Question

A radius of a cylinder is 1/3 times its height. The height is increasing at a rate of 2 cm per minute. Determine the change of volume to the time of the cylinder when h = 6.

Answer 1

The volume of a cylinder is
\[V=\frac{1}{3}\pi r^2h\]
You're told that the radius of the cylinder is 1/3 of the height, so you know
\[r=\frac{1}{3}h,\]
which gives you
\[\begin{align*}V&=\frac{1}{3}\pi \left(\frac{1}{3}h\right)^2h\\
&=\frac{1}{27}\pi h^3\end{align*}\]
And, you also know that
\[\frac{dh}{dt}=2\frac{\text{cm}}{\text{min}}\]
Try differentiating the last volume equation with respect to h.

Answer 2

These questions are essentially always asking you to find a formula for the quantities described and then do implicit differentiation with respect to time.

@SithsAndGiggles The volume of a cylinder is plain old V=pi*r^2*h. There's no 1/3 involved.

Answer 3

Wait, I still think you've gotten something wrong here.

Answer 4

Let me correct myself again...
\[V=\frac{1}{9}\pi h^3\]
Thanks again.

Answer 5

So, I would have to take the derivative of that formula @SithsAndGiggles? >_<; Word problems on top of implicit differentiation.... my poor brain, hah.

Answer 6

Yes, you would have to do that. You shouldn't be surprised, though. Calculus concepts are everywhere in a dynamic world.

Answer 7

Doesn't make them easy. :p

Answer 8

So the derivative of that: \[(\pi h^2)/3\]

Answer 9

Yes. You have to find the rate of change of volume (dV/dt) when h = 6. And now that I see another mistake, I should say that I meant to say "differentiate with respect to t," not h. So, you have
\[V'(t)=\frac{dV}{dt}=\frac{1}{3}\pi h^2\frac{dh}{dt}\]
All you need to do now is find V'(6).

Answer 10

Would the answer be: \[24\pi\]