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

I got -9 units/min

Answer 2

Let's see your implicit derivative.

Answer 3

4xx' + 3y2 y' = 0
4xx' + 9y2 = 0
4*1*x' +9*2^2 = 0
=-9

Answer 4

@tkhunny ?

Answer 5

It might be more clear to write (dx/dt) and (dy/dt), rather than (x') and (y'), respectively.

When x = 1, we have: 2(1)^2 + y^3 =10 ==> y^3 = 8 ==? y = 2

When x = 1, y = 2, and dy/dt = 3, we have:

4(1)(dx/dt) + 3(2)^2 (3) = 0 ==> 4(dx/dt) + 36 = 0 ==> dx/dt = -9

Okay. Why do you doubt?

Answer 6

thank you I just wanted to make sure that was correct :)

Answer 7

Let's work on that confidence. :-)