# Applications of DIfferentiation - Cylinder

• Sep 6th 2006, 07:43 AM
classicstrings
Applications of DIfferentiation - Cylinder
I keep getting the wrong answer for this question, help appreciated.

A cylinder is designed so that it is always as high as its base diameter. It is expanding.

What is the rate of change of its surface area when radius is 1cm?

What is the rate of change of its volume for the same radius?
• Sep 6th 2006, 07:48 AM
ThePerfectHacker
Quote:

Originally Posted by classicstrings

A cylinder is designed so that it is always as high as its base diameter. It is expanding.

What is the rate of change of its surface area when radius is 1cm?

What is the rate of change of its volume for the same radius?

When the radius is 1 then the diamter is 2 thus the height is 2.
You have,
$\displaystyle V=\pi r^2 h$
Thus,
$\displaystyle \frac{dV}{dt}=2rh\pi \frac{dr}{dt}+ \pi r^2\frac{dh}{dt}$
There is not enough infromation, i.e. what are the rates of change in radius and height. No wonder you cannot solve it.
• Sep 6th 2006, 08:45 AM
Glaysher
But $\displaystyle h=2r$

So $\displaystyle V=2 \pi r^3$

and $\displaystyle \frac{dV}{dr} = 6 \pi r^2$

and $\displaystyle \frac{dV}{dt}=\frac{dV}{dr} \frac{dr}{dt}$

Only need rate of change of r
• Sep 6th 2006, 09:36 AM
classicstrings
Thanks! ---> worked all the answers out
• Sep 6th 2006, 09:54 AM
ThePerfectHacker
Quote:

Originally Posted by Glaysher
But $\displaystyle h=2r$

So $\displaystyle V=2 \pi r^3$

and $\displaystyle \frac{dV}{dr} = 6 \pi r^2$

and $\displaystyle \frac{dV}{dt}=\frac{dV}{dr} \frac{dr}{dt}$

Only need rate of change of r

You are right, I missed that.
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$\displaystyle \frac{dh}{dt}=2\frac{dr}{dt}$