The side of a square is equal to the diameter of a circle. If the side and radius change at the same rate, then the ratio of the change of their areas, is

A
$1 : π$
B
$π : 1$
C
$2 : π$
D
$1 : 2$

Solution:

Let x be the length of a side of the square and r be the radius of the circle.

Then, $x = 2 r$ [Given] and $\frac{d x}{d t} = \frac{d r}{d t}$

Let $A_{1}$ and $A_{2}$ be the areas of the square and circle respectively. Then,

$A_{1} = x^{2}$ and $A_{2} = π r^{2}$

$\Rightarrow \frac{d A_{1}}{d t} = 2 x \frac{d x}{d t}$ and $\frac{d A_{2}}{d t} = 2 π r \frac{d r}{d t}$

$\Rightarrow \frac{\frac{d A_{1}}{d t}}{\frac{d A_{2}}{d t}} = \frac{2 x \frac{d x}{d t}}{2 π r \frac{d r}{d t}} = \frac{2}{π} [∵ x = 2 r and \frac{d x}{d t} = \frac{d r}{d t}]$

Solution:

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