The radius of a circle is increasing uniforml | vedclass

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(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}

Vedclass · Accepted Answer

$60 \pi 	ext{ cm}^2/	ext{s}$

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:$\frac{dA}{dt} = \frac{d}{dt}(\pi r^2) = 2 \pi r \frac{dr}{dt}$ (using the chain rule).Given that the rate of change of the radius is $\frac{dr}{dt} = 3 	ext{ cm/s}$.Substituting the values,we get:$\frac{dA}{dt} = 2 \pi r (3) = 6 \pi r$.When the radius $r = 10 	ext{ cm}$,the rate of change of the area is:$\frac{dA}{dt} = 6 \pi (10) = 60 \pi 	ext{ cm}^2/	ext{s}$.Thus,the area of the circle is increasing at the rate of $60 \pi 	ext{ cm}^2/	ext{s}$.

The radius of a circle is increasing uniformly at the rate of $3 \text{ cm/s}$. Find the rate at which the area of the circle is increasing when the radius is $10 \text{ cm}$.

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