Question

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

Answer 1

Since $x$ and $y$ are the sides of the two squares such that $y=x-x^2$
$\therefore$ area of the first square $\left(A_1\right)=x^2$
And area of the second square $\left(A_2\right)=y^2=(x-x^2{)}^2$
$\therefore \frac{d{A}_2}{dt}=\frac{d}{dt}(x-x^2{)}^2=2(x-x^2\left)\right(\frac{dx}{dt}-2x.\frac{dx}{dt})$
$=\frac{dx}{dt}(1-2x)2(x-x^2)$
and $\frac{d{A}_1}{dt}=\frac{d}{dt}{x}^2=2x\frac{dx}{dt}$
$\therefore \frac{d{A}_2}{d{A}_1}=\frac{d{A}_2/dt}{d{A}_1/dt}=\frac{\frac{dx}{dt}(1-2x)(2x-2x^2)}{2x\frac{dx}{dt}}$
$=\frac{(1-2x)2x(1-x)}{2x}$
$=(1-2x)(1-x)$
$=1-x-2x+2x^2$
$=2x^2-3x+1$