A stone is dropped into a quiet lake and wave | vedclass

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Question

(A) The area of a circle $A$ with radius $r$ is given by $A = \pi r^2$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = \fr

Vedclass · Accepted Answer

$80 \pi 	ext{ cm}^2/	ext{s}$

Vedclass · Answer

(A) The area of a circle $A$ with radius $r$ is given by $A = \pi r^2$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2\pi r \frac{dr}{dt}$.Given that the rate of change of the radius is $\frac{dr}{dt} = 5 	ext{ cm/s}$.At the instant when $r = 8 	ext{ cm}$,we substitute the values into the derivative:$\frac{dA}{dt} = 2 \pi (8 	ext{ cm}) (5 	ext{ cm/s}) = 80 \pi 	ext{ cm}^2/	ext{s}$.Thus,the enclosed area is increasing at the rate of $80 \pi 	ext{ cm}^2/	ext{s}$.

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

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