1. ## Differentiation ( Parametric)

Given x = t - ( 1/ t) , y = 2t + ( 1/t) with t is a non-zero parameter.
a) Show that dy/dx = 2 - [ 3 / (t^2 + 1 ) ] ...... Done
b) Hence, deduce that -1 < dy/dx < 2 .............. Undone

I would be grateful if you can help me with the part B with explanations.

2. Originally Posted by ose90
Given x = t - ( 1/ t) , y = 2t + ( 1/t) with t is a non-zero parameter.
a) Show that dy/dx = 2 - [ 3 / (t^2 + 1 ) ] ...... Done
b) Hence, deduce that -1 < dy/dx < 2 .............. Undone

I would be grateful if you can help me with the part B with explanations.

$\displaystyle \frac{dy}{dx} = 2 - \frac{3}{t^2+1}$
as $\displaystyle t$ approach $\displaystyle \pm \infty$, then $\displaystyle \frac{3}{t^2+1}$ approach $\displaystyle 0^+$ but not zero.. thus, $\displaystyle \frac{dy}{dx} = 2 - 0^+ < 2$
next, if $\displaystyle t$ approach $\displaystyle 0$, then $\displaystyle \frac{3}{t^2+1}$ approach $\displaystyle 3^-$.. hence, $\displaystyle \frac{dy}{dx} = 2 - 3^- > -1$..