Question

given the relationship 2x^2 + y^3 =10, with y > 0 and dy/dt = 3 units/min., find the value of dx/dt at the instant x = 1 unit.

Answer 1

$\dfrac{\mathrm d}{\mathrm dt}\left[2x^2+y^3\right]=\dfrac{\mathrm d}{\mathrm dt}[10]\implies 4x\dfrac{\mathrm dx}{\mathrm dt}+3y^2\dfrac{\mathrm dy}{\mathrm dt}=0$

When $x=1$, the original equation tells you that you have

$2\times1^2+y^3=10\implies y^3=8\implies y=2$

You also know that when $x=1$, you have $\dfrac{\mathrm dy}{\mathrm dt}=3$. Substituting everything you know into the differentiated equation, you get

$4\times1\times\dfrac{\mathrm dx}{\mathrm dt}+3\times2^2\times3=0\implies\dfrac{\mathrm dx}{\mathrm dt}=-9$

So $x$ is changing at a rate of $-9\text{ units/min}$.