Question

A particle moves along the parabolic path given by y^2=4x in such a way that when it is at the point (1,-2), its horizontal velocity (in the direction of the x-axis) is 3ft/s. What is its vertical velocity (in the direction of the y-axis) at this instant? PLEASE SHOW WORK AND ANSWER IN THE COMMENTS!

Answer 1

The particle's position as a function of time is parameterized by the ordered pair of functions $(x(t),y(t))$, with

$y(t)^2=4x(t)$

If we differentiate both sides of this equation with respect to time $t$, we have

$2y(t)\dfrac{\mathrm dy(t)}{\mathrm dt}=4\dfrac{\mathrm dx(t)}{\mathrm dt}$

where the derivatives of $x(t)$ and $y(t)$ represent change in horizontal and vertical position, respectively, over time. In other words, these are the particle's horizontal/vertical velocities.

We're told that the horizontal velocity at a certain point in time - namely, when the particle's position is (1, -2) - is 3 ft/s, which is to say that at some time $t$, we have

$2(-2)\dfrac{\mathrm dy(t)}{\mathrm dt}=4(3)\implies\dfrac{\mathrm dy(t)}{\mathrm dt}=-3$

In other words, under the given conditions, the particle has a downward vertical velocity of 3 ft/s.