Question

A particle is traveling on the curve, R, where R = {(x, y) = x² +9y² = 25} At time, t(o), the particle is at (-2,1); given that X' (to) = -1/2, what is y'(to) ?​

Answer 1

so is in essence implicit differentiation

`R(x,y)\implies x^2+9y^2=25\implies \stackrel{ chain~rule }{2x\cdot \cfrac{dx}{dt}}+9\cdot \stackrel{ chain~rule }{2y\cdot \cfrac{dy}{dt}}=0 \\\\\\ x\cdot \cfrac{dx}{dt}+9y\cdot \cfrac{dy}{dt}=0\implies 9y\cdot \cfrac{dy}{dt}=-x\cdot \cfrac{dx}{dt}\implies \cfrac{dy}{dt}=-x\cdot \cfrac{dx}{dt}\cdot \cfrac{1}{9y} \\\\[-0.35em] ~\dotfill`

`\textit{we also know that at t(0)}\hspace{5em} x'(t(0))=-\cfrac{1}{2}\hspace{5em}(\stackrel{x}{-2}~~,~~\stackrel{y}{1}) \\\\[-0.35em] ~\dotfill\\\\ \left. \cfrac{dy}{dt}=-x\cdot \cfrac{dx}{dt}\cdot \cfrac{1}{9y}\right|_((-2,1))\implies -(-2)\left( -\cfrac{1}{2} \right)\cdot \cfrac{1}{9(1)}\implies \boxed{-\cfrac{1}{9}}`