If the area of an expanding circular region increases at a constant rate with respect to time, then the rate of increase of the perimeter with respect to the time

A
Varies inversely as radius
B
Varies directly as radius
C
Remains constant
D
Varies directly as square of the radius

Solution:

Let $A$ be the area and $P$ be the perimeter of the

circular region of radius $r$ Then,

$A = π r^{2}$ and $P = 2 π r$

$\Rightarrow \frac{d A}{d t} = 2 π r \frac{d r}{d t}$ and $\frac{d P}{d t} = 2 π \frac{d r}{d t}$

It is given that $\frac{d A}{d t} = k (constant)$

$\begin{array}{l} \Rightarrow \frac{d r}{d t} = \frac{k}{2 π r} \\ ∴ \frac{d P}{d t} = 2 π \times \frac{k}{2 π r} = \frac{k}{r} \propto \frac{1}{r} \end{array}$

Solution:

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