Question

`A function is represented parametrically by the equation $x=\frac{1+t}{t^3},y=\frac{3}{2t^2}+\frac{2}{t}$ then $\frac{dy}{dx}-x{\left(\frac{dy}{dx}\right)}^2$ has the value equal to

Answer 1

$x=\frac{1+t}{t^3},y=\frac{3}{2t^2}+\frac{2}{t}$

$\frac{dx}{dt}=\frac{d}{dt}[\frac{1}{t^2}+\frac{1}{t^2}]$

$\frac{dx}{dt}=[\frac{-3}{t^4}-\frac{2}{t^3}]$

$\frac{d^{2}x}{d{t}^2}=[\frac{12}{t^5}+\frac{6}{t^4}]$

$\frac{dy}{dt}=\frac{-3}{t^3}-\frac{2}{t^2}=-\left[\frac{3+2t}{t^3}\right]$

$\frac{d^{2}y}{dt}-\frac{9}{t^4}+\frac{4}{t^3}=\left[\frac{9+4t}{t^4}\right]$

$\frac{dy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=?$
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}$=$\frac{-\left(\frac{3+2t}{t^3}\right)}{-\left(\frac{3+2t}{t^4}\right)}=t$

$\frac{dy}{dx}-x(\frac{dy}{dx}{)}^2=t-\frac{(t+t)}{t^3}\times t^2$

$=t-\frac{(1+t)}{t}$

$\frac{dy}{dx}-x{\left(\frac{dy}{dx}\right)}^2=t-\frac{1}{t}-1$