Question

Solve the problem. Determine dy/dt for x^3 + y^3 = 9 given x = 1, y = 2, and dx/dt= -3

Answer 1

Differentiate both sides with respect to $t$, treating both $x$ and $y$ as functions of $t$.

$x^3 + y^3 = 9 \implies 3x^2 \dfrac{dx}{dt} + 3y^2 \dfrac{dy}{dt} = 0$

Then when $x=1$, $y=2$ and $\frac{dx}{dt}=-3$, we get

$3\times1^2\times(-3) + 3\times2^2\dfrac{dy}{dt} = 0 \implies \dfrac{dy}{dt} = \dfrac9{12} = \boxed{\frac34}$