MHT CET · Maths · Differential Equations
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
- A \(10 \pi \mathrm{cm}^{2} / \mathrm{min}\)
- B \(12 \pi \mathrm{cm}^{2} / \mathrm{min}\)
- C \(14 \pi \mathrm{cm}^{2} / \mathrm{min}\)
- D \(16 \pi \mathrm{cm}^{2} / \mathrm{min}\)
Answer & Solution
Correct Answer
(B) \(12 \pi \mathrm{cm}^{2} / \mathrm{min}\)
Step-by-step Solution
Detailed explanation
(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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