document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1055332>Question 1055332</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The spread of a contaminant on the surface of a lake is increasing in a circular pattern. The radius of the contaminant can be modeled by f(t)=6 square root t ( just 6 squart of t no parenthesis) where f(t) is in meters and t is hours since contamination. Reminder: The area of a circle is pie r^2. (piesign r to the power of 2)\r<BR>\n" );
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document.write( "A. Find a function (A of F)(t) that gives the area of the circular leak in terms of the time t since the spread began. Simplify this function as much as possible.\r<BR>\n" );
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document.write( "B.) Find the size of the contaminated area after 20 hours. Include units and round to the nearest whole number.\r<BR>\n" );
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document.write( "c.) When will the size of the contaminated area be 850 square meters? Include units and round to one decimal place.\r<BR>\n" );
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document.write( "I'm in desperate need of help with this problem. I am really lost. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #670567 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=670567\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> f(t)=6 sqrt(t).  That is the radius<BR>\n" );
document.write( "The area is pi*r^2=pi*36t.  6 sqrt(t) squared is 6 squared* sqrt(t) squared,or 36 t.<BR>\n" );
document.write( "A(t)=36t*pi m^2.<BR>\n" );
document.write( "after 20 hours, the area is 720 pi m^2=2262 m^2.  Check with the radius, which is 6sqrt(20)=26.83 m<BR>\n" );
document.write( "The area is 26.83^2*pi=2261.47, close enough with rounding.<BR>\n" );
document.write( "850 m^2=36*pi*t<BR>\n" );
document.write( "divide 850 by 36*pi=7.5 hours.<BR>\n" );
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document.write( "Check<BR>\n" );
document.write( "radius after 7.5 hours is 6*sqrt(7.5)=16.43 m<BR>\n" );
document.write( "The area is pi*(16.43^2) m^2=848.05 m^2, close enough with rounding. </SPAN>\n" );
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