Question

A function is represented parametrically by the equations $x=\frac{1+t}{t^{3}}$ and $y=\frac{3}{2 t^{2}}+\frac{2}{t}$. Then, $\frac{d y}{d x}-x \cdot \frac{d y^{3}}{d x}$ has the absolute value equal to:

Answer 1

Given, $x=\frac{1+t}{t^{3}} \ldots 1$
and $y=\frac{3}{2 t^{2}}+\frac{2}{t} \ldots 2$
From equation $1, x=\frac{1}{t^{3}}+\frac{1}{t^{2}}$.
Differentiating the above equation 1 with respect to $t$ we get,
$\frac{d x}{d t}=\frac{d}{d t} \frac{1}{t^{3}}+\frac{1}{t^{2}}$
$\Rightarrow \frac{d x}{d t}=\frac{d}{d t} \frac{1}{t^{3}}+\frac{d}{d t} \frac{1}{t^{2}}$
$\Rightarrow \frac{d x}{d t}=\frac{-3}{t^{4}}+\frac{-2}{t^{3}}$
$\Rightarrow \frac{d x}{d t}=-\frac{3+2 t}{t^{4}}$
Differentiating equation $2$ with respect to $t$ we get,
$\frac{d y}{d t}=\frac{3}{2} \frac{-2}{t^{3}}+2 \frac{-1}{t^{2}} $
$\Rightarrow \frac{d y}{d t}=-\frac{3+2 t}{t^{3}}$
Hence,
$\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}} .$
$\frac{d y}{d x}=t$
Hence,
$\frac{d y}{d x}-x \frac{d y}{d x}^{3}=t-x t^{3} $
$\Rightarrow \frac{d y}{d x}-x \frac{d y^{3}}{d x}=t-\frac{1+t}{t^{3}} t^{3}$
$\{$ from equation $1\}$
$\frac{d y}{d x}-x \frac{d y}{d x}^{3}=-1$
$\frac{d y}{d x}-x \frac{d y}{d x}^{3}=-1=1$