Question

Given the relationship x2 + 3y2 =12, with y > 0 and dx, dt = 2 units/min., find the value of dy, dt at the instant y = 1 unit.

Answer 1

$\bf x^2+3y^2=12\implies \stackrel{chain~rule}{2x\cfrac{dx}{dt}}+3 \stackrel{chain~rule}{\left(2y\cfrac{dy}{dt}\right)}=0 \\\\\\ 2x\cfrac{dx}{dt}+6y\cfrac{dy}{dt}=0 \implies 6y\cfrac{dy}{dt}=-2x\cfrac{dx}{dt}\implies \cfrac{dy}{dt}=\cfrac{-2x\frac{dx}{dt}}{ 6y\frac{dy}{dt}} \\\\\\ \boxed{\cfrac{dy}{dt}=-\cfrac{x\frac{dx}{dt}}{3y}}\\\\ -------------------------------$

$\bf \textit{now, when y = 1, what is \underline{x}?}\qquad x^2+3y^2=12\implies x^2+3(1)^2=12 \\\\\\ x^2=9\implies x=\sqrt{9}\implies x=3\\\\ -------------------------------\\\\ \begin{cases} y=1\\ x=3\\ \frac{dx}{dt}=2 \end{cases}\implies \cfrac{dy}{dt}=-\cfrac{3\cdot 2}{3\cdot 1}\implies \cfrac{dy}{dt}=-2$