SOLUTION: A spherical balloon is being inflated. Estimate the rate at which its surface area is changing with respect to the radius when the radius measures 20 cm. Answer: 160pi cm^2/cm

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Question 160801: A spherical balloon is being inflated. Estimate the rate at which its surface area is changing with respect to the radius when the radius measures 20 cm.
Answer: 160pi cm^2/cm
How do I get it though?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
SA=4%2Api%2AR%5E2
You're looking for the rate of change of SA with respect to R.
That's the same as the derivative.
Take a small step in R, calculate the SA, subtract the original SA, and divide by the small step.
SA%28R%2BdR%29=4%2Api%2A%28R%2BdR%29%5E2=4%2Api%2A%28R%5E2%2B2RdR%2BdR%5E2%29
SA%28R%29=4%2Api%2AR%5E2
%28SA%28R%2BdR%29-S%28R%29%29%2FdR=%284%2Api%2A%28R%5E2%2B2RdR%2BdR%5E2-R%5E2%29%29%2FdR
Assume that dR is small, then dR%5E2 is even smaller (=0)
%28SA%28R%2BdR%29-S%28R%29%29%2FdR=%284%2Api%2A%282RdR%2Bcross%28dR%5E2%29%29%29%2FdR
%28SA%28R%2BdR%29-S%28R%29%29%2FdR=8%2Api%2AR
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.
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That's the hard way to find the derivative of SA with respect to R.
You can also differentiate.
SA=4%2Api%2AR%5E2
d%28SA%29%2FdR=8%2Api%2AR
So when R=20
d%28SA%29%2FdR=8%2Api%2A20=160%2Api