Question

x and y are both variables that are differentiable with respect to t, but not to each other. however, there is a relationship between them: x^2=3y^2+7. we also know that x changes with respect to t at a steady rate of square root of 12 (dx/dt=x'= square root 12). when x is 4 how fast does y change?

Answer 1

x^2 = 3y^2 + 7
2x dx/dt = 6y dy/dt

You know dx/dt
find y when x=4 from the original, that leaves
dy/dt to solve for

Answer 2

okay so after i find y and its 8 square root of 12 = 6(1/3)dy/dt how do you get rid of the square root?

Answer 3

is 4 square root of 12 right?

Answer 4

when x=4
16 = 3y^2 + 7
9 = 3y^2
y^2 = 3
y = ± √3

using x=4, y = +√3 , dx/dt = √12

2(4)√12 = 6(√3) dy/dt
dy/dt = 8√12/(6√3) = 4√4/3 = 8/3

repeat for y = - √3