document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=187283>Question 187283</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A trough is 2 m long, and its ends are triangles with sides of length 1 m, 1 m, and 1.2 m.  \r<BR>\n" );
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document.write( "a) Find the volume V of the water in the trough as a function of the water level h. \r<BR>\n" );
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document.write( "b) If water is pumped into the empty trough at the rate of 6 L/min, find the water level h as a function of the time t after the pumping begins. (1 m^3 = 1000L)  </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #140425 by </B><A HREF=/tutors/aboutme.mpl?userid=solver91311><B>solver91311(24713)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=solver91311><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=solver91311><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=140425\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <font face=\"Times New Roman\" size=\"+2\">\r<BR>\n" );
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document.write( "You don't say, but I have to presume that the sides of the trough are 1 meter and the measure across the top is 1.2 meters.\r<BR>\n" );
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document.write( "Part a.\r<BR>\n" );
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document.write( "Since the triangle is isoceles, the height segment is a perpendicular bisector of the base, forming two right triangles with sides 1, .6, and <IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\Large sqrt{1 - (0.6)^2} = 0.8\">\r<BR>\n" );
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document.write( "That means the base varies as the height in the proportion 1.2 to 0.8, or:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \  \frac{b}{h}\ =\ \frac{1.2}{0.8} \ \ \Rightarrow\ \ b\ =\ \frac{3h}{2}\">\r<BR>\n" );
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document.write( "Hence, the area of the triangle as a function of the height of the water is:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ A(h)\ =\ \frac{\frac{3h^2}{2}}{2}\ =\ \frac{3h^2}{4}m^2\">\r<BR>\n" );
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document.write( "And therefore the volume is just the area times the length of the trough:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ V(h)\ =\ 2\,\cdot\,\frac{3h^2}{4}\ =\ \frac{3h^2}{2}m^3\">\r<BR>\n" );
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document.write( "Part b.\r<BR>\n" );
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document.write( "We need to find volume per unit time, or\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \  \frac{V(h)}{t} = 6 L/min = .006 m^3/min\">\r<BR>\n" );
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document.write( "So:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ \frac{3h^2}{2t}m^3/min\ =\ .006 m^3/min\"> \r<BR>\n" );
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document.write( "Solve for <i>h</i>:\r<BR>\n" );
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document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE \ \ \ \ \ \ \ \ \ \ h^2\ =\ \frac{.006\,\cdot\,2t}{3}\ =\ .004t \ \ \Rightarrow\ \ h(t) = sqrt{.004t}\ m\">\r<BR>\n" );
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document.write( "John<BR>\n" );
document.write( "<IMG STYLE=\"max-width:80%;\" SRC=\"/cgi-bin/rendertex.cgi?\LARGE e^{i\pi} + 1 = 0\"><BR>\n" );
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