The path of a particle is described by y^2 + 3x^2 = 10. If dx/dt = 2 and x = 1, what is dy/dt
The path of the particle is described by y^2 + 3x^2 = 10. If x = 1,
y^2 + 3*1 = 10
=> y^2 = 7
=> `y = +-sqrt 7`
Take the derivative of y^2 + 3x^2 = 10 with respect to t,
`2y*(dy/dt) + 6x*(dx/dt) = 0`
=> `dy/dt = (-3x)/y*(dx/dt)`
At x = 1, `dx/dt = 2` and y can take on the values `-sqrt 7` and `+sqrt 7`
This gives `dy/dt = -6/sqrt 7` and `dy/dt = 6/sqrt 7` .
The position of the particle satisfies the given relation at two different points. The value of `dy/dt` is different for each of them.
At `y = sqrt 7` , `dy/dt = -6/sqrt 7` and at `y = -sqrt 7` , `dy/dt = 6/sqrt 7`