If the radius of a circle increases at the ra | vedclass

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(A) Given that the rate of change of radius is $\frac{dr}{dt} = 7 	ext{ cm/sec}$.The area of a circle is given by $A = \pi r^2$.Differentiating both

Vedclass · Accepted Answer

$1,84,800 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(A) Given that the rate of change of radius is $\frac{dr}{dt} = 7 	ext{ cm/sec}$.The area of a circle is given by $A = \pi r^2$.Differentiating both sides with respect to time $t$,we get $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.After $10 	ext{ minutes}$,the time $t = 10 	imes 60 = 600 	ext{ seconds}$.Since the radius increases at a constant rate of $7 	ext{ cm/sec}$,the radius after $600 	ext{ seconds}$ is $r = 7 	imes 600 = 4200 	ext{ cm}$.Substituting the values into the derivative formula: $\frac{dA}{dt} = 2 	imes \frac{22}{7} 	imes 4200 	imes 7$.$\frac{dA}{dt} = 2 	imes 22 	imes 600 	imes 7 = 1,84,800 	ext{ cm}^2/	ext{sec}$.

If the radius of a circle increases at the rate of $7 \text{ cm/sec}$,then the rate of increase of its area after $10 \text{ minutes}$ is:

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