document.write( "Question 1163171: please help\r
\n" ); document.write( "\n" ); document.write( "A circular plate is cooled and contracts so its radius is shrinking by 0.01cm/s. How fast is the area of one face changing when the radius is 2.5cm?\r
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Algebra.Com's Answer #787224 by ikleyn(52797) $\"\"$ $\"About$

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document.write( "The area of the plate is  A = ,\r\n" );
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document.write( "or, since the radius depends on time  r = r(t),  the area is\r\n" );
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document.write( "    A(t) = .      (1)\r\n" );
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document.write( "The derivative A'(t) is\r\n" );
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document.write( "    A'(t) = 2*pi*r(t)*r'(t).     (2)\r\n" );
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document.write( "We are given  r'(t) = 0.01 sm/s  and  r(t) = 2.5 cm.  Substitute these values into the formula (2) and get the\r\n" );
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document.write( "ANSWER.  A'(t) = 2*pi*2.5*0.01 = 2*3.14159*2.5*0.01 = 0.157 cm^2/s.\r\n" );
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