document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1163278>Question 1163278</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The level of water in a rain gutter is given by the following<BR>\n" );
document.write( "V(t) =  <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=+10t+sqrt%28+4-t%5E2+%29+ ALIGN=MIDDLE  ALT=\"+10t+sqrt%28+4-t%5E2+%29+\"   DISABLED_style= \"max-width:90%; height:auto;\" >. Note : The square root contains 4-t^2\r<BR>\n" );
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document.write( "a. Determine dV/dt\r<BR>\n" );
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document.write( "b. Is the water level rising or falling at t = 1.5 seconds?\r<BR>\n" );
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document.write( "Thank you.\r<BR>\n" );
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document.write( "10t(√4-t^2)\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #787331 by </B><A HREF=/tutors/aboutme.mpl?userid=Boreal><B>Boreal(15235)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Boreal><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=Boreal><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=787331\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> V=10t*(4-t^2)^(1/2)<BR>\n" );
document.write( "dV/dt=10t *(1/2)(4-t^2)^(-1/2)*(-2t)+10*sqrt(4-t^2)<BR>\n" );
document.write( "This is -10t^2/(sqrt(4-t^2))+10 sqrt(4-t^2).  (the 1/2 and 2 go away)<BR>\n" );
document.write( "when t=1.5<BR>\n" );
document.write( "dV/dt=-22.50/sqrt(1.75)+10*sqrt(1.75)=-17.01+13.23<BR>\n" );
document.write( "=-3.78, which is negative\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=graph%28300%2C300%2C-2%2C3%2C-10%2C30%2C10x%2Asqrt%284-x%5E2%29%29 ALIGN=MIDDLE  ALT=\"graph%28300%2C300%2C-2%2C3%2C-10%2C30%2C10x%2Asqrt%284-x%5E2%29%29\"   DISABLED_style= \"max-width:90%; height:auto;\" >\r<BR>\n" );
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document.write( "-10t^2/sqrt(4-t^2)=-10*sqrt(4-t^2) when derivative equals 0<BR>\n" );
document.write( "t^2=4-t^2<BR>\n" );
document.write( "2t^2=4<BR>\n" );
document.write( "t^2=2<BR>\n" );
document.write( "t=1.414 where the derivative is 0. It was rising until then but it is falling after that, which is where 1.5 seconds is. </SPAN>\n" );
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