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Please solve RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 9 maths textbook solution

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Answer: \frac{d A}{d t}=80 \: \pi \: \mathrm{cm}^{2} / \mathrm{sec}

Hint: As we know that the area of circle is \pi r^{2} .

Given: As a stone is dropped into a quiet lake and waves move in circle at a speed of  4 cm/sec

Solution: Suppose r  be the radius of the circle and A  be the area of the circle.

Whenever stone is dropped into the lake waves in circle at the speed of 4 cm/sec.

That is the radius of the circle increases at the rate of 4 cm/sec.

\frac{dr}{dt}=4 cm/sec                                ……………. (i)


\frac{d A}{d t}=\frac{d\left(\pi r^{2}\right)}{d t}=\frac{\pi d\left(r^{2}\right)}{d t}=\pi \times 2 r \frac{d r}{d t}=2 \pi r \times 4

So, when the radius of circle is 10 cm,

\frac{d A}{d t}=2 \pi \times 10 \times 4=80 \pi\; cm^{2}/sec

Thus the enclosed area is increasing at the rate of 80\pi  cm2/sec.

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