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

Question

x and y are the sides of two squares such that y = x - x2. The rate of change of the area of the second square with respect to the area of the first square is (A) x2 - x + 1 (B) 2 x2 + 2 x - 1 (C) 2x2 - 3x + 1  (D)  x2 + x - 1

Tardigrade Admin · Accepted Answer

A2 = (x-x2)2 , A1 =x2 (dA2/dx) = 2(x-x2) (1-2x) ⇒ (dA1/dx) = 2x ⇒ (dA2/dA1) = 1+2x2 - 3x

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

$x^{2} - x + 1$

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

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

$x^{2} + x - 1$

$A_{2} = (x - x^{2})^{2}, A_{1} = x^{2}$
$\frac{d A _{2}}{d x} = 2 (x - x^{2}) (1 - 2 x)$
$\Rightarrow \frac{d A _{1}}{d x} = 2 x \Rightarrow \frac{d A _{2}}{d A _{1}} = 1 + 2 x^{2} - 3 x$

Q. x and y are the sides of two squares such that y=x−x2. The rate of change of the area of the second square with respect to the area of the first square is

x2−x+1

2x2+2x−1

2x2−3x+1

x2+x−1