# Chain rule homework

Does anyone know how to do this with chain rule??? :surprised

If a cone has height 1 m and radius 30 cm, and the height is increasing at a rate of 1 cm/s, whereas the radius is decreasing at a rate of 1 cm/s, what is the rate of change of the cones volume? Solve the problem using the chain rule.

Thanks!!!

## Answers and Replies

dr/dt = .1 r = .1t + .3
dh/dt = .1 h= .1t + 1

v = 1/3 pi r^2 h
v = 1/3 pi (.1t+.3)^2 (.1t +1)
dv/dt = 1/3 pi 2(.1t+3) (.1) (.1)
notice chain rule used

dv/dt = 2pi/3 (.001t + .003)

HallsofIvy
Or: Since $V= \frac{\pi}{3}r^h$, $\frac{dV}{dt}= \frac{2\pi}{3}rh\frac{dr}{dt}+ \frac{\pi}{3}r^2$
We are told that $\frac{dr}{dt}= -1 cm/sec$ and $\frac{dh}{dt}= 1 cm/sec$
so $\frac{dV}{dt}=\frac{\pi}{3}r^2- \frac{2\pi}{3}rh cm^3/sec$