Let there be the function f(alpha)= (L( v_a - v_0 sin alpha))/(v_0 cos alpha) Where L ,v_a and v_o are constants. Determine alpha such that f(alpha) is minimal ?

Jul 5, 2018

$f \left(\alpha\right) = \frac{L \left({v}_{a} - {v}_{0} \sin \alpha\right)}{{v}_{0} \cos \alpha}$

$= L \left({v}_{a} / {v}_{o} \sec \alpha - \tan \alpha\right)$

$L$ is just a scaling factor so it can be removed, thus writing:

• $F \left(\alpha\right) = f \frac{\alpha}{L} q \quad F ' \left(\alpha\right) = \frac{f ' \left(\alpha\right)}{L} q \quad \text{ etc }$

To optimise $F \left(\alpha\right)$, differentiate wrt $\alpha$:

• ${F}_{\alpha} = {v}_{a} / {v}_{o} \sec \alpha \tan \alpha - {\sec}^{2} \alpha = 0$

$\implies {\underbrace{{\sec}^{2} \alpha}}_{> 0 , \forall \alpha} \left({v}_{a} / {v}_{o} \sin \alpha - 1\right) q \quad = \square q \quad = 0$

Critical points can, therefore, only occur under these rules:

• $\sin \alpha = {v}_{o} / {v}_{a} q \quad q \quad \implies \left\{\begin{matrix}{v}_{o} < = {v}_{a} \\ \cos \alpha = \frac{\sqrt{{v}_{a}^{2} - {v}_{o}^{2}}}{v} _ a \\ \tan \alpha = {v}_{o} / \sqrt{{v}_{a}^{2} - {v}_{o}^{2}}\end{matrix}\right.$

To determine the max or min nature of the Critical Point, grab the second derivative of $\square$, using the product rule:

${F}_{\alpha \alpha} = \left(2 {\sec}^{2} \alpha \tan \alpha\right) \left({\underbrace{{v}_{a} / {v}_{o} \sin \alpha - 1}}_{= 0}\right) + {\sec}^{2} \alpha \left({v}_{a} / {v}_{o} \cos \alpha\right)$

$= {\sec}^{2} \alpha \left({v}_{a} / {v}_{o} \cos \alpha\right) = {v}_{a} / {v}_{o} \sec \alpha$

$\therefore {F}_{\alpha \alpha} = \textcolor{b l u e}{{v}_{a} / {v}_{o}} \cdot \frac{1}{\textcolor{g r e e n}{\sqrt{1 - {v}_{o}^{2} / {v}_{a}^{2}}}}$

With assumption ${v}_{o} < = {v}_{a}$, then the green term is positive; whole expression is positive provided ${v}_{a} , {v}_{o} > 0$.

Accordingly, there is a min when this occurs:

• $\sin \alpha = {v}_{o} / {v}_{a}$

This min will be periodic.

I have solved this way. See the answer below: 