Monday, May 26, 2008

Introduction to Anti-Derivatives (sounds like some sort of medicine)

So, on the previous class, we were introduced to anti-derivatives..

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, f ' (x) = x 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 f ' (x) = x. 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:

f ' (x) = xn -------> f (x) = xn+1/n+1 + C, where C is the constant. I'll explain later.

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.

Now this is where I explain the C part in the rule:

f ' (x) = xn -------> f (x) = xn+1/n+1 + C, where C is the constant

We can't really find C not unless we're given a point in the graph. C is 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 + C, not unless we're given a point in the graph.

Then we were just given a practice problem, which was pretty straightforward.

The next scribe is etimz, or Ethan.

Tuesday, May 20, 2008


Well, 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.

Good luck.

Monday, May 19, 2008


What 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.

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!

Wednesday, May 14, 2008

Putting the pieces together

Today'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.

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 CRITICAL NUMBERS instead of stating the CRITICAL POINTS. To find the critical points, just plug in the critical numbers at the parent function, and you should have it.

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.

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. f is increasing where f ' is positive, and f is decreasing where f ' is negative. We then test it by using the number line test, and the

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.

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.

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. =)

That's it! We have a pre-test next class and a test the class after.

The next scribe is... *drum roll please*

haven't decided yet. lol

Today's Slides: May 14

Here they are ...

Monday, May 12, 2008

Student Voices Episode 3: Chris, Craig, and Graeme

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 Developing Expert Voices project. It ends with a challenge, the result of which will be featured in a future podcast.

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 on their class blog.

(Download File 31.8Mb, 26 min. 30 sec.)

The video mentioned near the end of the podcast is called Daft Hands. Here it is:

Photo Credit: Shadow singer by flickr user EugeniusD80

Scribe: Related Rates!


Hello, I'm known as Tim-Math-Y 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.

Key points to solving Related Rates questions:
  • Find a way to relate all the variables within a given question
  • Differentiate implicitly using chain rule, with respect to time


Well, I completed the scribe as soon as I could because it seemed that many people were missing today. I hope this helped!

Notice that most of these solutions include the following:
  • A labelled diagram to help you visualize
  • Listing given information, and finding information that will be required in the solution
  • Differentiating implicitly using chain rule with respect to time
  • A sentence answer including appropriate units
That sums it up folks! Thanks for reading =)

The Scribe for next class will be, John D. !

Today's Slides: May 12

Here they are ...

Wednesday, May 7, 2008

More optimization problems

Hello everyone this is the scribe post for May 6, sorry it is late but I had to write the AP calc 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).

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) =
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 reject 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.

Remember to do the other two questions and the homework given in class, the next scribe is Ethan.

Friday, May 2, 2008

Optimization problems

Today, we were doing optimization problems.
In order to describe how we are solving these problems, I will be using an example from the slide 5.
It is the open box question where you have to cut squares from the corners to maximize the volume of the box.

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.

Next we know that the volume of the box is going to be V=L*W*H.
We know that the height(H) is x and each side (L & W) is the same and equal to (24-2x), so the equation is going to be
V(x)= (24-2x)*(24-2x)*x
which is
V(x)=((24-2x)^2) * x

Third step is to simplify the equation and you will get
V(x)=(4x^3) - (96x^2) + 576x.

We take the derivative of that to find the max.
V'(x) =(12x^2) - 192x + 576
V'(x) =12((x^2) - 16x + 48)

The derivative at the max is equal to zero.
So we let V'(x) = 0
0 = 12((x^2) - 16x + 48)
0 = ((x^2) - 16x + 48)
0 = (x - 4) (x - 12)
x = 4, 12
x can't equal to 12 because that would make the side of the box equal to zero.
Therefore x = 4.

The last step is to test it.

4 by 4 squares should be cut out to maximize the volume of the box.

The next scribe is Dino.

Today's Slides: May 2

Here they are ...

Optimization Problems

Hey 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.

Slide 1

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.

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:

Step 1: Find what you are trying to maximize or minimize. This will be stated excplicitly (in the question).

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.

Step 3: Try to get the equation into a two variable form, so you can take the derivative.
- 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.
- Step 3b: ISolate one of the two variables in the equation drawn in step 3a.
- Step 3c: In the original equation that you are trying to minimize or maximize, replace the variable you isolated in step 3b.

Step 4: Take the derivative of your two-variable equation.

Step 5: Set the derivative to 0, and solve for the value of the remaining variable.

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.

NOTE: This is just an outline so follow it loosely. There is really no chronological way to answer this problems.

Thats it , the next scribe is haiyan.