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If the radius of a circular blot of oil is increasing at the rate of $2 \mathrm{~cm} / \mathrm{min}$, then the rate
of change of its area when its radius is $3 \mathrm{cms}$ is
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of change of its area when its radius is $3 \mathrm{cms}$ is
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Verified Answer
The correct answer is:
$12 \pi \mathrm{cm}^{2} / \mathrm{min}$
(A)
Given $\frac{\mathrm{dr}}{\mathrm{dt}}=2 \mathrm{~cm} / \mathrm{min}$ and $\mathrm{r}=3 \mathrm{cms}$
Area $=\pi r^{2}$
Differentiating w.r.t. $t$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\pi \cdot 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=\pi \times 2(3) \times(2)=12 \pi \mathrm{cm}^{2} / \mathrm{min}$
Given $\frac{\mathrm{dr}}{\mathrm{dt}}=2 \mathrm{~cm} / \mathrm{min}$ and $\mathrm{r}=3 \mathrm{cms}$
Area $=\pi r^{2}$
Differentiating w.r.t. $t$
$\frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}}=\pi \cdot 2 \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}}=\pi \times 2(3) \times(2)=12 \pi \mathrm{cm}^{2} / \mathrm{min}$
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