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Chain rule derivative applied to an ice cube

  1. Oct 21, 2010 #1
    1. The problem statement, all variables and given/known data

    A cubical block of ice is melting in such a way that each edge decreases steadily by 9.8 cm every hour. At what rate is its volume decreasing when each edge is 10 meters long?

    2. Relevant equations

    V(t) = (l(t))^3 m^3
    l'(t) = 0.098 m/h

    3. The attempt at a solution


    V'(t) = f'(g)*g(t) chain rule formula, f(g) = g^3, g(t) = l = 10
    V'(t) = 3g^2 * 0.098
    V'(t) = 3(10)^2 * 0.098
    V'(t) = 300 * 0.098
    V'(t) = 29.4 m^3/h
     
  2. jcsd
  3. Oct 21, 2010 #2

    Mark44

    Staff: Mentor

    My only real quibble is that V'(t) should be negative, since the block of ice is melting.
     
  4. Oct 21, 2010 #3
    that was the answer, thanks. i don't like doing online assignments because there is no feedback. the website kept telling me i was wrong and i had no idea where
     
  5. Oct 22, 2010 #4

    Mark44

    Staff: Mentor

    This is how I would do this problem.

    V(t) = (x(t))3
    V'(t) = 3(x(t))2 * x'(t)

    x'(t) = -0.098 m/hr is constant.
    At some time t = t0, x(t0) = 10,
    so V'(t0) = 3 * 102 * -0.098 = -29.4 m3/hr.
     
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