The radius of a circle is increasing at the rate of 0.02cms^-1. Find the rate at which the area is increasing when the radius of the circle is 7cm.

0.75cm²S^-1

0.53cm²S^-1

0.35cm²S^-1

0.88cm²S^-1

Correct answer is D

A = \(\pi\)r², \(\frac{\delta A}{\delta r}\) = 2πr

So, using \(\frac{\delta A}{\delta t}\) = \(\frac {\delta A}{\delta r}\) x \(\frac {\delta A}{\delta t}\)

= 2\(\pi\)r x 0.02

= 2\(\pi\) x 7 x 0.02

= 2 x \(\frac{22}{7}\) x 0.02

= 0.88cm²s^-1

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The radius of a circle is increasing at the rate of 0.02cms^-1. Find the rate at which the area is increasing when the radius of the circle is 7cm.