document.write( "Question 1163173: A spherical snow ball is melting at 120cm3/min. At what rate is the surface area decreasing at when the radius is 15cm?
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Algebra.Com's Answer #787235 by ikleyn(52788) $\"\"$ $\"About$

You can put this solution on YOUR website!
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document.write( "The volume of a sphere is  V = .\r\n" );
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document.write( "Since both the volume and the radius depend on time, the formula becomes  V(t) = .\r\n" );
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document.write( "The derivative over time is   = \r\n" );
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document.write( "Hence,   =  = .   (1)\r\n" );
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document.write( "The surface area of a sphere is  A = ,  or  A(t) = .\r\n" );
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document.write( "The derivative over time is   =  = \r\n" );
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document.write( "Substitute here  the expression   =   from (1)  and r(t) = 15 cm.  You will get\r\n" );
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document.write( "     =  =  =  = 2*8 = 16 cm^2/min.\r\n" );
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