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(D) Let $x$ be the thickness of the ice layer. The radius of the sphere including the ice is $r = (10 + x) \, cm$.The volume of the ice layer is $V =

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$1 / (18\pi) \, cm/min$

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(D) Let $x$ be the thickness of the ice layer. The radius of the sphere including the ice is $r = (10 + x) \, cm$.The volume of the ice layer is $V = \frac{4}{3}\pi (10 + x)^3 - \frac{4}{3}\pi (10)^3$.Differentiating with respect to time $t$,we get $\frac{dV}{dt} = 4\pi (10 + x)^2 \frac{dx}{dt}$.Given that the ice melts at a rate of $50 \, cm^3/min$,we have $\frac{dV}{dt} = -50 \, cm^3/min$ (since volume is decreasing).Substituting $x = 5$ and $\frac{dV}{dt} = -50$ into the equation:$-50 = 4\pi (10 + 5)^2 \frac{dx}{dt}$$-50 = 4\pi (15)^2 \frac{dx}{dt}$$-50 = 4\pi (225) \frac{dx}{dt}$$-50 = 900\pi \frac{dx}{dt}$$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \, cm/min$.The rate at which the thickness decreases is $\frac{1}{18\pi} \, cm/min$.

$A$ spherical iron ball of radius $10 \, cm$ is coated with a layer of ice of uniform thickness that melts at a rate of $50 \, cm^3/min$. When the thickness of the ice is $5 \, cm$,find the rate at which the thickness of the ice decreases.

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