document.write( "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 ?
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Algebra.Com's Answer #797124 by math_helper(2461)\"\" \"About 
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Assuming that \"1 in 2/min\" is supposed to be \"+1+%28in%5E2%2Fmin%29+\"...\r
\n" ); document.write( "\n" ); document.write( "\"+V+\" = \"+%284%2F3%29pi%2Ar%5E3+\"
\n" ); document.write( "\"+A+\" = \"++4%2Api%2Ar%5E2+\"
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\n" ); document.write( "\n" ); document.write( "V in terms of A:
\n" ); document.write( "\"+V+\" =
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\n" ); document.write( "\n" ); document.write( "Taking the derivative of V wrt A:
\n" ); document.write( "\"+dV%2FdA+\" = \"+sqrt%28A%29%2F%284%2Asqrt%28pi%29%29+\"
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\n" ); document.write( "\n" ); document.write( "Need an expression for dV/dt in terms of information given:
\n" ); document.write( "\"+dV%2Fdt+\" = \"+%28dV%2FdA%29+%2A+%28dA%2Fdt%29+\"
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\n" ); document.write( "\n" ); document.write( "At r=3, A=\"4%2Api%2A3%5E2\" = \"36pi\" ---> \"sqrt%28A%29\" = \"+6%2Asqrt%28pi%29+\"

\n" ); document.write( "dA/dt is given as \"1+%28in%5E2%2Fmin%29+\"
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\n" ); document.write( "\n" ); document.write( "So, finally, \"+dV%2Fdt+\" = \"+%286%2Asqrt%28pi%29+%2F+%284%2Asqrt%28pi%29%29%29+%2A+1+\" = \"highlight%283%2F2%29\"\"in%5E3%2Fmin\"\r
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\n" ); document.write( "Another approach would be to use 'r' instead of 'A': dV/dt = (dV/dr)*(dr/dt)
\n" ); document.write( "It is about equally messy to go this route. Because you are given dA/dt, you
\n" ); document.write( " have the additional step dr/dt = (dA/dt)*(dr/dA), so dV/dt = (dV/dr)*(dA/dt)*(dr/dA) \r
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\n" ); document.write( "Dear student: Please do not use the \"thank you\" message system to post additional problems. Please post your questions (one at a time) using
\n" ); document.write( "the normal process. Thank you.\r
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