If the rate of increase of the area of a circ | vedclass

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(C) Let $P$ be the perimeter and $A$ be the area of the circle with radius $r$.$P = 2\pi r$ and $A = \pi r^2$.Differentiating both with respect to tim

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As the radius

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(C) Let $P$ be the perimeter and $A$ be the area of the circle with radius $r$.$P = 2\pi r$ and $A = \pi r^2$.Differentiating both with respect to time $t$:$\frac{dP}{dt} = 2\pi \frac{dr}{dt}$ and $\frac{dA}{dt} = 2\pi r \frac{dr}{dt}$.From the first equation,$\frac{dr}{dt} = \frac{1}{2\pi} \frac{dP}{dt}$.Substituting this into the expression for $\frac{dA}{dt}$:$\frac{dA}{dt} = 2\pi r \left( \frac{1}{2\pi} \frac{dP}{dt} \right) = r \frac{dP}{dt}$.Since the rate of increase of the perimeter $\frac{dP}{dt}$ is constant,let $\frac{dP}{dt} = k$ (where $k$ is a constant).Then $\frac{dA}{dt} = k \cdot r$.Therefore,the rate of increase of the area varies directly as the radius $r$.

If the rate of increase of the area of a circle is not constant but the rate of increase of the perimeter is constant,then the rate of increase of the area varies:

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