# Paths

Similar to parametric surfaces, paths are typically defined by using a parametric function mapping a single input, $t$ to any number of dimensions through a vector valued function. However, instead of visualizing $t$ as time, we instead visualize it by tracing out the entire path along the domain of $t$

$r\left(t\right)=\left(\mathrm{cos}\left(2\pi t\right),\mathrm{sin}\left(2\pi t\right),t\right)$

One question we might have about a path is how long it is. We can use integrals to work out how long a path is by breaking the problem into smaller sub-problems.

Say for instance we take a very small section of that path like $dt$. You could imagine approximating that path using a vector, or a series of vectors in their place.

Because we can reduce the problem into a sum of vector lengths over a defined interval, we can also express the problem as an integral! To do this we integrate over the length of each vector for $dt$. We determine what each vector is by taking the gradient vector of the path at each point $t$. Then measure the length.

${\int }_{0}^{t}\sqrt{{\frac{\mathrm{\partial }x}{\mathrm{\partial }t}}^{2}+{\frac{\mathrm{\partial }y}{\mathrm{\partial }t}}^{2}+{\frac{\mathrm{\partial }z}{\mathrm{\partial }t}}^{2}}\phantom{\rule{thinmathspace}{0ex}}dt$

So for instance, on our function above, we would have:

${\int }_{0}^{t}\sqrt{\left(-2\pi \mathrm{sin}\left(2\pi t\right){\right)}^{2}+\left(2\pi \mathrm{cos}\left(2\pi t\right){\right)}^{2}+1}\phantom{\rule{thinmathspace}{0ex}}dt$

Now considering that a negative squared is a positive, we can rewrite as follows:

${\int }_{0}^{t}\sqrt{\left(2\pi \mathrm{sin}\left(2\pi t\right){\right)}^{2}+\left(2\pi {\mathrm{cos}}^{2}\left(2\pi t\right){\right)}^{2}+1}\phantom{\rule{thinmathspace}{0ex}}dt$

And remembering that ${\mathrm{sin}}^{2}\left(\theta \right)+{\mathrm{cos}}^{2}\left(\theta \right)=1$ we can simplify:

${\int }_{0}^{t}\sqrt{4{\pi }^{2}si{n}^{2}\left(2\pi t\right)+4{\pi }^{2}co{s}^{2}\left(2\pi t\right)+1}\phantom{\rule{thinmathspace}{0ex}}dt$${\int }_{0}^{t}\sqrt{4{\pi }^{2}\left(si{n}^{2}\left(2\pi t\right)+co{s}^{2}\left(2\pi t\right)\right)+1}\phantom{\rule{thinmathspace}{0ex}}dt$${\int }_{0}^{t}\sqrt{4{\pi }^{2}+1}\phantom{\rule{thinmathspace}{0ex}}dt=6.36$