Question

If a function is represented parametrically by $x=\frac{2}{t^2}+\frac{5}{t}$ and $y=\frac{8t^2+15t^3}{6t^5}$, then the value of $|x\frac{dy}{dx}-\frac{dy}{dx}|$ at $t=-1$is

Answer 1

$y=\frac{8t^2+15t^3}{6t^5}=\frac{4}{3t^3}+\frac{5}{2t^2}$
$\Rightarrow \frac{dy}{dt}=\frac{4}{3}\left(\frac{-3}{t^4}\right)+\frac{5}{2}\left(\frac{-2}{t^3}\right)\Rightarrow \frac{dy}{dx}=-(\frac{4}{t^4}+\frac{5}{t^3})$

$x=\frac{2}{t^2}+\frac{5}{t}$
$\Rightarrow \frac{dx}{dt}=\frac{-4}{t^3}-\frac{5}{t^2}=-(\frac{4}{t^3}+\frac{5}{t^2})$

$\frac{dy}{dx}=\frac{1}{t}$
$|x\frac{dy}{dx}-\frac{dy}{dx}|=|(\frac{2}{t^2}+\frac{5}{t})\frac{1}{t}-\frac{1}{t}|$
$\Rightarrow |x\frac{dy}{dx}-\frac{dy}{dx}|=|(\frac{2}{t^3}+\frac{5}{t^2})-\frac{1}{t}|$
Putting $t=-1$
$\Rightarrow |x\frac{dy}{dx}-\frac{dy}{dx}|=|-2+5+1|=4$