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Math Help - related rates question?

  1. #1
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    related rates question?

    Assume the volume V of a cube is increasing at a constant rate of 3cm^3 per second. Let t0 be the instant (t>0) when the rate of change of the volume (cm^3/sec) is numerically equal to the rate of change of the surface area (cm^2/sec) for the cube. Assume V=0 when t=0.

    a. Find the rate of change of the length of a side when t=t0
    b. Find the rate of change of the surface area when t=t0
    c. Find the value of t0.


    Thanks a lot everyone.
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  2. #2
    Senior Member apcalculus's Avatar
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    Quote Originally Posted by GlobalCooling View Post
    Assume the volume V of a cube is increasing at a constant rate of 3cm^3 per second. Let t0 be the instant (t>0) when the rate of change of the volume (cm^3/sec) is numerically equal to the rate of change of the surface area (cm^2/sec) for the cube. Assume V=0 when t=0.

    a. Find the rate of change of the length of a side when t=t0
    b. Find the rate of change of the surface area when t=t0
    c. Find the value of t0.

    Thanks a lot everyone.
    V = r^3 and A= 6r^2 where r depends on t.

    dV/dt = 3 r^2 dr/dt (by the chain rule)

    dA/dt = 12 r dr/dt


    part c) first:

    Setting the two equal to each other, we get 4r = r^2, with solutions r=0 and r = 4. So when the side is 4 units the two rates (dV/dt and dA/dt) are equal.

    a) Use the fact that dV/dt is constant and r = 2, to find dr/dt:
    3 r^2 dr/dt = 3
    3 4^2 dr/dt = 3
    dr/dt = 1/16 cm/s

    part b) should be straight-forward now.

    Good luck!!
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