If x and y are sides of two squares such that $y = x - x^{2}$ . Find the rate of change of area of second square (side y) with respect to area of the first square (side x) when x = 1 cm

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Solution

The correct option is A $0$
$y = x - x^{2}$
$A_{x} = x^{2} \Rightarrow$ $\frac{d A_{x}}{d x} = 2 x$
$A_{y} = y^{2} \Rightarrow \frac{d A_{y}}{d x} = 2 (x - x^{2}) (1 - 2 x)$
$\Rightarrow \frac{d A_{y}}{d A_{x}} = (1 - x) (1 - 2 x)$
$∴ \frac{d A_{y}}{d A_{x}} = 0$ , when $x = 1$ cm

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If x and y are sides of two squares such that y=x−x2. Find the rate of change of area of second square (side y) with respect to area of the first square (side x) when x = 1 cm

The correct option is A 0y=x−x2 Ax=x2⇒ dAxdx=2xAy=y2⇒dAydx=2(x−x2)(1−2x)⇒dAydAx=(1−x)(1−2x)∴dAydAx=0, when x=1cm

If x and y are sides of two squares such that $y = x - x^{2}$ . Find the rate of change of area of second square (side y) with respect to area of the first square (side x) when x = 1 cm

The correct option is A $0$
$y = x - x^{2}$
$A_{x} = x^{2} \Rightarrow$ $\frac{d A_{x}}{d x} = 2 x$
$A_{y} = y^{2} \Rightarrow \frac{d A_{y}}{d x} = 2 (x - x^{2}) (1 - 2 x)$
$\Rightarrow \frac{d A_{y}}{d A_{x}} = (1 - x) (1 - 2 x)$
$∴ \frac{d A_{y}}{d A_{x}} = 0$ , when $x = 1$ cm