I know how to find extrema of a function, however, I'm wondering how calculus would approach finding the maximum x-intercept achieved given a parametric function representing a parabola (or at least the first derivative so that the max distance achieved has a derivative of zero).
I was using a parametric equation for representing the path of a projectile.
$$x=v_0\cos(\theta)t$$ $$y=v_0\sin(\theta)t+\frac{1}{2}(-9.8)\cos(t)$$
I thought initially to do parametric differentiation, but that wouldn't help find max x-intercepts. The goal would be to show that the max range is when $\theta$ is $45\unicode{xb0}$ through showing that the max x-intercept of the parabola is when $\theta$ equals $45\unicode{xb0}$. Is this achievable through calculus? Maybe does it have to do with the max x-coordinate of the vertex?
That should be
$$x=v_0\cos(\theta)t$$
$$y=v_0\sin(\theta)t+\frac{1}{2}(-9.8) t^2$$
$\dot x=v_0\cos(\theta)$ velocity makes projectile to travel to x $\rightarrow \infty $
To find $ x_{max}$ at starting height just let $y=0$ in the equation of parabolic trajectory.
To find maximum height reached set $\dot y = 0 $ treating $\theta$ temporarily constant and then next treat it as a variable for differentiation, getting $45^0.$