SOLUTION: A spherical snowball is melting in such a way that its surface area decreases at the rate of 1 in 2 /min . How fast is its volume shrinking when its radius is 3 in ?

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Question 1172158: A spherical snowball is melting in such a way that its surface area decreases at the rate of 1 in 2 /min . How fast is its volume shrinking when its radius is 3 in ?
Found 2 solutions by Alan3354, math_helper:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
the rate of 1 in 2 /min
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What does that mean?

Answer by math_helper(2461) About Me  (Show Source):
You can put this solution on YOUR website!
Assuming that "1 in 2/min" is supposed to be +1+%28in%5E2%2Fmin%29+...
+V+ = +%284%2F3%29pi%2Ar%5E3+
+A+ = ++4%2Api%2Ar%5E2+

V in terms of A:
+V+ =

Taking the derivative of V wrt A:
+dV%2FdA+ = +sqrt%28A%29%2F%284%2Asqrt%28pi%29%29+

Need an expression for dV/dt in terms of information given:
+dV%2Fdt+ = +%28dV%2FdA%29+%2A+%28dA%2Fdt%29+

At r=3, A=4%2Api%2A3%5E2 = 36pi ---> sqrt%28A%29 = +6%2Asqrt%28pi%29+

dA/dt is given as 1+%28in%5E2%2Fmin%29+

So, finally, +dV%2Fdt+ = +%286%2Asqrt%28pi%29+%2F+%284%2Asqrt%28pi%29%29%29+%2A+1+ = highlight%283%2F2%29in%5E3%2Fmin
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Another approach would be to use 'r' instead of 'A': dV/dt = (dV/dr)*(dr/dt)
It is about equally messy to go this route. Because you are given dA/dt, you
have the additional step dr/dt = (dA/dt)*(dr/dA), so dV/dt = (dV/dr)*(dA/dt)*(dr/dA)
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