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.

Question

asked May 15, 2018 195k views

2 Answers

The Bearded Llama · Answer 1 · 2018-05-20T21:11:42+0000

Answer:

(dx)/(dt) = -9 when
x = 1

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(dx)/(dt) + 3y^(2)(dy)/(dt) = 0

We have that:

x = 1, y = 2, (dy)/(dt) = 3

4(dx)/(dt) + 36 = 0

(dx)/(dt) = -9

(dx)/(dt) = -9 when
x = 1