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

`(\mathrm d)/(\mathrm dt)\left[2x^2+y^3\right]=(\mathrm d)/(\mathrm dt)[10]\implies 4x(\mathrm dx)/(\mathrm dt)+3y^2(\mathrm dy)/(\mathrm dt)=0`

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


`2*1^2+y^3=10\implies y^3=8\implies y=2`

You also know that when
`x=1`, you have
`(\mathrm dy)/(\mathrm dt)=3`. Substituting everything you know into the differentiated equation, you get


`4*1*(\mathrm dx)/(\mathrm dt)+3*2^2*3=0\implies(\mathrm dx)/(\mathrm dt)=-9`

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