document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "Answer: 160pi cm^2/cm
\n" ); document.write( "How do I get it though?
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Algebra.Com's Answer #118565 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
\"SA=4%2Api%2AR%5E2\"
\n" ); document.write( "You're looking for the rate of change of SA with respect to R.
\n" ); document.write( "That's the same as the derivative.
\n" ); document.write( "Take a small step in R, calculate the SA, subtract the original SA, and divide by the small step.
\n" ); document.write( "\"SA%28R%2BdR%29=4%2Api%2A%28R%2BdR%29%5E2=4%2Api%2A%28R%5E2%2B2RdR%2BdR%5E2%29\"
\n" ); document.write( "\"SA%28R%29=4%2Api%2AR%5E2\"
\n" ); document.write( "\"%28SA%28R%2BdR%29-S%28R%29%29%2FdR=%284%2Api%2A%28R%5E2%2B2RdR%2BdR%5E2-R%5E2%29%29%2FdR\"
\n" ); document.write( "Assume that \"dR\" is small, then \"dR%5E2\" is even smaller (=0)
\n" ); document.write( "\"%28SA%28R%2BdR%29-S%28R%29%29%2FdR=%284%2Api%2A%282RdR%2Bcross%28dR%5E2%29%29%29%2FdR\"
\n" ); document.write( "\"%28SA%28R%2BdR%29-S%28R%29%29%2FdR=8%2Api%2AR\"
\n" ); document.write( ".
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\n" ); document.write( ".
\n" ); document.write( "That's the hard way to find the derivative of SA with respect to R.
\n" ); document.write( "You can also differentiate.
\n" ); document.write( "\"SA=4%2Api%2AR%5E2\"
\n" ); document.write( "\"d%28SA%29%2FdR=8%2Api%2AR\"
\n" ); document.write( "So when R=20
\n" ); document.write( "\"d%28SA%29%2FdR=8%2Api%2A20=160%2Api\"
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