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Section 3 Arc lenght: Curve properties.

One of the most fundamental characteristics of a curve is its length. As curves defined by vector functions are being introduced now in this material, we need to define the length of such a curve means, and to do that in a way that is compatible with the lengths of polygonal lines (which we know from euclidean geometry). We do this by approximating the length of the curve by polygonal lines and taking the limit of the lengths of these polygonal lines.

Recall that the length of the segment joining the points \(P=(P_x, P_y, P_z)\) and \(Q=(Q_x, Q_y, Q_z)\) can be calculated by

\begin{equation*} \|Q-P\|=\sqrt{(Q_x-P_x)^2+(Q_y-P_y)^2+(Q_z-P_z)^2}. \end{equation*}

Consider (formally) a curve defined by a vector function \(\vec{r}:[a,b] \to \mathbb{R}^n\) (we are mostly concerned with \(n=2\) ou \(3\)). Then let us partitionate the interval \([a,b]\) into \(n\) parts of equal length \(P=\{a, a+\Delta t, a+2\Delta t, \cdots, a+n\Delta t=b\}\text{,}\) with \(\Delta t = (b-a)/n\text{,}\) and approximate the length of the curve by the polygonal joining consecutive points of the curve defined by the partition \(\vec{r}(a+i\Delta t)\text{.}\)

Please note (and try the code) that the polygonal line gets indistinguishable from the curve as we increase the number of points in the partition. The length of the curve can be approximated by the sums of the lengths of the line segments in the polygonal line:

\begin{align*} L \approx \amp\sum_{i=0}^{n-1} \|\vec{r}(a+(i+1)\Delta t) - \vec{r}(a+i\Delta t)\| \\ \approx\amp \sum_{i=0}^{n-1} \left\|\frac{\vec{r}(a+(i+1)\Delta t) - \vec{r}(a+i\Delta t)}{\Delta t}\right\|\Delta t. \end{align*}

Numerical calculations for lengths are often done by such sums with fine partitions.

This motivates us to define the length as

\begin{equation*} L =\lim_{n\to \infty} \sum_{i=0}^{n-1} \left\|\frac{\vec{r}(a+(i+1)\Delta t) - \vec{r}(a+i\Delta t)}{\Delta t}\right\|\Delta t = \int_a^b\|\vec{r}'(t)\|dt. \end{equation*}
Definition 3.1.

If \(\vec{r}: I \to \mathbb{R}^n\) is a smooth parametrization (piecewise 1 ) of a curve that does not cover arcs of the curve more than once, then we define its length as

\begin{equation*} L = \int_a^b\|\vec{r}'(t)\|dt, \end{equation*}

if the limit exists.

This means that there is, at most, a finite number of points, where the function is not differentiable.
Remark 3.2.
It is possible to show that the length defined like this does not depend on the chosen parametrization (although the formula explicitly shows a parametrization), if the parametrization only passes through each section of curve once. This is important, then the notion of length cannot depend on the vector function that describes the curve. The following theorem shows this.

If \(\vec{r}\) and \(\vec{q}\) are in one to one correspondence with \(C\text{,}\) we can change from the parameter \(t\) to the parameter \(s\) through \(s=(\vec{q}^{-1} \circ \vec{r})(t)\text{.}\) Let us assume that both parametrizations run through the curve in the same direction, so that \((\vec{q}^{-1} \circ \vec{r})(a)=c\) and \((\vec{q}^{-1} \circ \vec{r})(b)=d\text{.}\)  2  As the functions are smooth, \((\vec{q}^{-1} \circ \vec{r})'(t)\) exists and as the parametrizations run the curve in the same direction \((\vec{q}^{-1} \circ \vec{r})'(t)\geq 0\) for all \(t\text{.}\) Then we can change variables in the following integral and calculate:

\begin{align*} \int_c^d\|\vec{q}'(s)\|ds =\amp \int_a^c\|\vec{q}'((\vec{q}^{-1} \circ \vec{r})(t))\|(\vec{q}^{-1} \circ \vec{r})'(t)dt\\ =\amp \int_a^c\|(\vec{q}^{-1} \circ \vec{r})'(t)\vec{q}'((\vec{q}^{-1} \circ \vec{r})(t))\|dt \\ =\amp \int_a^b \|(\vec{q}\circ \vec{q}^{-1} \circ \vec{r})'(t)\|dt = \int_a^b \|\vec{r}'(t)\|dt, \end{align*}

where we used the chain rule from the second to the third line. Thus the same length is obtained using either of the parametrizations.

If that is not the case, the proof is also possible, but with extra care to the signs and limits of the integrals.
Remark 3.4.
If \(\vec{r}(t)=(x(t),y(t))\) is a smooth parametrization that does not cover multiple times any arc of a curve \(C \subset \mathbb{R}^2\text{,}\) then:
\begin{equation*} L = \int_a^b\|\vec{r}'(t)\|dt = \int_a^b\sqrt{(x'(t))^2+(y'(t))^2}dt. \end{equation*}
If \(\vec{r}(t)=(x(t),y(t),z(t))\) is a smooth parametrization that does not cover multiple times any arc a curve \(C \subset \mathbb{R}^3\text{,}\) then:
\begin{equation*} L = \int_a^b\|\vec{r}'(t)\|dt = \int_a^b\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt. \end{equation*}

Using the definition of length, find an expression for the length of the graph of a function \(y=f(x)\text{,}\) with \(a\leq x \leq b\text{.}\)

Solution.
The graph of a function has a natural parametrization, namely \(\vec{r}(t)=(t,f(t))\text{,}\) with \(a\leq t \leq b\text{.}\) Let us calculate the length using this parametrization. To that we calculate \(\vec{r}'(t)=(1,f'(t))\text{,}\) so that the length is
\begin{equation*} L = \int_b^a \sqrt{1 +(f'(t))^2} dt, \end{equation*}
which agrees with length calculated in single variable calculus.

The following code calculates the length of the curve numerically (using scipy) \(C\subset \mathbb{R}^3\) parametrized by \(r(t)\text{,}\) with \(t \in [a,b]\text{.}\)

To each of the vector functions, calculate \(\vec{r}'\text{,}\) \(\|\vec{r}'\|\) and the length \(L\) of the curve defined by the vector function. Do not forget to check if the parametrization is covering arcs of the curve multiple times (plotting the curves and changing the domain might help). Then check your solutions with the code above

  1. \(\vec{r}(t) = (t, \cosh(t)), \,\,\, t \in [-1,1] \text{;}\) (hint: \(cosh^2(t) - sinh^2(t) = 1\))

  2. \(\vec{r}(t) = (\cos(t), \sin(t), 3\sin(t)), \,\,\, t \in [0,4\pi] \text{;}\)

  3. \(\vec{r}(t) = (t^2, t^3), \,\,\, t \in [-2,2] \text{;}\)

Solution.

  1. \(\vec{r}'(t) = (1, \sinh(t))\text{,}\) \(\|\vec{r}'\|= \sqrt{1 + \sinh^2(t)} = \sqrt{\cosh^2(t)}= |\cosh(t)| = \cosh(t)\) and

    \begin{equation*} L = \int_{-1}^1 \cosh(t) dt = \sinh(t)|_{-1}^1 = \sinh(1) - \sinh(-1) = 2\sinh(1) \approx 2.35 \end{equation*}

  2. \(\vec{r}'(t) = (-\sin(t), \cos(t), 3\cos(t))\text{,}\) \(\|\vec{r}'\|= \sqrt{(-\sin(t))^2 + (\cos(t))^2+ (3\cos(t))^2} = \sqrt{1+9\cos^2(t)}\) and

    \begin{equation*} L = \int_{0}^{2\pi} \sqrt{1+9\cos^2(t)} dt \approx 13.97 \end{equation*}

  3. \(\vec{r}'(t) = (2t, 3t^2)\text{,}\) \(\|\vec{r}'\|= \sqrt{4t^2 + 9t^4} = |t|\sqrt{4+9t^2}\) and

    \begin{align*} L =\amp \int_{-2}^2 \|\vec{r}'\| dt = \int_{-2}^0 -t\sqrt{4+9t^2} dt + \int_{0}^2 t\sqrt{4+9t^2} dt\\ \amp + \frac{-1}{18}\int_{40}^4 \sqrt{u} dt + \frac{1}{18}\int_{4}^{40} \sqrt{u} dt\\ \amp = \frac{-1}{18}\int_{40}^4 \sqrt{u} dt + \frac{1}{18}\int_{4}^{40} \sqrt{u} dt = \frac{-1}{27} u^{3/2}|_{40}^4 + \frac{1}{27} u^{3/2}|_{4}^{40} \approx 9.07 \end{align*}

Change the parameter in the vector of functions of the last exercise by doing \(t=s^3\) and changing the domains to \([\sqrt[3]{a},\sqrt[3]{b}]\text{.}\) Then verify, with the code, that the lengths obtained by using the new parametrizations \(\vec{l}(s)=\vec{r}(s^3)\) are the same as the ones obtained before.

Solution.
Replace the appropriate functions in the code (see video solution).

If we have a trajectory given by some smooth vector function \(\vec{r}(t)\text{,}\) we might be interested into calculating the distance traveled from a starting time \(t_0\) (often 0) until time \(t\text{.}\) Letting \(t\) vary, we obtain the arc length function:

\begin{equation*} L(t)=\int_{t_0}^t \|\vec{r}'(u)\|du. \end{equation*}

The following code writes an expression for the arc length function.

Given a parametrization \(\vec{r}(t)\) of a spacial curve, \(C\text{,}\) we might use the arc length function in order to obtain a special parametrization of \(C\text{,}\) called parametrization by arc length, witch we shall denote by \(\vec{r}_0(s)\text{,}\) such that the tangent vector has always length 1: \(\|\vec{r}'_0(s)\| = 1\text{.}\) This parametrization is found by making \(s=L(t)\) and replacing \(t=L^{-1}(s)\) in \(\vec{r}(t)\text{,}\) so that \(\vec{r}_0(s)=\vec{r}(L^{-1}(s))\text{.}\) 3 

We can calculate that \(\vec{r}_0(s)\) has the desired property, \(\|\vec{r}'_0(s)\|^2=1\text{,}\) using the chain rule and the inverse function theorem in \((L^{-1}(s))'=1/L'(t) = \|\vec{r}'(t)\|.\)

Let \(\vec{r}(t)=\cos t\vec{i}+\sin t\vec{j}+3t\vec{k}\text{,}\) be a parametrization of a helix, defined in \([0,\infty)\text{.}\) Then the derivative of \(\vec{r}(t)\) is:

\begin{equation*} \vec{r'}(t)=-\sin t\vec{i}+\cos t\vec{j}+3\vec{k}, \end{equation*}

so that

\begin{equation*} \|\vec{r'}(t)\|^2=(\sin t)^2+(\cos t)^2+9=10 \end{equation*}

and the arc length function is

\begin{equation*} L(t)=\int_0^t \|\vec{r}'(u)\|du = \int_0^t \sqrt{10} du =\sqrt{10} t. \end{equation*}

Doing \(s=L(t)=\sqrt{10} t\text{,}\) we isolate \(t\) in terms of \(s\text{:}\)

\begin{equation*} t=\frac{s}{\sqrt{10}}. \end{equation*}

So the arc length parametrization of this helix is

\begin{align*} \vec{r}_0(s) =\amp \vec{r}(t) = \vec{r}\left(\frac{s}{\sqrt{10}}\right) \\ =\amp \cos \left(\frac{s}{\sqrt{10}}\right)\vec{i}+\sin \left(\frac{s}{\sqrt{10}}\right)\vec{j}+\left(\frac{3s}{\sqrt{10}}\right)\vec{k}. \end{align*}
Remark 3.9.
The arc length parametrization of a curve is particularly important to study the geometry of the curve, then the scalar speed of the curve is constant (and we do not risk mixing trajectory properties with the geometrical ones). However, it might be hard to calculate the inverse of \(L(t)\) explicitly and even computer methods might find difficult to find an expression. So we might prefer to do this numerically, as done in the next code:

The following code can be use to animate trajectories defined by some parametrizations (by arc length or not):

For each of the vector functions below use the codes above to animate the defined trajectories:

  1. \(\vec{r}(t) = (\ln(t), \cos(t),0), \,\,\,\,\, t \in [1,10]\text{;}\)

  2. \(\vec{r}(t) = (\cos(2t), sin(t), 3t), \,\,\,\,\, t \in [0,10]\text{;}\)

  3. \(\displaystyle \vec{r}(t) = (t, t^2, t^3/3), \,\,\,\,\, t \in [0,3]\)

Solution.
Replace the appropriate functions in the code (see video solution).

A notion of how much a curve \(C\) defined by some smooth vector function \(r:[a,b] \to \mathbb{R}^3\)is bending is important, then it can indicate points of higher deformation and how tight a curve is in a road. Such a notion is called curvature in math and it is defined as second derivative of the arc length parametrization of \(C\text{:}\) \(||\vec{r}_0''(s)||\text{.}\) We use the arc length parametrization 4  because the second derivative corresponds to the acceleration. The expression \(k=||\vec{r}_0''(s)||\) is prone to numerical error, and the formula

\begin{equation*} k=\frac{\sqrt{(z''y'-y''z')^2+(x''z'-z''x')^2+(y''x'-x''y')^2}}{(x'^2+y'^2+z'^2)^{3/2}} \end{equation*}

is better suited (although its proof is beyond the scope of this material). The following code calculates the curvature and plots it against the parameter \(t\text{:}\)

That has constant scalar speed.

For each of the vector functions below use the code above to determine approximately (by the graph) points of maxima and minima for the curvature of the curves defined by the vector functions:

  1. \(\vec{r}(t) = (\ln(t), \cos(t),0), \,\,\,\,\, t \in [1,10]\text{;}\)

  2. \(\vec{r}(t) = (\cos(2t), sin(t), 3t), \,\,\,\,\, t \in [0,10]\text{;}\)

  3. \(\displaystyle \vec{r}(t) = (t, t^2, t^3/3), \,\,\,\,\, t \in [0,3]\)

Solution.

a)

We find points of maxima for the curvature at \(\pi, 2\pi, 3\pi\text{.}\) That correspond to the three high bending points in the plot:

b)

We find points two different types of local maxima: the lower ones at \(0, \pi, 2\pi, 3\pi\text{,}\) that correspond to the bends "in the middle" of the plot, and higher ones at \(\pi/2, 3\pi/2, 5\pi/2, 7\pi/2\text{,}\) that correspond to the bends "in the corners" of the plot. The minima show the flatter parts, in between consecutive local maxima:

c)

We find a point of maxima at \(0\text{,}\) with curvature 2, and the curvature seems to tend to 0 as \(t\) increases: