Solving Cone Volume Change with Chain Rule

[Monsu]

Does anyone know how to do this with chain rule?

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!

 

[Phymath]

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
(both product rule and chain rule used!)
We are told that \frac{dr}{dt}= -1 cm/sec and \frac{dh}{dt}= 1 cm/sec
(Phymath: you missed the fact that r is decreasing! Also you do not state the units, which is crucial.)
so \frac{dV}{dt}=\frac{\pi}{3}r^2- \frac{2\pi}{3}rh cm^3/sec