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

$\frac{d A}{d t} = - 0.02 c m^{2} / sec$ $(dA)/dt= - 0.02 """"" "cm^2"/"sec"$

The figure is a sphere
Volume decreases
$\frac{d V}{d t} = - 0.05 c m^{3} / sec$ $(dV)/dt=-0.05""""" "cm^3"/"sec"$

$V = (\frac{4}{3}) π r^{3}$ $V=(4/3)pir^3$
Differentiate with respect to time

$\frac{d V}{d t} = (\frac{4}{3}) \cdot π \cdot 3 r^{2} \cdot \frac{d r}{d t}$ $(dV)/dt=(4/3)*pi*3r^2*(dr)/dt$

Substitute $\frac{d V}{d t} = - 0.05$ $(dV)/dt=-0.05$ and $r = \frac{D}{2} = \frac{10}{2} = 5$ $r=D/2=10/2=5$

$- 0.05 = (\frac{4}{3}) \cdot π \cdot 3 {(5)}^{2} \cdot \frac{d r}{d t}$ $-0.05=(4/3)*pi*3(5)^2*(dr)/dt$

Solve for $\frac{d r}{d t}$ $(dr)/dt$

$\frac{d r}{d t} = - \frac{0.0005}{π} c m / sec$ $(dr)/dt=-0.0005/pi""""" "cm"/"sec"$

Now, the surface Area

$A = 4 π r^{2}$ $A=4pir^2$
Differentiate with respect to time

$\frac{d A}{d t} = 4 π \cdot 2 r \cdot \frac{d r}{d t}$ $(dA)/dt=4pi*2r*(dr)/dt$

Substitute now $r = 5$ $r=5$ and $\frac{d r}{d t} = - \frac{0.0005}{π}$ $(dr)/dt=-0.0005/pi$

$\frac{d A}{d t} = 4 π \cdot 2 r \cdot \frac{d r}{d t}$ $(dA)/dt=4pi*2r*(dr)/dt$

$\frac{d A}{d t} = 4 π \cdot 2 \cdot 5 \cdot (- \frac{0.0005}{π})$ $(dA)/dt=4pi*2*5*(-0.0005/pi)$

$\frac{d A}{d t} = - 0.02 c m^{2} / sec$ $(dA)/dt=-0.02""""" "cm^2"/"sec"$

Negative sign means decreasing surface area.

I hope the explanation is useful...God bless