This section is aimed at students in upper secondary education in the Danish
school system, some objects will be simplified and some details omitted.
Curve Length
To determine the curve length of the graph of a (vector)function we can
make the following observation
$$v(t)=s'(t)\implies s(t)\in\int s'(t)dt=\int v(t)dt$$
so if I want to determine distances, I can integrate its derivative.
This yields the following result
Curve Length of a Vector Function
For a vector function
$$\underline{r}(t)=\begin{pmatrix}x(t)\\y(t)\end{pmatrix}$$
its curve length from \(t=a\) to \(t=b\) is calculated as such
$$L_{\underline{r}}(a,b)=\int_a^b\sqrt{x'(t)^2+y'(t)^2}dt$$
Proof.
Consider moving along the curve of this vector function, and denote by
\(s(t)\) the distance you've moved from some point \(\underline{r}(a)\).
Then we just need to find the speed at which we're moving, but this can
be achieved by finding the tangent vector to the vector function. So
\begin{align}
v(t)=&|\underline{v}(t)|=|\underline{r}'(t)|\\
=&\left|\begin{pmatrix}x'(t)\\y'(t)\end{pmatrix}\right|\\
=&\sqrt{x'(t)^2+y'(t)^2}
\end{align}
This immediately yields
$$L_{\underline{r}}(a,b)=\int_a^b\sqrt{x'(t)^2+y'(t)^2}dt$$
Single Valued Functions
If we consider a real function \(f\) that is appropriately
differentiable, then we can find its curve length by
$$L_f(a,b)=\int_a^b\sqrt{1+f'(x)^2}dx$$
which is just a consequence of taking
$$\underline{r}(t)=\begin{pmatrix}t\\f(t)\end{pmatrix}$$
and "substituting" the variable \(t\) for \(x\).