Answer:
[tex]\dfrac{dx}{d\theta}=\dfrac{2v_0^2\cos2\theta}{g}[/tex]
[tex]x=\dfrac{v_0^2}{g}[/tex] at [tex]\theta=45[/tex]
Step-by-step explanation:
[tex]x=\dfrac{v_0^2\sin2\theta}{g}[/tex]
Differentiating with respect to [tex]\theta[/tex] means [tex]v_0[/tex] and [tex]g[/tex] are constants. So we differentiate only the [tex]\sin2\theta[/tex] term.
Assume [tex]y=\sin2\theta[/tex].
Let [tex]u=2\theta[/tex].
[tex]\dfrac{du}{d\theta}=2[/tex]
Also, [tex]y[/tex] becomes
[tex]y=\sin u[/tex]
[tex]\dfrac{dy}{du}=\cos u[/tex]
By chain rule,
[tex]\dfrac{dy}{d\theta}=\dfrac{dy}{du}\cdot \dfrac{du}{d\theta}[/tex]
[tex]\dfrac{dy}{d\theta}=\cos u\cdot2=2\cos u=2\cos 2\theta[/tex]
Adding back the constants,
[tex]\dfrac{dx}{d\theta} = \dfrac{2v_0^2\cos2\theta}{g}[/tex]
Equating this to 0,
[tex]\dfrac{dx}{d\theta}=\dfrac{2v_0^2\cos2\theta}{g}=0[/tex]
[tex]\cos2\theta=0[/tex]
[tex]2\theta = \cos^{-1}0=90[/tex]
[tex]\theta=45[/tex]
Using this value of [tex]\theta[/tex] in the expression for [tex]x[/tex],
[tex]x=\dfrac{v_0^2\sin(2\times145)}{g}=\dfrac{v_0^2\sin90}{g}=\dfrac{v_0^2}{g}[/tex] (since [tex]\sin90=1[/tex])
