A stone is dropped into a still pond,creating | vedclass

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(A) Let $r$ be the radius of the ripple and $A$ be the area of the circle enclosed by the ripple.Given,the rate of change of the radius is $\frac{dr}{

Vedclass · Accepted Answer

$52.5 \pi 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(A) Let $r$ be the radius of the ripple and $A$ be the area of the circle enclosed by the ripple.Given,the rate of change of the radius is $\frac{dr}{dt} = 3.5 	ext{ cm/sec}$.The area of the circle is given by $A = \pi r^2$.Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$.At the instant when $r = 7.5 	ext{ cm}$,we substitute the values:$\frac{dA}{dt} = 2 	imes \pi 	imes 7.5 	imes 3.5$.$\frac{dA}{dt} = 15 	imes 3.5 	imes \pi$.$\frac{dA}{dt} = 52.5 \pi 	ext{ cm}^2/	ext{sec}$.Thus,the area is increasing at a rate of $52.5 \pi 	ext{ cm}^2/	ext{sec}$.

$A$ stone is dropped into a still pond,creating ripples that move at a speed of $3.5 \text{ cm/sec}$. At the instant when the radius of the ripples is $7.5 \text{ cm}$,how fast is the enclosed area increasing?

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