If $x$ and $y$ are the sides of two squares such that $y = x - x^2$, then find the rate of change of the area of second square with respect to the area of first square.

$2x^2 - 3x + 1$

The correct answer is Option (2) → $2x^2 - 3x + 1$ ##

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

$∴$ Area of the first square $(A_1) = x^2$

and area of the second square $(A_2) = y^2 = (x - x^2)^2$

$∴\quad \frac{dA_2}{dt} = \frac{d}{dt} (x - x^2)^2 = 2(x - x^2) \left( \frac{dx}{dt} - 2x \cdot \frac{dx}{dt} \right)$

$\quad \quad \quad = \frac{dx}{dt} (1 - 2x) \cdot 2(x - x^2)$

and $\quad \frac{dA_1}{dt} = \frac{d}{dt} x^2 = 2x \cdot \frac{dx}{dt}$

$∴ \quad \frac{dA_2}{dA_1} = \frac{dA_2 / dt}{dA_1 / dt} = \frac{\frac{dx}{dt} \cdot (1 - 2x)(2x - 2x^2)}{2x \cdot \frac{dx}{dt}}$

$\quad \quad \quad = \frac{(1 - 2x) \cdot 2x(1 - x)}{2x}$

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

$\quad \quad \quad = 1 - x - 2x + 2x^2$

$\quad \quad \quad = 2x^2 - 3x + 1$