Tuesday, January 19, 2010

Post 22

Ok, so today is Tuesday. I totally forgot to do my blog over the weekend. Although my lateness might affect my grade on the blog, I hope it makes up for it in content. If I would have done my blog on Sunday, I probably would have looked up something old from my notes and just wrote it on here, but between yesterday night and today in class, I learned a few new things. The concepts of the things I learned may not exactly be new, but I learned new ways of looking at them. For the last couple of weeks, we have been working on the AP tests. We have covered mostly all the material given on the test, but many of the questions are difficult to answer; especially the calculator portion (I got a zero).

Anyway, a few things that have refreshed my memory include infinity rules, limits that go to zero, second implicit derivatives, average rate of change, relative maximum, vertical tangents, and some other stuff.

Some people think infinity rules are very easy, but many forget. For a limit approaching infinity you have three rules:

1. If degree of the top is greater than the degree of the bottom, the limit is approaching infinity

2. If the degree of the top is less than the degree of the bottom, the limit is approaching zero

3. If the degrees of the top and bottom are the same, you just make a fraction out of the coefficients in front of the variables.

Limits that go to zero:
I used to forget what to do with these. Today I learned that if a limit is going to zero, you are thinking about taking a derivative. If you see a +h, this means to definitely take a derivative.

A second derivative is just taking two derivatives, but taking a second implicit derivative can be tricky. When you are taking the first derivative of an implicit, you solve for dy/dx and that’s your answer. For the second derivative, you take the derivative of what dy/dx is equal to. When you further have more dy/dx’s in your second derivative, you plug the first dy/dx in to get your answer.

Average rate of change is not 1/b-a. It is f(b)-f(a)/b-a.

A vertical tangent is only found when a slope is undefined. First you need to find the derivative of the function and set the bottom equal to zero. You set the bottom equal to zero because zero at the bottom of any fraction means it is undefined.

These things I understand very well. These things are also on the non-calculator portion of the AP. I’m still a little confused oh now to use functions in my calculator. I forgot how to take a derivative in my calculator. So someone can help me with that.

3 comments:

  1. To take a derivative in your calculator, type in the function in the y= as if you were going to graph it. Press the [SECOND] key then press [CALC] . Press the dy/dx option and it will go to the graph and ask you for an x-value. Type in the x value that you want to know the derivative at and then press [ENTER]. It will give you the value of the derivative or the slope at that particular x-value on the lower left hand corner of the screen. You can also use the nDeriv function...which is in the math list. You would input it as, for example, nDeriv(x^2+2x+2,x,5) to find the derivative at x=5 for the function x^2 + 2x + 2.

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  2. Hmm...John already explained how to take a Deriv in your calc, but whenever I get confused, I try to remember it (what they're asking, I mean) and play around with my calc later on when I have no life. Though I'm usually afraid I'll break something, it's still kinda fun to see what you've got hidden on these machines :):)

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  3. well, i was going to take this one..but john has it. I figured out that if you get an arguement error message it means something doesn't add up with your parenthesis..so that's what you need to change.

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