# Thread: Calc III: Partial Derivatives

1. ## Calc III: Partial Derivatives

The radius of a right circular cone is increasing at a rate of 5 inches per second and its height is decreasing at a rate of 3 inches per second. At what rate is the volume of the cone changing when the radius is 50 inches and the height is 40 inches?

I went to get help with this problem and I was advised to differentiate the Volume equation in terms of r and h and plug in the numbers. But I didn't get the right answer. I did this:

$V=\pi r^2(t)h(t)$
$dV/dt = 2\pi r(t)r'(t)h(t)+h'(t)\pi r^2(t)$
Plugged in the numbers and got $12500\pi$, which isn't right!

What am I doing wrong?

2. Hello, saxyliz!

Did you forget the denominator?

The radius of a right circular cone is increasing at a rate of 5 in/sec
and its height is decreasing at a rate of 3 in/sec.
At what rate is the volume of the cone changing
when the radius is 50 inches and the height is 40 inches?
Your derivative is correct . . .

We have: . $V \;=\;\frac{\pi}{3}r^2h$

Then: . $\frac{dV}{dt} \;=\;\frac{2\pi}{3}rh\!\cdot\!\frac{dh}{dt} + \frac{\pi}{3}r^2\!\cdot\!\frac{dh}{dt}$

We are given: . $r = 50,;\;h=40,\;\;\frac{dr}{ht} = 5,\;\;\frac{dh}{dt} = \text{-}3$

Therefore: . $\frac{dV}{dt} \;=\;\frac{2\pi}{3}(50)(40)(5) + \frac{\pi}{3}(50^2)(-3) \;=\;\frac{12,500}{3}\pi$ in³/sec.

3. Oh my goodness! It's a cone!! Wow. Neither me nor the person who helped me caught that, we read it as a cylinder for some reason... I feel dumb now.

Thank you so much!