A stone is dropped in a quiet lake and it is | vedclass

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(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 2 	ext{ cm/sec}$.We need to find the rate of change of the area $A$ when the radi

Vedclass · Accepted Answer

$40$

Vedclass · Answer

(A) Given that the rate of change of the radius is $\frac{dr}{dt} = 2 	ext{ cm/sec}$.We need to find the rate of change of the area $A$ when the radius $r = 10 	ext{ cm}$.The area of a circle is given by $A = \pi r^2$.Differentiating both sides with respect to time $t$,we get:$\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.Substituting the given values $r = 10 	ext{ cm}$ and $\frac{dr}{dt} = 2 	ext{ cm/sec}$:$\frac{dA}{dt} = 2 \pi 	imes 10 	imes 2 = 40 \pi 	ext{ cm}^2/	ext{sec}$.

$A$ stone is dropped in a quiet lake and it is observed that waves move in circles. If the radius of a circular wave increases at the rate of $2 \text{ cm/sec}$,then the rate of increase in its area at the instant when its radius is $10 \text{ cm}$,is (in $\text{cm}^2/\text{sec}$): (in $\pi$)

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