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.
Showing posts with label John D.. Show all posts
Showing posts with label John D.. Show all posts
Tuesday, May 20, 2008
Monday, April 21, 2008
Application of Derivatives
On Friday, our class started talking about the application of derivatives. To start off, Mr. K showed us this slide:
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.
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 f in this graph to find out where is f ' is positive. First we have to find out where the tangent lines in f are zero. Looking at the function f, we can clearly see that the tangent lines are zero in -2 and +2. This means that f ' will have roots at -2 and +2. Then, we look when the tangent lines of f are approaching zero. There are 2 instances in this graph. The first instance is from (-∞, -2]. The slope from that domain is increasing, since it's coming from a negative value to zero. But that still doesn't solve our problem since f ' is still negative at that point. Then, from (-2, 2) in f, 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, f ' 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 f ' positive, which is from (-2, o) to (2,0).
Now to find out where f '' is positive. From the graph, we can see that f 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 f '' is a straight line. Looking at the shape of f ' (say it is -2x2), it is definitely a negative quadratic function because it has a maximum, not a minimum. We then have to find out the derivative of f '' 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 f ', which is f '', is -4x. From this we can see that the function is positive at quadrant 2.
The critical numbers (which is another term for the roots) of f are -2 and 2.
This is a little difficult...
For a), f is increasing wherever f ' is positive.
For b), f(0) is negative.
Well, that was all we talked about last class. Mr. K tried to squeeze in one more slide but we ran out of time.
The next scribe is Kristin.
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.
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 f in this graph to find out where is f ' is positive. First we have to find out where the tangent lines in f are zero. Looking at the function f, we can clearly see that the tangent lines are zero in -2 and +2. This means that f ' will have roots at -2 and +2. Then, we look when the tangent lines of f are approaching zero. There are 2 instances in this graph. The first instance is from (-∞, -2]. The slope from that domain is increasing, since it's coming from a negative value to zero. But that still doesn't solve our problem since f ' is still negative at that point. Then, from (-2, 2) in f, 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, f ' 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 f ' positive, which is from (-2, o) to (2,0).
Now to find out where f '' is positive. From the graph, we can see that f 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 f '' is a straight line. Looking at the shape of f ' (say it is -2x2), it is definitely a negative quadratic function because it has a maximum, not a minimum. We then have to find out the derivative of f '' 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 f ', which is f '', is -4x. From this we can see that the function is positive at quadrant 2.
The critical numbers (which is another term for the roots) of f are -2 and 2.
This is a little difficult...
For a), f is increasing wherever f ' is positive.
For b), f(0) is negative.
Well, that was all we talked about last class. Mr. K tried to squeeze in one more slide but we ran out of time.
The next scribe is Kristin.
Tuesday, April 15, 2008
BOB
Almost forgot to put up my BOB before the test. Whew.
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.
Night.
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.
Night.
Tuesday, March 4, 2008
BOBby McFerrin
I 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.
Anyway, good luck on the test tomorrow guys!
Anyway, good luck on the test tomorrow guys!
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