document.write( "Question 1056385: Help me solve the problem\r
\n" ); document.write( "\n" ); document.write( "The rate of change of a quantity u with respect to t is du/dt = pt + c, p and c being constants. find u in terms of t given that the quantity has a maximum value of 5.5 when t=1 and that its rate of change when t=2 is -3
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Algebra.Com's Answer #671542 by Fombitz(32388)\"\" \"About 
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\"du%2Fdt=pt%2Bc\"
\n" ); document.write( "\"u=%28pt%5E2%29%2F2%2Bct%2Bd\" where \"d\" is a constant.
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\n" ); document.write( "In vertex form,
\n" ); document.write( "\"u=a%28t-1%29%5E2%2B5.5\"
\n" ); document.write( "\"u=a%28t%5E2-2t%2B1%29%2B5.5\"
\n" ); document.write( "\"u=at%5E2-2at%2B%28a%2B5.5%29\"
\n" ); document.write( "Comparing terms,
\n" ); document.write( "\"a=p%2F2\"
\n" ); document.write( "\"c=-2a=-p\"
\n" ); document.write( "\"d=a%2B5.5\"
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\n" ); document.write( "Using the rate of change,
\n" ); document.write( "\"-3=p%282%29%2Bc\"
\n" ); document.write( "\"2p%2Bc=-3\"
\n" ); document.write( "\"2%28-c%29%2Bc=-3\"
\n" ); document.write( "\"-2c%2Bc=-3\"
\n" ); document.write( "\"-c=-3\"
\n" ); document.write( "\"c=3\"
\n" ); document.write( "So then,
\n" ); document.write( "\"p=-3\"
\n" ); document.write( "\"a=-3%2F2\"
\n" ); document.write( "and
\n" ); document.write( "\"d=-3%2F2%2B11%2F2\"
\n" ); document.write( "\"d=8%2F2\"
\n" ); document.write( "\"d=4\"
\n" ); document.write( "So,
\n" ); document.write( "\"u=-%283%2F2%29%28t-1%29%5E2%2B5.5\"
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