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Q.
The rate of change of the area of a circle with respect to its radius $r$ at
Application of Derivatives
Solution:
Let $A$ denotes the area of the circle when its radius is $r$, then $A=\pi r^2$.
Now, the rate of change of the area with respect to its radius is given by,
$\frac{d A}{d r}=\frac{d}{d r}\left(\pi r^2\right)=2 \pi r$
(a) When $r=3\, cm ,\left(\frac{d A}{d r}\right)_{r=3}=2 \pi(3)=6 \pi cm ^2 / cm$.
Hence, the area of the circle is changing at the rate of $6 \pi cm ^2 / cm$ when its radius is $3 cm$.
(b) When $r=4 \,cm ,\left(\frac{d A}{d t}\right)_{r=4}=2 \pi(4)=8 \pi cm ^2$ per $cm$.
Hence, the area of the circle is changing at the rate of $8 \pi cm ^2 / cm$ when its radius is $4 \,cm$.