MHT CET · Maths · Application of Derivatives
The radius of a circle is increasing at the rate \(2 \mathrm{~cm} / \mathrm{sec}\). The rate at which its area is
increasing when the radius of the circle is 5 decimeters is
- A \(100 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
- B \(200 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
- C \(2000 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
- D \(20 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
Answer & Solution
Correct Answer
(B) \(200 \pi \mathrm{cm}^{2} / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
Here \(\frac{\mathrm{dr}}{\mathrm{dt}}=2\) and \(\mathrm{r}=5\) decimeter \(=50 \mathrm{~cm}\)
Now Area of circle \(=\mathrm{A}=\pi \mathrm{r}^{2}\).
\(\begin{aligned}
\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}} &=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \\
&=2 \times \pi \times 50 \times 2=200 \pi \mathrm{cm}^{2} / \mathrm{sec}
\end{aligned}\)
Now Area of circle \(=\mathrm{A}=\pi \mathrm{r}^{2}\).
\(\begin{aligned}
\therefore \frac{\mathrm{d} \mathrm{A}}{\mathrm{dt}} &=2 \pi \mathrm{r} \frac{\mathrm{dr}}{\mathrm{dt}} \\
&=2 \times \pi \times 50 \times 2=200 \pi \mathrm{cm}^{2} / \mathrm{sec}
\end{aligned}\)
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