Question

x and y are the sides of two squares such that y = x – x². Find the rate of change of the area of second square with respect to the area of first square.

Answer 1

Let area of the first square A₁ = x²

And area of the second square A₂ = y²

Now A₁= x² and A₂ = y² = (x – x²)²

Differentiating both A₁ and A₂ w.r.t. t, we get

`("dA"_1)/"dt" = 2x * "dx"/"dt"` and `("dA"_2)/"dt" = 2(x - x^2)(1 - 2x) * "dx"/"dt"`

∴ `("dA"_2)/("dA"_1) = ("dA"_2/"dt")/("dA"_1/"dt")`

= `(2(x - x^2)(1 - 2x) * "dx"/"dt")/(2x * "dx"/"dt")`

= `(x(1 - x)(1 - 2x))/x`

= (1 – x)(1 – 2x)

= 1 – 2x – x + 2x²

= 2x² – 3x + 1

Hence, the rate of change of area of the second square with respect to first is 2x² – 3x + 1.