A circular plate of radius 5 text{ cm} is hea | vedclass

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(C) Let $r$ be the radius and $A$ be the area of the circular plate at any time $t$. We know that the area of a circle is given by $A = \pi r^2$. Diff

Vedclass · Accepted Answer

$0.52 \pi 	ext{ cm}^2/	ext{s}$

Vedclass · Answer

(C) Let $r$ be the radius and $A$ be the area of the circular plate at any time $t$. We know that the area of a circle is given by $A = \pi r^2$. Differentiating both sides with respect to $t$,we get: $\frac{dA}{dt} = 2 \pi r \frac{dr}{dt}$. Given that the rate of change of radius is $\frac{dr}{dt} = 0.05 	ext{ cm/s}$. We need to find the rate of change of area when $r = 5.2 	ext{ cm}$. Substituting the values into the derivative formula: $\frac{dA}{dt} = 2 	imes \pi 	imes 5.2 	imes 0.05$. $\frac{dA}{dt} = 10.4 	imes 0.05 	imes \pi = 0.52 \pi 	ext{ cm}^2/	ext{s}$. Thus,the rate at which the area is increasing is $0.52 \pi 	ext{ cm}^2/	ext{s}$.

$A$ circular plate of radius $5 \text{ cm}$ is heated. Due to expansion,its radius increases at the rate of $0.05 \text{ cm/s}$. The rate at which its area is increasing when the radius is $5.2 \text{ cm}$ is:

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