Question

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

Answer 1

Answer: 1

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