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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.
Solution:
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Verified Answer
Area of the first square $\left(\mathrm{A}_1\right)=\mathrm{x}^2$ and area of the second square
$$
\left(A_2\right)=y^2=\left(x-x^2\right)^2
$$
$$
\therefore \quad \frac{\mathrm{dA}_2}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{x}-\mathrm{x}^2\right)^2=2\left(\mathrm{x}-\mathrm{x}^2\right)\left[\frac{\mathrm{dx}}{\mathrm{dt}}-2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}\right]
$$
$=\frac{\mathrm{dx}}{\mathrm{dt}}(1-2 \mathrm{x}) 2\left(\mathrm{x}-\mathrm{x}^2\right)$ and $\frac{\mathrm{dA} \mathrm{A}_1}{\mathrm{dt}}=2 \mathrm{x} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}$
$$
\therefore \frac{\mathrm{dA}_2}{\mathrm{dA}_1}=\frac{\mathrm{dA}{ }_2 / \mathrm{dt}}{\mathrm{dA_{1 } / \mathrm { dt }}}=\frac{\frac{\mathrm{dx}}{\mathrm{dt}} \cdot(1-2 \mathrm{x})\left(2 \mathrm{x}-2 \mathrm{x}^2\right)}{2 x \cdot \frac{\mathrm{dx}}{\mathrm{dt}}}
$$
$$
=\frac{(1-2 x) 2 x(1-x)}{2 x}
$$
$$
\Rightarrow \frac{\mathrm{dA}_2}{\mathrm{dA}_1}=2 \mathrm{x}^2-3 \mathrm{x}+1
$$
$$
\left(A_2\right)=y^2=\left(x-x^2\right)^2
$$
$$
\therefore \quad \frac{\mathrm{dA}_2}{\mathrm{dt}}=\frac{\mathrm{d}}{\mathrm{dt}}\left(\mathrm{x}-\mathrm{x}^2\right)^2=2\left(\mathrm{x}-\mathrm{x}^2\right)\left[\frac{\mathrm{dx}}{\mathrm{dt}}-2 \mathrm{x} \frac{\mathrm{dx}}{\mathrm{dt}}\right]
$$
$=\frac{\mathrm{dx}}{\mathrm{dt}}(1-2 \mathrm{x}) 2\left(\mathrm{x}-\mathrm{x}^2\right)$ and $\frac{\mathrm{dA} \mathrm{A}_1}{\mathrm{dt}}=2 \mathrm{x} \cdot \frac{\mathrm{dx}}{\mathrm{dt}}$
$$
\therefore \frac{\mathrm{dA}_2}{\mathrm{dA}_1}=\frac{\mathrm{dA}{ }_2 / \mathrm{dt}}{\mathrm{dA_{1 } / \mathrm { dt }}}=\frac{\frac{\mathrm{dx}}{\mathrm{dt}} \cdot(1-2 \mathrm{x})\left(2 \mathrm{x}-2 \mathrm{x}^2\right)}{2 x \cdot \frac{\mathrm{dx}}{\mathrm{dt}}}
$$
$$
=\frac{(1-2 x) 2 x(1-x)}{2 x}
$$
$$
\Rightarrow \frac{\mathrm{dA}_2}{\mathrm{dA}_1}=2 \mathrm{x}^2-3 \mathrm{x}+1
$$
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