Question

Find $\frac{dy}{dx}$ of each of the functions expressed in parametric form.
$x=t+\frac{1}{t},y=t-\frac{1}{t},$

Answer 1

$x=t+\frac{1}{t}andy=t-\frac{1}{t},\therefore \frac{dx}{dt}=\frac{d}{dt}(t+\frac{1}{t})and\frac{dy}{dt}=\frac{d}{dt}(t-\frac{1}{t})\Rightarrow \frac{dx}{dx}=1+(-1)t^{-2}and\frac{dy}{dt}=1-(-1)t^{-2}\to \frac{dx}{dt}=1-\frac{1}{t^2}and\frac{dy}{dt}=1+\frac{1}{t^2}\Rightarrow \frac{dx}{dt}=\frac{t^2-1}{t^2}and\frac{dy}{dt}=\frac{t^2+1}{t^2}\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{t^2+1}{t^2-1}$