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:
\n" ); document.write( "a) find V(r), the volume of the water remaining
\n" ); document.write( "b) find the rate at which the surface radius is changing at the instant when the water is 5m deep.
\n" ); document.write( "The diameter is 6m. Height 8m.
\n" ); document.write( "

Algebra.Com's Answer #721138 by greenestamps(13200)\"\" \"About 
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( "\"r+=+%283%2F8%29h\"; \"h+=+%288%2F3%29r\"

\n" ); document.write( "(a) The volume remaining at any time is
\n" ); document.write( "

\n" ); document.write( "\"V%28r%29+=+%288%2F9%29%28pi%29%28r%5E3%29\"
\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( "\"dV%2Fdt+=+%28dV%2Fdr%29%2A%28dr%2Fdt%29\"
\n" ); document.write( "\"dr%2Fdt+=+%28dV%2Fdt%29%2F%28dV%2Fdr%29\"

\n" ); document.write( "We know dV/dt; and from part (a) we can find dV/dr to be \"%288%2F3%29%28pi%29%28r%5E2%29\"

\n" ); document.write( "So

\n" ); document.write( "\"dr%2Fdt+=+%28-0.2%29%2F%28%288%2F3%29%28pi%29%28r%5E2%29%29\"

\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( "\"dr%2Fdt+=+%28-0.2%29%2F%28%288%2F3%29%28pi%29%28%2815%2F8%29%5E2%29%29\"
\n" ); document.write( "\"dr%2Fdt+=+%28-1%2F5%29%2F%28%2875%2F8%29%28pi%29%29\"
\n" ); document.write( "\"dr%2Fdt+=+%28-8%29%2F%28375%28pi%29%29\"
\n" ); document.write( "
\n" );