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Calculus 3 (Math 221) Notes and Study Guide

Brenton LeMesurier

Section 3.3 Arc Length and Curvature

References.

Topics.

Subsection 3.3.1 Arc Length

The arc length of parameterized curves in the plane was computed in Section 1.2, Calculus of Parametric Curves as
\begin{equation} L = \int_{t=\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt = \int_a^b \sqrt{[f'(t)]^2+[g'(t)]^2} dt,\tag{3.3.1} \end{equation}
(see Equation (1.2.6)) and the same procedure can be applied to compute the length of a smooth curve in space as
\begin{equation} \begin{split} L \amp= \int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2 + \left(\frac{dz}{dt}\right)^2} \, dt \\ \amp= \int_a^b \sqrt{[f'(t)]^2 + [g'(t)]^2 + [h'(t)]^2} \, dt. \end{split}\tag{3.3.2} \end{equation}
This has the more compact form \(L=\int_a^b \| \vec{r}'(t) \| dt\text{,}\) valid in either two or three dimensions.

Other Parametrizations and the Arc Length Function.

A curve can be parameterized in many ways. For example
\begin{equation*} \vec{r}(t)=\vector{t,t^2,t^3},\; 1 \leq t \leq 2 \end{equation*}
describes the same curve as
\begin{equation*} \vec{r}(t)=\vector{e^u,e^{2u},e^{3u}},\; 0 \leq u \leq \ln 2. \end{equation*}
One particular useful and elegant parametrization is based on arc length, starting with the arc length function \(s\text{,}\) given by
\begin{equation} s(t) = \int_a^t \|\vec{r}'(t)\| dt\tag{3.3.3} \end{equation}
which is the length of the part of the curve from the initial point \(\vec{r}(a)\) to the point \(\vec{r}(t)\text{.}\) This is equivalent to \(\ds\frac{ds}{dt}= \|\vec{r}'(t)\|.\) If a curve is parameterized as \(\vec{r}(s)\) with arclength as its parameter then \(s(t)=t-a\text{.}\)

Parametrizing with Arc Length.

Any smooth curve can be expressed with arc length as parameter: \(ds/dt = \| \vec{r}'(t) \| >0\text{,}\) so \(s(t)\) is increasing and has an inverse, \(t(s)\text{,}\) and this inverse gives \(\vec{r} = \vec{r}(t(s))\) as the arc length parametrization. However it is often not possible to get an explicit formula for \(t(s)\) or for the arc length parametrization.

Subsection 3.3.2 Curvature

A smooth curve \(C\) described by vector function \(\vec{r}\) has a tangent direction at each point (except possibly at its end points) given by the unit tangent vector.
\begin{equation} \T(t) = \frac{d\vec{r}/dt}{\| d\vec{r}/dt \|} = \frac{d\vec{r}/dt}{ds/dt} = \frac{d\vec{r}}{ds}\tag{3.3.4} \end{equation}
Change in \(\T\) as one moves along the curve measures change in direction, so its rate of change measures curvature:

Definition 3.3.1.

The curvature of a curve with unit tangent \(\T(t)\) is
\begin{equation} \kappa = \kappa(t) = \left\| \frac{d \T}{ds} \right\|, = \left\| \frac{d \T/dt}{ds/dt} \right\| = \frac{\| \T'(t) \|}{\| \vec{r}'(t) \|}.\tag{3.3.5} \end{equation}

Exercise 3.3.2.

Show that the curvature of a circle of radius \(a\) is \(1/a\) at every point.

Proof.

Computing the Curvature of Plane Curves.

For the special case of a curve in the plane given by \(y=f(x)\text{,}\) the parametric form \(\vec{r}=\vector{x, f(x),0}\) gives
\begin{equation} \kappa(x) = \frac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}\tag{3.3.7} \end{equation}
so there is a relation to the second derivative, but not as simple as you might have guessed.
Exercise 3.3.4.
Compute \(y''\) and then the curvature for
  1. the parabola \(y=x^2\) and
  2. the semi-circle \(y=\sqrt{1-x^2}\text{.}\)
Discuss why these results suggests that the curvature is a more reasonable geometrical measure than the second derivative.

Subsection 3.3.3 Normal and Binormal Vectors, and the Osculating Plane

The Principle Unit Normal Vector.

With arc-length parametization, \(\kappa(s) = \|\vec{r}'(s) \times \vec{r}''(s)\|\) and \(\vec{r}'(s) = \T(s)\text{,}\) so
\begin{equation*} \kappa(s) = \left\|\T(s) \times \frac{d\T}{ds}(s)\right\|. \end{equation*}
The directions of these three vectors \(\T(s)\text{,}\) \(\T'(s)\) and \(\T(s) \times \T'(s)\) help describe the direction, curvature and "bending" of a curve in more detail.
Firstly, \(\T'\) is orthogonal (or normal) to \(\T\) and thus is normal to the curve (unless it is zero). It is natural to normalize it, giving the principal unit normal vector or unit normal.
\begin{equation} \N(t) = \frac{\T'(t)}{\|\T'(t)\|}\tag{3.3.8} \end{equation}
Note: In the special case that the curvature is zero a point, \(\T' = \vec{0}\) there, and so the unit normal vector and normal direction are undefined.

The Binormal Vector.

In the plane, the tangent and normal vectors forrm a basis that can be used to describe all directions in the plane. In space (\(\reals^3\)) instead, there is a whole plane normal to \(\T\) and thus to the curve, and a third dimension not cover by these two vectors. Then we can get a second normal direction, the binormal vector with
\begin{equation} \B(t) = \T(t) \times \N(t)\tag{3.3.9} \end{equation}
which is automatically a unit vector, normal to both \(\T(t)\) and \(\N(t)\text{.}\)
Question 3.3.5.
Why?
Note: At a point where the curvature is zero, the binormal vector is also undefined.

The Normal and Osculating Planes of a Curve.

At any point \(P\) given by \(\vec{r}(t)\) on a smooth curve \(C\text{,}\) the three vectors \(\T\text{,}\) \(\N\text{,}\) and \(\B\) form a right-handed set of orthogonal directions, and in some sense specify a natural set of coordinates for looking at the world with \(P\) as the origin.
Two planes through the point \(P\) are important in this view:
  • The plane through \(P\) with normal \(\T\) (and thus containing the two "normal" directions \(\N\) and \(\B\)) is the normal plane of \(C\) at \(P\text{.}\)
  • The plane through \(P\) containing directions \(\T\) and \(\N\text{,}\) and so with normal \(\B\text{,}\) is in some sense the plane that the curve is momentarily moving in, and is called the osculating plane of \(C\) at \(P\). In particular, if the curve lines entirely in a plane, \(\B\) is always normal to that plane, and so the osculating plane is that plane.
Note: At a point where the curvature is zero, the osculating plane is undefined, but the normal plane makes sense at any point on any smooth curve.

Tangent Lines and Osculating Circles.

If we look closely enough at a curve near a point it is approximated by a straight line, the tangent line, determined by \(\vec{r}(t)\) and \(\T(t)\text{:}\) geometrically, this is the straight line that best fits the curve near that point.
A more precise picture is given by noting the curvature too, and looking instead for the circle that best fits the curve at that point. This circle turns out to have radius \(\rho = 1/\kappa\text{,}\) lying in the osculating plane, to the side of the curve given by the principal normal direction \(\N(t)\text{.}\)
Note: At a point where the curvature is zero, this circle effectively becomes a straight line: the tangent line.
Exercise 3.3.6.
Find the osculating circle at the origin on the parabola \(y=x^2\text{.}\)
Exercise 3.3.7.
Verify that when the curve is a circle, the osculating circle at each point of the curve is that same circle (as it should be!)
For a visualization of the three directions \(\T\text{,}\) \(\N\) and \(\B\text{,}\) and the related osculating circle, see this YouTube video:
Figure 3.3.8. The \(\T\)-\(\N\)-\(\B\) frame and osculating circle.
(From the printed version of these notes or if the above does not work, the link is https://www.youtube.com/watch?v=JZGFcwipHYY
 8 
www.youtube.com/watch?v=JZGFcwipHYY
.)

A Relationship Between \(\kappa\text{,}\) \(\T\) and \(\N\).

The formula for curvature in terms of arc-length parameter \(s\) is \(\kappa(s)=\|\T'(s)\|\text{,}\) and \(\ds\N(s) = \frac{\T'(s)}{\left\|\T'(s)\right\|}\text{,}\) so
\begin{equation} \frac{d \T}{ds} = \kappa\N.\tag{3.3.10} \end{equation}
A geometrical interpretation if this is that the direction of a curve is changing at a rate \(\kappa\) in the principal normal direction.
This perhaps helps to explain why over a short period of time, the curve lies roughly in the osculating plane.
Remark 3.3.9.
This formula gives sometimes easier alternatives for computing the unit normal vector: either compute the curvature \(\kappa\) and then
\begin{equation} \N = \frac{1}{\kappa}\frac{d \T}{ds} = \frac{1}{\|\vec{r}'\| \kappa}\frac{d \T}{dt}\tag{3.3.11} \end{equation}
or get it directly as
\begin{equation} \N = \frac{\|\vec{r}'\|^2 }{\|\vec{r}'(t) \times \vec{r}''(t)\|}\frac{d \T}{dt}\tag{3.3.12} \end{equation}

Study Guide.

Study Section 3.3 of Calculus Volume 3
 9 
openstax.org/books/calculus-volume-3/pages/3-3-arc-length-and-curvature
; in particular
  • All the Definitions, Theorems, Examples and Checkpoints.
  • One or several exercises from each of the following ranges: 102–109, 110, 130, 133–134, and several from 113–126.
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