# rate of change

• Feb 6th 2010, 08:16 AM
discobob
rate of change
Hey, I'm doing some homework and I'm a little confused with this question.

Find the rate of change of the volume V of a cube with respect to
(a) the length w of a diagonal on one of the faces.
(b) the length z of the one of the diagonals of the cube.

So i'm trying to find $\frac{dv}{dw}$
I know that the Volume of a cube is $v=x^3$

What i need to do is find a way of expressing V in terms of w.

I know that $x^2 + x^2 = w^2$ and solving for x I get $x= \frac{w}{2}$. Unless my math is incorrect (which is very possible).

The next step is the plug that into $v=x^3$ so i get $v=(\frac{w}{2})^3$

Do i just find the derivative of that to get answer? I know that's wrong somewhere since the answer is $(\frac{3 (2^(\frac{1}{2})}{4})w^2$

Can anyone tell me what i'm doing wrong? thank you very much
• Feb 6th 2010, 08:44 AM
skeeter
Quote:

Originally Posted by discobob
Hey, I'm doing some homework and I'm a little confused with this question.

Find the rate of change of the volume V of a cube with respect to
(a) the length w of a diagonal on one of the faces.
(b) the length z of the one of the diagonals of the cube.

$V = x^3$

$w^2 = x^2 + x^2$

$w^2 = 2x^2$

$w = \sqrt{2} \cdot x$

$x = \frac{w}{\sqrt{2}}$

$V = \frac{w^3}{2\sqrt{2}}$

find $\frac{dV}{dw}$

$z^2 = w^2 + x^2$

$z^2 = w^2 + \frac{w^2}{2}$

$z^2 = \frac{3w^2}{2}$

$w = \sqrt{\frac{2}{3}} \cdot z$

use the chain rule ...

$\frac{dV}{dz} = \frac{dV}{dw} \cdot \frac{dw}{dz}$