document.write( "Question 1106186: Water exits a conical tank at a constant rate of o.2 m3/minutes. If the surface of the water has radius r:
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document.write( "a) find V(r), the volume of the water remaining
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document.write( "b) find the rate at which the surface radius is changing at the instant when the water is 5m deep.
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document.write( "The diameter is 6m. Height 8m. \n" );
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Algebra.Com's Answer #721138 by greenestamps(13200)![]() ![]() You can put this solution on YOUR website! \n" ); document.write( "The conical tank has a radius of 3m and a height (depth) of 8m. The ratio of the radius of the surface of the water to the depth of the water will always be 3:8. So \n" ); document.write( " \n" ); document.write( "(a) The volume remaining at any time is \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "(b) We want to find the rate at which the radius of the surface of the water is changing (dr/dt) when the depth is 5m. We are given that the volume of water in the tank is decreasing at a rate of 0.2 m^3/minute: dV/dt = -0.2. \n" ); document.write( " \n" ); document.write( " \n" ); document.write( "We know dV/dt; and from part (a) we can find dV/dr to be \n" ); document.write( "So \n" ); document.write( " \n" ); document.write( "When the depth is 5m, the radius is (3/8)*5 = (15/8)m. So dr/dt when the depth is 5 is \n" ); document.write( " \n" ); document.write( " \n" ); document.write( " |