Application of Derivatives — Class 12 Maths Solution

Since, $x$ and $y$ are the sides of two squares such that $y = x - {x^2}$.

Therefore the area of the first square $\left( {{A_1}} \right) = {x^2}$
and area of the second square $\left( {{A_2}} \right) = {y^2} = {\left( {x - {x^2}} \right)^2}$

Therefore,$\frac{{d{A_2}}}{{dt}} = \frac{d}{{dt}}{\left( {x - {x^2}} \right)^2} = 2\left( {x - {x^2}} \right)\left( {\frac{{dx}}{{dt}} - 2x \cdot \frac{{dx}}{{dt}}} \right)$

$= \frac{{dx}}{{dt}}(1 - 2x)2\left( {x - {x^2}} \right)$

And $\frac{{d{A_1}}}{{dt}} = \frac{d}{{dt}}{x^2} = 2x \cdot \frac{{dx}}{{dt}}$

Therefore, $\frac{{d{A_2}}}{{d{A_1}}} = \frac{{d{A_2}/dt}}{{d{A_1}/dt}} = \frac{{\frac{{dx}}{{dt}} \cdot (1 - 2x)\left( {2x - 2{x^2}} \right)}}{{2x \cdot \frac{{dx}}{{dt}}}}$
$= \frac{{(1 - 2x)2x(1 - x)}}{{2x}}$

$= (1 - 2x)(1 - x)$

$= 1 - x - 2x + 2{x^2}$

$= 2{x^2} - 3x + 1$