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(D) Let $r$ be the thickness of the ice layer at any time $t$ and $V$ be the volume of the ice layer.Given that the rate of change of volume is $\frac

Vedclass · Accepted Answer

$1/(18\pi)$

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(D) Let $r$ be the thickness of the ice layer at any time $t$ and $V$ be the volume of the ice layer.Given that the rate of change of volume is $\frac{dV}{dt} = -50 	ext{ cm}^3/	ext{min}$.We need to find $\frac{dr}{dt}$ when $r = 5 	ext{ cm}$.The radius of the sphere including the ice layer is $(r + 10) 	ext{ cm}$.The volume of the ice layer $V$ is the volume of the sphere with ice minus the volume of the iron sphere:$V = \frac{4}{3}\pi(r + 10)^3 - \frac{4}{3}\pi(10)^3$.Differentiating both sides with respect to $t$:$\frac{dV}{dt} = 4\pi(r + 10)^2 \frac{dr}{dt}$.Substituting the given values $\frac{dV}{dt} = -50$ and $r = 5$:$-50 = 4\pi(5 + 10)^2 \frac{dr}{dt}$.$-50 = 4\pi(15)^2 \frac{dr}{dt}$.$-50 = 4\pi(225) \frac{dr}{dt}$.$-50 = 900\pi \frac{dr}{dt}$.$\frac{dr}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} 	ext{ cm/min}$.Since the question asks for the rate at which the thickness is decreasing,the rate is $1/(18\pi) 	ext{ cm/min}$.

An ice layer of uniform thickness is deposited on an iron sphere of radius $10 \text{ cm}$. The ice melts at a rate of $50 \text{ cm}^3/\text{min}$. When the thickness of the ice layer is $5 \text{ cm}$,the rate at which the thickness of the ice layer is decreasing is ...... $\text{cm/min}$.

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