TS EAMCET · Maths · Application of Derivatives
If the area of a circle increases at the rate of \(\frac{1}{\sqrt{\pi}}\) sq. units/sec, then the rate (in units/sec) at which the perimeter of the circle changes, when perimeter is \(\sqrt{\pi}\) units, is
- A 2
- B 4
- C \(\frac{1}{\sqrt{\pi}}\)
- D \(\sqrt{\pi}\)
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
Let \(A\) and \(P\) are area and perimeter of circle. We have, \(\frac{d A}{d t}=\frac{1}{\sqrt{\pi}} \Rightarrow \frac{d}{d t}\left(\pi r^2\right)=\frac{1}{\sqrt{\pi}}\)…
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