Chain rule homework

  • Thread starter Monsu
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  • #1
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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

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