# Section11.5The Arc Length Parameter and Curvature¶ permalink

In normal conversation we describe position in terms of both time and distance. For instance, imagine driving to visit a friend. If she calls and asks where you are, you might answer “I am 20 minutes from your house,” or you might say “I am 10 miles from your house.” Both answers provide your friend with a general idea of where you are.

Currently, our vector–valued functions have defined points with a parameter $t\text{,}$ which we often take to represent time. Consider Figure 11.5.1(a), where $\vrt = \la t^2-t,t^2+t\ra$ is graphed and the points corresponding to $t=0,\ 1$ and $2$ are shown. Note how the arc length between $t=0$ and $t=1$ is smaller than the arc length between $t=1$ and $t=2\text{;}$ if the parameter $t$ is time and $\vec r$ is position, we can say that the particle traveled faster on $[1,2]$ than on $[0,1]\text{.}$

Now consider Figure 11.5.1(b), where the same graph is parametrized by a different variable $s\text{.}$ Points corresponding to $s=0$ through $s=6$ are plotted. The arc length of the graph between each adjacent pair of points is 1. We can view this parameter $s$ as distance; that is, the arc length of the graph from $s=0$ to $s=3$ is 3, the arc length from $s=2$ to $s=6$ is 4, etc. If one wants to find the point 2.5 units from an initial location (i.e., $s=0$), one would compute $\vec r(2.5)\text{.}$ This parameter $s$ is very useful, and is called the arc length parameter.

How do we find the arc length parameter?

Start with any parametrization of $\vec r\text{.}$ We can compute the arc length of the graph of $\vec r$ on the interval $[0,t]$ with \begin{equation*} \text{ arc length } = \int_0^t\norm{\vec r\,'(u)}\ du. \end{equation*}

We can turn this into a function: as $t$ varies, we find the arc length $s$ from $0$ to $t\text{.}$ This function is $$s(t) = \int_0^t \norm{\vec r\,'(u)}\ du. \label{eq_vvfarc}\tag{11.5.1}$$

This establishes a relationship between $s$ and $t\text{.}$ Knowing this relationship explicitly, we can rewrite $\vec r(t)$ as a function of $s\text{:}$ $\vec r(s)\text{.}$ We demonstrate this in an example.

##### Example11.5.4Finding the arc length parameter

Let $\vec r(t) = \la 3t-1,4t+2\ra\text{.}$ Parametrize $\vec r$ with the arc length parameter $s\text{.}$

Solution

Things worked out very nicely in Example 11.5.4; we were able to establish directly that $s=5t\text{.}$ Usually, the arc length parameter is much more difficult to describe in terms of $t\text{,}$ a result of integrating a square–root. There are a number of things that we can learn about the arc length parameter from Equation (11.5.1), though, that are incredibly useful.

First, take the derivative of $s$ with respect to $t\text{.}$ The Fundamental Theorem of Calculus (see Theorem 5.4.6) states that $$\frac{ds}{dt}=s\,'(t) = \norm{\vrp(t)}. \label{eq_vvfarc3}\tag{11.5.2}$$

Letting $t$ represent time and $\vec r(t)$ represent position, we see that the rate of change of $s$ with respect to $t$ is speed; that is, the rate of change of “distance traveled” is speed, which should match our intuition.

The Chain Rule states that \begin{align*} \frac{d\vec r}{dt} \amp = \frac{d\vec r}{ds}\cdot\frac{ds}{dt}\\ \vrp(t) \amp = \vrp(s)\cdot \norm{\vrp(t)}. \end{align*}

Solving for $\vrp(s)\text{,}$ we have $$\vrp(s) = \frac{\vrp(t)}{\norm{\vrp(t)}} = \vec T(t), \label{eq_vvfarc2}\tag{11.5.3}$$ where $\vec T(t)$ is the unit tangent vector. Equation (11.5.3) is often misinterpreted, as one is tempted to think it states $\vrp(t) = \vec T(t)\text{,}$ but there is a big difference between $\vrp(s)$ and $\vrp(t)\text{.}$ The key to take from it is that $\vrp(s)$ is a unit vector. In fact, the following theorem states that this characterizes the arc length parameter.

# Subsection11.5.1Curvature

Consider points $A$ and $B$ on the curve graphed in Figure 11.5.7(a). One can readily argue that the curve curves more sharply at $A$ than at $B\text{.}$ It is useful to use a number to describe how sharply the curve bends; that number is the curvature of the curve.

We derive this number in the following way. Consider Figure 11.5.7(b), where unit tangent vectors are graphed around points $A$ and $B\text{.}$ Notice how the direction of the unit tangent vector changes quite a bit near $A\text{,}$ whereas it does not change as much around $B\text{.}$ This leads to an important concept: measuring the rate of change of the unit tangent vector with respect to arc length gives us a measurement of curvature.

##### Definition11.5.10Curvature

Let $\vec r(s)$ be a vector–valued function where $s$ is the arc length parameter. The curvature $\kappa$ of the graph of $\vec r(s)$ is \begin{equation*} \kappa = \snorm{\frac{d\vec T}{ds}} = \snorm{\vec T\,'(s)}. \end{equation*}

If $\vec r(s)$ is parametrized by the arc length parameter, then \begin{equation*} \vec T(s) = \frac{\vrp(s)}{\norm{\vrp(s)}} \text{ and } \vec N(s) = \frac{\vec T\,'(s)}{\norm{\vec T\,'(s)}}. \end{equation*}

Having defined $\norm{\vec T\,'(s)} =\kappa\text{,}$ we can rewrite the second equation as $$\vec T\,'(s) = \kappa\vec N(s). \label{eq_curvature}\tag{11.5.4}$$

We already knew that $\vec T\,'(s)$ is in the same direction as $\vec N(s)\text{;}$ that is, we can think of $\vec T(s)$ as being “pulled” in the direction of $\vec N(s)\text{.}$ How “hard” is it being pulled? By a factor of $\kappa\text{.}$ When the curvature is large, $\vec T(s)$ is being “pulled hard” and the direction of $\vec T(s)$ changes rapidly. When $\kappa$ is small, $T(s)$ is not being pulled hard and hence its direction is not changing rapidly.

We use Definition 11.5.10 to find the curvature of the line in Example 11.5.4.

##### Example11.5.11Finding the curvature of a line

Use Definition 11.5.10 to find the curvature of $\vrt = \la 3t-1,4t+2\ra\text{.}$

Solution

While the definition of curvature is a beautiful mathematical concept, it is nearly impossible to use most of the time; writing $\vec r$ in terms of the arc length parameter is generally very hard. Fortunately, there are other methods of calculating this value that are much easier. There is a tradeoff: the definition is “easy” to understand though hard to compute, whereas these other formulas are easy to compute though it may be hard to understand why they work.

We practice using these formulas.

##### Example11.5.13Finding the curvature of a circle

Find the curvature of a circle with radius $r\text{,}$ defined by $\vec c(t) = \la r\cos(t) ,r\sin(t) \ra\text{.}$

Solution

Example 11.5.13 gives a great result. Before this example, if we were told “The curve has a curvature of 5 at point $A\text{,}$” we would have no idea what this really meant. Is 5 “big” — does is correspond to a really sharp turn, or a not-so-sharp turn? Now we can think of 5 in terms of a circle with radius 1/5. Knowing the units (inches vs. miles, for instance) allows us to determine how sharply the curve is curving.

Let a point $P$ on a smooth curve $C$ be given, and let $\kappa$ be the curvature of the curve at $P\text{.}$ A circle that:

• passes through $P\text{,}$

• lies on the concave side of $C\text{,}$

• has a common tangent line as $C$ at $P$ and

• has radius $r=1/\kappa$ (hence has curvature $\kappa$)

is the osculating circle, or circle of curvature, to $C$ at $P\text{,}$ and $r$ is the radius of curvature. Figure 11.5.14 shows the graph of the curve seen earlier in Figure 11.5.7 and its osculating circles at $A$ and $B\text{.}$ A sharp turn corresponds to a circle with a small radius; a gradual turn corresponds to a circle with a large radius. Being able to think of curvature in terms of the radius of a circle is very useful.

(The word “osculating” comes from a Latin word related to kissing; an osculating circle “kisses” the graph at a particular point. Many beautiful ideas in mathematics have come from studying the osculating circles to a curve.)

##### Example11.5.15Finding curvature

Find the curvature of the parabola defined by $y=x^2$ at the vertex and at $x=1\text{.}$

Solution
##### Example11.5.17Finding curvature

Find where the curvature of $\vec r(t) = \la t, t^2, 2t^3\ra$ is maximized.

Solution

# Subsection11.5.2Curvature and Motion

Let $\vec r(t)$ be a position function of an object, with velocity $\vvt = \vrp(t)$ and acceleration $\vec a(t)=\vrp'(t)\text{.}$ In Section 11.4 we established that acceleration is in the plane formed by $\vec T(t)$ and $\vec N(t)\text{,}$ and that we can find scalars $a_\text{T}$ and $a_\text{N}$ such that \begin{equation*} \vat = a_\text{T} \vec T(t) + a_\text{N} \vec N(t). \end{equation*}

Theorem 11.4.13 gives formulas for $a_\text{T}$ and $a_\text{N}\text{:}$ \begin{equation*} a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) \text{ and } a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}}. \end{equation*}

We understood that the amount of acceleration in the direction of $\vec T$ relates only to how the speed of the object is changing, and that the amount of acceleration in the direction of $\vec N$ relates to how the direction of travel of the object is changing. (That is, if the object travels at constant speed, $a_\text{T} =0\text{;}$ if the object travels in a constant direction, $a_\text{N} =0\text{.}$)

In Equation (11.5.2) at the beginning of this section, we found $s\,'(t) = \norm{\vvt}\text{.}$ We can combine this fact with the above formula for $a_\text{T}$ to write \begin{equation*} a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) = \frac{d}{dt}\big( s\,'(t)\big) = s\,''(t). \end{equation*}

Since $s\,'(t)$ is speed, $s\,''(t)$ is the rate at which speed is changing with respect to time. We see once more that the component of acceleration in the direction of travel relates only to speed, not to a change in direction.

Now compare the formula for $a_\text{N}$ above to the formula for curvature in Theorem 11.5.12: \begin{equation*} a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}} \text{ and } \kappa = \frac{\norm{\vrp(t)\times\vrp'(t)}}{\norm{\vrp(t)}^3}=\frac{\norm{\vvt\times \vat}}{\norm{\vvt}^3} . \end{equation*}

Thus \begin{align*} a_\text{N} \amp = \kappa \norm{\vvt}^2\\ \amp = \kappa\Big(s\,'(t)\Big)^2 \end{align*}

This last equation shows that the component of acceleration that changes the object's direction is dependent on two things: the curvature of the path and the speed of the object.

Imagine driving a car in a clockwise circle. You will naturally feel a force pushing you towards the door (more accurately, the door is pushing you as the car is turning and you want to travel in a straight line). If you keep the radius of the circle constant but speed up (i.e., increasing $s\,'(t)$), the door pushes harder against you ($a_\text{N}$ has increased). If you keep your speed constant but tighten the turn (i.e., increase $\kappa$), once again the door will push harder against you.

Putting our new formulas for $a_\text{T}$ and $a_\text{N}$ together, we have \begin{equation*} \vat = s\,''(t)\vec T(t) + \kappa\norm{\vvt}^2\vec N(t). \end{equation*}

This is not a particularly practical way of finding $a_\text{T}$ and $a_\text{N}\text{,}$ but it reveals some great concepts about how acceleration interacts with speed and the shape of a curve.

The minimum radius of the curve in a highway cloverleaf is determined by the operating speed, as given in the table in Figure 11.5.22. For each curve and speed, compute $a_\text{N}\text{.}$

Solution

We end this chapter with a reflection on what we've covered. We started with vector–valued functions, which may have seemed at the time to be just another way of writing parametric equations. However, we have seen that the vector perspective has given us great insight into the behavior of functions and the study of motion. Vector–valued position functions convey displacement, distance traveled, speed, velocity, acceleration and curvature information, each of which has great importance in science and engineering.

# Subsection11.5.3Exercises

Terms and Concepts

In the following exercises, a position function \vrt is given, where $t=0$ corresponds to the initial position. Find the arc length parameter $s\text{,}$ and rewrite \vrt in terms of $s\text{;}$ that is, find $\vec r(s)\text{.}$

In the following exercises, a curve $C$ is described along with 2 points on $C\text{.}$

1. Using a sketch, determine at which of these points the curvature is greater.

2. Find the curvature $\kappa$ of $C\text{,}$ and evaluate $\kappa$ at each of the 2 given points.

In the following exercises, find the value of $x$ or $t$ where curvature is maximized.

In the following exercises, find the radius of curvature at the indicated value.

In the following exercises, find the equation of the osculating circle to the curve at the indicated $t$-value.