Question

Using calculus, you can find a function’s maximum or minimum by differentiating and setting the result to zero. Do this for equation x=v20gsin2θ0x=v02 sin⁡2θ0, differentiating with respect to θθ, and thus find the maximum range for θθ.

Answer 1

Answer:

$\dfrac{dx}{d\theta}=\dfrac{2v_0^2\cos2\theta}{g}$

$x=\dfrac{v_0^2}{g}$ at $\theta=45$

Step-by-step explanation:

$x=\dfrac{v_0^2\sin2\theta}{g}$

Differentiating with respect to $\theta$ means $v_0$ and $g$ are constants. So we differentiate only the $\sin2\theta$ term.

Assume $y=\sin2\theta$.

Let $u=2\theta$.

$\dfrac{du}{d\theta}=2$

Also, $y$ becomes

$y=\sin u$

$\dfrac{dy}{du}=\cos u$

By chain rule,

$\dfrac{dy}{d\theta}=\dfrac{dy}{du}\cdot \dfrac{du}{d\theta}$

$\dfrac{dy}{d\theta}=\cos u\cdot2=2\cos u=2\cos 2\theta$

Adding back the constants,

$\dfrac{dx}{d\theta} = \dfrac{2v_0^2\cos2\theta}{g}$

Equating this to 0,

$\dfrac{dx}{d\theta}=\dfrac{2v_0^2\cos2\theta}{g}=0$

$\cos2\theta=0$

$2\theta = \cos^{-1}0=90$

$\theta=45$

Using this value of $\theta$ in the expression for $x$,

$x=\dfrac{v_0^2\sin(2\times145)}{g}=\dfrac{v_0^2\sin90}{g}=\dfrac{v_0^2}{g}$ (since $\sin90=1$)