Question

Given the relationship 2x2 + y3 =10, with y > 0 and dy, dt = 3 units/min., find the value of dx, dt at the instant x = 1 unit.

Answer 1

I believe it is -1/9. 

Answer 2

Answer:

$\frac{dx}{dt} = -9$ when $x = 1$

Step-by-step explanation:

The instant x = 1.

We have to find the value of y at this instant.

We have that:

$2x^{2} + y^{3} = 10$

$2 + y^{3} = 10$

$y^{3} = 8$

$y = 2$

Now we find the implicit derivative

The derivative of a constant is 0. So:

$4x\frac{dx}{dt} + 3y^{2}\frac{dy}{dt} = 0$

We have that:

$x = 1, y = 2, \frac{dy}{dt} = 3$

$4\frac{dx}{dt} + 36 = 0$

$\frac{dx}{dt} = -9$

$\frac{dx}{dt} = -9$ when $x = 1$