Answer to Question #349090 in Calculus for gogo

Question

A particle moves along the parabola y²Â = 4x with a constant horizontal component velocity of 2 m/s. Find the vertical component of the velocity at the point (1, 2).

Answer 1

We know that:

$v_x=\frac{dx}{dt},$ $v_y=\frac{dy}{dt},$ and given $|\vec{v_x}|=2$ m/s.

Let's differentiate the function $y^2=4x$ :

$2y\frac{dy}{dt}=4\frac{dx}{dt},$

$2yv_y=4v_x,$

$v_y=\frac{4v_x}{2y}=\frac{2v_x}{y}=\frac{2\cdot2}{y}=\frac{4}{y}.$

At the point (1, 2) we have y=2, so

$v_y=\frac{4}{2}=2$ m/s.

Answer: 2 m/s.