I solved the following problem myself:
"The radius r(t) of a cone is increasing at a rate of 3 centimeters per second and the height h(t) of the cone is decreasing at a rate of 4 centimeters per second. At a certain instant t_0, the radius is 8 centimeters and the height is 10 centimeters. What is the rate of change of the volume V(t) of the cone at that instant?"
and have come up with the answer -64π. Can someone verify this?
Here is my attempt:
Since: \(V_{cone}=\frac{1}{3}\pi r^2h\) (*)
and given: \(\dfrac{dr}{dt}=3\), and \(\dfrac{dh}{dt}=-4\)
Differentiate (*) with respect to t
\(\dfrac{d}{dt}V(t)=\frac{1}{3}\pi(2r\frac{dr}{dt}h+r^2\frac{dh}{dt})\)
We want to find the rate at \(t_0\) and we are given r=8 and h=10:
\(V'(t)=\frac{1}{3}\pi(2(8)(3)+(8)^2(-4))=\frac{224}{3}\pi\)
For the differentiation part: notice that r(t) and h(t) are functions of time and they are varying. So when we differentiate r, we do the normal differentiation (r -->1) but then applying the chain rule, we multiply by r' which is dr/dt, and similarly for h). So to differentiate (*) you need to know the product rule and chain rule.
