document.write( "Question 893873: A particle is moving along the curve y= 3 \sqrt{5 x + 6}. As the particle passes through the point (2, 12), its x-coordinate increases at a rate of 5 units per second. Find the rate of change of the distance from the particle to the origin at this instant. \n" ); document.write( "

Algebra.Com's Answer #541615 by Fombitz(32388) $\"\"$ $\"About$

You can put this solution on YOUR website!
$\"y=3%2F%28sqrt%285x%2B6%29%29=3%2A%285x%2B6%29%5E%28-1%2F2%29\"$
\n" ); document.write( " $\"dy%2Fdx=-%2815%2F2%29%285x%2B6%29%5E%28-3%2F2%29\"$
\n" ); document.write( "At (2,12),
\n" ); document.write( " $\"dx%2Fdt=2\"$
\n" ); document.write( " $\"dy%2Fdt=%28dy%2Fdx%29%28dx%2Fdt%29=-%2815%2F2%29%285%282%29%2B6%29%5E%28-3%2F2%29%2A2\"$
\n" ); document.write( " $\"dy%2Fdt=%28dy%2Fdx%29%28dx%2Fdt%29=-15%2816%29%5E%28-3%2F2%29\"$
\n" ); document.write( " $\"dy%2Fdt=%28dy%2Fdx%29%28dx%2Fdt%29=-15%2F64\"$
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( ".
\n" ); document.write( " $\"V=sqrt%28%28dx%2Fdt%29%5E2%2B%28dy%2Fdt%29%5E2%29\"$
\n" ); document.write( " $\"V=sqrt%282%5E2%2B%2815%2F64%29%5E2%29\"$
\n" ); document.write( " $\"V=sqrt%284%2B225%2F4096%29\"$
\n" ); document.write( " $\"V=sqrt%2816384%2F4096%2B225%2F4096%29\"$
\n" ); document.write( " $\"V=sqrt%2816609%29%2F64%29\"$ \n" ); document.write( "

\n" );