If f(x) = $\displaystyle \frac {1} {\sqrt {x} } $and g(x) = 1 -$\displaystyle x^2$,find f(g(x)).

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- Jul 3rd 2007, 04:54 AM #1

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- Jul 3rd 2007, 05:21 AM #2

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- Jul 3rd 2007, 05:26 AM #3

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Actually, I was just making sure that I'm right...

I have a problem with word problems

The volume of a right circular cone of radius r and height r is given by V= $\displaystyle \frac {\pi} {3} r^3$. How fast is the volume changing with respect to changes in r when the radius is equal to 2 feet?.

- Jul 3rd 2007, 05:39 AM #4
Oh, you have a different formula.

$\displaystyle \frac {\pi} {3} r^3$

now $\displaystyle \frac {\pi} {3} $ is a constant.

So $\displaystyle \frac {dV} {dr} = (3)( \frac {\pi} {3} r^2) $

$\displaystyle = \pi r^2 $

Set r = 2

$\displaystyle = \pi 2^2 $

$\displaystyle = 4 \pi $

- Jul 3rd 2007, 05:39 AM #5

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- Jul 3rd 2007, 05:41 AM #6

- Jul 3rd 2007, 05:43 AM #7

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- Jul 3rd 2007, 05:46 AM #8

- Jul 3rd 2007, 05:50 AM #9

- Jul 3rd 2007, 06:07 AM #10

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$\displaystyle

(\frac \pi {3})

$ = $\displaystyle \mathbb{V}_{1}$ = $\displaystyle \frac \pi {3}$ * $\displaystyle r^2$

$\displaystyle

2^2 = 4

$

$\displaystyle \mathbb{V}_{2}$ = $\displaystyle \frac \pi {3} $ * r$\displaystyle ({r}_{2})^2 $ =$\displaystyle \frac \pi {3}$ * 1 $\displaystyle (2^2)$ = 4 $\displaystyle \frac \pi {3}$

- Jul 3rd 2007, 06:12 AM #11

- Jul 3rd 2007, 06:33 AM #12

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- Jul 3rd 2007, 06:37 AM #13

- Jul 3rd 2007, 06:49 AM #14

- Jul 3rd 2007, 07:03 AM #15
I see what's the point of confusion here.

The volume of a right-cylinder is given by $\displaystyle V = \frac { \pi}{3} r^2 h$, however, in this particular cone, $\displaystyle h = r$. This is why Samantha made a distinction between the $\displaystyle r$'s and used $\displaystyle r \cdot r^2$. realizing that the first $\displaystyle r$ is for the height, she decided not to differentiate that one. However, this is incorrect.**Janvdl is right**. if $\displaystyle h = r$ it is perfectly valid to treat the $\displaystyle h$ as $\displaystyle r$ in every way. in fact, this is a prevalent related rates technique. a lot of questions like this require you to solve for one variable in terms of another so you can differentiate a single variable function, we do not make distinctions between the same variable despite the fact it came from two different quantities.

even if your method was right Samantha, you would still get it wrong, why did you plug in 1 for the r that represents the height? if h is equal to r, then when r = 2, the height would be the same. and looking at it again, it seems you didn't differentiate at all. i'm sorry but what you did is confusing (i am looking at your last attempt at the question). can you explain exactly what you were doing so that we can clear up any misconceptions you have?