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

Vedclass · Accepted Answer

$\frac{1}{18\pi} \ cm/min$

Vedclass · Answer

(D) Let $x$ be the thickness of the ice. The total radius of the ball 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 $\frac{dV}{dt} = -50 \ cm^3/min$ (since the ice is melting,the volume is decreasing).At $x = 5 \ cm$,we have $-50 = 4\pi(10 + 5)^2 \frac{dx}{dt}$.$-50 = 4\pi(15)^2 \frac{dx}{dt} = 4\pi(225) \frac{dx}{dt} = 900\pi \frac{dx}{dt}$.Therefore,$\frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} \ cm/min$.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. The ice melts at a rate of $50 \ cm^3/min$. When the thickness of the ice is $5 \ cm$,the rate at which the thickness of the ice decreases is:

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