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Chain rule

  1. Sep 27, 2004 #1
    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.

  2. jcsd
  3. Sep 27, 2004 #2
    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)
  4. Sep 28, 2004 #3


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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]
    Last edited: Sep 28, 2004
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