What is the rate of change for the radius of a growing sphere after six weeks?

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SUMMARY

The discussion focuses on determining the rate of change of the radius of a growing sphere, specifically an orange, after six weeks, given a constant volume increase of 4 cm³/day. The volume formula for a sphere, V = 4/3(pi*r³), is utilized alongside the relationship dV/dt = dV/dr * dr/dt. After calculating the volume at 42 days, the radius is derived, leading to the conclusion that the radius change rate can be accurately determined by first finding the volume at that time and then applying the volume formula.

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Homework Statement



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

Homework Equations



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

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 don't 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 doesn't seem to make sense. Also it would then seem as though the orange started with a radius of 0 which is dumb.
 
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Ok I've 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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