The radius of a circle is increasing at the r | vedclass

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Question

(B) Given that the rate of change of the radius is $\frac{dr}{dt} = 2 	ext{ cm/sec}$.Since $1 	ext{ decimeter} = 10 	ext{ cm}$,the radius $r = 5

Vedclass · Accepted Answer

$200 \pi 	ext{ cm}^2/	ext{sec}$

Vedclass · Answer

(B) Given that the rate of change of the radius is $\frac{dr}{dt} = 2 	ext{ cm/sec}$.Since $1 	ext{ decimeter} = 10 	ext{ cm}$,the radius $r = 5 	ext{ decimeters} = 50 	ext{ cm}$.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 values $r = 50 	ext{ cm}$ and $\frac{dr}{dt} = 2 	ext{ cm/sec}$:$\frac{dA}{dt} = 2 	imes \pi 	imes 50 	imes 2 = 200 \pi 	ext{ cm}^2/	ext{sec}$.

The radius of a circle is increasing at the rate of $2 \text{ cm/sec}$. The rate at which its area is increasing when the radius of the circle is $5 \text{ decimeters}$ is:

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