AP EAMCET · Maths · Differential Equations
The radius of a circle is increasing at a rate of \(0.1 \mathrm{~cm} \mathrm{~s}^{-1}\). Then the rate of change of area, when its radius is \(5 \mathrm{~cm}\), is .........
- A \(\pi^2 \mathrm{~cm}^2 \mathrm{~s}^{-1}\)
- B \(\pi \mathrm{cm}^2 \mathrm{~s}^{-1}\)
- C \(2 \pi \mathrm{cm}^2 \mathrm{~s}^{-1}\)
- D \(\frac{\pi}{2} \mathrm{~cm}^2 \mathrm{~s}^{-1}\)
Answer & Solution
Correct Answer
(B) \(\pi \mathrm{cm}^2 \mathrm{~s}^{-1}\)
Step-by-step Solution
Detailed explanation
Let the circle have radius \(r \mathrm{~cm}\) and area \(A \mathrm{~cm}^2\) and it is given that \(\frac{d r}{d t}=0.1 \mathrm{~cm} . \mathrm{s}^{-1}\) and as we know area \(A=\pi r^2 \Rightarrow \frac{d A}{d t}=2 \pi r \frac{d r}{d t}\) \(\therefore\) Rate of change of area,…
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