Sunday, October 18, 2009

Post # 9

Okay, other than taking those exams... this week we learned explicit derivatives!!! Alright, to begin, Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx or y' (I prefer y').
3: Solve for dx/dy or y'
(if you want to find slope plug in an x and y value)

example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(y')+2y(y')-5(y')-2x=0

Then you have to solve for y', so you get:
y'(3y^2+2y-5)=2x which then is further solved for to get y'=2x/(3y^2+2y-5)

And that's all!!! this problem is finished.

Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)

First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(y'))=100(y+x(y'))

then, you need to foil it n stuff, so you get:
12x^3+12x^2(y')+12xy^2+12y^2(y'))=100y+100x(y')

then, as usual, you would have to solve for y':
y'=(-12^3-12xy^2+100y)/(12x^2+12y-100x)

after you solved for y', you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)

and I finally learned how to do optimization!!! I got that problem right on the free response part of the exam, but the thing I don't understand, was the second problem on that page. I forgot what it's called, and I don't remember it. All I know is that when I saw it, I didn't know what to do. If anyone remembers try to explain it please. Thanks.

1 comment:

  1. are u talking about the problem where they give you the function and you take the derivative and solve for x. if so this is how you do it.

    take the derivative, solve for x and those are your critical values.

    then to find max and min, you plug the x's into the derivative to get another point and which point is highest is the max and the lowest is the min.

    and finally to find points of inflection, take the second derivative of the function and solve for x again. set up intervals and find where the function is concave up or down, where there is a change in concavity there is a point of inflection...

    hope this helps you!

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