(B) Given that the rate of change of the radius is $ \frac{dr}{dt} = 5 \text{ cm s}^{-1} $. The area of a circle is given by $ A = \pi r^2 $. Differen

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$ 80 \pi \text{ cm}^2 \text{ s}^{-1} $

(B) Given that the rate of change of the radius is $ \frac{dr}{dt} = 5 \text{ cm s}^{-1} $. The area of a circle is given by $ A = \pi r^2 $. Differentiating both sides with respect to time $ t $,we get: $ \frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt} $. Substituting the given values $ r = 8 \text{ cm} $ and $ \frac{dr}{dt} = 5 \text{ cm s}^{-1} $: $ \frac{dA}{dt} = 2 \pi (8) (5) = 80 \pi \text{ cm}^2 \text{ s}^{-1} $. Thus,the enclosed area is increasing at a rate of $ 80 \pi \text{ cm}^2 \text{ s}^{-1} $.

Class 12 Mathematics Applications of Derivatives Rate of Change of Quantities

Easy

$A$ stone is dropped into a quiet lake and waves move in circles at the speed of $ 5 \text{ cm s}^{-1} $. At that instant,when the radius of the circular wave is $ 8 \text{ cm} $,how fast is the enclosed area increasing?

KCET 2014 All KCET

A
$ 8 \pi \text{ cm}^2 \text{ s}^{-1} $
B
$ 80 \pi \text{ cm}^2 \text{ s}^{-1} $
C
$ 6 \pi \text{ cm}^2 \text{ s}^{-1} $
D
$ 800 \pi \text{ cm}^2 \text{ s}^{-1} $

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Applications of Derivatives

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Rate of Change of Quantities

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$A$ stone is dropped into a quiet lake and waves move in circles at the speed of $ 5 \text{ cm s}^{-1} $. At that instant,when the radius of the circular wave is $ 8 \text{ cm} $,how fast is the enclosed area increasing?

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