12.5 The Arc Length Parameter and Curvature

Consider points $A$ and $B$ on the curve graphed in Figure 12.5.3(a). One can readily argue that the curve curves more sharply at $A$ than at $B$. 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 12.5.3(b), where unit tangent vectors are graphed around points $A$ and $B$. Notice how the direction of the unit tangent vector changes quite a bit near $A$, whereas it does not change as much around $B$. 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.

If $\overrightarrow{r}(s)$ is parameterized by the arc length parameter, then

Having defined $\parallel {\overrightarrow{T}}^{\prime }(s)\parallel =\kappa$, we can rewrite the second equation as

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

We use Definition 12.5.1 to find the curvature of the line in Example 12.5.1.

While the definition of curvature is a useful mathematical concept, it is nearly impossible to use most of the time; writing $\overrightarrow{r}$ 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 trade-off: 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.

• Proof

  
  We’ll prove statement 3; statements 1 and 2 are applications that we leave to the exercises. By the chain rule and then Equation (12.3),

  $$\frac{d\overrightarrow{T}(t)}{dt}=\frac{d\overrightarrow{T}(s)}{ds}\frac{ds}{dt}=\frac{d\overrightarrow{T}(s)}{ds}\parallel {\overrightarrow{r}}^{\prime }(t)\parallel ,$$

  so that

  $$\kappa =\parallel \frac{d\overrightarrow{T}(s)}{ds}\parallel =\parallel \frac{\frac{d\overrightarrow{T}(t)}{dt}}{\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }\parallel =\frac{\parallel \frac{d\overrightarrow{T}(t)}{dt}\parallel }{\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }=\frac{\parallel {\overrightarrow{T}}^{\prime }(t)\parallel }{\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }.$$

  Now, ${\overrightarrow{r}}^{\prime }(t)=\parallel {\overrightarrow{r}}^{\prime }(t)\parallel \overrightarrow{T}(t)$ so that

  	${\overrightarrow{r}}^{\prime \prime }(t)$	$={\left(\parallel {\overrightarrow{r}}^{\prime }(t)\parallel \right)}^{\prime }\overrightarrow{T}(t)+\parallel {\overrightarrow{r}}^{\prime }(t)\parallel {\overrightarrow{T}}^{\prime }(t)\mathit{\hspace{1em}\hspace{1em}}\text{and}$
  	${\overrightarrow{r}}^{\prime }(t)\times {\overrightarrow{r}}^{\prime \prime }(t)$	$=\parallel {\overrightarrow{r}}^{\prime }(t)\parallel \overrightarrow{T}(t)\times \parallel {\overrightarrow{r}}^{\prime }(t)\parallel {\overrightarrow{T}}^{\prime }(t)={\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }^2\overrightarrow{T}(t)\times {\overrightarrow{T}}^{\prime }(t)$

  Because $\parallel \overrightarrow{T}\parallel =1$, Theorem 12.2.5 tells us that $\overrightarrow{T}$ and ${\overrightarrow{T}}^{\prime }$ are orthogonal. This means that

  	$\parallel {\overrightarrow{r}}^{\prime }(t)\times {\overrightarrow{r}}^{\prime \prime }(t)\parallel$	$={\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }^2\parallel \overrightarrow{T}(t)\parallel \parallel {\overrightarrow{T}}^{\prime }(t)\parallel ={\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }^2\parallel {\overrightarrow{T}}^{\prime }(t)\parallel ,\text{and}$
  	$\kappa$	$={\displaystyle \frac{\parallel {\overrightarrow{T}}^{\prime }(t)\parallel }{\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }}={\displaystyle \frac{\parallel {\overrightarrow{r}}^{\prime }(t)\times {\overrightarrow{r}}^{\prime \prime }(t)\parallel }{{\parallel {\overrightarrow{r}}^{\prime }(t)\parallel }^3}}.$

  Theorem 12.4.2 tells us that ${\overrightarrow{r}}^{\prime \prime }(t)=a_T\overrightarrow{T}(t)+({\overrightarrow{r}}^{\prime \prime }(t)\cdot \overrightarrow{N}(t))\overrightarrow{N}(t)$. Since ${\overrightarrow{r}}^{\prime }(t)=\overrightarrow{v}(t)$ and $T(t)$ are parallel, their cross product is zero and

  $$\kappa =\frac{\parallel \overrightarrow{v}(t)\times [a_T\overrightarrow{T}(t)+({\overrightarrow{r}}^{\prime \prime }(t)\cdot \overrightarrow{N}(t))\overrightarrow{N}(t)]\parallel }{{\parallel \overrightarrow{v}(t)\parallel }^3}=\frac{\parallel \overrightarrow{v}(t)\times (\overrightarrow{a}(t)\cdot \overrightarrow{N}(t))\overrightarrow{N}(t)\parallel }{{\parallel \overrightarrow{v}(t)\parallel }^3}.$$

  Since $\overrightarrow{v}(t)$ and $\overrightarrow{N}(t)$ are orthogonal, the norm of their cross product is the product of their norms, and

  $$\kappa =\frac{\parallel \overrightarrow{v}(t)\parallel (\overrightarrow{a}(t)\cdot \overrightarrow{N}(t))\parallel \overrightarrow{N}(t)\parallel }{{\parallel \overrightarrow{v}(t)\parallel }^3}=\frac{\overrightarrow{a}(t)\cdot \overrightarrow{N}(t)}{{\parallel \overrightarrow{v}(t)\parallel }^2}.\mathit{∎}$$

We practice using these formulas.

Example 12.5.3 gives a great result. Before this example, if we were told “The curve has a curvature of 5 at point $A$,” 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 $P$. A circle that:

• •

  passes through $P$,

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  lies on the concave side of $C$,

• •

  has a common tangent line as $C$ at $P$ and ^††margin: Figure 12.5.4: Illustrating the osculating circles for the curve seen in Figure 12.5.3.

• •

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

is the osculating circle, or circle of curvature, to $C$ at $P$, and $r$ is the radius of curvature. Figure 12.5.4 shows the graph of the curve seen earlier in Figure 12.5.3 and its osculating circles at $A$ and $B$. 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 useful ideas in mathematics have come from studying the osculating circles to a curve.)

Let $\overrightarrow{r}(t)$ be a position function of an object, with velocity $\overrightarrow{v}(t)={\overrightarrow{r}}^{\prime }(t)$ and acceleration $\overrightarrow{a}(t)={\overrightarrow{r}}^{\prime \prime }(t)$. In Section 12.4 we established that acceleration is in the plane formed by $\overrightarrow{T}(t)$ and $\overrightarrow{N}(t)$, and that we can find scalars $a_{\text{T}}$ and $a_{\text{N}}$ such that

Theorem 12.4.2 gives formulas for $a_{\text{T}}$ and $a_{\text{N}}$:

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

In Equation (12.3) at the beginning of this section, we found $s^{\prime }(t)=\parallel \overrightarrow{v}(t)\parallel$. We can combine this fact with the above formula for $a_{\text{T}}$ to write

Since $s^{\prime }(t)$ is speed, $s^{\prime \prime }(t)$ 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 12.5.2:

Thus

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 $s^{\prime }(t)$), 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

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

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.