MHT CET · Maths · Application of Derivatives
The radius of a circular plate is increasing at the rate of 0.01 \(\mathrm{cm} / \mathrm{sec}\), when the radius is \(12 \mathrm{~cm}\). Then the rate at which the area increases is
- A \(60 \pi\) sq. cm \(/ \mathrm{sec}\)
- B \(0.24 \pi\) sq. cm \(/ \mathrm{sec}\)
- C \(1.2 \pi\) sq. cm \(/ \mathrm{sec}\)
- D \(24 \pi\) sq. cm \(/ \mathrm{sec}\)
Answer & Solution
Correct Answer
(B) \(0.24 \pi\) sq. cm \(/ \mathrm{sec}\)
Step-by-step Solution
Detailed explanation
We have \(\frac{\mathrm{dr}}{\mathrm{dt}}=0.01\)
\(\mathrm{A}=\pi \mathrm{r}^2\)
\(\therefore \frac{\mathrm{dA}}{\mathrm{dt}}=\pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=(2 \pi)(12)(0.01)=0.24 \pi \mathrm{sq}\) \(\cdot \mathrm{cm} / \mathrm{sec}>\)
\(\mathrm{A}=\pi \mathrm{r}^2\)
\(\therefore \frac{\mathrm{dA}}{\mathrm{dt}}=\pi(2 \mathrm{r}) \frac{\mathrm{dr}}{\mathrm{dt}}=(2 \pi)(12)(0.01)=0.24 \pi \mathrm{sq}\) \(\cdot \mathrm{cm} / \mathrm{sec}>\)
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