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Related rate with a sphere

  1. Jun 7, 2007 #1
    1. The problem statement, all variables and given/known data

    An orange is spherical. Suppose it grows so that its volume increases at an average rate of 4cm^3/day. Determine the rate at which the radius of the orange is changing six weeks after it begins growing.

    So we are given dV/dt = 4 and looking at t = 42 days

    2. Relevant equations

    V = 4/3(pi*r^3) for vol of sphere

    3. The attempt at a solution

    Im looking for dr/dt at 42 days.

    dV/dt = dV/dr * dr/dt

    dV/dt = 4 and dV/dr = 4pi*r^2 (derivative of V = 4/3(pi*r^3))

    dV/dt = dV/dr * dr/dt
    4 = 4pi*r^2 * dr/dt
    dr/dt = 1/(pi*r^2) (divided 4pi*r^2 and reduce 4)

    But I dont know where to go from here. I need a radius at 42 days but if I just use the 4cm^3/day then i end up with 1.13x10^-5 as the rate which doesnt seem to make sense. Also it would then seem as though the orange started with a radius of 0 which is dumb.
  2. jcsd
  3. Jun 7, 2007 #2
    Ok Ive got it. Just had to find volume at 42 days using dV/dt and then find r with the volume formula. Forgot that 4cm^3/day was volume and not radius (doh).
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