A snowball melts at a rate of 0.05 [cm3 / s]. At what speed does the surface area decrease when the sphere is 10 [cm] in diameter?

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Answer 1 · 2018-06-10T20:57:25

#(dA)/dt= - 0.02 """"" "cm^2"/"sec"#

The figure is a sphere
Volume decreases
#(dV)/dt=-0.05""""" "cm^3"/"sec"#

#V=(4/3)pir^3#
Differentiate with respect to time

#(dV)/dt=(4/3)*pi*3r^2*(dr)/dt#

Substitute #(dV)/dt=-0.05# and #r=D/2=10/2=5#

#-0.05=(4/3)*pi*3(5)^2*(dr)/dt#

Solve for #(dr)/dt#

#(dr)/dt=-0.0005/pi""""" "cm"/"sec"#

Now, the surface Area

#A=4pir^2#
Differentiate with respect to time

#(dA)/dt=4pi*2r*(dr)/dt#

Substitute now #r=5# and #(dr)/dt=-0.0005/pi#

#(dA)/dt=4pi*2r*(dr)/dt#

#(dA)/dt=4pi*2*5*(-0.0005/pi)#

#(dA)/dt=-0.02""""" "cm^2"/"sec"#

Negative sign means decreasing surface area.

I hope the explanation is useful...God bless