# Optimizing Projectile Arclength

I was running through some old Putnam problems and came across one from the 1940 exam that asked the following:

A stone is thrown from the ground with speed $v$ at an angle $θ$ to the horizontal. There is no friction and the ground is flat. Find the total distance it travels before hitting the ground. Show that the distance is greatest when $\sin θ \cdot\ln (\sec θ + \tan θ) = 1$.

My work:

We can describe the motion with the following vector:

$$\vec r(t)=\langle v\cos\theta \cdot t,v\sin\theta \cdot t-gt^2/2\rangle$$

We know the vector hits the ground when the $y$ component is equal to $0$. Solving for $t_0$, we get that the ball reaches the ground again at $t_0=\frac{2v\sin\theta}{g}$.

Let's set up an integral for the arc length $s$:

$$s=\int_0^{t_0}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}dt$$

Some quick differentiation and we can replace $\frac{dx}{dt}$ and $\frac{dy}{dt}$ with the following identities:

$$s=\int_0^{t_0}\sqrt{(-v\sin\theta)^2+(v\cos\theta-gt)^2}dt$$

Since we are trying to optimize $\theta$, if we take the derivative of the arc length, set it equal to $0$, and find the maximum, then we can solve for optimal $\theta$. We'll also replace $t_0$ with its identity in terms of $\theta$.

$$\frac{ds}{d\theta}=\frac{d}{d\theta}\int_0^{\frac{2v\sin\theta}{g}}\sqrt{v^2-2v\cos\theta\cdot gt+g^2t^2}dt$$

$$\frac{ds}{d\theta}=\frac{2v\cos\theta}{g}\sqrt{v^2-2v\cos\theta\cdot gt+g^2t^2}$$

Here I am unsure how to relate $t$ to $v$ and $\theta$. Do I use the identity for $t_0$, a kinematics equation, or simply treat $t$ as a constant? This, of course, assumes my work thus far has been valid.

Next I would either show that the identity holds after finding $\theta$ or use some relation involving $\theta$ and show that it's an identity to the listed equation.

Thanks for taking the time to read/respond.

• It looks like you've made an error in the very last step you've written, because the expression should now be independant of t, which was simply a variable for definite integration. In particular, you'd need to use the Leibniz integral rule to get the correct final step. Everything else looks good so far though! Feb 22 '17 at 3:08
• So replace $t$ with $\frac{2v\sin\theta}{g}$ per 2nd Fundamental Theorem and simplify, no? Feb 22 '17 at 3:12
• That'll fix one of the terms, but there should be another term where you'd have to compute a new integral. Feb 22 '17 at 3:13
• Why is that? @B.Mehta Feb 22 '17 at 4:01
• qc.edu.hk/math/Resource/AL/Trajectory%20of%20Projectile.pdf Feb 22 '17 at 5:02

WLOG, $v=1$ and $g=1$ (you can rescale time and space independently), the trajectory is

$$x=t\cos\theta,\\y=t\sin\theta-\frac{t^2}2,$$ and the total travel time is

$$2\sin\theta.$$

Then you want to maximize

$$L=\int_0^{2\sin\theta}\sqrt{(t-\sin\theta)^2+\cos^2\theta}\,dt=\int_{-\sin\theta}^{\sin\theta}\sqrt{t^2+\cos^2\theta}\,dt\\ =\sin^2\theta\int_{-1}^{1}\sqrt{u^2+\tan^2\theta}\,du\\ =\frac12\sin^2\theta\left.\left(u\sqrt{u^2+\tan^2\theta}+\tan^2\theta\log(u+\sqrt{u^2+\tan^2\theta})\right)\right|_{u=-1}^1\\ =\frac12\sin^2\theta\left(2\sec\theta+\tan^2\theta\log\frac{\sec\theta+1}{\sec\theta-1}\right).$$

The claim should follow by differentiation and simplification.

Instruction is as follows

1) Evaluate $s(\theta, v)$

2) Take the gradient of the expression and equate to zero $\nabla_{v, \theta}s(v, \theta)=0$

3) This gives two equations

4) Solve them

5) Find the Hessian $$\begin{bmatrix} \partial^{2}_{v, v}s(v, \theta) & \partial^{2}_{v, \theta}s(v, \theta) \\ \partial^{2}_{v, \theta}s(v, \theta) & \partial^{2}_{\theta, \theta}s(v, \theta) \end{bmatrix}$$ 6) Evaluate the Hessain at the extremum's you've found in 4. Find the eigenvalues of the Hessian.

7) If all the eigenvalues of Hessian are positive then we have a minimum. If all the eigenvalues of Hessian are negavive then we have a minimum.

• Your mistake was obviously in that you've optimised by the direction only, but not the magnitude. Feb 22 '17 at 9:49
• Obviously not. $v$ is a constant, otherwise the distance is maximized by $v=\infty$.
– user65203
Feb 22 '17 at 10:09