Question

A particle moves alongthecurve $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1, with constant speed v. Express its velocity vectorially as a function of (x,y).

Answer 1

Given $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1
$\Rightarrow$ $\frac{2x}{9}\left[\frac{dx}{dt}\right]$ + $\frac{2y}{4}\left[\frac{dx}{dt}\right]$ = 1 $[\frac{dz}{dt}=v_x;\frac{dy}{dt}=v_y]$
$v_x$ = $-\frac{9}{4}\frac{y}{x}{v}_y$ (i)
Also $v_x^2+v_y^2=v^2$ $\Rightarrow$ ${(-\frac{9}{4}\frac{y}{x}{v}_y)}^2$ + $v_y^2=v^2$
$\Rightarrow$ $v_y^2=\frac{16x^2{v}^2}{16x^2+81y^2}$ $\Rightarrow$ $v_y=\frac{\pm 4xv}{\sqrt{16x^2+81y^2}}$ (ii)
From (i) and (ii), we get $v_x=\frac{\mp 9yv}{\sqrt{16x^2+81y^2}}$
So, velocity is given by $v_x\hat{i}+v_y\hat{j}=\frac{(\mp 9y\hat{i}\pm 4x\hat{j})v}{\sqrt{16x^2+81y2}}$