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**Samantha** · July 3rd 2007, 04:54 AM

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

**Soroban** · July 3rd 2007, 05:21 AM

Hello, Samantha!

Exactly where is your difficulty?

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

We have: . $f(g(x)) \;=\;f\left(1-x^2\right) \;=\;\frac{1}{\sqrt{1-x^2}}$

**Samantha** · July 3rd 2007, 05:26 AM

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= $\frac {\pi} {3} r^3$ . How fast is the volume changing with respect to changes in r when the radius is equal to 2 feet?.

**janvdl** · July 3rd 2007, 05:39 AM

Originally Posted by Samantha

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= $\frac {\pi} {3} r^3$ . How fast is the volume changing with respect to changes in r when the radius is equal to 2 feet?.

Oh, you have a different formula.

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

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

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

$= \pi r^2$

Set r = 2

$= \pi 2^2$

$= 4 \pi$

**Samantha** · July 3rd 2007, 05:39 AM

$ (\frac (\pi) {3}) = V = \frac (\pi) {3} * r * r^2 $

r^2 = 4

v = \frac (\pi) {3} * 2 * (r)^2 = \frac (\pi) {3}*1(2^2)= 4 \frac (\pi) {3}
[/tex]

Am I right?

**janvdl** · July 3rd 2007, 05:41 AM

Originally Posted by Samantha

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

Am I right?

Hmm, seems you are having problems with latex.
Just check my latest reply. I think it's right.

**Samantha** · July 3rd 2007, 05:43 AM

Some how I got 4 $\frac \pi {3}$

**janvdl** · July 3rd 2007, 05:46 AM

Originally Posted by Samantha

Some how I got 4 $\frac \pi {3}$

I cant see how you got it.
Im struggling to read your reply.

**janvdl** · July 3rd 2007, 05:50 AM

Originally Posted by Samantha

$ (\frac (\pi) {3}) = V = \frac (\pi) {3} * r * r^2 $

r^2 = 4

v = \frac (\pi) {3} * 2 * (r)^2 = \frac (\pi) {3}*1(2^2)= 4 \frac (\pi) {3}
[/tex]

Am I right?

No, why did you say $r * r^2$ ????

And why is it $\frac{3}{ \pi }$ in your calculations there?

**Samantha** · July 3rd 2007, 06:07 AM

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

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

**janvdl** · July 3rd 2007, 06:12 AM

Samantha, i really think your answer is wrong...

I'll ask Jhevon or someone to check mine though.

**Samantha** · July 3rd 2007, 06:33 AM

Can u look at my answer again? It's on the first page =)

**janvdl** · July 3rd 2007, 06:37 AM

Originally Posted by Samantha

Can u look at my answer again? It's on the first page =)

I checked it again, and no, i dont agree with it, sorry.

**janvdl** · July 3rd 2007, 06:49 AM

Originally Posted by Samantha

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

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

Explain $\frac \pi {3}$ * $r^2$

Why did you use $r^2$ ????

**Jhevon** · July 3rd 2007, 07:03 AM

Originally Posted by Samantha

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= $\frac {\pi} {3} r^3$ . How fast is the volume changing with respect to changes in r when the radius is equal to 2 feet?.

Originally Posted by janvdl

Explain $\frac \pi {3}$ * $r^2$

Why did you use $r^2$ ????

I see what's the point of confusion here.

The volume of a right-cylinder is given by $V = \frac { \pi}{3} r^2 h$ , however, in this particular cone, $h = r$ . This is why Samantha made a distinction between the $r$ 's and used $r \cdot r^2$ . realizing that the first $r$ is for the height, she decided not to differentiate that one. However, this is incorrect. Janvdl is right. if $h = r$ it is perfectly valid to treat the $h$ as $r$ 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?

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