Arc Length of Projectile

We will show that:

The arc length of any projectile subject to no forces other than gravity can be expressed by the formula:

\(\ell = \frac{(v_0\cos(\phi))^2}{2g} \left[ 2\sec(\phi)\tan(\phi) + \ln \lvert\frac{1+\sin(\phi)}{1-\sin(\phi)} \rvert \right]\)

Where:

$\phi$ is the angle the projectile initially makes with the horizontal axis

$v_0$ is the initial velocity

$g$ is the force of gravity

I suppose before we start, I should explain why I decided to put such a disgusting looking formula up here. Arc length integrals are not particularly difficult (usually), but they are often times very annoying to actually compute, at least in my opinion. With this formula, arc length can now be considered a function of any of the above variables, even $g$. For instance, suppose we wished to study the effect of gravitational fields with differing magnitude (but always constant direction) on the arc length of a projectile. We would simply make the above formula a function of the variable $g$ and hold everything else constant (as long as all other factors are the same). This is why I felt this formula has its advantages and is thus worthy of sharing.

Proof:

Consider a massive particle projected from the origin of a Cartesian Coordinate system with initial velocity $v_0$ and angle $\phi$. There is negligible air resistance and the only force acting on said particle is gravity. Then, by Newton, we have:

$\displaystyle \sum \vec{F} = m\vec{a} = \vec{W}$

$\vec{W}$ is just the weight of the particle:

$\vec{W} = -mg \hat{j}$

Then:

$\vec{a} = -g \hat{j}$

Acceleration is clearly constant, hence we may proceed to derive vector-valued functions for our velocity and position:

$\vec{v} = \displaystyle \int \vec{a} \,dt = (v_0\cos(\phi))\hat{i} + (v_0\sin(\phi) - gt)\hat{j}$

$\vec{r} = \displaystyle \int \vec{v}\,dt = (v_0\cos(\phi)t)\hat{i} +(v_0\sin(\phi)t - \frac{gt^2}{2})\hat{j}$

(This assumes that we zero-reference our position from the initial position)

Now, keep these in mind as we derive an integral formula for arc length:

By the Pythagorean Theorem, if we consider a triangle with infinitesimal hypotenuse on the trajectory of the particle, we have:

$d\ell^2 = dx^2 +dy^2$

Now, our position equations are parameterized in time. Then:

$\frac{d\ell^2}{dt^2} = \frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}$

$\Rightarrow$ $\frac{d\ell}{dt} = \sqrt{\frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}}$

$\Rightarrow$ $\ell = \displaystyle \int_{0}^{t_0} \sqrt{\frac{dx^2}{dt^2} + \frac{dy^2}{dt^2}}\,dt$

Where $t_0$ is the time at which the particle has exactly zero height.

(As a quick side note, this integral should make sense. The integrand is simply the magnitude of the velocity as a function of time, hence we are summing all infinitesimal velocity elements along the curve at all times $0\leq t \leq t_0$ to obtain our length)

Now we have our position functions for $x$ and $y$, these are just the components of our position vector. Then:

$x(t) = v_0\cos(\phi)t$ $\Rightarrow$ $\frac{dx}{dt} = v_0\cos(\phi)$

$y(t) = v_0\sin(\phi)t - \frac{gt^2}{2}$ $\Rightarrow$ $\frac{dy}{dt} = v_0\sin(\phi) - gt$

Then our integral is:

$\ell = \displaystyle \int_{0}^{t_0} \sqrt{ v_0^2\cos^2(\phi) + \frac{dy^2}{dt^2}}\,dt$

Let: $u= \frac{dy}{dt} = v_0\sin(\phi) - gt$ $\Rightarrow$ $-\frac{du}{g} = dt$

Then:

$\ell =-\frac{1}{g} \displaystyle \int_{v_0\sin(\phi)}^{v_0\sin(\phi) -gt_0} \sqrt{ v_0^2\cos^2(\phi) + u^2}\,du$

Recall that: $\sin^2(\phi) + \cos^2(\phi) = 1$

Dividing the entire equation by $\cos^2(\phi)$ we obtain:

$\tan^2(\phi) + 1 = \sec^2(\phi)$

Let: $u=v_0\cos(\phi)\tan(\theta)$ $\Rightarrow$ $du = v_0\cos(\phi) \sec^2(\theta) d\theta$

Before plugging this in to our integral, we shall find an expression for $t_0$. This will help us change our bounds:

$0 = v_0\sin(\phi) t_0 - \frac{gt_0^2}{2}$ $\Rightarrow$ $t_0 = 0$

or

$t_0 = \frac{2v_0\sin(\phi)}{g}$

$t_0=0$ is simply the initial time, hence we neglect this solution in favor of the latter.

Now let us change our bounds:

$\displaystyle \lim_{u \to v_0\sin(\phi)} \theta = \displaystyle \lim_{u \to v_0\sin(\phi)} \tan^{-1} \left(\frac{u}{v_0 \cos(\phi)} \right) = \phi$

$\displaystyle \lim_{u \to v_0\sin(\phi)-gt_0} \theta = \displaystyle \lim_{u \to v_0\sin(\phi)-gt_0} \tan^{-1} \left(\frac{u}{v_0 \cos(\phi)} \right) = -\phi$

Then:

$\ell = -\frac{1}{g} \displaystyle \int_{\phi}^{-\phi} \sqrt{ v_0^2\cos^2(\phi) +v_0^2\cos^2(\phi)\tan^2(\theta)}(v_0\cos(\phi)\sec^2(\theta))\,d\theta$

$\Rightarrow$ $\ell = \frac{(v_0\cos(\phi))^2}{g} \displaystyle \int_{-\phi}^{\phi} \sec^3(\theta) \,d\theta = \frac{(v_0\cos(\phi))^2}{g} \displaystyle \int_{-\phi}^{\phi} \sec(\theta)\frac{d}{d\theta} \tan(\theta)\,d\theta$

$\Rightarrow$ $\ell =\frac{(v_0\cos(\phi))^2}{g} \left[ \sec(\theta)\tan(\theta) |_{-\phi}^{\phi} - \displaystyle \int_{-\phi}^{\phi} \sec(\theta)\tan^2(\theta) \,d\theta \right]$

$\Rightarrow$ $\frac{(v_0\cos(\phi))^2}{g} \displaystyle \int_{-\phi}^{\phi} \sec^3(\theta) \,d\theta =\frac{(v_0\cos(\phi))^2}{g} \left[\sec(\theta)\tan(\theta) |_{-\phi}^{\phi} - \displaystyle \int_{-\phi}^{\phi} \sec^3 (\theta) \,d\theta + \displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta \right]$

$\Rightarrow$ $\ell = \frac{(v_0\cos(\phi))^2}{2g}\left[\sec(\theta)\tan(\theta) |_{-\phi}^{\phi} + \displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta \right]$

Now, focus on $\displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta$. We'll deal with everything else later, we need to evaluate this integral first:

$\displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta = \displaystyle \int_{-\phi}^{\phi} \sec(\theta) \frac{\frac{1+\sin(\theta)}{\cos(\theta)}}{\frac{1+\sin(\theta)}{\cos(\theta)}} \,d\theta$

If we let $u= \frac{1+\sin(\theta)}{\cos(\theta)}$, then:

$du = \sec(\theta) \frac{1+\sin(\theta)}{\cos(\theta)}d\theta$

Then we have:

$\displaystyle \int_{-\phi}^{\phi} \sec(\theta) \,d\theta = \displaystyle \int \frac{du}{u} |_{-\phi}^{\phi} = \ln \lvert \frac{1+\sin(\theta)}{\cos(\theta)} \rvert |_{-\phi}^{\phi}$

$\Rightarrow$ $\ell = \frac{(v_0\cos(\phi))^2}{2g} \left [ \sec(\theta)\tan(\theta) + \ln \lvert \sec(\theta) +\tan(\theta) \rvert \right]_{-\phi}^{\phi}$

$\Rightarrow$ $\ell = \frac{(v_0\cos(\phi))^2}{2g} \left [ 2 \sec(\phi) \tan(\phi) + \ln \lvert\frac{1+\sin(\phi)}{1-\sin(\phi)} \rvert \right]$

Since $\sec(-\phi) = \sec(\phi)$ and $\tan(-\phi) = - \tan(\phi)$

This is the intended result.

QED