A spherical iron ball of radius 10 text{ cm} | vedclass

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Question

(A) Let $x$ be the thickness of the ice layer. The total radius of the ball including the ice is $R = 10 + x$.The volume of the ice layer is $V = \fra

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $x$ be the thickness of the ice layer. The total radius of the ball including the ice is $R = 10 + x$.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 $\frac{dV}{dt} = -50 	ext{ cm}^3/	ext{min}$ (since it melts) and $x = 5 	ext{ cm}$.Substituting these values: $-50 = 4\pi(10+5)^2 \frac{dx}{dt}$.$-50 = 4\pi(15)^2 \frac{dx}{dt} \implies -50 = 4\pi(225) \frac{dx}{dt}$.$-50 = 900\pi \frac{dx}{dt} \implies \frac{dx}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi} 	ext{ cm/min}$.Thus, the rate at which the thickness decreases is $\frac{1}{18\pi} 	ext{ cm/min}$.

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

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