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explain solution RD Sharma class 12 chapter Derivative As a Rate Measure exercise 12.2 question 4 maths

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Answer: The rate of increasing of the circle’s circumference will be 1.4\pi cm/sec

Hint: The circumference of circle at any time  t  will be C=2\pi r  cm

Given: The radius of a circle is increasing at the rate of 0.7 cm/sec.

Solution: Suppose the radius of the given circle be  r  cm at any instant time.

Now according to the question,

The rate of radius of a circle is increasing is \frac{dr}{dt}=0.7 cm/sec            ...… (i)

Now the circumference of the circle at any time  t   will be C=2\pi r cm

By applying derivative on circumference,

\frac{d C}{d t}=\frac{d(2 \pi r)}{d t}=2 \pi \frac{d r}{d t}=2 \pi \times 0.7=1.4 \pi cm/sec

\left\{\frac{d r}{d t}=0.7\left(\text { from } e q^{n}(i)\right) \text { and } C=2 \pi r(\text { Given })\right\} 

Thus the rate of increasing of circle circumference will be 1.4\pi  cm/sec.

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