document.write( "Question 1082937: Two botas starts at the same point. One sail due east starting 10 A.M. at a constant rate of 20 kph. The other sail due south starting 11 A.M. at constant rate of 9 kph. How fast are they separating at noon?
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Algebra.Com's Answer #701297 by Fombitz(32388)\"\" \"About 
You can put this solution on YOUR website!
The distance between them is the hypotenuse of a right triangle who's legs are measured by the rate of each boat.
\n" ); document.write( "\"A%5E2%2BB%5E2=H%5E2\"
\n" ); document.write( "Let's measure time starting at 11 AM,
\n" ); document.write( "\"A=20t%2B20=20%28t%2B1%29\"
\n" ); document.write( "\"B=9t\"
\n" ); document.write( "So then,
\n" ); document.write( "\"400%28t%2B1%29%5E2%2B%289t%29%5E2=H%5E2\"
\n" ); document.write( "\"400%28t%5E2%2B2t%2B1%29%2B81t%5E2=H%5E2\"
\n" ); document.write( "\"H%5E2=481t%5E2%2B800t%2B400\"
\n" ); document.write( "So at noon, \"t=2\",
\n" ); document.write( "\"H%5E2=481%284%29%2B800%282%29%2B400\"
\n" ); document.write( "\"H%5E2=1924%2B1600%2B400\"
\n" ); document.write( "\"H%5E2=3924\"
\n" ); document.write( "\"H=sqrt%283924%29\"
\n" ); document.write( "\"H=6sqrt%28109%29\"
\n" ); document.write( "So now to find the rate of separation, implicitly differentiate the equation above with respect to time.
\n" ); document.write( "\"2A%28dA%2Fdt%29%2B2B%28dB%2Fdt%29=2H%28dH%2Fdt%29\"
\n" ); document.write( "\"A=20%282%2B1%29=60\"
\n" ); document.write( "\"B=9%282%29=18\"
\n" ); document.write( "\"dA%2Fdt=20\"
\n" ); document.write( "\"dB%2Fdt=9\"
\n" ); document.write( "Substituting,
\n" ); document.write( "\"2%2860%29%2820%29%2B2%2818%29%289%29=2%286sqrt%28109%29%29%28dH%2Fdt%29\"
\n" ); document.write( "\"12sqrt%28109%29%28dH%2Fdt%29=2724%29\"
\n" ); document.write( "\"dH%2Fdt=227%2Fsqrt%28109%29\"
\n" ); document.write( "\"highlight%28dH%2Fdt=%28227%2F109%29sqrt%28109%29%29\"\"kph\"\r
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