document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=269008>Question 269008</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the depth of fluid,H cm, in a vessel at time t minutes is given by<BR>\n" );
document.write( "H = 16+4t+2t^2-(t^3/16)\r<BR>\n" );
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document.write( "determine the rate at which the depth is changing after 4 minutes.\r<BR>\n" );
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document.write( "if the cross section of the vessel is circular,of diameter 2 m, determine the rate of filling in m3 min-1,at this time. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #197258 by </B><A HREF=/tutors/aboutme.mpl?userid=scott8148><B>scott8148(6628)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=scott8148><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=scott8148><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=197258\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> this is a differential calculus problem using the first derivative of depth with respect to time\r<BR>\n" );
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document.write( "dH/dt = 4 + 4t - (3/16)t^2\r<BR>\n" );
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document.write( "substituting 4 for t gives a rate of 17 cm/min\r<BR>\n" );
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document.write( "the volume rate is ___ .17 pi m^3/min\r<BR>\n" );
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