Answer
The proof is below.
Work Step by Step
We know that equation 3.14 is given by:
$$ y = x tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x^2$$
From this, we find that dy/dx is:
$$ \frac{dy}{dx} = tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x$$
We know that dy/dx gives the slope, along with $ tan \theta$. Thus, we know:
$$ tan\theta = \frac{dy}{dx} = tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x$$
We now consider equations 3.10 and 3.11:
$$ v_y = v_{y0}-gt$$
$$v_x = v_{x0}$$
We simplify to find:
$$ v_y = v_{0}sin \theta_0-gt$$
$$v_x = v_{0}cos \theta_0$$
Since $v_y = dy$ and $v_x=dx$, we find:
$$\frac{v_y}{v_x}=\frac{dy}{dx}=tan \theta = \frac{ v_{0}sin \theta_0-gt}{v_{0}cos \theta_0}$$
We finally simplify to:
$$ tan\theta = \frac{dy}{dx} = tan \theta_0 - \frac{g}{ v_o^2cos^2 \theta}x$$
Thus, we conclude that $dy/dx$ is, indeed, the velocity.
