SOLUTION: GENERAL: Hailstones A hailstone (a small sphere of ice) is forming in the clouds so that its radius is growing at the rate of 1 millimeter per minute. How fast is its volume growin

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Question 1125312: GENERAL: Hailstones A hailstone (a small sphere of ice) is forming in the clouds so that its radius is growing at the rate of 1 millimeter per minute. How fast is its volume growing at the moment when the radius is 2 millimeters? [Hint: The volume of a sphere of radius r is V = 4/3*pi r^3.]
Found 3 solutions by Alan3354, ikleyn, rothauserc:
Answer by Alan3354(69443)   (Show Source): You can put this solution on YOUR website!
Hailstones A hailstone (a small sphere of ice) is forming in the clouds so that its radius is growing at the rate of 1 millimeter per minute. How fast is its volume growing at the moment when the radius is 2 millimeters?
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dV/dt = 4pi*r^2*dr/dt
dV/dt = 4pi*4*1 = 16pi cubic mms/minute
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Hint: We don't need hints.

Answer by ikleyn(52805)   (Show Source): You can put this solution on YOUR website!
.
You are given  V(t) = V(r(t)) = ,  where the radius  r  varies  such that   = 1 millimeter per minute.


Then   = . = . = 4*3.14*4 =  = 50.24 mm^3 per minute.


Thus the instantaneous volume growing rate is  50.24 mm^3/minute.



Interesting, that if you calculate the averaged volume growing rate between  t= 2 seconds  and  t= 3 seconds, you will get another value


 =  -  =  =  =  = 79.55  mm^3/minute.


But it only demonstrates the difference between these two conceptions: instantaneous and averaged rates.


Answer.   The instantaneous volume growing rate at this moment is  50.24 mm^3 per minute.



Answer by rothauserc(4718)   (Show Source): You can put this solution on YOUR website!
take first derivative of V and r with respect to time
:
dV/dt = 3 * (4/3) * pi * r^2 * dr/dt = 4 * pi * r^2 * dr/dt
:
dV/dt = 4 * pi * 4 * 1 = 16 * pi is approximately 50.27 mm^3 per minute
:

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