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?
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
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?
is 4 square root of 12 right?
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
To find how fast y changes with respect to t when x is 4, we first need to find dy/dt, which represents the rate of change of y with respect to t.
Given the relationship x^2 = 3y^2 + 7, we can differentiate both sides of the equation with respect to t using implicit differentiation.
d/dt (x^2) = d/dt (3y^2 + 7)
Using the chain rule, we can differentiate each term:
2x(dx/dt) = 6y(dy/dt)
Since we are given that dx/dt (x') is the square root of 12, we can substitute that value into the equation:
2(4)(√12) = 6y(dy/dt)
Simplifying, we have:
8√12 = 6y(dy/dt)
Now we can solve for dy/dt by substituting the given values:
dy/dt = (8√12) / (6y)
Since we are looking for the rate of change of y when x is 4, we need to find the value of y when x is 4. We can obtain this by substituting x = 4 into the initial relationship x^2 = 3y^2 + 7:
(4)^2 = 3y^2 + 7
16 = 3y^2 + 7
9 = 3y^2
Dividing both sides by 3:
3 = y^2
Taking the square root of both sides:
√3 = y
Substituting this value into the equation for dy/dt, we have:
dy/dt = (8√12) / (6√3)
Simplifying further:
dy/dt = (4√3) / (3√3)
The square root of 3 cancels out, leaving us with:
dy/dt = 4/3
Therefore, when x is 4, y changes at a rate of 4/3.