tag:blogger.com,1999:blog-32176108590114916872012-02-16T12:28:11.041-06:00A window through the walls of our classroom. This is an interactive learning ecology for students and parents in my Calculus 45S class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.comBlogger831100tag:blogger.com,1999:blog-3217610859011491687.post-53539318427397875722008-06-26T15:30:00.000-05:002008-06-26T15:30:01.022-05:00<img src="http://static.flickr.com/73/161775713_4ed2bb3663_m_d.jpg" align="right" hspace="8">We had our graduation exercises today. A gentle push into the world for all of you. I hope you're leaving with the keys to your future in your hand.<br /><br /><b><i>I'm so glad we've had this time together,<br /><br />Just to have a laugh or learn some math,<br /><br />Seems we've just got started and before you know it,<br /><br />Comes the time we have to say, "So Long!"</b></i><br /><br />So long everybody! Watch this space in the fall for pointers to new blogs for each of my classes. <br /><br />Farewell, Auf Wiedersehen, Adieu, and all those good bye things. ;-)<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5353931842739787572?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-68045434578681052162008-06-15T23:17:00.001-05:002008-06-15T23:20:50.460-05:00<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://farm2.static.flickr.com/1162/1390706332_8572d1c1f8_d.jpg"><img style="margin: 0pt 0pt 10px 10px; float: right; cursor: pointer; width: 320px;" src="http://farm2.static.flickr.com/1162/1390706332_8572d1c1f8_d.jpg" alt="" border="0" /></a>First an update on this podcast: While we have received few comments on this or any of our class blogs the number of times the audio files have been downloaded is remarkable ...<br /><br /><blockquote><a href="http://adifference.blogspot.com/2008/04/student-voices-episode-1-jessie.html">Episode 1: Jessie</a> 2440 downloads<br /><br /><a href="http://adifference.blogspot.com/2008/04/student-voices-episode-2-timmathy.html">Episode 2: Tim_MATH_y</a> 1766 downloads<br /><br /><a href="http://adifference.blogspot.com/2008/05/student-voices-episode-3-chris-craig.html">Episode 3: Chris, Craig, Graeme</a> 1367 downloads</blockquote><br /><br />Thanks to all our listeners. We might get one more published during this school year but this may be the last until September. In any case feel free to let us know your thoughts about what you heard; every comment is appreciated.<br /><br />In this episode of <a href="http://adifference.blogspot.com/search/label/Student%20Voices">Student Voices</a> Justice, Lawrence, and Richard talk about how they put together <a href="http://expertvoices08.blogspot.com/2008/06/team-lrj-studios-present.html">their Developing Expert Voices project</a> and what they learned in the process: how they they best learn math, how it can best be taught, and many other incidental things like team work and organizational skills.<br /><br />They have titled their project with one of my favourite reminders to all my students: Mathematics is the Science of Patterns. If you watch any of the video content they created you'll hear several "in jokes", listen for them. Without any further ado, here is the podcast. A copy of the poster they made for their work is below.<br /><br /><center><embed src="http://www.archive.org/download/DarrenKuropatwaStudentVoices_Episode4Justice_Lawrence_andRichard/SV04JusticeLaurenceRichardPC40SW08.mp3" type="audio/mpeg" autostart="false" loop="false" bgcolor="#DCDE97" height="74" width="144"></embed><br />(<a href="http://www.archive.org/download/DarrenKuropatwaStudentVoices_Episode4Justice_Lawrence_andRichard/SV04JusticeLaurenceRichardPC40SW08.mp3">Download File</a> 12.2Mb, 25 min. 30 sec.)<br /></center><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_tlF3qs3v9cw/SFXm56L5XhI/AAAAAAAAAXY/6BZ6-EQewm8/s1600-h/Poster.jpg"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_tlF3qs3v9cw/SFXm56L5XhI/AAAAAAAAAXY/6BZ6-EQewm8/s400/Poster.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5212326026297302546" /></a><br /><br /><div align="right">Photo Credit: <a href="http://www.flickr.com/photos/zhzheka/1390706332/">Shadow singer</a> by flickr user EugeniusD80</div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6804543457868105216?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-5644827496202041062008-06-04T15:56:00.002-05:002008-06-04T15:59:11.926-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_447562"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080604-1212612823429205-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080604-1212612823429205-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-june-4-2008?src=embed" title="View Calculus 45S Slides June 4, 2008 on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-564482749620204106?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-27831011563045792052008-05-26T21:49:00.008-05:002008-05-26T23:22:11.346-05:00So, on the previous class, we were introduced to anti-derivatives..<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_AECKI0ALk0E/SDt4VKSkTrI/AAAAAAAAAGU/FG8mKwbf51U/s1600-h/scribeslide301.jpg"><img style="cursor: pointer;" src="http://2.bp.blogspot.com/_AECKI0ALk0E/SDt4VKSkTrI/AAAAAAAAAGU/FG8mKwbf51U/s320/scribeslide301.jpg" alt="" id="BLOGGER_PHOTO_ID_5204886099291557554" border="0" /></a><br />At the beginning of the class, we were asked to find the derivatives of the above functions. Needless to say, we found it difficult... NOT. It's 2x for all functions. And then we were given the derivative, <span style="font-style: italic;">f ' (x) = x</span> and were asked to find the parent function. We were quite stumped at first, but then managed to figure it out. We had to go back to a previous lesson since we were asked whether if given a constant and a function and if we had to find the derivative of the function, does the constant play a role? No, it doesn't. Anyway, we were asked to find the parent function of the derivative <span style="font-style: italic;">f ' (x) = x</span>. If x is the derivative, then obviously the parent function is x². But the derivative of x² is 2x and the derivative we want is x. So we figured, we have to multiply 2x by 1/2 to get x. From that, we got the parent function, which is 1/2x². Mr. K then asked us, what about if I give you the derivative x²? What would be its parent function? We then thought about it. If x² is the derivative, obviously the parent function is x³. But the derivative of that function will give us 3x², which we don't want since we're given the derivative x². So we thought multiplying 3x² by 1/3 would give us x². We reckon that the parent function is 1/3x³. From that we came up with a rule for all power functions, which is:<br /><br /><span style="font-style: italic; color: rgb(255, 0, 0);">f ' (x) = xⁿ -------> f (x) = xⁿ⁺¹/n+1 + <span style="color: rgb(51, 204, 0);">C, where C is the constant. I'll explain later.<br /><br /></span></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_AECKI0ALk0E/SDt-V6SkTsI/AAAAAAAAAGc/6t9kX_Cdyiw/s1600-h/scribeslide302.jpg"><img style="cursor: pointer;" src="http://1.bp.blogspot.com/_AECKI0ALk0E/SDt-V6SkTsI/AAAAAAAAAGc/6t9kX_Cdyiw/s320/scribeslide302.jpg" alt="" id="BLOGGER_PHOTO_ID_5204892709246226114" border="0" /></a><br />We then made anti-derivative rules for almost all derivative rules we remember. Mr. K said that the anti-derivative rules for the product and quotient rule are much more complicated and we won't be taking it until First Year University Calculus, so yeah.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_AECKI0ALk0E/SDuK2aSkTtI/AAAAAAAAAGk/iNY6RK9o1TE/s1600-h/scribeslide303.jpg"><img style="cursor: pointer;" src="http://3.bp.blogspot.com/_AECKI0ALk0E/SDuK2aSkTtI/AAAAAAAAAGk/iNY6RK9o1TE/s320/scribeslide303.jpg" alt="" id="BLOGGER_PHOTO_ID_5204906461731507922" border="0" /></a><br />Now this is where I explain the <span style="color: rgb(51, 204, 0);">C <span style="color: rgb(0, 0, 0);">part in the rule:<br /><br /></span></span><span style="font-style: italic; color: rgb(255, 0, 0);">f ' (x) = xⁿ -------> f (x) = xⁿ⁺¹/n+1 + <span style="color: rgb(51, 204, 0);">C, where C is the constant<br /><br /></span></span><span style="color: rgb(255, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);">We can't really find</span></span></span><span style="font-style: italic; color: rgb(255, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"><span style="font-style: italic;"><span style="font-style: italic;"> <span style="color: rgb(51, 204, 0);">C </span></span></span></span></span></span><span style="color: rgb(255, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);">not unless we're given a point in the graph.</span></span></span></span></span><span style="font-style: italic; color: rgb(255, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"><span style="font-style: italic;"><span style="font-style: italic;"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"> <span style="color: rgb(51, 204, 0);">C</span> i</span></span></span></span></span></span></span><span style="color: rgb(255, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);">s basically the y-intercept, but if we're just given the derivative and asked to solve for the parent function, all we can do is find the parent function and then</span></span></span></span></span><span style="font-style: italic; color: rgb(255, 0, 0);"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"><span style="font-style: italic;"><span style="font-style: italic;"><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"> + <span style="color: rgb(51, 204, 0);">C</span>, not unless we're given a point in the graph.</span></span></span></span></span><br /></span></span><span style="color: rgb(51, 204, 0);"><span style="color: rgb(0, 0, 0);"><br /></span></span>Then we were just given a practice problem, which was pretty straightforward.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_AECKI0ALk0E/SDuMM6SkTuI/AAAAAAAAAGs/z0ydoSFT9G8/s1600-h/scribeslide304.jpg"><img style="cursor: pointer;" src="http://1.bp.blogspot.com/_AECKI0ALk0E/SDuMM6SkTuI/AAAAAAAAAGs/z0ydoSFT9G8/s320/scribeslide304.jpg" alt="" id="BLOGGER_PHOTO_ID_5204907947790192354" border="0" /></a><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_AECKI0ALk0E/SDuMVKSkTvI/AAAAAAAAAG0/jaRy3j0ddbM/s1600-h/scribeslide305.jpg"><img style="cursor: pointer;" src="http://2.bp.blogspot.com/_AECKI0ALk0E/SDuMVKSkTvI/AAAAAAAAAG0/jaRy3j0ddbM/s320/scribeslide305.jpg" alt="" id="BLOGGER_PHOTO_ID_5204908089524113138" border="0" /></a><br /><br />The next scribe is etimz, or Ethan.<br /><span style="font-style: italic; color: rgb(255, 0, 0);"><br /><br /></span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2783101156304579205?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-74189241131012055912008-05-23T13:28:00.001-05:002008-05-23T13:29:21.116-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_424310"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080523-1211559338880347-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080523-1211559338880347-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-may-23-2008?src=embed" title="View Calculus 45S Slides May 23, 2008 on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7418924113101205591?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-27582879237205637402008-05-20T21:16:00.003-05:002008-05-20T21:20:24.285-05:00Well, another test. This unit hasn't been so bad for me, I understand most of it. Hopefully I do well in the test. I do have a couple of problems with differentiating implicitly but it's nothing practice cannot solve. Otherwise, I'm good to go.<br /><br />Good luck.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2758287923720563740?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-70376760880606573612008-05-19T21:52:00.002-05:002008-05-19T21:54:41.864-05:00What can I say? Well for one, this unit was simple the second time around =). I really think that I'm going to do well on this next test and yeah, hopefully I don't stumble on something that I haven't yet seen or tried to do. As always, the solution is practice! Hmm, I think the best place to start review would be to go through the past slides, and then see if I have any muddiest points.<br /><br />This bob was short and straight to the point. I hope everyone remembers to bob and and and! Good luck on the test Wednesday I believe!<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7037676088060657361?l=cal45sw08.blogspot.com' alt='' /></div>Tim_MATH_yhttp://www.blogger.com/profile/09929458936647070771noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-2553987465057716042008-05-16T11:17:00.002-05:002008-05-16T16:28:31.059-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_410618"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080516-1210954543437459-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080516-1210954543437459-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-may-16-2008?src=embed" title="View 'Calculus 45S Slides May 16, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-255398746505771604?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-81984363753129776272008-05-14T22:48:00.014-05:002008-05-15T00:43:29.908-05:00Today's lesson was about putting the pieces together, so to speak. Mr. K taught us how we should be able to come up with a graph of a function simply by using derivatives and critical points and whatnot.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_AECKI0ALk0E/SCuz0Pr3dvI/AAAAAAAAAFk/50rVwRCr37s/s1600-h/scribeslide201.jpg"><img style="cursor: pointer;" src="http://1.bp.blogspot.com/_AECKI0ALk0E/SCuz0Pr3dvI/AAAAAAAAAFk/50rVwRCr37s/s320/scribeslide201.jpg" alt="" id="BLOGGER_PHOTO_ID_5200447904874460914" border="0" /></a><br />This question was pretty straightforward. It just asks us to determine the critical points of the function, which means we have to find out where the derivative = 0. First we have to find the derivative of the certain function, as shown above, factor it if possible. However, don't make the mistake I did, which is state the <span style="font-weight: bold;">CRITICAL NUMBERS</span> instead of stating the <span style="font-weight: bold;">CRITICAL POINTS. </span>To find the critical points, just plug in the critical numbers at the parent function, and you should have it.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_AECKI0ALk0E/SCu03fr3dwI/AAAAAAAAAFs/As6jVfjFgsk/s1600-h/scribeslide202.jpg"><img style="cursor: pointer; width: 369px; height: 230px;" src="http://2.bp.blogspot.com/_AECKI0ALk0E/SCu03fr3dwI/AAAAAAAAAFs/As6jVfjFgsk/s320/scribeslide202.jpg" alt="" id="BLOGGER_PHOTO_ID_5200449060220663554" border="0" /></a><br /><br />This is another straightforward question. Mr. K said that we should be able to solve these kinds of questions for 30-60 seconds. To find the relative max and min of an equation, we simply have to find out where the derivative of the parent function = 0. We then find the derivative of the parent function, as seen above, and then we use the first derivative test to find out if we have a maximum or a minimum. Then, we solve for the second derivative and use the second derivative test to find out again the maximum or a minimum of a function. An interesting tidbit: the second derivative function's main purpose is to find out the maximum or a minimum of a parent function. If we're only given the second derivative of a function, then yes, the 2nd derivative function is useful. Then Mr. K said, but what if we're given the parent function? Isn't that kind of a waste of time? The first derivative test already tells us the maximum or a minimum of a function, so why bother with the second derivative test? If it's main purpose is to find out the concavity of a parent function, then there we go.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_AECKI0ALk0E/SCu7lPr3dxI/AAAAAAAAAF0/qAmbgjHncJA/s1600-h/scribeslide203.jpg"><img style="cursor: pointer;" src="http://1.bp.blogspot.com/_AECKI0ALk0E/SCu7lPr3dxI/AAAAAAAAAF0/qAmbgjHncJA/s320/scribeslide203.jpg" alt="" id="BLOGGER_PHOTO_ID_5200456443269445394" border="0" /></a><br />For this question, we had to find out several things; intervals in which y is increasing or decreasing, the coordinates of any local extrema (minimum and maximum), intervals of concavity, inflection points, and the graph of the function. To find out where y is increasing or decreasing, we have to find out its derivative first and test the answers that we get. As you can see on the picture above, the critical numbers are x = 1 and 2. <span style="font-style: italic;">f is increasing where f ' is positive, and f is decreasing where f ' is negative. </span>We then test it by using the number line test, and the<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_AECKI0ALk0E/SCu7t_r3dyI/AAAAAAAAAF8/992k4Tb0u1o/s1600-h/scribeslide204.jpg"><img style="cursor: pointer;" src="http://4.bp.blogspot.com/_AECKI0ALk0E/SCu7t_r3dyI/AAAAAAAAAF8/992k4Tb0u1o/s320/scribeslide204.jpg" alt="" id="BLOGGER_PHOTO_ID_5200456593593300770" border="0" /></a><br />To find the coordinates of the local extrema, we use the first derivative test. From the number line test we did, we can see that we have a max at x = 1 and a min at x = 2. We then plug in those critical numbers on the parent function to find out the y coordinate of the maximum and the minimum points.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_AECKI0ALk0E/SCu78vr3dzI/AAAAAAAAAGE/bLbZ7xp_i00/s1600-h/scribeslide205.jpg"><img style="cursor: pointer;" src="http://3.bp.blogspot.com/_AECKI0ALk0E/SCu78vr3dzI/AAAAAAAAAGE/bLbZ7xp_i00/s320/scribeslide205.jpg" alt="" id="BLOGGER_PHOTO_ID_5200456846996371250" border="0" /></a><br /> To find the intervals of concavity, we simply use the second derivative test. We find the derivative of the first derivative, which of course gives us the second derivative. We then find its critical point. We then use the number line test to find out where the parent function is concave up and concave down. From the solution above, we find out that the parent function is concave down from (-∞, 3/2) and concave up at (3/2, ∞). Therefore, we can say that we have a point of inflection at x = 3/2.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_AECKI0ALk0E/SCu89Pr3d0I/AAAAAAAAAGM/FwE_kKFbcS4/s1600-h/scribeslide206.jpg"><img style="cursor: pointer;" src="http://1.bp.blogspot.com/_AECKI0ALk0E/SCu89Pr3d0I/AAAAAAAAAGM/FwE_kKFbcS4/s320/scribeslide206.jpg" alt="" id="BLOGGER_PHOTO_ID_5200457955097933634" border="0" /></a><br />Well this is pretty straightforward. We just gather what we found out from the previous solutions and, shall I say, "put the pieces together." Then we sketch the graph. =)<br /><br />That's it! We have a pre-test next class and a test the class after.<br /><br />The next scribe is... *drum roll please*<br /><br />haven't decided yet. lol<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8198436375312977627?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-18883346751964360292008-05-14T15:35:00.001-05:002008-05-14T15:37:54.609-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_405916"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080514-1210797112725929-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080514-1210797112725929-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-may-14-2008?src=embed" title="View 'Calculus 45S Slides May 14, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1888334675196436029?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-59233430629649342362008-05-12T23:22:00.002-05:002008-06-15T23:20:40.802-05:00<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://farm2.static.flickr.com/1162/1390706332_8572d1c1f8_d.jpg"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 320px;" src="http://farm2.static.flickr.com/1162/1390706332_8572d1c1f8_d.jpg" border="0" alt="" /></a><br />In this episode of Student Voices three Advanced Placement Calculus students, Chris, Craig, and Graeme, talk about a wiki assignment they did to prepare for the exam. Then the conversation transitions to a discussion of the many things they learned while doing their <a href="http://expertvoices08.blogspot.com">Developing Expert Voices</a> project. It ends with a challenge, the result of which will be featured in a future podcast.<br /><p><br />Let Chris, Craig, and Graeme know what you thought about the podcast by leaving a comment here on this post or on the mirror of this post <a href="http://apcalc07.blogspot.com/2008/05/student-voices-episode-3-chris-craig.html">on their class blog</a>.<br /><p><br /><center><br /> <embed src="http://www.archive.org/download/DarrenKuropatwaStudentVoices_Episode3Chris_Craig_Graeme/SV03ChrisCraigGraemeapcalc07.mp3" width="144" height="74" type="audio/mpeg" autostart="false" loop="false" bgcolor="#DCDE97"> </embed><br /><br><br />(<a href="http://www.archive.org/download/DarrenKuropatwaStudentVoices_Episode3Chris_Craig_Graeme/SV03ChrisCraigGraemeapcalc07.mp3">Download File</a> 31.8Mb, 26 min. 30 sec.)<br /></center><br /><p><br />The video mentioned near the end of the podcast is called <a href="http://www.youtube.com/watch?v=K2cYWfq--Nw">Daft Hands</a>. Here it is:<br /><p><br /><center><br /><object width="425" height="355"><param name="movie" value="http://www.youtube.com/v/K2cYWfq--Nw&hl=en"></param><param name="wmode" value="transparent"></param><embed src="http://www.youtube.com/v/K2cYWfq--Nw&hl=en" type="application/x-shockwave-flash" wmode="transparent" width="425" height="355"></embed></object><br /></center><br /><p><br /><div align="right">Photo Credit: <a href="http://www.flickr.com/photos/zhzheka/1390706332/">Shadow singer</a> by flickr user EugeniusD80</div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5923343062964934236?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-33205090297707694062008-05-12T15:57:00.004-05:002008-05-12T17:32:24.059-05:00<span style="font-weight: bold;font-size:130%;" >Introduction:</span><br /><br /><span style="font-size:100%;">Hello, I'm known as <span style="color: rgb(0, 0, 153);">Tim-Math-Y</span> on our class blog and I will be the scribe for today's lessons! Today's class started with an image that introduced our content. The image was a combination of three that portrayed related rates with spheres, water ripples, and shadows. It was the precursor to the questions that we were to solve.<br /><br /><span style="color: rgb(0, 0, 153);">Key</span> points to solving <span style="color: rgb(0, 0, 153);">Related Rates</span> questions:<br /></span><ul><li><span style="font-size:100%;">Find a way to relate all the variables within a given question</span></li><li><span style="font-size:100%;">Differentiate implicitly using chain rule, with respect to time</span></li></ul><span style="font-weight: bold;font-size:130%;" >Content:</span><br /><br /><div style="width:425px;text-align:left" id="__ss_401300"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=junk-scribe-1210631405645742-8"><param name="allowFullScreen" value="true"><param name="allowScriptAccess" value="always"><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=junk-scribe-1210631405645742-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare" /></a> | <a href="http://www.slideshare.net/Timmathy/scribe1?src=embed" title="View 'Scribe1' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /><br /><span style="font-size:130%;"><span style="font-weight: bold;">Conclusion:</span><br /></span><br />Well, I completed the scribe as soon as I could because it seemed that many people were missing today. I hope this helped!<br /><br />Notice that most of these solutions include the following:<br /><ul><li>A labelled diagram to help you visualize</li><li>Listing given information, and finding information that will be required in the solution<br /></li><li>Differentiating implicitly using chain rule with respect to time</li><li>A sentence answer including appropriate units</li></ul>That sums it up folks! Thanks for reading =)<br /><br />The Scribe for next class will be, <span style="font-weight: bold; color: rgb(0, 0, 153);">John D.</span> !<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3320509029770769406?l=cal45sw08.blogspot.com' alt='' /></div>Tim_MATH_yhttp://www.blogger.com/profile/09929458936647070771noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-7959832407397330962008-05-12T15:49:00.001-05:002008-05-12T15:52:49.282-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_401110"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080512-1210625314869103-9"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080512-1210625314869103-9" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-may-12-2008?src=embed" title="View 'Calculus 45S Slides May 12, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-795983240739733096?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-86976877025410940902008-05-08T23:07:00.014-05:002008-05-08T23:48:40.099-05:00<a href="http://4.bp.blogspot.com/_fvK_Af4KmnA/SCPWeX4VBFI/AAAAAAAAAJA/numvwifcQ3E/s1600-h/scribe+into+cal_1.jpeg"><img id="BLOGGER_PHOTO_ID_5198234212210377810" style="WIDTH: 412px; CURSOR: hand; HEIGHT: 319px" height="321" alt="" src="http://4.bp.blogspot.com/_fvK_Af4KmnA/SCPWeX4VBFI/AAAAAAAAAJA/numvwifcQ3E/s400/scribe+into+cal_1.jpeg" width="423" border="0" /></a><br /><br /><div><a href="http://2.bp.blogspot.com/_fvK_Af4KmnA/SCPWZ34VBEI/AAAAAAAAAI4/zQe6kqbPScE/s1600-h/scribe+into+cal_2.jpeg"><img id="BLOGGER_PHOTO_ID_5198234134900966466" style="WIDTH: 411px; CURSOR: hand; HEIGHT: 325px" height="322" alt="" src="http://2.bp.blogspot.com/_fvK_Af4KmnA/SCPWZ34VBEI/AAAAAAAAAI4/zQe6kqbPScE/s400/scribe+into+cal_2.jpeg" width="425" border="0" /></a><br /><br /><div><a href="http://3.bp.blogspot.com/_fvK_Af4KmnA/SCPWWH4VBDI/AAAAAAAAAIw/3XtXabcrUn0/s1600-h/scribe+into+cal_3.jpeg"><img id="BLOGGER_PHOTO_ID_5198234070476457010" style="WIDTH: 411px; CURSOR: hand; HEIGHT: 313px" height="323" alt="" src="http://3.bp.blogspot.com/_fvK_Af4KmnA/SCPWWH4VBDI/AAAAAAAAAIw/3XtXabcrUn0/s400/scribe+into+cal_3.jpeg" width="424" border="0" /></a><br /><div><a href="http://4.bp.blogspot.com/_fvK_Af4KmnA/SCPWTX4VBCI/AAAAAAAAAIo/FcY005ucPWU/s1600-h/scribe+into+cal_4.jpeg"><img id="BLOGGER_PHOTO_ID_5198234023231816738" style="WIDTH: 410px; CURSOR: hand; HEIGHT: 331px" height="332" alt="" src="http://4.bp.blogspot.com/_fvK_Af4KmnA/SCPWTX4VBCI/AAAAAAAAAIo/FcY005ucPWU/s400/scribe+into+cal_4.jpeg" width="424" border="0" /></a><br /><div><a href="http://2.bp.blogspot.com/_fvK_Af4KmnA/SCPWP34VBBI/AAAAAAAAAIg/-eqYgML4Dyo/s1600-h/scribe+into+cal_5.jpeg"><img id="BLOGGER_PHOTO_ID_5198233963102274578" style="WIDTH: 411px; CURSOR: hand; HEIGHT: 345px" height="341" alt="" src="http://2.bp.blogspot.com/_fvK_Af4KmnA/SCPWP34VBBI/AAAAAAAAAIg/-eqYgML4Dyo/s400/scribe+into+cal_5.jpeg" width="423" border="0" /></a><br /><div><a href="http://1.bp.blogspot.com/_fvK_Af4KmnA/SCPWLn4VBAI/AAAAAAAAAIY/Nk_vjSClI_c/s1600-h/scribe+into+cal_6.jpeg"><img id="BLOGGER_PHOTO_ID_5198233890087830530" style="WIDTH: 410px; CURSOR: hand; HEIGHT: 329px" height="333" alt="" src="http://1.bp.blogspot.com/_fvK_Af4KmnA/SCPWLn4VBAI/AAAAAAAAAIY/Nk_vjSClI_c/s400/scribe+into+cal_6.jpeg" width="424" border="0" /></a><br /><div><a href="http://1.bp.blogspot.com/_fvK_Af4KmnA/SCPWIn4VA_I/AAAAAAAAAIQ/06y20-vpFys/s1600-h/scribe+into+cal_7.jpeg"><img id="BLOGGER_PHOTO_ID_5198233838548222962" style="WIDTH: 408px; CURSOR: hand; HEIGHT: 330px" height="347" alt="" src="http://1.bp.blogspot.com/_fvK_Af4KmnA/SCPWIn4VA_I/AAAAAAAAAIQ/06y20-vpFys/s400/scribe+into+cal_7.jpeg" width="424" border="0" /></a> </div></div></div></div></div></div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8697687702541094090?l=cal45sw08.blogspot.com' alt='' /></div>etimzhttp://www.blogger.com/profile/09626175119223517919noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-72529712057728366152008-05-08T15:39:00.001-05:002008-05-08T15:41:11.180-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_395160"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080508-1210279148249123-9"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080508-1210279148249123-9" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-may-8-2008?src=embed" title="View 'Calculus 45S Slides May 8, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7252971205772836615?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-51783044533089399152008-05-07T14:29:00.003-05:002008-05-07T15:47:47.509-05:00Hello everyone this is the scribe post for May 6, sorry it is late but I had to write the AP <span class="blsp-spelling-error" id="SPELLING_ERROR_0">calc</span> exam and I was not able to post it earlier. Well in Tuesday's class we only did one question, but there were two others for homework. The first question was about a man who was driving his car through the desert, and wanted to know at what distance he should drive through the desert and the paved road to get to the city B 5km south from city A. He can drive 15mph in the desert and 35mph on the paved road, he is 5 km east of city A. So the maximum distance the man could travel is 5km in the desert and 5km of pavement to get to city B (The diagram is on the slides). So that would be unrealistic, so in order to get the minimum amount of time the car must cut time through the desert and pavement alike. So the pavement equation will become 5km subtract x, x the amount of distance that is going to be cut by driving through the desert. Then the amount of distance driving through the desert is found by a^2 + b^2 = c^2. Where a and b is 5-x. So c is (25+x^2)^.5 (square root is ^.5).<br /><br />As you know Rate, Distance, and Time is all related. Time is Distance over rate. So the total time is (5-x)/35 + ((25+x^2)^.5)/15 =t(x). Solving for the derivative (solved on the slides) , t'(x) =<br />7x-3((25+x^2)^0.5)/105((25-x^2)^0.5). Now solving for the zero's you get 40x^2-225. So the roots are -15/2((10)^0.5) and 15/2((10)^0.5). We <span class="blsp-spelling-corrected" id="SPELLING_ERROR_1">reject</span> the negative value because its not possible in the context of the question. So the man should go and head for the point x = 15/2((10)^0.5), as that point is the minimum value of the First Derivative Function.<br /><br />Remember to do the other two questions and the homework given in class, <span style="color: rgb(255, 0, 0);">the next scribe is Ethan. </span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5178304453308939915?l=cal45sw08.blogspot.com' alt='' /></div>Dinohttp://www.blogger.com/profile/16720903626604404687noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-51150667021640273692008-05-06T12:28:00.001-05:002008-05-06T12:34:33.809-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_390753"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080506-1210090495307494-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080506-1210090495307494-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-may-6-2008?src=embed" title="View 'Calculus 45S Slides May 6, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5115066702164027369?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-6153144572837360642008-05-02T17:40:00.004-05:002008-05-02T18:40:12.019-05:00Today, we were doing optimization problems.<br />In order to describe how we are solving these problems, I will be using an example from the slide 5.<br />It is the open box question where you have to cut squares from the corners to maximize the volume of the box.<br /><br />First step is to draw a picture of what you want to do. We are cutting squares from the corners and we label the square's side with x. We know that the side of the entire piece is 24 inches per side. Therefore the equation is (24-2x) for the space between the two corner squares you have to cut. Everything is a square so all sides are the same.<br /><br />Next we know that the volume of the box is going to be V=L*W*H.<br />We know that the height(H) is x and each side (L &amp; W） is the same and equal to (24-2x), so the equation is going to be<br />V<span style="font-size:78%;">(x)</span>= （24-2x)*(24-2x)*x<br />which is<br />V<span style="font-size:78%;">(x)</span>=((24-2x)^2) * x<br /><br />Third step is to simplify the equation and you will get<br />V<span style="font-size:78%;">(x)</span>=(4x^3) - (96x^2) + 576x.<br /><br />We take the derivative of that to find the max.<br />V'<span style="font-size:78%;">(x) <span style="font-size:100%;">=(12x^2) - 192x + 576</span></span><br /> V'<span style="font-size:78%;">(x)</span> =12((x^2) - 16x + 48)<br /><br />The derivative at the max is equal to zero.<br />So we let V'<span style="font-size:78%;">(x)</span> = 0<br />0 = 12((x^2) - 16x + 48)<br />0 = ((x^2) - 16x + 48)<br />0 = (x - 4) (x - 12)<br />x = 4, 12<br />x can't equal to 12 because that would make the side of the box equal to zero.<br />Therefore x = 4.<br /><br />The last step is to test it.<br />___+___4__-___12_____<br /><br />4 by 4 squares should be cut out to maximize the volume of the box.<br /><br /><br />The next scribe is Dino.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-615314457283736064?l=cal45sw08.blogspot.com' alt='' /></div>haiyanhttp://www.blogger.com/profile/04253570568706636473noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-77540228435700022942008-05-02T14:14:00.002-05:002008-05-02T14:16:39.645-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_385174"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080502-1209755569904131-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080502-1209755569904131-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-may-2-2008?src=embed" title="View 'Calculus 45S Slides May 2, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7754022843570002294?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-57289299258840545442008-05-02T06:38:00.003-05:002008-05-02T07:06:18.555-05:00Hey everyone this is m@rk and i'm scribing for Wednesday's class. First of we took a quiz on visual calculus about what we've learned from the first and second derivative test. Second , Mr. K introduced a new topic called Optimization Problems. Optimization problems are problems that deal with either finding the maximum or minimum of the function in the problem.<br /><br />Slide 1<br /><br />First of all you need to do is draw a diagram to help you visualize the problem. Second, construct the constraint equation. The constraint equation is the equation is something that limits the problem. In this case it is 320=2L+2W. Third, find what you are trying to optimize and construct an optimization equation. In this case you are trying to maximize area, so you need to use the area formula A=LW. Fourth, you need to solve for one of the unknown variables on the constraint equation and then plug it in on the optimization equation. We choose to solve for L in this case and we plug it in on the optimization equation. Fifth, solve for the derivative of our new optimization equation. Sixth, solve for the roots of the derivative of our optimization equation. We found out that the root is located at W=80. Seventh, use the first derivative test and find out whether it is a maximum or a minimum. Lastly, answer the question in a complete sentence.<br /><br />I'm not going to go over the other question since it is just almost the same problem but instead I'm going to give you an outline how to solve an optimization problem, so after digging up the apcalc blog here it is:<br /><br /><div>Step 1: Find what you are trying to maximize or minimize. This will be stated excplicitly (in the question).</div><div><br />Step 2: Write an equation for it. Use V = for volume and M = for material, A for area, etc. I must insist on using descriptive variables, because in optimization, if you are sloppy, you lose track of what's going on.</div><div><br />Step 3: Try to get the equation into a two variable form, so you can take the derivative.<br />- Step 3a: To do step 1, you will often have to create a second equation from additional information given in the problem. This may require ingenuity, but it should become natural.<br />- Step 3b: ISolate one of the two variables in the equation drawn in step 3a.<br />- Step 3c: In the original equation that you are trying to minimize or maximize, replace the variable you isolated in step 3b.</div><div><br />Step 4: Take the derivative of your two-variable equation.</div><div><br />Step 5: Set the derivative to 0, and solve for the value of the remaining variable.</div><div><br />Step 6: Plus the value of that variable into the first two equations to find all dimensions, including the final goal, such as the amount of material, cost, or volume.<br /><br />NOTE: This is just an outline so follow it loosely. There is really no chronological way to answer this problems.<br /><br />Thats it , the next scribe is haiyan.<br /></div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5728929925884054544?l=cal45sw08.blogspot.com' alt='' /></div>m@rkhttp://www.blogger.com/profile/00509544958003356512noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-13303215679184397002008-04-30T13:47:00.001-05:002008-04-30T13:48:59.648-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_382010"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080430-1209581157114495-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080430-1209581157114495-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-30-2008?src=embed" title="View 'Calculus 45S Slides April 30, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1330321567918439700?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-79792078139373518852008-04-28T21:04:00.004-05:002008-04-28T21:34:12.958-05:00Why hello there, it's Kristin_R here, one of the few non-AP students left in the calculus class.<br /><br /><br />Today in class we learned about the <span style="color:#ff0000;">first and second derivative tests.<br /><br /></span><strong>The first derivative test:<br /></strong><br />c is a critical point (meaning the inputs where the derivative is zero or undefined, which is a horizontal tangent line) somewhere over the interval (a,b).<br /><br />If f'(x) > 0 over the interval (a,c) and f'(x) < 0 for all values of x in the interval (c, b), then that function has a maximum over that interval.<br /><br />If f'(x) ,0 over the interval (a,c) and f'(x) . 0 over the interval (c,b), then that function has a minimum over that interval.<br /><br />Here's an example...<br /><br />f(x) = x^2-8x=4<br /><br />therefore, f'(x) = 2x-8<br /><br />If we let the derivative equal zero, the root is at x=4.<br /><br />When we plug four back into the original function, we get a y-value of -12. .<br /><br />This means that the root is at (4, -12).<br /><br />- ----- -12 ------ - --------4-------- +<br /><br />By the first derivative test, x=4 is a minimum because to the left of 4, f is decreasing , and to the right of four, f is increasing.<br /><br /><br /><strong>The second derivative test:<br /></strong><br />When the parent function is <span style="color:#ff0000;">concave up (valley),</span> f'' is postive.<br /><br />When the parent function is <span style="color:#ff0000;">concave down (hill),</span> f'' is negative.<br /><br />When f' has a root, the parent function has a <span style="color:#ff0000;">point of inflection</span>, which is where the function changes concavity.<br /><br />So, the first derivative test shows where the function is increasing or decreasing, as well as maximum or minimum extrema.<br /><br />The second derivative test shows us the concavity of the graph.<br /><br />"<em>Just finding a root of your second derivative is not enough! You have to check your intervals!"</em><br /><br />Just a quick little note - in grade eleven, you should have been taught about how to graph certain expontential functions. Remember that the degree of the given function is the same amount of possible roots in the graph.<br /><br />So, thats it, thats the whole ball of wax...y'all dig?<br /><br />Homework is up to and including page 64 in the white and orange book, as well as up to and including page 42 in the blue book.<br /><br />The scribe for next class is....<span style="color:#ff0000;">John. D!</span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7979207813937351885?l=cal45sw08.blogspot.com' alt='' /></div>Kristin_Rhttp://www.blogger.com/profile/13674690777406107391noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-5544044236823879072008-04-28T16:31:00.003-05:002008-04-28T16:42:44.504-05:00Yes me again. There actually isn't a lot of information to include on this scribe.<br /><br />Due to the fact, Mr. K thought it was a lot of information to absorb for the past few classes.<br /><br />So we....<br /><br />basically reviewed Tuesday's class.<br /><br />We got into groups, did the second slide question.<br /><br />And critical numbers of f, are when f' and f'' are equal to zero.<br /><br />Next scribe... Kristen.<br /><br />Sorry it's so short...<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-554404423682387907?l=cal45sw08.blogspot.com' alt='' /></div>Vanhttp://www.blogger.com/profile/07679810325423377354noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-68966792537657122252008-04-28T16:01:00.002-05:002008-04-28T16:30:12.000-05:00Sorry it's so late. Kept forgetting and procrastinating. Yes, it was supposed to be me... but, I wasn't even participating during the class, because of APcalc in the corner of the room. But, I can at least explain the slides.<br /><br />Slide 1, 2<br /><br />Picture and the question.<br /><br />Slide 3,4,5<br /><br />Velocity is the change of distance over the change in time.<br /><br />a. The average velocity, is the slope of the line, from the interval 0 to 4 of the given position function.<br /><br />b. Velocity is zero when, the derivative of x(t) is equal to zero.<br /><br />c. When the point is moving to the right (positive) meaning, the velocity is positive. (Derivative of x is positive)<br /><br />d. Same thing as c, but moving left (negative)<br /><br />e. When the derivative is equal to 3<br /><br />f. Taking the derivative of the velocity function. Or the second derivative of the position function, called the acceleration function is equal to 3.<br /><br />q. (Not sure why q, but okay) Draw what x(t) looks like. Red graph is the derivative of x(t)<br /><br />Slide 6,7<br /><br />The graph, and line test means, you are trying to find from the derivative, where there may be local minimums or maximums on the parent function.<br /><br />So, when the derivative has values that crosses the x axis(zero) there is a local maximum or minimum. To determine whether it is a maximum or a minimum, on the left side of the zero, if it is a positive value then crosses the x-axis and then is negative, it is a local maximum. Vice versa, it is a local minimum.<br /><br />Slide 8<br /><br />The extreme value theorem, says, on a differentiated function, within a closed interval, there exists a maximum and minimum value on the parent function.<br /><br />Slide 9 and 10 practice what we just learned.<br /><br />That's it for that scribe. I'm onto Thursday's scribe now. Ciao.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6896679253765712225?l=cal45sw08.blogspot.com' alt='' /></div>Vanhttp://www.blogger.com/profile/07679810325423377354noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-73166675046566115052008-04-28T11:41:00.001-05:002008-04-28T11:43:20.226-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_377034"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080428-1209400779686370-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080428-1209400779686370-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-28-2008?src=embed" title="View 'Calculus 45S Slides April 28, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7316667504656611505?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-30346499890098952362008-04-27T19:25:00.002-05:002008-06-15T23:21:16.009-05:00<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://farm2.static.flickr.com/1162/1390706332_8572d1c1f8_d.jpg"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;width: 320px;" src="http://farm2.static.flickr.com/1162/1390706332_8572d1c1f8_d.jpg" border="0" alt="" /></a><br />In this episode Timothy came back to school on Friday afternoon to talk about his week attending the miniUniversity program at the University of Winnipeg. He talks about the differences he finds between teaching and learning at high school and university and describes learning in the university classroom using a thought provoking metaphor, listen for it. Also, we have a cameo appearance by two very special people at the very end.<br /><p><br />Please feel free to leave Tim_MATH_y your comments here on this post.<br /><p><br /><center><br /> <embed src="http://ia360905.us.archive.org/1/items/DarrenKuropatwaStudentVoices_Episode2Tim_math_y/SV02TimMathYapcalc07_64kb.mp3" width="144" height="74" type="audio/mpeg" autostart="false" loop="false" bgcolor="#DCDE97"> </embed><br /><br><br />(<a href="http://ia360905.us.archive.org/1/items/DarrenKuropatwaStudentVoices_Episode2Tim_math_y/SV02TimMathYapcalc07_64kb.mp3">Download File</a> 7.2Mb, 15 min. 3 sec.)<br /></center><br /><p><br /><div align="right">Photo Credit: <a href="http://www.flickr.com/photos/zhzheka/1390706332/">Shadow singer</a> by flickr user EugeniusD80</div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3034649989009895236?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-898223529187716802008-04-24T13:18:00.001-05:002008-04-24T13:23:59.041-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_370856"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080424-1209061146190925-9"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080424-1209061146190925-9" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-24-2008?src=embed" title="View 'Calculus 45S Slides April 24, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-89822352918771680?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-80983435505602301932008-04-22T14:00:00.003-05:002008-04-22T14:56:02.114-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_367574"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080422-1208890534169345-9"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080422-1208890534169345-9" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-22-2008?src=embed" title="View 'Calculus 45S Slides April 22, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8098343550560230193?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-74551292256405198292008-04-21T16:52:00.009-05:002008-04-21T19:38:52.638-05:00On Friday, our class started talking about the application of derivatives. To start off, Mr. K showed us this slide:<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_AECKI0ALk0E/SA0Zlk0ySrI/AAAAAAAAAFM/H4P8pEDZnXA/s1600-h/scribeslide01.JPG"><img style="cursor: pointer; width: 372px; height: 178px;" src="http://4.bp.blogspot.com/_AECKI0ALk0E/SA0Zlk0ySrI/AAAAAAAAAFM/H4P8pEDZnXA/s320/scribeslide01.JPG" alt="" id="BLOGGER_PHOTO_ID_5191834078758980274" border="0" /></a><br />Mr. K then asked us when was the car moving fastest. The class decided that the car was moving fastest during t2, because the slope of the tangent line was the steepest at that point. During the interval [t1,t2], it is clear that the car is speeding up during that interval because the slope of the tangent lines are getting steeper going from t1 to t2. The car is definitely slowing down on the interval [t2,t3] as we can see that the slope of the tangent lines are decreasing.<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_AECKI0ALk0E/SA0abk0ySsI/AAAAAAAAAFU/ftfjGwVBAnI/s1600-h/scribeslide02.jpeg"><img style="cursor: pointer;" src="http://4.bp.blogspot.com/_AECKI0ALk0E/SA0abk0ySsI/AAAAAAAAAFU/ftfjGwVBAnI/s320/scribeslide02.jpeg" alt="" id="BLOGGER_PHOTO_ID_5191835006471916226" border="0" /></a><br />Now this was a tough graph to decipher. The AP Calculus students had no problem understanding this one, but to those of us who aren't taking AP, this was kind of difficult. I myself found it quite confusing at first, but after clarification, I understand it. Now, you have to find the derivative of <span style="font-style: italic;">f</span> in this graph to find out where is <span style="font-style: italic;">f ' </span>is positive. First we have to find out where the tangent lines in <span style="font-style: italic;">f</span> are zero. Looking at the function <span style="font-style: italic;">f,</span> we can clearly see that the tangent lines are zero in -2 and +2. This means that <span style="font-style: italic;">f '</span> will have roots at -2 and +2. Then, we look when the tangent lines of <span style="font-style: italic;">f </span>are approaching zero. There are 2 instances in this graph. The first instance is from (-<!--[if gte msEquation 12]><m:omathpara><m:omath><i style="'mso-bidi-font-style:normal'"><span style="'line-height:;font-size:11.0pt;"><m:r>∞</m:r></span></i></m:oMath></m:oMathPara><![endif]--><!--[if !msEquation]--><span style="line-height: 115%;"><!--[if gte vml 1]><v:shapetype id="_x0000_t75" coordsize="21600,21600" spt="75" preferrelative="t" path="m@4@5l@4@11@9@11@9@5xe" filled="f" stroked="f"> <v:stroke joinstyle="miter"> <v:formulas> <v:f eqn="if lineDrawn pixelLineWidth 0"> <v:f eqn="sum @0 1 0"> <v:f eqn="sum 0 0 @1"> <v:f eqn="prod @2 1 2"> <v:f eqn="prod @3 21600 pixelWidth"> <v:f eqn="prod @3 21600 pixelHeight"> <v:f eqn="sum @0 0 1"> <v:f eqn="prod @6 1 2"> <v:f eqn="prod @7 21600 pixelWidth"> <v:f eqn="sum @8 21600 0"> <v:f eqn="prod @7 21600 pixelHeight"> <v:f eqn="sum @10 21600 0"> </v:formulas> <v:path extrusionok="f" gradientshapeok="t" connecttype="rect"> <o:lock ext="edit" aspectratio="t"> </v:shapetype><v:shape id="_x0000_i1025" type="#_x0000_t75" style="'width:9pt;"> <v:imagedata src="file:///C:\DOCUME~1\John\LOCALS~1\Temp\msohtmlclip1\01\clip_image001.png" title="" chromakey="white"> </v:shape><![endif]--><!--[if !vml]--><!--[endif]--></span><!--[endif]-->∞, -2]. The slope from that domain is <span style="font-style: italic;">increasing</span>, since it's coming from a negative value to zero. But that still doesn't solve our problem since <span style="font-style: italic;">f '</span> is still negative at that point. Then, from (-2, 2) in <span style="font-style: italic;">f</span>, we can see that the slope is increasing until (0,0) and it starts to decrease until it reaches (2,3), where it becomes zero. At this point, <span style="font-style: italic;">f '</span> is positive as the slopes of the tangent lines from (-2, -1) to (0,0) are increasing from zero and reaches its maximum at (0,0), and then it starts to decrease from (0,0) until it reaches (2,3) where the slope is zero. This then solves the first question which asks where is <span style="font-style: italic;">f ' </span>positive, which is from (-2, o) to (2,0).<br /><br />Now to find out where <span style="font-style: italic;">f ''</span> is positive. From the graph, we can see that <span style="font-style: italic;">f</span> is some kind of a cubic function, so we know that its derivative will be a quadratic function. Now, the derivative of a quadratic function is a straight line, so now we know that <span style="font-style: italic;">f '' </span>is a straight line. Looking at the shape of <span style="font-style: italic;">f ' (say it is -2x²), </span>it is definitely a negative quadratic function because it has a maximum, not a minimum. We then have to find out the derivative of <span style="font-style: italic;">f ''</span> which is represented as -2x² (However, it IS NOT -2x², it's just a representation). Using the power rule, we can determine that the derivative of <span style="font-style: italic;">f ', </span>which is <span style="font-style: italic;">f ''</span>, is -4x. From this we can see that the function is positive at quadrant 2.<br /><br />The critical numbers (which is another term for the roots) of <span style="font-style: italic;">f </span>are -2 and 2.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_AECKI0ALk0E/SA0qagmhe3I/AAAAAAAAAFc/i8wYqG0QsS4/s1600-h/scribeslide03.JPG"><img style="cursor: pointer;" src="http://3.bp.blogspot.com/_AECKI0ALk0E/SA0qagmhe3I/AAAAAAAAAFc/i8wYqG0QsS4/s320/scribeslide03.JPG" alt="" id="BLOGGER_PHOTO_ID_5191852580344527730" border="0" /></a><br />This is a little difficult...<br />For a), <span style="font-style: italic;">f </span>is increasing wherever <span style="font-style: italic;">f '</span> is positive.<br />For b),<span style="font-style: italic;"> f(0)</span> is negative.<br /><br />Well, that was all we talked about last class. Mr. K tried to squeeze in one more slide but we ran out of time.<br /><br />The next scribe is Kristin.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7455129225640519829?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-58472981587289690642008-04-19T13:25:00.003-05:002008-06-15T23:21:38.586-05:00<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_tlF3qs3v9cw/SAo3CljRiII/AAAAAAAAAWw/ATUn9MTojWw/s1600-h/SVKidsOfConversation.jpg"><img style="float:right; margin:0 0 10px 10px;cursor:pointer; cursor:hand;" src="http://4.bp.blogspot.com/_tlF3qs3v9cw/SAo3CljRiII/AAAAAAAAAWw/ATUn9MTojWw/s320/SVKidsOfConversation.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5191022038077245570" /></a><br />I was talking to Jessie, one of my Applied Math students, earlier this week while helping her review over the lunch hour. I found her comments so compelling I asked her (and later her parents) if I could record and publish her comments so other students could hear what she had to say. I've long thought students need to hear from other students how they best learn to help them all learn.<br /><p><br />This is the first in what I hope will be a series of podcasts called <b>Student Voices</b>. I'm hoping to have one of these short conversations with a student published each week. If you'd like to volunteer to be featured in one of these just let me know.<br /><p><br />In this episode Jessie shares how she uses <a href="http://am40sw08.blogspot.com">her class blog</a> to learn and describes her personal "tipping point" from being confused to understanding Statistics very well. She also discusses the value of learning conversations and how sometimes being a "teacher" and sometimes a student helps her learn.<br /><p><br />Please feel free to leave Jessie your comments here or <a href="http://am40sw08.blogspot.com/2008/04/student-voices-podcast-episode-1-jessie.html">on this post on her class blog</a>.<br /><p><br /><center><br /> <embed src="http://www.archive.org/download/DarrenKuropatwaStudentVoices_Episode1Jessie/SV01JessieAM40SW00_64kb.mp3" width="144" height="74" type="audio/mpeg" autostart="false" loop="false" bgcolor="#DCDE97"> </embed><br /><br><br />(<a href="http://www.archive.org/download/DarrenKuropatwaStudentVoices_Episode1Jessie/SV01JessieAM40SW00_64kb.mp3">Download File</a> 5.6Mb, 11 min. 40 sec.)<br /></center><br /><p><br /><div align="right">Photo Credit: <a href="http://flickr.com/photos/11739182@N03/1263985679/">Kids of conversation</a> by flickr user Kris Hoet</div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5847298158728969064?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-59945611073720084542008-04-18T13:40:00.001-05:002008-04-18T13:43:21.727-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_360428"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080418-1208542546856586-9"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080418-1208542546856586-9" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-18-2008?src=embed" title="View 'Calculus 45S Slides April 18, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5994561107372008454?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-75593894952798206992008-04-16T07:23:00.003-05:002008-04-16T07:30:40.427-05:00Something that I found that I thought was cool was when Mr. K showed us of something "brilliant" where we added terms - like adding zero.<br /> This unit I thought was funner than the first unit - Limits. When you get to cancel or reduce terms and create a whole new equation different from the beginning it creates a feeling of completion if it is the derivative of that equation. I feel that I need some more work towards this unit to fully understand it - overall I like this unit out of the two: Limits and Derivatives.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7559389495279820699?l=cal45sw08.blogspot.com' alt='' /></div>1http://www.blogger.com/profile/05403602652691839017noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-49947095868114363662008-04-15T23:46:00.002-05:002008-04-15T23:48:35.935-05:00Almost forgot to put up my BOB before the test. Whew.<br /><br />Anyway, I thought this unit was pretty fun and stuff. There are some things I still have to work on, which is knowing when to apply the product rule and stuff. Other than that I think I'm pretty much know what's going on. A good sleep and breakfast should do the trick, and a good review too.<br /><br />Night.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-4994709586811436366?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-24597041414083717952008-04-15T22:18:00.005-05:002008-04-15T22:57:32.108-05:00Well, as I sit here and watch the hockey game, I will take some time (during an intermission of course) to reflect upon our last unit:<br /><br /><span style="color: rgb(0, 102, 0);">DERIVATIVES<br /><br /><span style="color: rgb(0, 0, 0);">Now I have already learned the content of this unit, and applied it many a time in the AP Calculus class. This means that everything on the test should be fairly easy for myself, but then again, it is one of Mr. K.'s tests.<br />As I saw with the Pre-Test, small mistakes are easy to make. Forgetting to take the derivate of a constant, or the other side of an equation can result in a major headache.<br />I think a good sleep, a hearty breakfast, and a little extra concentration should do the trick.<br /><br />Good luck to all on the test.<br /><br /><span style="color: rgb(51, 51, 255);">Remember the derivative rules as well as the limit definition of a derivative!</span><br /></span></span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2459704141408371795?l=cal45sw08.blogspot.com' alt='' /></div>«Craig»http://www.blogger.com/profile/08860268378924842699noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-28396516583037407982008-04-15T22:03:00.006-05:002008-04-15T23:38:17.725-05:00Hey, MrSiwWy here for Monday's scribe! Now, all that we really did during class was write the pre-test, though Graeme, Mark and I all decided to rush through it and discuss it in our own group so that we could leave early to practice for the Literacy presentation in the afternoon. I know that it was my fault that we lost a few silly marks because of the fact that we rushed through everything on Monday, but since I really haphazardly rushed and finished way too early some of my solutions were flawed (and we ended up handing in my paper). Well, that's in the past, so on to the scribe. Now, usually a scribe for a pre-test class wouldn't really be that detailed, but I feel obligated to try and explain the solution process for each of the questions on the pre-test as well as I can. I'll accompany each with the corresponding slide, to show Mr. K's work (if that question was brought up in class) and to show the question itself.<br /><br /><span style="font-weight: bold;">1. What is </span><span style="font-style: italic; font-weight: bold;">d</span><span style="font-weight: bold;">c/</span><span name="intelliTxt" id="intelliTxt"><span style="font-style: italic;"></span><span style="font-style: italic; font-weight: bold;">dx</span></span><span style="font-weight: bold;">, where</span><span style="font-style: italic; font-weight: bold;"> c</span><span style="font-weight: bold;"> is a constant?</span><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_dvbjJeJb4LE/SAV-4k_9gYI/AAAAAAAAAQM/XD5wcR-Di1E/s1600-h/pic1.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_dvbjJeJb4LE/SAV-4k_9gYI/AAAAAAAAAQM/XD5wcR-Di1E/s400/pic1.PNG" alt="" id="BLOGGER_PHOTO_ID_5189693656084087170" border="0" /></a><br />Well, this is simply testing to see if you remember the constant rule for differentiation. Thus the answer is simply (b) which is 0. It might help if I reviewed with you the basic idea of each of the rules we have learned for differentiation, since the test is just about 12 hours away from now.<br /><span style="font-weight: bold; color: rgb(204, 0, 0);">Constant Rule:</span> The constant rule basically dictates that any function that is solely just a constant, such as f(x) = 4, or g(x) = 13, has a derivative that is 0, or f'(x) = 0 and g'(x) = 0.<br /><span style="font-weight: bold; color: rgb(204, 0, 0);">Coefficient Rule:</span> When differentiating a first-order function with a coefficient, such as f(x) = 2x, or g(x) = 215x, the derivative will simply be that coefficient. In the examples given, f(x) = 2x yields f'(x) = 2, while g(x) = 215x yields g'(x) = 215.<br /><span style="font-weight: bold; color: rgb(204, 0, 0);">Sum/Difference Rule:</span> When a function is composed of more than one term, then you can break that function up into it's constituent parts and differentiate each part separately. Examples include f(x) = 388x + 6, or g(x) = 56x² - 21x, where f'(x) = 388 + 0 = 388, or g'(x) = 102x - 21.<br /><span style="font-weight: bold; color: rgb(204, 0, 0);">Power Rule:</span> For any function where x is raised to an exponent n, or f(x) = xⁿ, the derivative of that function will be x raised to the exponent (n-1) all multiplied by n, or f'(x) = nx<span class="blsp-spelling-error" id="SPELLING_ERROR_5"></span>^n-1.<br /><span style="font-weight: bold; color: rgb(204, 0, 0);">Product Rule:</span> For any function that is composed of two functions being multiplied together, the derivative is <span style="font-weight: bold;">not</span> the derivative of their product, nor is it the product of their respective derivatives. To find the derivative of a product, the derivative will always be given by h'(x) = f(x)g'(x) + g(x)f'(x), assuming h(x) = f(x)g(x). It's basically the derivative of the first function multiplied by the other function, then add on the derivative of the second function multiplied by the first function.<br /><span style="font-weight: bold;"><span style="color: rgb(204, 0, 0);">Quotient Rule:</span> </span>To find the derivative of a quotient, all you have to do is recall the quotient rule song! Assuming that h(x) = f(x) / g(x) then h'(x) = [g'(x)f(x) - f'(x)g(x)]/g²(x). Here's the song in case you forgot: "High de low minus low dehigh, all over low low"<br /><span style="font-weight: bold; color: rgb(204, 0, 0);">Chain Rule:</span> This must be used whenever a function is actually buried deep inside another function. This can be given by h(x) = f(g(x)), such as h(x) = (2x² + 3x)³. First you must find what g(x) is, or what the inner function is, then it should be easy to determine what is the outer function. The derivative of any composition of functions (when a function has another function inside of it) is given by h'(x) = g'(x)f'(g(x)). In the given example, the derivative would be h'(x) = (4x + 3)[3(2x² + 3x)²], or (12x + 9)(2x² + 3x)² since g(x) = 2x² + 3x and f(x) = x³.<br /><br /><span style="font-weight: bold;">2. What is </span><span style="font-style: italic; font-weight: bold;">d</span><span style="font-weight: bold;">x</span><sup style="font-weight: bold;">n</sup><span style="font-weight: bold;">/</span><span name="intelliTxt" id="intelliTxt"><span style="font-style: italic;"></span><span style="font-style: italic; font-weight: bold;">dx</span></span><span style="font-weight: bold;">?</span><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_dvbjJeJb4LE/SAV-5E_9gZI/AAAAAAAAAQU/l5xOJj7_vGk/s1600-h/pic2.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_dvbjJeJb4LE/SAV-5E_9gZI/AAAAAAAAAQU/l5xOJj7_vGk/s400/pic2.PNG" alt="" id="BLOGGER_PHOTO_ID_5189693664674021778" border="0" /></a><br />Again, this is just testing the knowledge of a rudimentary rule given above.<br /><br /><span style="font-weight: bold;">3. Given f(x) = </span><span name="intelliTxt" id="intelliTxt"><span style="font-weight: bold;"> √(1-x</span><sup style="font-weight: bold;">2</sup><span style="font-weight: bold;">) find and simplify f'(x).</span></span><span name="intelliTxt" id="intelliTxt"><br /></span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_dvbjJeJb4LE/SAV-5E_9gaI/AAAAAAAAAQc/JqvwHdhdJxg/s1600-h/pic3.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_dvbjJeJb4LE/SAV-5E_9gaI/AAAAAAAAAQc/JqvwHdhdJxg/s400/pic3.PNG" alt="" id="BLOGGER_PHOTO_ID_5189693664674021794" border="0" /></a><br /><span name="intelliTxt" id="intelliTxt">Well, this is basically just applying the rules of differentiating I reviewed above in a very straight forward manner. You must be able to recognize that you must use the chain rule with the inner function g(x) = 1 - x², while the outer function f(x) = </span><span name="intelliTxt" id="intelliTxt"> √x. Then the derivative becomes f'(x) = -2x / 2</span><span name="intelliTxt" id="intelliTxt"> √(1-x²), which simplifies to </span><span name="intelliTxt" id="intelliTxt">f'(x) = -x / </span><span name="intelliTxt" id="intelliTxt">√(1-x²).</span><br /><span name="intelliTxt" id="intelliTxt"><br /><span style="font-weight: bold;">4. Find the equation of the line tangent to the graph of 8xy<span class="blsp-spelling-error" id="SPELLING_ERROR_9"></span></span><sup style="font-weight: bold;">2</sup><span style="font-weight: bold;"> = (x + y)</span><sup style="font-weight: bold;">4</sup><span style="font-weight: bold;"> at the point (1/2, 1/2).</span><br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_dvbjJeJb4LE/SAV-5U_9gbI/AAAAAAAAAQk/UIOdMeF1SyY/s1600-h/pic4.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://2.bp.blogspot.com/_dvbjJeJb4LE/SAV-5U_9gbI/AAAAAAAAAQk/UIOdMeF1SyY/s400/pic4.PNG" alt="" id="BLOGGER_PHOTO_ID_5189693668968989106" border="0" /></a><span name="intelliTxt" id="intelliTxt"><br /><span style="font-style: italic;"></span><span style="font-style: italic;"></span>This was a rather tricky question, but I remember first determining how I forgot to take the derivative of x + y as 1 + <span class="blsp-spelling-error" id="SPELLING_ERROR_10"></span></span><span name="intelliTxt" id="intelliTxt"><span style="font-style: italic;">dy</span>/<span style="font-style: italic;">dx</span></span><span name="intelliTxt" id="intelliTxt"><span class="blsp-spelling-error" id="SPELLING_ERROR_10"></span><span class="blsp-spelling-error" id="SPELLING_ERROR_11"></span>, instead of just<span class="blsp-spelling-error" id="SPELLING_ERROR_12"></span> </span><span name="intelliTxt" id="intelliTxt"><span style="font-style: italic;">dy</span>/<span style="font-style: italic;">dx</span></span><span name="intelliTxt" id="intelliTxt">, and I also forgot to distribute the 8 through the left side and didn't read the question and gave him only a slope for an answer. I really need to take my time answering these questions =/. Well anyways...</span><br /><br /><span name="intelliTxt" id="intelliTxt">The solution is shown above, but the main idea behind solving this question is just realizing that you must differentiate the equation implicitly, since y might represent any one of a variety of possible functions. If you can see that you must differentiate implicitly, it's pretty just grunt work applying the above rules and keeping in mind the fact that any time you differentiate the variable y, you must use the chain rule since it has some unknown buried inside of it, meaning that that term must be multiplied by y'. In the work you can see this being done. Say a function x + y = 4 was to be differentiated, then afterwards it would transform into 1 + 1<span style="font-weight: bold;">y'</span> = 0, or y' = -1.<br /></span><br /><span name="intelliTxt" id="intelliTxt">Don't forget, as I did, that the question is asking for the equation to the tangent line at (1/2, 1/2), not just the slope of that tangent line. So, once you can solve for y' as shown above through implicit differentiation, you can use the point given (1/2, 1/2) to find the equation of the tangent line, thus completing the question.<br /><br /><span style="font-weight: bold;">5. (a) If f(x) = </span></span><span name="intelliTxt" id="intelliTxt"><span style="font-weight: bold;">√(x</span><sup style="font-weight: bold;">2</sup><span style="font-weight: bold;"> + 9), use the limit definition of the derivative to find f'(0). You must show all work and use the limit definition properly to receive any credit.</span><br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_dvbjJeJb4LE/SAV-5k_9gcI/AAAAAAAAAQs/dsD19wCCLTU/s1600-h/pic5.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_dvbjJeJb4LE/SAV-5k_9gcI/AAAAAAAAAQs/dsD19wCCLTU/s400/pic5.PNG" alt="" id="BLOGGER_PHOTO_ID_5189693673263956418" border="0" /></a><br /><span name="intelliTxt" id="intelliTxt">The work for this question was once again fully shown by Mr. K, but the key to this question is realizing that it's asking for the derivative at a point, and not the derivative function of f(x). Using this knowledge, we can apply the limit definition at f(0) instead of f(x), and f(0+h) instead of at f(x+h). What Mr. K did with his work was that he determined what f(0) was by simply plugging 0 into x inside f(x), and then determined what f(0+h) would be by plugging in (0 + h) into f(x). He did this so that he could use the values he arrived at in these initial steps, and simply apply them to the limit definition as shown near the bottom of the slide.<br /><br /><span style="font-weight: bold;">5 (b) If g(x) = x2 - 3x + 4, use the definition of the derivative to find an equation or formula for the derivative of x. Again no credit will be awarded unless you demonstrate competent use of the limit definition and show all your work.<br /><br /></span></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_dvbjJeJb4LE/SAV_yk_9gdI/AAAAAAAAAQ0/bZ69vXYiCew/s1600-h/pic6.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_dvbjJeJb4LE/SAV_yk_9gdI/AAAAAAAAAQ0/bZ69vXYiCew/s400/pic6.PNG" alt="" id="BLOGGER_PHOTO_ID_5189694652516499922" border="0" /></a><br /><span name="intelliTxt" id="intelliTxt">Once again you can see the solution to this question in the slide above. What you had to realize to solve this question is that you have to know what g(x+h) will be when expanded, as Mr. K has shown at the top portion of the slide. Once this is known, you must plug it into the fundamental limit definition of the derivative and things should start to cancel quickly and elegantly. In the final steps of the solution, once you plug in 0 into h, then the lim part goes away and any term with an h in it leaves along with it. Thus leaving 2x - 3, which gives the derivative of g(x), or rather gives g'(x).<br /><br />Well that's it for my scribe post, I think that this might be one of my shortest scribe posts since way back in grade 12 pre-cal ^^. Too bad I wasn't present for the entire class, but I guess I tried to cover everything to the best of my ability since I was only present for maybe 20 minutes. Don't forgot to study tonight folks, since the test is tomorrow. I wish everyone the best of luck!<br />Oh yes, almost forgot, the next scribe will be: <span style="color: rgb(0, 102, 0);"><span style="font-weight: bold;">John D.</span></span><br /></span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2839651658303740798?l=cal45sw08.blogspot.com' alt='' /></div>MrSiwWyhttp://www.blogger.com/profile/18072494079396275311noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-6286317165982338152008-04-15T21:56:00.002-05:002008-04-15T22:03:35.317-05:00MrSiwWy here for my second bob in intro to calculus. Oh wow, I just got home and I still have to scribe tonight =/ I don't expect to be faced with a terribly difficult test tomorrow, since we've already covered this unit in great depth in our AP calc class. Though, I must make sure to take my time this time and not to rush during any portion of the test, as I did on the pre-test (which cost me 2 rather easy marks). I'll probably explain that in more detail in my scribe post later tonight, but with that said, I don't think that there's anything I'm too worried about now. I usually make a longer bob than this, but I'll probably just integrate what I would usually say in my scribe post. Good luck on the test everyone ! Don't forget to practice those derivations!<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-628631716598233815?l=cal45sw08.blogspot.com' alt='' /></div>MrSiwWyhttp://www.blogger.com/profile/18072494079396275311noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-8454547909002984572008-04-15T21:13:00.003-05:002008-04-15T21:21:00.634-05:00Well, this unit about derivatives isn't too difficult considering that i learned this before. The only thing that i need to be careful on is working too fast for my own good. I usually make most of my mistakes on being too overconfident, but just to make sure I'll go over the orange and blue book tonight. Hopefully, i manage to get a good mark. Good luck everyone and study hard.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-845454790900298457?l=cal45sw08.blogspot.com' alt='' /></div>m@rkhttp://www.blogger.com/profile/00509544958003356512noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-83938316674006855532008-04-15T18:36:00.003-05:002008-04-15T19:17:08.104-05:00Tomorrow is the test! Wow it's quite surprising how so many people have dropped this course throughout the minor portion of what has passed. Well, tomorrow is the test on Derivatives! It shouldn't be too difficult to be honest, especially for the AP Calc Students who have already learned this unit. Though I understand the most part of this unit, I know that I may still stumble somewhere, as questions are always different: there are always new wrinkles that you can run into.<br /><br /><span style="font-weight: bold;">Summary of Differentiation:</span><br /><ul><li style="color: rgb(0, 0, 153);">Power Rule</li><li style="color: rgb(0, 0, 153);">Product Rule</li><ul><li><span style="color: rgb(0, 102, 0);">F'(x) = f'(x) g(x) + f(x) g'(x)</span><br /></li></ul><li style="color: rgb(0, 0, 153);">Quotient Rule</li><ul><li>Low D High, minus High D Low, All over Low Low (The SONG!)</li><li style="color: rgb(0, 102, 0);">F'(x) = [ g(x) f'(x) - f(x) g'(x) ] / [ g(x) ]^2</li></ul><li style="color: rgb(0, 0, 153);">Chain Rule</li><ul style="color: rgb(0, 102, 0);"><li>F'(x) = f'( g(x) ) g'(x)</li></ul></ul>Including these rules, remember to understand the <span style="color: rgb(0, 0, 153);">Definition of a Derivative</span> and how to find the derivative using this definition.<br /><br />Good luck everyone!<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8393831667400685553?l=cal45sw08.blogspot.com' alt='' /></div>Tim_MATH_yhttp://www.blogger.com/profile/09929458936647070771noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-68282774485232296232008-04-14T22:35:00.004-05:002008-04-14T23:09:12.982-05:00Well I dropped the ball here, haven't really been paying too much attention to this particular blog of late. So by Murphy's law I was picked as scribe. Alright this isn't my best work I can tell you now, but this is how those two classes I have to scribe for go:<br /><br />Day 1:<br /><br />The point of this class was to come up with the equations of tangent lines and normal lines.<br />-Tangent lines are lines touch that touch the graph only once on a certain interval (the graph may wiggle and the tangent line can cross the graph at another point that does not matter)<br />-Normal lines are lines that are perpendicular to a tangent line.<br />-Recall that the first derivative of any funtion gives the slope of the graph at any point<br />-To find the equation of a tangent line at a point on a graph you need a couple things<br />-You need a point and you need slope<br />-You may use slope-intercept formula if you are given the y intercept y=mx+b<br />-But you'll probably be using point-slope formula (y-y1) = m(x - x1)<br />-So the general method to finding the formula of a tangent line to a point goes something like this:<br /> -Find the derivative of your function<br /> -Evaluate at your point<br /> -Plug it into point-slope formula<br />and presto! There's your answer!<br /><br />To find the equation of a normal line (a line that is perpendicular to a tangent line) there is only one more step. When you find the slope you just need to take the negative reciprocal of that and then plug it into our formula.<br /><br />The last question we had that class asks us to find when the tangent line is horizontal. Well that is when the derivative equals zero. So you come up with the formula of the derivative then solve for the zeroes. Then you plug those x-co's back into the original function to find the points where the tangent line equals zero.<br /><br />Day 2:<br /><br />This class is a little more complicated.<br />-Mr.K started off with a talk about questions that the answers to were that just because we don't know for sure doesn't mean that something is or isn't there or happening.<br />-The shadow of the balcony, we would infer that there is a balcony casting that shadow, doesn't have to be, could be something that looks like a balcony.<br />-Then is that roof insulated? Well there is melted snow in places so we could assume no.<br />-Then do those people agree with one another? Through the story of Mr.K since their body language is all somewhat the same they probably are, but we don't know for sure.<br /><br />-We then found the derivative of a semicircle.<br />-Mr.K showed us that we could define this function in other ways to get an infinite amount of other circle bits-and-pieces functions (see the slides)<br />-This was to show us that there could be many functions buried within another and that even though we don't know which one there may be we have to treat it as if there is another function within it and therefor when we differentiate we have to use the chain rule.<br />-One thing to notice about this is that the derivative might be in terms of y and x so you may need a set of coordinates to solve for the slope at a certain point instead of just an x-co.<br /><br /><br />Bob<br /><br />I feel fairly confident for the upcoming test. No worries at all really. Schedule should die down a little so I wont' miss any more scribes or bobs!<br /><br />G'night! (might edit this sometime to add some examples but I am *really* tired right now so sleep is a must... after bio...)<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6828277448523229623?l=cal45sw08.blogspot.com' alt='' /></div>Grey-Mhttp://www.blogger.com/profile/10237241798100727258noreply@blogger.com1tag:blogger.com,1999:blog-3217610859011491687.post-85157266359865541022008-04-14T10:45:00.003-05:002008-04-15T13:01:25.331-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_354825"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080414-1208282353867848-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080414-1208282353867848-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-14-2008-354825?src=embed" title="View 'Calculus 45S Slides April 14, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8515726635986554102?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com2tag:blogger.com,1999:blog-3217610859011491687.post-62056178009557855282008-04-10T15:54:00.003-05:002008-04-10T16:40:19.236-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_346713"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080410-1207862829092684-9"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080410-1207862829092684-9" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-10-2008?src=embed" title="View 'Calculus 45S Slides April 10, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6205617800955785528?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-22449701660832026412008-04-08T15:43:00.003-05:002008-04-08T15:49:48.158-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_342753"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080408-1207687175294930-8"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=cal45sw0820080408-1207687175294930-8" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-april-8-2008?src=embed" title="View 'Calculus 45S Slides April 8, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2244970166083202641?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-68873486713219916762008-04-07T22:34:00.002-05:002008-04-07T23:14:19.199-05:00Hey everyone! I apologize for the super super late scribe post. I know you are all waiting for it so here it is.<br /><br />March 19.<br />The class started off with a discussion about Article 13 and the students' comments on their class blog. Moving on from that, we reviewed the sum/difference rule and then moved on to the new topic. The class topic was mostly about the product rule. All the information about the class work and the rule and how to apply it is in the slides below. Also we learned about the quotient rule but we ran out of time so it was continued in the other class.<br /><br />March 28.<br />This was our last class before the spring break. During this class we reviewed the quotient rule and Mr K. taught the class the quotient rule song. The song goes like this: "<span style="font-style: italic;">high de low minus low de high all over low low</span>" which translate to: <span style="font-weight: bold;">[f(x)^1 * g(x) - g(x)^1 * f(x)]/g(x)^2.</span><br />But this class' real topic was... an introduction to the chain rule! The chain rule basically states that the whole function is called F(x). This F(x) is composed but different functions which are f(x) and g(x) which g(x) is the inner function and f(x) is the outer function. If examples are need it is in the slides below.<br /><br />Thank you for your time and I do apologize for the super super late blog post. Now that I finished this scribe post the next scribe shall be... <span style="font-weight: bold;">GREY-M</span>.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6887348671321991676?l=cal45sw08.blogspot.com' alt='' /></div>rusz.lnoreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-85084018885915033632008-03-28T14:24:00.003-05:002008-03-29T08:09:51.939-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_325963"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-28-2008-1206732208943039-5"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-28-2008-1206732208943039-5" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-march-28-2008?src=embed" title="View 'Calculus 45S Slides March 28, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8508401888591503363?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-12960052060122075912008-03-19T14:21:00.001-05:002008-03-19T14:23:29.517-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_313922"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-19-2008-1205947308313591-5"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-19-2008-1205947308313591-5" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-march-19-2008?src=embed" title="View 'Calculus 45S Slides March 19, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1296005206012207591?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-25087140024880869982008-03-17T23:04:00.004-05:002008-03-17T23:15:47.990-05:00If you'd like to leave a comment to Al Upton and the minilegends in Australia click on the picture below to get to their blog.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://alupton.edublogs.org/"><img style="display:block; margin:0px auto 10px; text-align:center;cursor:pointer; cursor:hand;" src="http://3.bp.blogspot.com/_tlF3qs3v9cw/R99AHlbTURI/AAAAAAAAAVQ/2zJ2q7-iB38/s320/minilegends.jpg" border="0" alt=""id="BLOGGER_PHOTO_ID_5178928595549245714" /></a><br /><br />You can also read more about Article 13 and the Convention on the Rights of the Child <a href="http://www.unicef.org/knowyourrights/convention_cartoons.php">here</a>. This movie illustrates what it's all about:<br /><br /><center><object width="425" height="355"><param name="movie" value="http://www.youtube.com/v/xKBu2B8xZYE&hl=en"></param><param name="wmode" value="transparent"></param><embed src="http://www.youtube.com/v/xKBu2B8xZYE&hl=en" type="application/x-shockwave-flash" wmode="transparent" width="425" height="355"></embed></object></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2508714002488086998?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-11721000899342908362008-03-17T20:46:00.002-05:002008-03-17T21:09:56.342-05:00Today's class went similar to the class last Friday. Mr. K assigned people into different groups of three or four students, each group having at least one AP Calculus student. He wants the non-AP students to learn from their peers and the AP students to apply what they've learned before. The first proof that was done was the derivative of the power function, but before that Mr. K refreshed our memory about the binomial theorem. Next we worked on the proof of the derivative of a constant times a function. Then, we applied what we know about limits to prove the derivative of sum and difference of functions. Thats about everything that happened in class today. The next scribe will be <span style="color: rgb(255, 0, 0);">rusz.l</span>.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1172100089934290836?l=cal45sw08.blogspot.com' alt='' /></div>m@rkhttp://www.blogger.com/profile/00509544958003356512noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-21134485143024638122008-03-17T16:14:00.001-05:002008-03-17T16:18:17.941-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_310893"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-17-2008-1205787981544594-5"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-17-2008-1205787981544594-5" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-march-17-2008?src=embed" title="View 'Calculus 45S Slides March 17, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2113448514302463812?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-4774295625201001512008-03-16T17:02:00.002-05:002008-03-16T17:26:53.918-05:00Sorry for the late scribe everyone, but this week has been very hectic for me. Well Tuesday's class began very differently than usual. The class began with Mr. K splitting everyone into groups of two or three students, with each group having at least one AP calculus student. Mr. K wanted the AP calc students to be the teachers and the intro calc students to be attentive and learn what the AP calc students were teaching them. The first proof that was to be solved was (F(x+h) - F(x)) /h where (mx+b) was plugged in for (x). The derivative of mx+b is m, as shown in the slides from Class below this blog post. The second proof was plugging in a constant (K) in for x. Where the derivative of the constant is always zero (0). The derivative of K is zero (0). That was the class for the shown proofs scroll down below this post to see the proof slides. <span style="color: rgb(255, 0, 0);">The next scribe is M@rk.</span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-477429562520100151?l=cal45sw08.blogspot.com' alt='' /></div>Dinohttp://www.blogger.com/profile/16720903626604404687noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-53332344829971435412008-03-14T08:58:00.001-05:002008-03-14T09:01:40.232-05:00<center>Happy <font size="+2">&pi;</font> Day Everyone!!<br /><br /><a href="http://flickr.com/search/show/?q=pi+pie">Have some <font size="+2">&pi;</font> Pie!</a></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5333234482997143541?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-26339358246056430592008-03-13T08:13:00.001-05:002008-03-13T08:13:58.816-05:00<div style="text-align: center;"><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_HiqPzpd3O9Y/R9knopn8vDI/AAAAAAAAAG8/F5_2_zP7PpA/s1600-h/pi+day.gif"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://2.bp.blogspot.com/_HiqPzpd3O9Y/R9knopn8vDI/AAAAAAAAAG8/F5_2_zP7PpA/s200/pi+day.gif" alt="" id="BLOGGER_PHOTO_ID_5177212825960758322" border="0" /></a><a href="http://www.123greetings.com/view/MJ30313060109599">Click here to view your e card! </a><br /></div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2633935824605643059?l=cal45sw08.blogspot.com' alt='' /></div>Lani Ritter Hallhttp://www.blogger.com/profile/16352862711544966770noreply@blogger.com1tag:blogger.com,1999:blog-3217610859011491687.post-75838734714261953622008-03-11T12:28:00.002-05:002008-03-11T12:31:05.091-05:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_302056"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-11-2008-120525632879862-3"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-11-2008-120525632879862-3" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-march-11-2008?src=embed" title="View 'Calculus 45S Slides March 11, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7583873471426195362?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-6278843844611790222008-03-10T21:37:00.004-05:002008-03-11T00:19:39.915-05:00Sorry for the lateness of this scribe post, I was planning on doing this over the weekend but laziness took over me, haha.. and I fell asleep when I got home today and woke up around 8:30. Anyways...<br /><br /><b><u>Derivatives</u></b><br /><br />The class started off by watching Mr. K find a picture in flickr. He was using a nifty program called PicLens. It's pretty cool in my opinion and I downloaded the extension for my Firefox browser. About five minutes later, he found a picture of three guys at the beach. He then copied the picture, with the link of course, to the SmartBoard and appropriately named the picture "Tan Gents". Get it? "Tan Gents?" Haha, the picture is at the slides so you can go check it out there.<br /><br />Mr. K then introduced us again to the concept of tangent and secant lines. A tangent line is a line that touches a curve at exactly one point. A secant line is a line that touches a curve at two or more points. Then we were introduced to a familiar concept; the slope of a line, as seen on slide 5/9.<br /><br />Zooming in on the graph on slide 6/9 or following the link, we can see that the slope of the function f(x)=5^x is becoming relatively smaller and equal to the slope of the tangent line f(x)=x+1 as the scale of the graph becomes smaller.<br /><br />Moving on to the next slide, we were asked to solve for the slope of the tangent line @ x=1 on f(x)=x². It was pretty straightforward and the procedure is shown at the slide.<br /><br />However, on the next slide, we were introduced to the definition of a derivative, as shown. We were then introduced to the basic-thingy of derivatives, which is 2x, shown in the final slide.<br /><br />For homework we were assigned pages 1-31 and page 15 on the blue book.<br /><br />Next scribe is Dino.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-627884384461179022?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-32111104263139862802008-03-07T14:18:00.003-06:002008-03-07T14:22:39.746-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_297191"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-7-2008-1204921044478341-4"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-7-2008-1204921044478341-4" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-march-7-2008?src=embed" title="View 'Calculus 45S Slides March 7, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3211110426313986280?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-54727344217718493212008-03-04T23:04:00.002-06:002008-03-05T02:59:47.252-06:00Bobbing for Limits. This was a short unit. When we started to talk about the graphs, I was kinda lost at first, but after a while I got it and it was okay.<br />Good luck on the test! o(∩_∩)o<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5472734421771849321?l=cal45sw08.blogspot.com' alt='' /></div>haiyanhttp://www.blogger.com/profile/04253570568706636473noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-15087647492579476182008-03-04T22:57:00.003-06:002008-03-04T23:07:31.588-06:00Its that time again... BOB time! Reflecting on the unit limits, at first i thought that concept of it was going to be extremely hard because its calculus and the state of mind that I'll have to learn new concepts. As the unit progress i started to understand more and more the concept became a bit more easier. Also I would like to thank my fellow classmate who helped me out to understand the unit more. Going in to the test i hope i do alright. Good luck to everyone tomorrow.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1508764749257947618?l=cal45sw08.blogspot.com' alt='' /></div>rusz.lnoreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-33605130933707815362008-03-04T22:21:00.003-06:002008-03-04T22:24:54.100-06:00I just realized that I haven't blogged in a long time! Anyway, I thought that this unit, Limits, was pretty simple and quite easy to understand, that is, if you do your homework. I admit I forget to do it sometimes because we only have this class every other day. I get confused on interpreting graphs sometimes but after going over everything today, everything became clear and I understand it.<br /><br />Anyway, good luck on the test tomorrow guys!<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3360513093370781536?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com1tag:blogger.com,1999:blog-3217610859011491687.post-60866545281066812932008-03-04T22:08:00.000-06:002008-03-04T22:14:37.073-06:00Hey It's MrSiwWy here my BOB vicariously through Grey-M since I haven't been added to the blog yet. Well, the first unit in this class didn't really present anything new to the calculus students in class or anything terribly difficult to the rest of the class. Though, I feel as if I'm bound to make tonnes of incredibly silly mistakes throughout the test tomorrow. If so, I'll probably spend a little more time with some practice problems or concept reviews during each night so everything doesn't seem so foreign anymore. Hopefully we all do well, though, it's not that difficult of a unit. I wish everyone good luck on the test tomorrow and a great night."<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6086654528106681293?l=cal45sw08.blogspot.com' alt='' /></div>Grey-Mhttp://www.blogger.com/profile/10237241798100727258noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-47227851441189435052008-03-04T22:05:00.003-06:002008-03-04T22:12:22.028-06:00<div align="justify"><strong><span style="font-family:arial;color:#000000;">BOB-ing again?! WOW! Anyway, this unit took less time than I thought! It seems pretty easy so far, at least as far as I know. Who knows what I might encounter tomorrow during the test! Limits... Sounds like everything is easy because there IS a limit. Though, it gets tricky especially that part where we have to EVALUATE, and the part where we have to present evidence of continuity. Condition I. II.. III... It's interesting. We're exploring a different area in functions. I hope everyone does well tomorrow! Good luck to me... AND YOU! Study HARD. I should study now too! Goodnight everyone!</span></strong></div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-4722785144118943505?l=cal45sw08.blogspot.com' alt='' /></div>vincentrhttp://www.blogger.com/profile/17670701436682827532noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-67998668421519615212008-03-04T21:19:00.004-06:002008-03-04T21:40:12.117-06:00<div style="text-align: justify;"><span style="font-family:arial;">What can I say about this unit? We'll it was the first unit for Intro to Calculus and it was really boring and like bland. It was hard not to get distracted by other things. Limits isn't hard it was just writing everything out that I would forget to. Sometimes I wouldn't see the dots and thats where I lose marks. Other than the boring mechanical stuff limits was okay. That was a fast unit. We'll I guess I'm happy limits is over with.<br /><br /><br /></span></div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6799866842151961521?l=cal45sw08.blogspot.com' alt='' /></div>SAMUShttp://www.blogger.com/profile/00765502734004344768noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-20120021614692924742008-03-04T21:16:00.002-06:002008-03-04T21:24:37.953-06:00Hey it's me GreyM! Well gotta say for being in the other calc class I'm not terribly comfortable going into this test. I'm going to chalk it up to lack of doing any hw for this class as it was pretty much last priority for me every night. We'll see if I can move it up the priority list after this test.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2012002161469292474?l=cal45sw08.blogspot.com' alt='' /></div>Grey-Mhttp://www.blogger.com/profile/10237241798100727258noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-67429243107263461702008-03-04T19:57:00.002-06:002008-03-04T20:00:31.495-06:00Test time! Or tomorrow rather. Yep, it's that time again. This unit, on limits was a breeze. So far the unit hasn't gotten in depth yet so I don't believe that I have any problems, considering this is my second time around learning limits. I'm glad that I took this course cause despite the fact that much was a review, the second round trip opened my eyes to things that I couldn't catch during the first round. Best of luck to everyone else on the test tomorrow! Study? =)<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6742924310726346170?l=cal45sw08.blogspot.com' alt='' /></div>Tim_MATH_yhttp://www.blogger.com/profile/09929458936647070771noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-35541231116730537462008-03-04T16:19:00.004-06:002008-03-04T16:36:33.794-06:00<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://tbn0.google.com/images?q=tbn:Sg76H7oCkwKYEM:http://www-uxsup.csx.cam.ac.uk/%7Efanf2/hermes/doc/talks/2006-01-techlinks/speed_limit.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer; width: 109px; height: 87px;" src="http://tbn0.google.com/images?q=tbn:Sg76H7oCkwKYEM:http://www-uxsup.csx.cam.ac.uk/%7Efanf2/hermes/doc/talks/2006-01-techlinks/speed_limit.jpg" alt="" border="0" /></a><br /><br />This is 1's first Blogging on Blogging post (B.o.B. or B^2).<br />This unit was for "limits".<br /><br /><br /> I would like to comment how cool limits are. How cool is it like Mr. K said to ride on a roller coaster and all of a sudden get to a certain point and DROP straight down... This analogy was pretty cool as it related to limits. When you approach the limit of a function, the line goes into thin air! Amazingly, after this point on the graph it continued on like nothing happened at all. It is amazing in the fact that it just disappears as there is nothing in the spot...<br /><br />I think what might pose problems for me is the interpreting the graphs right but other than that I will continue practice some more questions in preparation for the test.<br /><br />I hope to gain some knowledge from this calculus class so that when I take it for real, I will be better prepared.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3554123111673053746?l=cal45sw08.blogspot.com' alt='' /></div>1http://www.blogger.com/profile/05403602652691839017noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-16955868592240771432008-03-04T00:07:00.004-06:002008-03-04T00:13:43.530-06:00Hello everyone, this is Craig here to do my first Blogging On Blogging (B.O.B.) post<br />for this Calculus 45 S course.<br /><br />This first unit has been a brief review for me because of my taking of the AP Calculus course, but it has been a nice refresher.<br /><br />It is actually still kind of tricky because I have to learn how to use the long, tedious, drawn-out methods of solving the problems I usually do very briefly. I guess it enhances my understanding of the concepts because we actually touch on the basics compared to the fast-paced AP Calculus' version of Limits.<br /><br />The pre-test was a clear example of how I need to use all of the techniques used in this class to get full marks. It is good practice and I look forward to more helpful hints from the class.<br /><br />Good luck to all on the test on Wednesday =D<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1695586859224077143?l=cal45sw08.blogspot.com' alt='' /></div>«Craig»http://www.blogger.com/profile/08860268378924842699noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-87779361039478350342008-03-03T22:32:00.002-06:002008-03-03T22:47:24.869-06:00Well, this unit about limits went by really fast. I didn't even feel that we already finished a unit. This unit has been really easy for me ( thanks to AP Calculus). This unit has been sort of a review of the section that i learned not too long ago from calculus but is just discussed in a greater depth. The only thing that will pose problems for me in the test is evaluating using the limit theorems. I particularly don't like writing the limit notation in every step of the process, other than that i think that i should be fine. Good luck everyone on the test and study hard!!<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8777936103947835034?l=cal45sw08.blogspot.com' alt='' /></div>m@rkhttp://www.blogger.com/profile/00509544958003356512noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-29563382238920631282008-03-03T20:38:00.010-06:002008-03-03T22:04:40.874-06:00<a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://flickr.com/photos/cromo/74284450/"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://4.bp.blogspot.com/_HLm_KkQlC58/R8y2aVb0V5I/AAAAAAAAAHM/uJUSFQZjtXE/s320/limits2.JPG" alt="" id="BLOGGER_PHOTO_ID_5173710635488532370" border="0" /></a><br />Hello! On our blog, I am known as <span style="color: rgb(204, 102, 0);">Tim-math-y</span>, and I will be your scribe for today's lessons.<br /><br /><span style="font-weight: bold;">Introduction:</span><br /><br />We started off the class with a brief discussion on our <span style="color: rgb(204, 102, 0);">del.icio.us</span> accounts and homework. We are to find, with effort, atleast one site that we can learn from and that can be leveled as a quality find. Then we 'tag' it with: <span style="color: rgb(204, 102, 0);">cal45sw08</span>, so that it will be added to our blog's bucket. Remember that it may not only aid in developing our learning outside the classroom but also, may prove to be great resources for others reading our blog.<br /><br />Sweeping that discussion aside, we started off our <span style="color: rgb(204, 102, 0);">pre-test</span> on the unit of <span style="color: rgb(204, 102, 0);">limits</span>! The pre-test consisted of 5 questions in total. For those who do not know the procedures of a pre-test, it is an effective practice worth marks where a short test is written. After a set test-writing duration, we are placed into even groups where we share our answers to compile the best solutions onto one paper, as a team. This individual test paper is handed in before Mr. K reveals and explains the correct solutions.<br /><br /><span style="font-weight: bold;">The Pre-test:</span><br /><br />As mentioned earlier, this Pre-test consisted of 5 questions: 2 multiple choice questions, 2 short answer questions (where work was required to show), and 1 long answer question.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_HLm_KkQlC58/R8y7A1b0V6I/AAAAAAAAAHU/Y_6cOsxxwF0/s1600-h/LimS1.JPG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://2.bp.blogspot.com/_HLm_KkQlC58/R8y7A1b0V6I/AAAAAAAAAHU/Y_6cOsxxwF0/s320/LimS1.JPG" alt="" id="BLOGGER_PHOTO_ID_5173715694960007074" border="0" /></a><br />The first question included an <span style="color: rgb(0, 0, 153);">error</span> that stumped everyone. The <span style="color: rgb(0, 0, 153);">x^4</span> in the numerator was supposed to be <span style="color: rgb(0, 0, 153);">x^2</span>. Because of this unintentional error, this question was <span style="color: rgb(0, 0, 153);">ommitted</span>, as far as marks go.<br /><br />However, this question could still be solved by exploring the function. This is shown on the slide. First, we notice that there is a <span style="color: rgb(0, 0, 153);">vertical asymptote at x = 4</span> (Remember that when a question is asking for a limit, it is essentially looking for a horizontal asymptote).<br /><br />By creating a <span style="color: rgb(0, 0, 153);">number line</span>, one will find that as 'x' approaches 4 from the <span style="color: rgb(0, 0, 153);">negative side</span>, the function goes to <span style="color: rgb(204, 102, 0);">positive infinity</span>. One would also find that as 'x' approaches 4 from the <span style="color: rgb(0, 0, 153);">positive side</span>, the function goes to <span style="color: rgb(204, 102, 0);">negative infinity</span>. Because of this occurrence, <span style="color: rgb(0, 0, 153);">the limit, as 'x' approaches 4 from the positive and negative side DOES NOT EXIST.<br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_HLm_KkQlC58/R8y9xlb0V7I/AAAAAAAAAHc/GZa7O2-pz_U/s1600-h/LimS2.JPG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_HLm_KkQlC58/R8y9xlb0V7I/AAAAAAAAAHc/GZa7O2-pz_U/s320/LimS2.JPG" alt="" id="BLOGGER_PHOTO_ID_5173718731501885362" border="0" /></a>This question is simply a give-away, as many may describe it. <span style="color: rgb(204, 102, 0);">As 'x' approaches the value of 1 from the positive and negative side, the value is 1.<br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_HLm_KkQlC58/R8zAf1b0V8I/AAAAAAAAAHk/FRf_9HMyAWI/s1600-h/LimS3.JPG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://2.bp.blogspot.com/_HLm_KkQlC58/R8zAf1b0V8I/AAAAAAAAAHk/FRf_9HMyAWI/s320/LimS3.JPG" alt="" id="BLOGGER_PHOTO_ID_5173721725094090690" border="0" /></a>A number of groups faltered on this question simply because of the nature of previous 'short answer' questions. In the past, short answer questions were marked based on the final answer only, with a chance to earn partial marks for work shown. However, this question stated: Evaluate using the <span style="color: rgb(204, 102, 0);">Limit Theorems</span>, upon which many did not. This question is extremely simple yet painful. As long as you know your limit theorems, you should be fine. Listed are the <span style="color: rgb(0, 102, 0);">limit theorems</span> from <span style="color: rgb(0, 102, 0);"><a href="http://archives.math.utk.edu/visual.calculus/1/limits.18/index.html">[visual calculus]</a>:<br /><br /></span><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_HLm_KkQlC58/R8zB5Vb0V9I/AAAAAAAAAHs/0BUe_YCsP0A/s1600-h/LT1.JPG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://4.bp.blogspot.com/_HLm_KkQlC58/R8zB5Vb0V9I/AAAAAAAAAHs/0BUe_YCsP0A/s320/LT1.JPG" alt="" id="BLOGGER_PHOTO_ID_5173723262692382674" border="0" /></a><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_HLm_KkQlC58/R8zChFb0V-I/AAAAAAAAAH0/n7E5sVEQsoQ/s1600-h/LT2.JPG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_HLm_KkQlC58/R8zChFb0V-I/AAAAAAAAAH0/n7E5sVEQsoQ/s320/LT2.JPG" alt="" id="BLOGGER_PHOTO_ID_5173723945592182754" border="0" /></a><br />These are the main limit theorems we are required to know.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_HLm_KkQlC58/R8zDDFb0V_I/AAAAAAAAAH8/p7pq2n8tklw/s1600-h/LimS4.JPG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_HLm_KkQlC58/R8zDDFb0V_I/AAAAAAAAAH8/p7pq2n8tklw/s320/LimS4.JPG" alt="" id="BLOGGER_PHOTO_ID_5173724529707735026" border="0" /></a>To start off this question, we chose to solve for the <span style="color: rgb(204, 102, 0);">horizontal asymptote</span> first. By <span style="color: rgb(0, 0, 153);">dividing each term by the variable with the highest degree</span>, we found that the <span style="color: rgb(204, 102, 0);">horizontal asymptote y=0</span>, when the value of <span style="color: rgb(0, 0, 153);">'x' approaches infinity</span> (any number divided by infinity is extremely close to zero, therefore in this method, terms are reduced substantially).<br /><br />Next we solved for the <span style="color: rgb(204, 102, 0);">vertical asymptotes</span>. This is done by <span style="color: rgb(0, 0, 153);">factoring the denominator</span> and <span style="color: rgb(0, 0, 153);">solving for restrictions</span> (the denominator can not equal to zero). We found the vertical asymptotes to be @ <span style="color: rgb(204, 102, 0);">x=-9, 0</span>.<br /><br />Finally, to help visualize the <span style="color: rgb(0, 0, 153);">graph and sketch</span> it, a simple method of finding out the positions around the asymptotes is by creating a <span style="color: rgb(204, 102, 0);">number line</span>:<br /><ul><li>As 'x' approaches -9 from the negative side, the limit is +infinity<br /></li><li>As 'x' approaches -9 from the positive side, the limit is -infinity</li><li>As 'x' approaches 0 from the negative side, the limit is -infinity</li><li>As 'x' approaches 0 from the positive side, the limit is +infinity</li></ul>Thus, the graph can be sketched.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_HLm_KkQlC58/R8zFdlb0WBI/AAAAAAAAAIM/DSILNVZf44M/s1600-h/LimS5.JPG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_HLm_KkQlC58/R8zFdlb0WBI/AAAAAAAAAIM/DSILNVZf44M/s320/LimS5.JPG" alt="" id="BLOGGER_PHOTO_ID_5173727183997523986" border="0" /></a>In the final question, we started off by running the <span style="color: rgb(204, 102, 0);">piece-wise</span> function through the <span style="color: rgb(204, 102, 0);">three steps of continuity testing</span>.<br /><ul style="color: rgb(0, 0, 153);"><li>Does f(a) exist?</li><li>Does the limit as 'x' approaches 'a' exist?</li><li>Does f(a) = L?</li></ul>If not, the function is discontinuous.<br /><ul style="color: rgb(204, 102, 0);"><li>f(a) = f(2)<br />f(2) = 2</li><li>the limit as 'x' approaches 2 is 5</li><li>f(2) does not equal L: 2 does not equal 5</li></ul>Therefore, this piece-wise function is not continuous. By discovering this, we found that this function is a <span style="color: rgb(204, 102, 0);">removable discontuity</span>.<br /><br />Finally, we had to sketch this piece-wise function. The graph maintains the <span style="color: rgb(0, 0, 153);">shape of (x+3)</span>. However, it has a <span style="color: rgb(0, 0, 153);">hole at x = 2</span> because there was a <span style="color: rgb(0, 0, 153);">reduction in the factors of (x-2)</span>. <span style="color: rgb(102, 51, 102);">Remember: when there is a reduction in factors, there is a hole at that point rather than an asymptote</span>. Because the function of f(x) has a value of 2 @ x = 2, there is a black dot at that location.<br /><br /><span style="font-weight: bold;">The Conclusion:</span><br /><br />Well that was our pre-test! To sum things up, there were multiple things that should be remembered.<br /><ul><li>A limit as 'x' approaches a value from both sides must meet at the same point, otherwise, the limit does not exist</li><li>Remember how to solve using the painful work of writing out all of the evaluation steps using the limit theorems</li><li>A number line really helps in determining the shape of the function</li><li>Remember the three steps to testing continuity</li><li>When factors reduce a restriction in the denominator, there is a hole at that value of 'x' rather than a vertical asymptote</li></ul>I hope this scribe helped any of the readers! There will be a test on <span style="color: rgb(204, 102, 0);">wednesday</span> so <span style="font-weight: bold;">DON'T DON'T DON'T DON'T DON'T FORGET TO <span style="color: rgb(204, 0, 0);"><span style="color: rgb(255, 102, 0);">"</span>B</span><span style="color: rgb(0, 0, 153);">O</span><span style="color: rgb(0, 102, 0);">B</span><span style="color: rgb(255, 0, 0);">" </span>!<br /><br /></span>Good luck everyone on the test! Do not forget to study either! =) Have a great night everyone.<br /><br />OoOoooOOo! And the scribe for the next class will be: (Give me a sec while I find the scribe list)<br /><br />.........<br />..........<br />...........<br /><br /><span style="text-decoration: underline;">John D. !!!!!</span><br /><br /><br /><span style="font-weight: bold;"></span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2956338223892063128?l=cal45sw08.blogspot.com' alt='' /></div>Tim_MATH_yhttp://www.blogger.com/profile/09929458936647070771noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-88923675524306886942008-03-03T13:55:00.001-06:002008-03-03T14:04:01.280-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_290630"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-3-2008-1204574303132406-3"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-march-3-2008-1204574303132406-3" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-march-3-2008?src=embed" title="View 'Calculus 45S Slides March 3, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-8892367552430688694?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-74763521144473589222008-03-02T22:01:00.008-06:002008-03-04T21:44:30.823-06:00<div style="text-align: justify; color: rgb(0, 0, 0);">THURSDAY'S CLASS<br /><br />We started off class with a quiz. It had two graphs and we had to find the limit of this and that. We marked them in class and went over some of them. We mainly focused on the different type of graphs that are discontinuity. The three different types of discontinuity are:<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_xI0D91rlrqE/R8uCsngIbSI/AAAAAAAAANM/9FinFaRXKdU/s1600-h/052.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://4.bp.blogspot.com/_xI0D91rlrqE/R8uCsngIbSI/AAAAAAAAANM/9FinFaRXKdU/s320/052.PNG" alt="" id="BLOGGER_PHOTO_ID_5173372299994754338" border="0" /></a><br />1. Removable Discontinuity:<br />A hole in a graph<span style="text-decoration: underline;"></span>. That is, a discontinuity that can be "repaired" by filling in a single point. In other words, a removable discontinuity is a point at which a graph is not connected but can be made connected by filling in a single point.<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_xI0D91rlrqE/R8uCs3gIbTI/AAAAAAAAANU/bP5WI3-LZzE/s1600-h/053.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_xI0D91rlrqE/R8uCs3gIbTI/AAAAAAAAANU/bP5WI3-LZzE/s320/053.PNG" alt="" id="BLOGGER_PHOTO_ID_5173372304289721650" border="0" /></a><br />2. Jump Discontinuity:<br />Jump discontinuities occur where the graph has a break in it is as this graph does. It can't be fixed or repaired so that the graph is continuous.<br /><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_xI0D91rlrqE/R8uCtHgIbUI/AAAAAAAAANc/e49aE07ImwU/s1600-h/054.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://2.bp.blogspot.com/_xI0D91rlrqE/R8uCtHgIbUI/AAAAAAAAANc/e49aE07ImwU/s320/054.PNG" alt="" id="BLOGGER_PHOTO_ID_5173372308584688962" border="0" /></a><br /><br />3. Infinite Discontinuity:<br />A discontinuity of a function for which the absolute value of the function can have arbitrarily large values arbitrarily close to the discontinuity. Can't be fixed either.<br /><br />The formal way of telling what kind of discontinuity it is:<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_xI0D91rlrqE/R8uFm3gIbVI/AAAAAAAAANk/Ur_uUCFPeqg/s1600-h/051.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_xI0D91rlrqE/R8uFm3gIbVI/AAAAAAAAANk/Ur_uUCFPeqg/s320/051.PNG" alt="" id="BLOGGER_PHOTO_ID_5173375499745389906" border="0" /></a><br />Then we worked on some questions in class.<br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_xI0D91rlrqE/R8uIw3gIbWI/AAAAAAAAANs/40GMxmGTryM/s1600-h/055.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_xI0D91rlrqE/R8uIw3gIbWI/AAAAAAAAANs/40GMxmGTryM/s320/055.PNG" alt="" id="BLOGGER_PHOTO_ID_5173378970078965090" border="0" /></a><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_xI0D91rlrqE/R8uIxXgIbXI/AAAAAAAAAN0/3720m4aTpVE/s1600-h/056.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_xI0D91rlrqE/R8uIxXgIbXI/AAAAAAAAAN0/3720m4aTpVE/s320/056.PNG" alt="" id="BLOGGER_PHOTO_ID_5173378978668899698" border="0" /></a><br />We also did some work on Mr. K's <span style="font-style: italic;font-size:130%;" ><a style="color: rgb(51, 153, 153);" href="http://archives.math.utk.edu/visual.calculus/1/index.html">favorite website for limits</a><span style="color: rgb(51, 153, 153);"> </span></span>( i think ). We worked together as a class to solve them and they're all on the slides that Mr. K posted up on the 28Th. It also has explanations and the homework assignments. And that was Thursday's class.<br />The next scribe is <span style="font-size:180%;"><span style="color: rgb(153, 51, 153);">KIM POSSIBLE</span></span>. Ha ha i totally ripped that off from your brother but he's not around.<br /><br /><br /></div><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7476352114447358922?l=cal45sw08.blogspot.com' alt='' /></div>SAMUShttp://www.blogger.com/profile/00765502734004344768noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-33766897488666132482008-03-02T18:27:00.011-06:002008-03-02T21:59:52.721-06:00<div style="text-align: justify;"><span style="font-family:arial;">TUESDAY'S MATH CLASS<br /><br />Someone forgot to scribe for Tuesday's math class and she's making up for it by scribing for Tuesday and Thursday. Sorry. Well, on Tuesday we mainly focused on limits in a symbolic approach. The first question was....</span><br /></div><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_xI0D91rlrqE/R8tWG3gIbMI/AAAAAAAAAMc/9xvwwWejmvU/s1600-h/046.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_xI0D91rlrqE/R8tWG3gIbMI/AAAAAAAAAMc/9xvwwWejmvU/s320/046.PNG" alt="" id="BLOGGER_PHOTO_ID_5173323272943070402" border="0" /></a><br /><ol style="text-align: justify;"><span style="font-family:arial;"><li>Factor both the numerator and denominator. If you don't factor the numerator and denominator and go straight to substituting in the value 2 for x, you'll end up with zero in the denominator making the whole thing undefined.<br /></li><li>Reduce<br /></li><li>After reducing the same terms in the numerator and denominator you substitute the value two in everywhere there is an x.<br /></li><li>Voila, you end up with the answer.</li></span></ol><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_xI0D91rlrqE/R8tiy3gIbOI/AAAAAAAAAMs/G8YCmAsvry0/s1600-h/047.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_xI0D91rlrqE/R8tiy3gIbOI/AAAAAAAAAMs/G8YCmAsvry0/s320/047.PNG" alt="" id="BLOGGER_PHOTO_ID_5173337222996847842" border="0" /></a><br /><br /><ol style="text-align: justify;"><span style="font-family:arial;"><ol><li>Rationalize the numerator because if you do then you can reduce 25 - x in the numerator and denominator.</li><li>After reducing, you're left with 1 over 5 + √x. You can now substitute the value 25 for x because it's in its most reduced form.<br /></li><li>Now simplify, the √25 is 5. What is left is 1 over 10.<br /></li></ol><br /></span><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://1.bp.blogspot.com/_xI0D91rlrqE/R8tmq3gIbPI/AAAAAAAAAM0/CZAWr_f7gaw/s1600-h/048.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://1.bp.blogspot.com/_xI0D91rlrqE/R8tmq3gIbPI/AAAAAAAAAM0/CZAWr_f7gaw/s320/048.PNG" alt="" id="BLOGGER_PHOTO_ID_5173341483604405490" border="0" /></a><br /><ol style="text-align: justify;"><span style="font-family:arial;">The whole thing is undefined because in the end your gonna have to substitute zero in for x and 3 over 0 is undefined. Any number over zero is undefined.<br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://4.bp.blogspot.com/_xI0D91rlrqE/R8tvlngIbQI/AAAAAAAAAM8/kaKgkoMrwuE/s1600-h/049.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://4.bp.blogspot.com/_xI0D91rlrqE/R8tvlngIbQI/AAAAAAAAAM8/kaKgkoMrwuE/s320/049.PNG" alt="" id="BLOGGER_PHOTO_ID_5173351289014742274" border="0" /></a><br /><ol style="text-align: justify;"><span style="font-family:arial;"><ol><li>The numerator can be factored so that something can be reduced from the bottom.</li><li>After reducing to the simplest form you now can substitute the value nine for where there's an x.<br /></li><li>The end result is negative 6.</li></ol><br /><br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://3.bp.blogspot.com/_xI0D91rlrqE/R8tyUXgIbRI/AAAAAAAAANE/tiyin08SPBY/s1600-h/050.PNG"><img style="margin: 0px auto 10px; display: block; text-align: center; cursor: pointer;" src="http://3.bp.blogspot.com/_xI0D91rlrqE/R8tyUXgIbRI/AAAAAAAAANE/tiyin08SPBY/s320/050.PNG" alt="" id="BLOGGER_PHOTO_ID_5173354291196882194" border="0" /></a><br /><ol style="text-align: justify;"><span style="font-family:arial;">That would have been marked wrong if it was on a test or exam. Why?<br />Well, in the end since it's just the notation and you've solved for it, you don't need the lim thing.<br /><br /><ol style="text-align: justify;"><span style="font-family:arial;">For the rest of the class, we briefly talked about horizontal and vertical asymptotes. Homework was posted in the slides.<br /></span></ol></span></ol></span></ol></span></ol></ol><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3376689748866613248?l=cal45sw08.blogspot.com' alt='' /></div>SAMUShttp://www.blogger.com/profile/00765502734004344768noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-48303760690829276612008-02-28T15:21:00.002-06:002008-02-28T15:24:24.720-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_285862"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-28-2008-1204233563537589-3"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-28-2008-1204233563537589-3" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-february-28-2008?src=embed" title="View 'Calculus 45S Slides February 28, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-4830376069082927661?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-76781847670773821532008-02-26T12:20:00.001-06:002008-02-26T12:23:12.139-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_282639"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-26-2008-1204050044875545-3"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-26-2008-1204050044875545-3" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-february-26-2008?src=embed" title="View 'Calculus 45S Slides February 26, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7678184767077382153?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-63790304433090515942008-02-20T22:29:00.002-06:002008-02-20T22:38:01.663-06:00Why hello there!<br /><br /><br /><br />Kristin here - reporting the events that took place in todays' Calculus class.<br /><br /><br /><br />The day started off with introductions to Mr.Rekrut- our substitute teacher for the class, with his first words being, "Looks like you have a quiz today" ....a little shocked, not going to lie.<br /><br /><br /><br />Luckily, it was only four questions involving the simplifying of problems with exponents, factoring with exponents, finding the equation of a line in general form given the slope and a point on the line, and finding the equation of a line given a point and a parallel line.<br /><br /><br /><br />Then we were assigned pages 45-58 in the orange and white booklet, as well as pages 6-8 in the blue booklet, which I'm assuming are due for next class, seeing as how we don't have one until next week..<br /><br /><br /><br />If there is anything I forgot, please let me know....<br /><br />The Scribe for next class is....SAMUS!!!<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6379030443309051594?l=cal45sw08.blogspot.com' alt='' /></div>Kristin_Rhttp://www.blogger.com/profile/13674690777406107391noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-72980746633708619062008-02-18T21:06:00.009-06:002008-02-18T22:48:59.623-06:00<div align="justify"><br /></div><div align="justify"><span style="font-size:85%;"><span style="font-family:trebuchet ms;color:#000000;"><span style="color:#000000;">Whoa! I can’t remember the last time I blogged for Math! But, yes, again,</span> <strong><span style="color:#ff9900;">blogging</span></strong> <span style="color:#000000;">is a very important instrument to enhance knowledge and develop self responsibility (remember the escalator video..). It is our utmost responsibility to do things ourselves even though we tend to lean on others – which means:</span> </span></span></div><span style="font-size:85%;"><span style="font-family:Trebuchet MS;"></span><ul><li><div align="justify"></span><span style="font-family:trebuchet ms;font-size:85%;color:#ff9900;"><em><strong>do our homework</strong></em></span></div></li><li><div align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#ff9900;"><em><strong>try to blog as extensive as possible for it determines how well you know the material</strong></em></span></div></li><li><div align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#ff9900;"><em><strong>read scribe post everyday</strong></em></span></div></li><li><div align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#ff9900;"><em><strong>ask questions if necessary</strong></em></span></div></li><li><div align="justify"><span style="font-family:trebuchet ms;color:#000000;"><em><strong><span style="font-size:85%;"><span style="color:#ff9900;">always check for misinterpretation by peers so that no one else will be confused and eventually mess up a test or exam in the future.</span> </span></strong></em></span></div></li></ul><span style="font-family:trebuchet ms;"><span style="font-size:85%;color:#000000;">So, have you read the blog-online rules? Have you watched the video? Have you finished your homework? If you haven’t done so, please do.</span><br /><br /><div align="justify"><span style="font-size:85%;color:#000000;">Talking about unlimited amount of math, Mr. K introduced a very interesting approach to</span> </div><div align="center"><br /><span style="font-family:verdana;font-size:180%;color:#33ccff;"><strong>UNIT 1: LIMITS</strong></span></div><br /><div align="justify"><span style="font-size:85%;color:#000000;">What is a limit? The first time I heard it was when I watched “Mean Girls” back then. I didn’t know what it was. Yes, 5 years after… here it comes, slowly making sense. Yes, AP CALC people, you can laugh since we’re only at the threshold of the house you have already explored three times! Here it goes.</span></div><div align="justify"><span style="font-size:85%;"></span></span> </div><p><strong><span style="font-family:trebuchet ms;font-size:130%;color:#000000;">This notation is used to express LIMITS, which means if you do not have this in every line, the entire thing is incorrect!</span></strong><br /><br /></p><p><img id="BLOGGER_PHOTO_ID_5168537942787792370" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://2.bp.blogspot.com/_kjxjlW0K2UU/R7pV3_f-nfI/AAAAAAAAAJU/K2idiYxyjiU/s400/pic7.jpg" border="0" /> </p><p><em><strong><span style="font-family:trebuchet ms;font-size:130%;color:#33ccff;">PROBLEM #1:</span></strong></em> </p><p><img id="BLOGGER_PHOTO_ID_5168537947082759682" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://3.bp.blogspot.com/_kjxjlW0K2UU/R7pV4Pf-ngI/AAAAAAAAAJc/_G7Ojzl86kQ/s400/pic8.jpg" border="0" /></p><div align="justify"><span style="font-family:trebuchet ms;color:#33ccff;">What do we first notice about this equation?</span></div><br /><ul><li><div align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><em><strong>a second degree function (x^2 : parabola) over a first degree (linear)</strong></em></span></div></li><li><div align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><em><strong>a difference of square in the numerator</strong></em></span></div></li><li><div align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><em><strong>if we graph this on our graphing calculator, it’s a straight line, opposing the fact that it should be some kind of a parabola. Hmmm, weird.</strong></em></span> </div></li></ul><img id="BLOGGER_PHOTO_ID_5168534468159249874" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://1.bp.blogspot.com/_kjxjlW0K2UU/R7pStvf-ndI/AAAAAAAAAJE/UIBM2B4lYp0/s400/pic5.jpg" border="0" /><br /><p><span style="font-family:trebuchet ms;"><span style="font-size:85%;"><span style="font-size:100%;color:#33ccff;">What happens if we factor the numerator?</span> </span></span></p><ul><li><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><strong><em>the (x-1) reduce </em></strong></span></li><li><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><strong><em>we’re left with f(x) = x + 1</em></strong></span></li><li><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><strong><em>graphically, it is identical to the graph we had earlier when we graphed the original equation</em></strong></span> </li></ul><img id="BLOGGER_PHOTO_ID_5168541563445222930" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://1.bp.blogspot.com/_kjxjlW0K2UU/R7pZKvf-nhI/AAAAAAAAAJk/c1w58q3fDPQ/s400/pic9.jpg" border="0" /><br /><p><a href="http://1.bp.blogspot.com/_kjxjlW0K2UU/R7pOUvf-naI/AAAAAAAAAIs/K32Z364iz8E/s1600-h/pic2.jpg"></a><span style="font-family:trebuchet ms;color:#33ccff;">What if x = 1?</span></p><ul><li><div align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><strong><em>Most of the students would say, “IT’S 2!” because of the equation: f(1) = 1 + 1 = 2</em></strong></span></div></li><li><div align="justify"><span style="font-family:trebuchet ms;"><span style="font-size:85%;"><span style="color:#000000;"><strong><em>However, some might disagree and say, “It’s undefined, buddy!”</em></strong> </span></span></span></div></li></ul><p align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;">The question now is WHY? Well, f(x) = x + 1 isn’t the original equation.Therefore, substituting 1 for all x gives us:</span> </p><img id="BLOGGER_PHOTO_ID_5168541593509994018" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://4.bp.blogspot.com/_kjxjlW0K2UU/R7pZMff-niI/AAAAAAAAAJs/1m5PA13Kv80/s400/pic10.jpg" border="0" /><br /><p align="justify"><br /><span style="font-family:trebuchet ms;font-size:85%;color:#000000;">This brought the discussion about the very round number called ZERO. Usually when we divide any number by zero, we say “YOU CAN’T!!!”. It’s very hard to explain. Actually it’s pretty simple. You can’t divide by NOTHING! This follows the same curvature of the ball of wax. Anything over ZERO is undefined… it’s not two… again, it’s undefined! </span><span style="font-family:trebuchet ms;"></p><p align="justify"><span style="font-size:85%;"><span style="color:#000000;">We also had a discussion about 0/0 is not 1, why is it so different from 2/2 = 1?, when both 0 and 2 are numbers? Isn’t a number divided by itself equal to 1? Why is zero such an exception? Well it could mean NOTHING, INFINITY, or ZERO. It all depends on the hwo you look at it. Interesting… Mr. K, took out his “block of wood” to further discuss how you can look at something at different ways but it still refers to the same thing.</span> </span></p><p align="justify"><span style="color:#ff9900;">Consider SLOPE. It can be represented in three ways:</span></p><ul><li><div align="justify"><strong><em><span style="font-size:85%;color:#000000;">m</span></em></strong></div></li><li><div align="justify"><strong><em><span style="font-size:85%;color:#000000;">y = rise / run</span></em></strong></div></li><li><div align="justify"><strong><em><span style="font-size:85%;color:#000000;">y = delta y / delta x</span></em></strong></div></li></ul><p align="justify"><span style="font-size:85%;color:#000000;">So, why do we have three ways to describe slope? ANSWER: because we have three ways to illustrate a function: </span></p><ul><li><div align="justify"><strong><em><span style="font-size:85%;color:#000000;">equation – where ‘m’ is present in the standard form of a line (y = mx +b) </span></em></strong></div></li><li><div align="justify"><strong><em><span style="font-size:85%;color:#000000;">numerically (table of values) – where we can take two ordered pairs and put them into the equation </span></em></strong></div></li><li><div align="justify"><span style="font-size:85%;"><strong><em><span style="color:#000000;">graphically – where we can locate two a point and use y = to find the slope.</span></em></strong> </span></div></li></ul><p align="justify"><span style="font-size:85%;">In this case, <em>f(1)</em> is in an indeterminate form, which means, when x = 1, it is <strong><span style="color:#ff9900;">undefined</span></strong>. To prove this, we can graph the equation one more time. But this time, hit ZOOM 4, which will provide you with a much closer scrutiny at the graph. </span><br /></p><br /><br /><p align="justify"><img id="BLOGGER_PHOTO_ID_5168535292792970722" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://1.bp.blogspot.com/_kjxjlW0K2UU/R7pTdvf-neI/AAAAAAAAAJM/6jpZBc-hQf4/s400/pic6.jpg" border="0" /><br /><span style="font-size:85%;"><span style="color:#000000;">Look at the gap on the linear equation. Isn’t it weird? Well that’s exactly what we had earlier. <em>f(1)</em> ix not 2. Because the point missing on the graph. Try tracing any integer greater or less than 1. It will give you the y-value but will not do it so if you enter x = 1. This squeezes out the value of 2 from both sides - >2 and <2.> </span></span><br /><br /></p><p><span style="font-family:trebuchet ms;font-size:85%;color:#000000;">Let's take a look at a very similar problem:</span><br /></p><p><img id="BLOGGER_PHOTO_ID_5168541602099928626" style="DISPLAY: block; MARGIN: 0px auto 10px; CURSOR: hand; TEXT-ALIGN: center" alt="" src="http://2.bp.blogspot.com/_kjxjlW0K2UU/R7pZM_f-njI/AAAAAAAAAJ0/xn8wfJtlW_s/s400/pic11.jpg" border="0" /></p><p align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;">As the end of class approaches its LIMIT, Mr. K, very quickly went through the laws of limits with all the mathematical operations (addition, subtraction, multiplication, and division). They are on the slides posted of Friday! It's pretty simple. It's somehow like logarithms but not really! <strong>AND AGAIN, MAKE SURE TO INCLUDE THE PROPER NOTATION FOR LIMITS OR YOUR WORK WILL LOSE A VERY FRUSTRATING AMOUNT OF MARKS!</strong></span></p><p align="justify"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;">We had a glimpse of the graphs of limits at the end of the period but since I was not sure about how it goes, I chose not to include it here. I hope Mr. K further go into details about that one in class next next class (he won't be here on Wednesday, which is the only class we have this week). This concludes my scribe post and I hope everyone had a great 3-day weekend! If you didn't, don't worry, there IS another one! YES! Anyway the next scribe will be...</span></p><p align="center"><span style="font-family:trebuchet ms;font-size:85%;color:#000000;"><span style="font-size:180%;color:#ff9900;"><strong>K r i s t i n !</strong></span></p><p align="justify"><br /></span><br /><br /><br /></p><div align="justify"><span style="font-family:trebuchet ms;"></span></div><br /></span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7298074663370861906?l=cal45sw08.blogspot.com' alt='' /></div>vincentrhttp://www.blogger.com/profile/17670701436682827532noreply@blogger.com1tag:blogger.com,1999:blog-3217610859011491687.post-74456529867946206462008-02-15T15:54:00.001-06:002008-02-15T15:55:52.998-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_267645"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-15-2008-120311233820829-5"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-15-2008-120311233820829-5" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-february-15-2008?src=embed" title="View 'Calculus 45S Slides February 15, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-7445652986794620646?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-2021881228289131802008-02-15T13:13:00.007-06:002008-03-25T22:36:54.170-05:00This is <b>The Scribe List</b>. Every possible scribe in our class is listed here. This list will be updated every day. If you see someone's name crossed off on this list then you CANNOT choose them as the scribe for the next class.<br /><br />This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.<br /><br /><blockquote>IMPORTANT: Make sure you label all your Scribe Posts properly or they will not be counted.</blockquote><br /><br /><b><u>Cycle 1</u></b><br /><table><tbody><tr><td width="120"><br /><s>m@rk</s><br />haiyan<br /><s>1</s><br /><s>SAMUS</s><br />MrSiwWy<br /></td><td width="120"><br />Grey-M<br />John D.<br /><s>«Craig»</s><br /><s>vincentr</s><br /><s>Dino</s><br /></td><td><br /><s>Tim_MATH_y</s><br />rusz.l<br /><s>Kristin_R</s> <b>(update your posts)</b><br /></td></tr></tbody></table><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-202188122828913180?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-63035096520654479722008-02-14T06:50:00.004-06:002008-02-16T12:29:15.924-06:00I will be taken Vincent's scribe duties as he requested.<br /><br />Mr. K was not present in class today so we had a sub. This class was a working period to do the review booklets. (<span style="color: rgb(51, 102, 255);">Blue</span>, <span style="color: rgb(204, 204, 204);">White</span>/<span style="color: rgb(255, 204, 0);">Orange</span>)<br /><br />The new assignments are: <span style="font-weight: bold;">Read pages<span style="font-style: italic;"> 30-45</span> (<span style="color: rgb(204, 204, 204);">White</span>/<span style="color: rgb(255, 204, 0);">Orange book</span>), Do <span style="font-style: italic;">page 5</span> (<span style="color: rgb(51, 102, 255);">blue</span> book).</span><br /><br />That concluded the class of February 13!<br /><br />The next scribe will be: <span style="font-weight: bold; color: rgb(255, 0, 0);">Vincentr!</span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6303509652065447972?l=cal45sw08.blogspot.com' alt='' /></div>1http://www.blogger.com/profile/05403602652691839017noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-53810526327701877732008-02-11T22:53:00.001-06:002008-02-11T23:11:02.494-06:00<span style="color: rgb(102, 102, 102); font-style: italic;">Oh Boy! Here we go, I'm scribing again.</span><br /><br /><span style="color: rgb(102, 102, 102); font-style: italic;">This is my fourth class with Mr. K., all of which I had to scribe for at least thrice, so I'm used to it.</span><br /><br />Anyway, today's class was basically our first official class of learning. However, it ended up being a review of grades 10 through 12. Mr. K. went over the Review Test we had the previous class, touching on the questions we seemed to have the most trouble with.<br /><br /><span style="color: rgb(0, 153, 0);">These questions included the concepts of:</span><br /><span style="color: rgb(0, 153, 0);">•Inequalities</span><br /><span style="color: rgb(0, 153, 0);">•Fraction Operations</span><br /><span style="color: rgb(0, 153, 0);">•</span><span style="font-weight: bold; color: rgb(0, 153, 0);">Factoring</span><br /><span style="color: rgb(0, 153, 0);">•</span><span style="font-weight: bold; color: rgb(0, 153, 0);">Rational Functions</span><br /><span style="color: rgb(0, 153, 0);">•Function Notation</span><br /><span style="color: rgb(0, 153, 0);">•Trigonometry</span><br /><br />As well, there is a list that may be seen <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-february-11-2008">here(slides 14 and 15)</a> that goes over the things we should already know.<br /><br />I believe Mr. K. did a sufficient job in discussing the problems we may have had in class, as no one seemed to ask questions.<br /><span style="color: rgb(255, 102, 0);"><span style="color: rgb(204, 51, 204);">***But remember, if you need any help in dealing with any of the content in the course, arrange a time with Mr.K. or even ask some other students to see if you can get the advice you need***</span><br /><br /><span style="color: rgb(0, 0, 0);">As well, I suggest if you got some questions wrong on this quiz that you should go back to the questions and try them again. It should help. Or, just open your booklets and do the homework; it provides tons of practice questions dealing with the <span style="color: rgb(0, 153, 0);">above concepts</span> and more.<br /><br /><span style="color: rgb(255, 0, 0); font-style: italic;"><span style="font-weight: bold;">Homework: </span></span><span style="color: rgb(255, 0, 0);"><span style="font-weight: bold;"><span style="color: rgb(255, 102, 0);">•</span><span style="color: rgb(255, 102, 0);">All exercises up to page #30 in the ORANGE BOOK</span></span></span><span style="color: rgb(255, 0, 0); font-style: italic;"><span style="font-weight: bold;"><span style="font-style: italic;"><br /> </span></span></span><span style="color: rgb(255, 0, 0);"><span style="font-weight: bold;"><span style="color: rgb(51, 51, 255);">•Pages 2, 3, 4, and 5 in the BLUE BOOK<br /><br /><span style="color: rgb(0, 0, 0);"><br />Last but not least, Wednesday's Scribe will be:......<br /><br /><br /><span style="color: rgb(255, 204, 0);"><span style="font-size:180%;">VINCENT!!! <span style="font-size:100%;">(he does awesome Scribes =D)</span></span></span><br /></span></span></span></span><span style="color: rgb(255, 0, 0); font-style: italic;"><span style="font-weight: bold;"><span style="font-style: italic;"></span></span></span><span style="color: rgb(255, 0, 0);"><span style="font-weight: bold;"></span></span></span></span><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-5381052632770187773?l=cal45sw08.blogspot.com' alt='' /></div>«Craig»http://www.blogger.com/profile/08860268378924842699noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-16320454168072183772008-02-11T13:59:00.000-06:002008-02-11T14:12:07.316-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_261807"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-11-2008-1202760128771042-5"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-11-2008-1202760128771042-5" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-february-11-2008?src=embed" title="View 'Calculus 45S Slides February 11, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1632045416807218377?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-34885966240804959822008-02-07T23:49:00.001-06:002008-02-07T23:52:38.776-06:00So.. this is what boredom does to you. Especially if you're LOLcat fan.<br /><br /><a onblur="try {parent.deselectBloggerImageGracefully();} catch(e) {}" href="http://2.bp.blogspot.com/_AECKI0ALk0E/R6vtjPH8mpI/AAAAAAAAAE0/l-OhTvsbWBM/s1600-h/funny-pictures-sisyphus-cat-watermelon-water.jpg"><img style="margin: 0pt 10px 10px 0pt; float: left; cursor: pointer;" src="http://2.bp.blogspot.com/_AECKI0ALk0E/R6vtjPH8mpI/AAAAAAAAAE0/l-OhTvsbWBM/s320/funny-pictures-sisyphus-cat-watermelon-water.jpg" alt="" id="BLOGGER_PHOTO_ID_5164482587321473682" border="0" /></a><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br /><br />I found it on http://icanhascheezburger.com/2008/01/22/funny-pictures-sisyphus-cat-tries-again/..<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3488596624080495982?l=cal45sw08.blogspot.com' alt='' /></div>John D.http://www.blogger.com/profile/12872067735171995032noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-67532891683220575332008-02-07T15:46:00.000-06:002008-02-07T15:50:44.921-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_256973"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-7-2008-1202420653529357-4"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-7-2008-1202420653529357-4" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-february-7-2008?src=embed" title="View 'Calculus 45S Slides February 7, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-6753289168322057533?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-20613214246715888052008-02-07T00:07:00.000-06:002008-02-07T00:10:40.928-06:00We were talking about exactly what sort of post you're supposed to make to get that one blogging mark on your test. The kind of post I'd like you to make should have one or more of these characteristics:<br /><ul><br /> <li>Write about what you understand the least in the unit so far; your personal "Muddiest Point."</li><br /> <li>A reflection on a particular class.</li><br /> <li>A reflective comment on your progress in the course.</li><br /> <li>A comment on something that you've learned that you thought was &quot;cool&quot;.</li><br /> <li>A comment about something that you found very hard to understand but now you get it! Describe what sparked that &quot;moment of clarity&quot; and what it felt like.</li><br /> <li>Have you come across something we discussed in class out there in the &quot;real world&quot; or another class? Describe the connection you made.</li><br /></ul><br />Your posts do not have to be long. I'm far more interested in the <b>quality</b> of what you write rather than the <b>quantity</b>. <br /><br />Make certain you always use 3 labels on your post: <b>[your name], [unit tag], BOB</b><br /><br />When you share where you are in your learning a few days before the unit test I can address those issues in class so, hopefully, you will get much more than one extra mark on the test. ;-)<br /><br /><b><i>Happy Blogging!</i></b><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-2061321424671588805?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-39859490009517249552008-02-05T23:47:00.000-06:002008-02-05T23:49:19.254-06:00Blogging is a <b>very public</b> activity. Anything that gets posted on the internet stays there. Forever. Deleting a post simply removes it from the blog it was posted to. Copies of the post may exist scattered all over the internet. I have come across posts from my students on blogs as far away as Sweden! That is why we are being so careful to respect your privacy and using first names only. We do not use pictures of ourselves. If you really want a graphic image associated with your posting use an avatar -- a picture of something that represents you but IS NOT of you.<br /><br />Here are a few videos that illustrate some of what I want you to think about:<br /><br /><center><embed src="http://web.splashcast.net/go/p/PGTA6216AC" wmode="transparent" width="400" height="300" allowFullScreen="true" type="application/x-shockwave-flash" /></embed></center><br /><br />Two teachers in the U.S.A. worked with their classes to come up with a list of guidelines for student bloggers.<br /><br />One of them, <a href="http://budtheteacher.typepad.com">Bud Hunt</a>, has these suggestions, among others:<br /><br /><blockquote><ol><li>Students using blogs are expected to treat blogspaces as classroom spaces. Speech that is inappropriate for class is not appropriate for our blog. While we encourage you to engage in debate and conversation with other bloggers, we also expect that you will conduct yourself in a manner reflective of a representative of this school.</li> <br /><br /><li>Never EVER EVER give out or record personal information on our blog. Our blog exists as a public space on the Internet. Don’t share anything that you don’t want the world to know. For your safety, be careful what you say, too. Don’t give out your phone number or home address. This is particularly important to remember if you have a personal online journal or blog elsewhere.</li><br /><br /><li>Again, your blog is a public space. And if you put it on the Internet, odds are really good that it will stay on the Internet. Always. That means ten years from now when you are looking for a job, it might be possible for an employer to discover some really hateful and immature things you said when you were younger and more prone to foolish things. Be sure that anything you write you are proud of. It can come back to haunt you if you don’t.</li> <br /><br /><li>Never link to something you haven’t read. While it isn’t your job to police the Internet, when you link to something, you should make sure it is something that you really want to be associated with. If a link contains material that might be creepy or make some people uncomfortable, you should probably try a different source.</li></ol></blockquote><br /><br />Another teacher, <a href="http://outsidethecave.blogspot.com/">Steve Lazar</a>, developed a set of guidelines in consultation with his students. You can read them <a href="http://budtheteacher.jot.com/WikiHome/studentblogpol">here</a>.<br /><br />Look over the guidelines and add the ones you like in the comments section below this post; either from one of Steve's students or one of your own. I think Bud's suggestions are excellent. We'll be using the one's I highlighted above as a basis for how we will use our blog.<br /><br />Cheers,<br />Mr. K.<div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-3985949000951724955?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0tag:blogger.com,1999:blog-3217610859011491687.post-10553919646940578782008-02-05T15:59:00.000-06:002008-02-05T16:04:28.438-06:00Here they are ...<br /><br /><center><br /><div style="width:425px;text-align:left" id="__ss_253721"><object style="margin:0px" width="425" height="355"><param name="movie" value="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-5-2008-120224505052821-4"/><param name="allowFullScreen" value="true"/><param name="allowScriptAccess" value="always"/><embed src="http://static.slideshare.net/swf/ssplayer2.swf?doc=calculus-45s-slides-february-5-2008-120224505052821-4" type="application/x-shockwave-flash" allowscriptaccess="always" allowfullscreen="true" width="425" height="355"></embed></object><div style="font-size:11px;font-family:tahoma,arial;height:26px;padding-top:2px;"><a href="http://www.slideshare.net/?src=embed"><img src="http://static.slideshare.net/swf/logo_embd.png" style="border:0px none;margin-bottom:-5px" alt="SlideShare"/></a> | <a href="http://www.slideshare.net/dkuropatwa/calculus-45s-slides-february-5-2008?src=embed" title="View 'Calculus 45S Slides February 5, 2008' on SlideShare">View</a> | <a href="http://www.slideshare.net/upload?src=embed">Upload your own</a></div></div><br /></center><div class="blogger-post-footer"><img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3217610859011491687-1055391964694057878?l=cal45sw08.blogspot.com' alt='' /></div>dkuropatwahttp://www.blogger.com/profile/08462283847470560887noreply@blogger.com0