Question

Calculus: implicit differentiation/particle's distance changing question?

A particle is moving along the parabola 4y = (x + 2)^2 in such a way that
its x-coordinate is increasing at a constant rate of 2 units per second. How
fast is the particle's distance to the point (-2, 0) changing at the moment
that the particle is at the point (2, 4)?

Answer 1

i would calculate the velocity vector at (2,4) and dot that with the *unit version* of direction vector from (-2,0) to (2,4).
that should give the velocity of the particle along that second vector ie the rate at which the distance is changing between the 2.

Answer 2

i would like try out irishboy's solution.
the way i would do it is
Given:
$$\Large { 4y = (x+2)^2\\~\\
4~\frac{dy}{dt} = 2(x+2)^1 ~ \frac{dx}{dt}
\\ ~\\\rm Plug ~in~\frac{dx}{dt}=2
\\

}
$$