document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1163173>Question 1163173</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A spherical snow ball is melting at 120cm3/min. At what rate is the surface area decreasing at when the radius is 15cm?<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 #787203 by </B><A HREF=/tutors/aboutme.mpl?userid=dkppathak><B>dkppathak(439)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dkppathak><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=dkppathak><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=787203\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  A spherical snow ball is melting at 120cm3/min. At what rate is the surface area decreasing at when the radius is 15cm?\r<BR>\n" );
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document.write( "solution<BR>\n" );
document.write( "dv/dt=120cm^3/min  given  <BR>\n" );
document.write( "v=4/3(πr3)<BR>\n" );
document.write( "dv/dt=4πr2 dr/dt   than dr/dt=120/4πr2<BR>\n" );
document.write( "Surface area of sphere  A=4πr2<BR>\n" );
document.write( "da/dt=A=4*2πr dr/dt=8πrdr/dt<BR>\n" );
document.write( "da/dt=8πrx120/4πr2=240/r<BR>\n" );
document.write( "putting r=15<BR>\n" );
document.write( "da/dt=240/15=16<BR>\n" );
document.write( "rate of change of decreasing  surface are 16cm^2/min </SPAN>\n" );
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