MHT CET · Maths · Application of Derivatives
A stone is dropped into a quiet lake and waves move in circles at speed of \(8 \mathrm{~cm} / \mathrm{sec}\). At the instant when the radius of the circular wave is 12 cm . how fast is the enclosed area increasing?
- A \(180 \pi \mathrm{~cm}^2 / \mathrm{sec}\)
- B \(196 \pi \mathrm{~cm}^2 / \mathrm{sec}\)
- C \(192 \pi \mathrm{~cm}^2 / \mathrm{sec}\)
- D \(200 \pi \mathrm{~cm}^2 / \mathrm{sec}\)
Answer & Solution
Correct Answer
(C) \(192 \pi \mathrm{~cm}^2 / \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
Given, the rate of increasing the radius
\(\begin{aligned}
& =\frac{\mathrm{dr}}{\mathrm{dt}}=8 \mathrm{~cm} / \mathrm{sec} \\
& \text { Area }=A=\pi r^2 \\
& \therefore \quad \frac{\mathrm{dA}}{\mathrm{dt}}=2 \pi \mathrm{dr} \\
& \Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}}=2 \pi(12)(8) \\
& \ldots[\because \mathrm{r}=12 \mathrm{~cm}] \\
& =192 \pi \mathrm{~cm}^2 / \mathrm{sec}
\end{aligned}\)
\(\begin{aligned}
& =\frac{\mathrm{dr}}{\mathrm{dt}}=8 \mathrm{~cm} / \mathrm{sec} \\
& \text { Area }=A=\pi r^2 \\
& \therefore \quad \frac{\mathrm{dA}}{\mathrm{dt}}=2 \pi \mathrm{dr} \\
& \Rightarrow \frac{\mathrm{dA}}{\mathrm{dt}}=2 \pi(12)(8) \\
& \ldots[\because \mathrm{r}=12 \mathrm{~cm}] \\
& =192 \pi \mathrm{~cm}^2 / \mathrm{sec}
\end{aligned}\)
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