Arc Length with Vector Functions

Determining the length of a vector function $\vec{r}(t)=⟨f(t),g(t),h(t)⟩$ on the interval $a\le t\le b$

We can first rewrite in parametric form:

$$x=f(t),y=g(t),z=h(t)$$

Recall the arc length of a 2D parametric curve:

$$L={\int }_a^b\sqrt{[f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2}dx$$

Extending this to 3D:

$$L={\int }_a^b\sqrt{[f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2+[h^{\prime }(t){]}^2}dt$$

This can be simplified based on the fact that the the integrand is actually the same as the magnitude of the tangent vector:

$$\mid \mid {\vec{r}}^{\prime }(t)\mid \mid =\sqrt{[f^{\prime }(t){]}^2+[g^{\prime }(t){]}^2+[h^{\prime }(t){]}^2}$$

Thus, we have:

Arc Length

$$L={\int }_a^b\mid \mid {\vec{r}}^{\prime }(t)\mid \mid dt$$

We can also have:

Arc length function

$$s(t)={\int }_0^t\mid \mid {\vec{r}}^{\prime }(u)\mid \mid du$$

Plugged into the original function, we can reparametrize the function into the form

$$\vec{r}(t(s))$$

This allows us to tell where we are on the curve after we’ve traveled a distance $t$ along the curve (start measurement for where we are at $t=0$)

Here’s an example of how this works:

Example: Arc length function

Where on the curve $\vec{r}(t)=⟨2t,3\sin (2t),3\cos (2t)⟩$ are we after traveling for a distance of $\frac{\pi \sqrt{10}}{3}$?

Finding the arc length function:

$$\begin{array}{rl}\mid \mid \vec{r}(t)\mid \mid & =\sqrt{4+36{\cos }^2(2t)+36{\sin }^2(2t)}=\sqrt{4+36}=2\sqrt{10}\\ s(t)& ={\int }_0^{t}2\sqrt{10}du=(2\sqrt{10}u{|}_0)=2\sqrt{10}t\end{array}$$

Solving for $t$ based on $s(t)$:

$$t=\frac{s}{2\sqrt{10}}$$

Thus, we have

$$\begin{array}{rl}\vec{r}(t(s))& =⟨\frac{s}{\sqrt{10}},3\sin \left(\frac{s}{\sqrt{10}}\right),3\cos \left(\frac{s}{\sqrt{10}}\right)⟩\\ \vec{r}\left(t\left(\frac{\pi \sqrt{10}}{3}\right)\right)& =⟨\frac{\pi }{3},3\sin \left(\frac{\pi }{3}\right),3\cos \left(\frac{\pi }{3}\right)⟩\\ & =⟨\frac{\pi }{3},\frac{3\sqrt{3}}{2},\frac{3}{2}⟩\end{array}$$

Therefore, after traveling a distance of $\frac{\pi \sqrt{10}}{3}$ along the curve, we are at the point $(\frac{\pi }{3},\frac{3\sqrt{3}}{2},\frac{3}{2})$.