Question

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.

Answer 1

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