document.write( "Question 383221: Hi, Please can somebody help with this,\r
\n" ); document.write( "\n" ); document.write( "The base radius r cm of a right-circular cone increases at 2 cm s^-1 and its height h cm at 3 cm s^-1. Show that the rate of increase in its volume V = 1/3 pi r^2h when r = 5 and h = 15 is 125 pi cm^3 s^-1.
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Algebra.Com's Answer #271390 by richard1234(7193)\"\" \"About 
You can put this solution on YOUR website!
We have\r
\n" ); document.write( "\n" ); document.write( "\"V+=+%28pi%2F3%29%28hr%5E2%29\" (separating all the constants)\r
\n" ); document.write( "\n" ); document.write( "h and r are both functions in terms of time, t. We can take dV/dt by the product rule:\r
\n" ); document.write( "\n" ); document.write( "\"dV%2Fdt+=+%28pi%2F3%29%28%28dh%2Fdt%29r%5E2+%2B+h%2A%282r%29%2A%28dr%2Fdt%29%29\"\r
\n" ); document.write( "\n" ); document.write( "At this instant, we have \"dh%2Fdt+=+3+cm%2Fs\", \"h+=+15+cm\", \"dr%2Fdt+=+2+cm%2Fs\", and \"r+=+5+cm\". Substituting all of the known variables in, we obtain\r
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