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Course Notes for Multivariable Calculus

Remkes Kooistra

Section 2.2 Calculus of Parametric Curves

Subsection 2.2.1 Tangents

Parametric curves are functions, but functions with several variables of output. What can calculus say about this kind of function? How can I use the tools of single variable calculus to analyze the movement of parametric curve? Since there is still only a single input, I can actually apply the tools of single-variable calculus quite directly by taking the ordinary, single variable derivative of each component of the parametric curve. This leads to the first key definition (stated in \(\RR^3\) as usual, but can be generalized).
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Definition 2.2.1.

Let \(\gamma(t): [a,b] \rightarrow \RR^3\) be a parametric curve. I can write \(\gamma(t)\) in it components.
\begin{equation*} \gamma(t) = (x(t), y(t), z(t)) \end{equation*}
The derivative of the parametric curve, \(\gamma^\prime(t)\text{,}\) is calculated by taking the ordinary single-variable derivative of each component.
\begin{equation*} \gamma^\prime(t) = (x^\prime(t), y^\prime(t), z^\prime(t)) \end{equation*}
This derivative, \(\gamma^\prime(t)\text{,}\) produces a vector for each value of \(t\text{.}\) This vector is called the tangent vector to the parametric curve.
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This tangent is no longer the slope of a tangent line from single-variable claculus. It is a vector; specifically, local direction vector along the curve. At \(\gamma(a)\text{,}\) \(\gamma^\prime(a)\) points in the local direction of movement. The notion of a derivative as a tangent direction is replaced by the notion of the derivative as the slope of the tangent line.
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Example 2.2.2.

Figure 2.2.3. A tangent vector to the logarithmic spiral
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Consider the logarithmic spiral \(\gamma(t) = (e^{\frac{t}{4}}\sin t, e^{\frac{t}{4}}\cos t)\) on the domain \(t \in \RR\text{.}\) I calculate its derivative.
\begin{equation*} \gamma^\prime(t) = \left( e^{\frac{t}{4}} \cos t + \frac{1}{4} e^{\frac{t}{4}} \sin t, -e^{\frac{t}{4}} \sin t + \frac{1}{4} e^{\frac{t}{4}} \cos t \right) \end{equation*}
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Evaluating at \(t=0\) at the point of the curve \(\gamma(0) = (0,1)\text{,}\) the derivative or tangent vector to the curve is \(\gamma^\prime(0) = (1,\frac{1}{4})\text{.}\)
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This spiral is moving outwards; I can see how the tangent vector points in the direction of movement as the curves goes through \((0,1)\text{.}\) Figure 2.2.3 shows how I treat the tangent as a local direction vector: it points from \(\gamma(0) = (0,1)\text{,}\) not from the origin.
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Example 2.2.4.

Figure 2.2.5. Length of tangent vectors depending on parametrization
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Consider the four parametric curves earlier in Example 2.1.16. All four of these followed the same parabola, but at different rates of movement. I’m know going to calculation all of the tangents to these four course at the point \((1,1)\text{.}\)
\begin{align*} \gamma_2^\prime (t) \amp = \left( 1, 2t \right) \amp \amp t=1 \text{ at } (1,1) \implies \gamma_2^\prime (1) = \left( 1, 2 \right) \\ \gamma_1^\prime (t) \amp = \left( 2t, 4t^3 \right) \amp \amp t=1 \text{ at } (1,1) \implies \gamma_1^\prime (1) = \left( 2, 4 \right) \\ \gamma_3^\prime (t) \amp = \left( \frac{1}{2\sqrt{t}}, 1 \right) \amp \amp t=1 \text{ at } (1,1) \implies \gamma_3^\prime (1) = \left( \frac{1}{2} , 1 \right) \\ \gamma_4^\prime (t) \amp = \left( 5, 50t \right) \amp \amp t = \frac{1}{5} \text{ at } (1,1) \implies \gamma_4^\prime \left( \frac{1}{5} \right) = \left( 5, 10 \right) \end{align*}
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All of these tangent vectors are multiples of the vector \((1,2)\text{,}\) but they have different lengths, as seen in Figure 2.2.5. The direction of movement is fixed at that momemnt, but the speed is different.
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The tangent vector indicates the instantaneous direction of motion. The tangent vector also has a length, independent of the direction. The length of the tangent vector measures the speed of the curve going through that point. If the movement along the curve is very fast, the tangent vector will have a large length; if the movement along the curve is slow, the tangent vector will be shorter.
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Definition 2.2.6.

If \(\gamma(t)\) is a parametric curve, then the length of the tangent vector, \(|\gamma^\prime(t)|\text{,}\) is called the speed of the curve. The speed is a scalar.
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For the parametrizations of the graph of the parabola in Figure 2.2.5, I can see this behaviour clearly: the direction of the tangent is independent of the parametrization, but the length of the tangent is entirely dependent on the parametrization. The parametrizations which move more quickly along the curve have longer tangents, hence higher speed through that point.
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Often it is useful to only consider the tangent direction.
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Definition 2.2.7.

If \(\gamma^\prime(t)\) is the tangent vector of a parametric curve, the unit tangent vector is the unique vector of length one in the same direction as \(\gamma^\prime(t)\text{.}\) It is often written \(T(t)\text{.}\) To calculate the unit tangent, I simply take the tangent vector and divide by its length.
\begin{equation*} T(t) = \frac{\gamma^\prime(t)}{|\gamma^\prime(t)|} \end{equation*}
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Example 2.2.8.

In the various parametrizations of the graph of the parabola in Example 2.1.16, the tangent direction was \((1,2)\text{,}\) so the unit tangent is \(\left( \frac{1}{\sqrt{5}}, \frac{2}{\sqrt{5}} \right)\text{.}\)
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Example 2.2.9.

As introducted in Section 2.1, the archimedian spiral is
\begin{equation*} \gamma(t) = (t \cos t, t \sin t) \text{.} \end{equation*}
The tangent vector is k
\begin{equation*} \gamma^\prime (t) = (\cos t - t \sin t, \sin t + t \cos t) \text{.} \end{equation*}
When \(t=\pi\) (at the point \((-\pi,0)\text{,}\) the tangent vector is
\begin{equation*} \gamma^\prime (\pi) = (-1,-\pi). \end{equation*}
The length of this vector is
\begin{equation*} |\gamma^\prime (\pi)| = \sqrt{1 + \pi^2}. \end{equation*}
Therefore the unit tangent at this point is
\begin{equation*} T (\pi) = \frac{1}{\sqrt{1+\pi^2}} (-1,-\pi). \end{equation*}
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Subsection 2.2.2 Arc Length

Figure 2.2.10. Arclength setup by approximation
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My goal in this section is to produce a formula describing the length of a parametric curve. I approximate this length by breaking up a curve \(\gamma\) into a series of straight lines. In order to visualize the process, I work in \(\RR^2\) for the moment, as in Figure 2.2.10.
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For each straight line segment, the length of the segment is given by using pythagorus on the triange formed by the segment and the changes in each variable. I know \(\Delta y\) and \(\Delta x\text{,}\) then the length of the segment is \(\sqrt{\Delta x^2 + \Delta y^2}\text{.}\) I approximate the total length of the curve by adding up the lengths of these segments.
\begin{equation*} L \cong \sum \sqrt{\Delta x^2 + \Delta y^2} \end{equation*}
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In a parametric curve, everything changes in time. I can adjust this approximation by considering that \(x\) and \(y\) change over time. If \(\Delta t\) is the change in time over one of these segments, then by multiplying and dividing by \(\Delta t\text{,}\) and taking the division into the square root, I get the following expression.
\begin{equation*} L \cong \sum \sqrt{ \frac{\Delta x^2}{\Delta t^2} + \frac{\Delta y^2}{\Delta t^2}} \Delta t \end{equation*}
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If I take the limit of this process, breaking the curve into smaller and smaller pieces, I produce a Riemann sum which defines a definite integral. In the limit, the \(\Delta\) terms become the expression \(dy\text{,}\) \(dx\) and \(dt\) and their ratios become derivatives.
\begin{equation*} L = \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } dt \end{equation*}
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The same arguments work in any \(\RR^n\text{,}\) leading to the general result. However, as has been my practice, I’ll state the general result in \(\RR^3\text{.}\)
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Example 2.2.12.

In \(\RR^3\text{,}\) consider the three-dimensional spiral \(\gamma(t) = (\cos t, \sin t, t)\text{.}\) On the domain \(t \in [0, 8\pi]\text{,}\) the spiral makes four revolutions. I calculate its arclength.
\begin{equation*} L = \int_0^{8\pi} \sqrt{ \sin^2 t + \cos^2 t + 1} dt = \int_0^{8\pi} \sqrt{2} dt = 8 \sqrt{2} \pi \end{equation*}
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Example 2.2.13.

The asteroid is the parametric curve \(\gamma(t) = (\cos^3 t, \sin^3 t)\) for \(t \in [0,2\pi]\text{,}\) shown in Figure 2.2.14. I want to calculate its arclength.
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Figure 2.2.14. The Asteroid
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\begin{align*} L \amp = \int_a^b \sqrt{ \left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2 } dt\\ \amp = \int_0^{2\pi} \sqrt{ (-3 \cos^2 t (\sin t))^2 + (3 \sin^2 t (\cos t))^2} dt\\ \amp = \int_0^{2\pi} \sqrt{ 9 \cos^4 t \sin^2 t + 9 \sin^4 t \cos^2 t} dt\\ \amp = \int_0^{2\pi} |3 \sin t \cos t | \sqrt{ \sin^2 + \cos^2 t} dt\\ \amp = \int_0^{2\pi} |3 \sin t \cos t | dt \end{align*}
The absolute value here causes trouble. A convenient way to drop it is to notice that both \(\sin t\) and \(\cos t\) are positive on \(\let[0, \frac{\pi}{2}\right]\text{.}\) That range covers a quarter of the asteroid and the asteroid is symmetric, so I can calculate the length of that quarter and multiply by 4.
\begin{equation*} L = 4\int_0^{\frac{\pi}{2}} 3 \sin t \cos t dt = 6 \int_0^{\frac{\pi}{2}} \sin 2t = -3\cos 2t \Big|_0^{\frac{\pi}{2}} = 6 \end{equation*}
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The symmetric argument in Example 2.2.13 deserves some attention. In parametric curves, this kind of symmetry argument can be very valuable. It’s often easier to work with a small section of a curve, particularly for integration. If I can argue that the whole curve is just a repetition of a small portion (perhaps reflected or rotated), then I can reasonably use the small portion to calculate information about the whole curve.
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Subsection 2.2.3 Parametrization by Arclength

The arclength calculation determins the length of the whole curve. However, I can also ask for the length of sections of the curve. I’ll state the definition in \(\RR^3\text{.}\)
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Definition 2.2.15.

Let \(\gamma(t): [a,b] \rightarrow \RR^3\) be a parametric curve with components \((\gamma_1(t), \gamma_2(t), \gamma_3(t))\text{.}\) The arclength function \(s(t): [a,b] \rightarrow [0, \infty)\) is defined by this integral.
\begin{equation*} s(t) = \int_a^t \sqrt{ \left( \frac{d\gamma_1(u)}{du} \right)^2 + \left( \frac{d\gamma_2(u)}{du} \right)^2 + \left( \frac{d\gamma_3(u)}{du} \right)^2 } du \end{equation*}
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The letter \(s\) is conventional notation for the arclength function.
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The arclength function measures the length of the curve as a function of the parameter; it is simply the distance travelled along the curve. For example, if \(t \in [0,10]\text{,}\) then \(s(3)\) is the distance along the curve from \(\gamma(0)\) to \(\gamma(3)\text{,}\) \(s(5)\) is the distance the curve has covered from \(\gamma(0)\) to \(\gamma(5)\text{,}\) \(s(8)\) is the distance along the curve from \(\gamma(0)\) to \(\gamma(8)\) and so on. Since \(s\) depends on \(t\) outside the integral, I have to choose a temporary variable \(u\) inside the integral; in the integral, \(t\) is simply repalced with \(u\text{.}\)
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I want to use the arclength function to reparametrize curve. The process, therefore, has three steps.
  1. I calculate the arclength fuction \(s(t)\) by integration.
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  2. I invert the arclength function. Arclength is always an increasing function, so it is always invertible. (I might not be able to write the inverse as an elementary function, but it does always exist.) I write the inverse as \(t(s)\text{.}\)
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  3. I use the inverse of the arclength function to reparametrize by replacing \(t\) with \(t(s)\text{.}\)
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Definition 2.2.16.

Let \(\gamma(t) : [a,b] \rightarrow \RR^n\) be a parametric curve. Let \(s(t)\) be the arclength function, with \(t(s)\) its inverse. The reparametrization \(\gamma(t(s))\) is called the parametrization by arclength. It is the unique parametrization where the parameter is the distance along the curve.
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Example 2.2.17.

Consider the helix \(\gamma(t) = (2 \cos t, 2 \sin t, 4t)\) defined for \(t > 0\text{,}\) shown in Figure 2.2.18.
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Figure 2.2.18. The helix \(\gamma(t) = (2 \cos t, 2 \sin t, 4t)\)
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I calculate the arclength function.
\begin{align*} s(t) \amp = \int_0^t \sqrt{x^\prime(u)^2 + y^\prime(u)^2 + z^\prime(u)^2} du\\ \amp = \int_0^t \sqrt{4\sin^2 u + 4 \cos^2 u + 4^2} du = \int_0^t \sqrt{20} du = t \sqrt{20} \end{align*}
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The arclength function is \(s = t \sqrt{20}\text{;}\) the inverse is \(t = \frac{s}{\sqrt{20}}\text{.}\) I replace \(t\) with \(t(s)\) in \(\gamma(t)\text{.}\)
\begin{equation*} \gamma(s) = \left( 2 \cos \left( \frac{s}{\sqrt{20}} \right), 2 \sin \left( \frac{s}{\sqrt{20}} \right), \frac{4s}{\sqrt{20}} \right) \end{equation*}
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Example 2.2.19.

\begin{equation*} \gamma(s) = \left( 2 \cos \left( \frac{s}{\sqrt{20}} \right), 2 \sin \left( \frac{s}{\sqrt{20}} \right), \frac{4s}{\sqrt{20}} \right) \end{equation*}
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I’ll calculate the speed in this parametrization by arclength. \(|\gamma^\prime(s)|\text{:}\)
\begin{align*} |\gamma^\prime(s)| \amp = \sqrt{ x^\prime(s)^2 + y^\prime(s)^2 + z^\prime(s)^2}\\ \amp = \sqrt{ \left( \frac{-2}{\sqrt{20}} \sin \left( \frac{s}{\sqrt{20}} \right) \right)^2 + \left( \frac{2}{\sqrt{20}} \cos \left( \frac{s}{\sqrt{20}} \right) \right)^2 + \left( \frac{4}{\sqrt{20}} \right)^2}\\ \amp = \sqrt{\frac{4}{20} \left( \cos^2 \left( \frac{5}{\sqrt{20}} \right) + \sin^2 \left( \frac{5}{\sqrt{20}} \right) \right) + \frac{16}{20} } = \sqrt{ \frac{4}{20} + \frac{16}{20}} = 1 \end{align*}
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I find that the speed is constantly one.
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The process of parametrization by arclenth is a lot of work. What is the point? The point is that the parametrization by arclength has several very pleasant properties, properties which will be useful in later constructions.
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These properties are very useful and convenient. In the remainder of this chapter, I will make frequent use of the parametrization by arclength; it serves as the default parametrization. I can do this since the parametrization by arclength always exists and is unique. Having a special, default parameter allows me to make good definitions which will be independent of the choice of parametrization. (In practice, the parametrization by arclength can be extremely difficult to actually calculate. For theoretical results, however, this difficulty isn’t relevant).
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The existence of multiple parametrization for parametric curves is an example of a very general trend in abstract mathematics. A mathematical structure is some kind of pattern or behaviour (such as a curve being a particular path through space). This pattern can be presented in a variety of ways (such as a variety of parametrizations). This always presents a fundamental difficulty: if you want to understand something about the general pattern -- about the mathematical structure in itself -- you need to make sure that your observations transcend the particularities of the presentation. Sometimes, there is a special presentation (such as the parametrization by arclength) which serves as the ideal model of the structure. Where there is such a presentation, mathematicians make heavy use of it. Where no such presentation exists (which is usually the case), mathematicians always have to struggle with the tension between pattern itself and how the pattern is presented.
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Subsection 2.2.4 Arclength and Tangents

There are some very natural connections between tangents and arclength which relate to the intepretation of parametric curves as movement through space. I’ll return to the arc length formula (in \(\RR^2\text{,}\) for simplicity).
\begin{equation*} L = \int_a^b \sqrt{x^\prime(t)^2 + y^\prime(t)^2} dt \end{equation*}
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The tangent vector is \(\gamma^\prime(t) = (x^\prime(t), y^\prime(t))\text{,}\) so this integrand in the arclength formula is nothing more than the length of the tangent vector. This lets me rewrite the arclength formula.
\begin{equation*} L = \int_a^b |\gamma^\prime(t)| dt \end{equation*}
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This hopefully makes intuitive sense — the length of the tangent vector measures the speed of the curve. To get length (the distance travelled), I integrate the speed of movement.
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An nice event occurs when I differentiate the arclength function \(s(t)\text{.}\) The fundamental theorem of calculus lets me differentiate an integral.
\begin{equation*} \frac{d}{dt} s(t) = \frac{d}{dt} \int_a^t |\gamma^\prime(u)| du = |\gamma^\prime (t)| \end{equation*}
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The derivative of the arclength function is the length of the tangent vector, which represents the speed of movement along the curve. To get speed, I differentiate length.
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Both of these observations generalize the relationship between distance and speed from single variable calculus. In that setting, the relationship was relatively direct: if \(d(t)\) was distance, then \(d^\prime(t)\) was speed and if \(v(t)\) was speed then \(\int v(t) dt\) was distance. The idea here is exactly the same, but since the movement is much more complicated, going through multi-dimensional space, I need more complicated definitions and notations to access the idea.
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Subsection 2.2.5 Curvature

The tangent vector for a parametric curve measures the speed and direction of movement along the curve. Speed and direction are good information, but not enough to completely describe motion in multiple dimensions. In particular, the tangent vector gives the direction but doesn’t say how direction is changing. This section builds up the notions of curvature and torsion, which will complete a full description of the changing directions of movement. In this section, I exclusively work in \(\RR^3\text{.}\)
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In \(\RR^3\text{,}\) I want to classify three types of movement.
  • Straight motion: the direction is fixed and only speed varies.
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  • Curving motion in a plane: motion is fixed to a plane, but the direction can change and curve in the plane.
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  • Twisting motion: the most general motion, where in addition to curving in a plane, a curve can also twist away from the plane.
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Throughout this section, let \(\gamma(t): [a,b] \rightarrow \RR^3\) be a curve in three dimensions. To start with (and to ensure that the following definition does not depend on parametrization), I will assume the curve is parametrized by arc length: I will use the parameter \(s\) and write \(\gamma(s)\text{.}\) In this parametrization, the tangent is a unit vector, so \(T(s) = \gamma^\prime(s)\text{.}\)
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Definition 2.2.21.

Let \(\gamma(s): [a,b] \rightarrow \RR^3\) be a parametric curve parametrized by arclength. The curvature of \(\gamma(s)\) is a scalar \(\kappa(s)\) which measure how quickly the curve is turning.
\begin{equation*} \kappa(s) = \left| \frac{dT(s)}{ds} \right| \end{equation*}
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There are a couple things to note about this definition. First, \(\kappa\) is the greek letter kappa, even though it looks like the Englith ‘k’. Second, \(T(s)\) is a vector, so the derivative \(\frac{dT(s)}{ds}\) is a vector derivative — the derivative is taken in each component of the vector. Curvature is the scalar length of the resulting vector. Finally, the unit tangent is the direction of motion. Therefore, the derivative of the unit tangent measure the change in direction. The length of this derivative is a scalar measure of how quickly direction is changing. (Note that thought \(|T(s)| = 1\text{,}\) there is no reason to expect that \(|T^\prime(s)| = 1\text{;}\) the magnitude of the change in a unit vector can be any possible magnitude.)
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The given definition of curvature only works for parametrization by arclength. I would like to calculate \(\kappa\) in terms of an arbitrary parametrization \(t\text{,}\) since arclength can be difficult to work with.
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Proof.

Recall the arclength function \(s(t)\text{.}\) I can write an arbitrary parametrization \(\gamma(s(t))\text{,}\) where \(s(t)\) is the specific arclength function for the parametrization. This composition also applies to the unit tangent: \(T(s(t)\text{.}\) Then to differentiate in \(t\) requires a chain rule.
\begin{equation*} \frac{dT(t)}{dt} = \frac{dT(s(t))}{ds} \frac{ds}{dt} \end{equation*}
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I can rearrange this equation and then take the lengths of the vectors.
\begin{equation*} \frac{dT}{ds} = \frac{\frac{dT}{dt}}{\frac{ds}{dt}} \implies \left| \frac{dT}{ds} \right| = \left| \frac{\frac{dT}{dt}}{\frac{ds}{dt}} \right| \implies \left| \frac{dT}{ds} \right| = \frac{\left|\frac{dT}{dt}\right|}{\left|\frac{ds}{dt}\right|} \end{equation*}
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The left hand side of this expression is exactly \(\kappa\text{,}\) the curvature. The numerator on the right hand side can be written \(|T^\prime(t)|\text{,}\) the change in the unit tangent for an arbitrary parametrization. The denominator \(\left| \frac{ds}{dt} \right|\) is just the length of the tangent vector \(|\gamma^\prime(t)|\text{.}\)
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Example 2.2.23.

Consider the curve in \(\RR^3\text{:}\) \(\gamma(t) = (at, bt, ct)\) for some constants \(a,b,c,\in \RR\text{.}\) This curve is the straight line through \((0,0,0)\) in the direction \((a,b,c)\text{.}\) I will calculate the curvature of this curve. I work in steps, calculating the various pieces in order.
\begin{align*} \gamma^\prime(t) \amp = (a,b,c)\\ |\gamma^\prime(t)| \amp = \sqrt{a^2 + b^2 + c^2}\\ T(t) \amp = \frac{\gamma^\prime(t)}{|\gamma^\prime(t)|} = \frac{1}{\sqrt{a^2 + b^2 + c^2}} (a,b,c)\\ T^\prime(t) \amp = (0,0,0)\\ \kappa(t) \amp = \frac{|T^\prime(t)|}{|\gamma^\prime(t)|} = \frac{0}{\sqrt{a^2 +b^2 + c^2}} = 0 \end{align*}
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Unsurprisingly, the straight line has zero curvature.
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Example 2.2.24.

Consider the unit circle in \(\RR^2\text{:}\) \(\gamma(t) = (\cos t, \sin t)\text{.}\)
\begin{align*} \gamma^\prime(t) \amp = (-\sin t, \cos t)\\ |\gamma^\prime(t)| \amp = \sqrt{\sin^2 t + \cos^2 t} = 1\\ T(t) \amp = \frac{\gamma^\prime(t)}{|\gamma^\prime(t)|} = (-\sin t, \cos t)\\ T^\prime(t) \amp = (-\cos t, - \sin t)\\ |T^\prime(t)| \amp = \sqrt{ \cos^2 t + \sin^2 t} = 1\\ \kappa(t) \amp = \frac{|T^\prime(t)|}{|\gamma^\prime(t)|} = \frac{1}{1} = 1 \end{align*}
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Again unsurprisingly, the unit circle has constant curvature of one; I would expect uniform curvature for a circle. This also give a reference for the scale of curvature: curvature of one is exactly the curvature of the unit circle.
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Example 2.2.25.

A similar calcluation can be done for the circle of radius \(a\text{,}\) which is \(\gamma(t) = (a \cos t, a \sin t)\text{.}\) If I repeat all the steps for the unit circle, I can conclude that \(\kappa(t) = \frac{1}{a}\text{.}\) Again, this is constant, which makes sense since the circle is equally curved at all points. However, it’s interesting to note that the curvature is inversely proportional to the radius. A curvature value of \(\kappa\) is interpreted as the curvature of a circle of radius \(\frac{1}{\kappa}\text{.}\)
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This hopefully makes sense — a circle with a very large radius doesn’t locally have to curve very much. I know this very well, living on the surface of the earth: since the radius of the earth is large, I don’t really see the curvature of its surface. However, for circle with very small radius, there is less distance to cover a whole revolution, so the curvature must be larger.
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Subsection 2.2.6 Normals

To complete the description of motion in \(\RR^3\text{,}\) I need to define the normal and binormal vectors. I apologize, on behalf of mathematicians everywhere, for the use of the word ‘normal’ in this definition (and, franktly, everywhere else it’s used in mathematics.) Unfortunately, it’s standard and we are stuck with it.
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Definition 2.2.26.

Let \(\gamma(t): [a,b] \rightarrow \RR^3\) be a parametric curve in an arbitrary parametrization. The normal vector to a curve is written \(N(t)\) and defined by the formula
\begin{equation*} N(t) = \frac{T^\prime(t)}{|T^\prime(t)|}\text{.} \end{equation*}
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The binormal vector to a curve is written \(B(t)\) and defined by the formula
\begin{equation*} B(t) = T(t) \times N(t)\text{.} \end{equation*}
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The following lemma helps to understand the direction of \(N(t)\) (and will be useful later in the course as well).
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Proof.

There are several ways to argue this result. First, I can think about the geometry of the sphere and its tangents (though, I confess, this argument uses some ideas from future weeks in the course.) Since \(\gamma(t)\) is always a unit vector, the path of \(\gamma(t)\) is a path on the unit sphere in \(\RR^3\text{.}\) The tangent direction of this path, therefore, must be a tangent to the unit sphere. Unit vector point directly out of the unit sphere; they are perpendicular to tangent vectors of the unit sphere. Therefore, \(\gamma\) and \(\gamma^\prime\) must be perpendicular vectors.
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Alternatively, I could argue more directly from the definition of a unit vector. If \(\gamma(t)\) is a unit vector, its length can never change -- the length is fixed at one. If \(\gamma^\prime(t)\) shares any small direction with \(\gamma(t)\) (more formally, if the projection was non-zero), then there would be a portion of the tangent vector pointing in the same direction of \(\gamma(t)\text{.}\) This would indicate that there is change in the direction as the vector. Such a change would necessarily change the length of the vector; an increase is growth aligned with the vector, and a decrease is growth against the vector. Therefore, \(\gamma(t)\) and \(\gamma^\prime(t)\) cannot share any such small direction. The projection of the latter onto the former must be zero, which is equivalent to them being perpendicular.
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Lemma 2.2.27 shows that \(N\) is perpendicular to \(T\text{.}\) The cross product of two vectors is perpedicular to both, to all of \(T\text{,}\) \(N\) and \(B\) are perpendicular to each other. In addition, \(T\) was already a unit vector and \(N\) was defined explicitly as a unit vector. The cross product of two perpendicular unit vectors is also a unit vector, so all three are unit vectors. An example of the arrangement of the three vrectors is show in Figure 2.2.28.
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Figure 2.2.28. The tangent, normal and binormal to a helix in \(\RR^3\)
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I want to complete these definitions with interpretation. I’ve already spent some time on the unit tangent vector \(T(t)\text{.}\) It represents the direction of movement of the curve at that point. The associated scalar is speed, measuring how fast the object is moving at this point in the curve. The normal vector \(N(t)\) represents the direction of curvature. Curvature was a scalar, measuring how curved the curve is at this point. The normal vector adds to this scalar a direction, telling you which way the curve is turning. The two vectors \(T\) and \(N\) determine a plane.
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Definition 2.2.29.

The plane determined by \(T\) and \(N\)is called the osculating plane.
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If a parametric curve isn’t twisted, the osculating plane is the plane on which the curve travels. The binormal vector \(B(t)\) is perpendicular to the osculating plane. Since planes are best described by their normals, I use \(B(t)\) to keep track of the osculating plane.
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Subsection 2.2.7 Torsion

The third type of movement I want to describe is twisting movement, where a curve can twist away from its plane of movement. Since the osculating plane is determined by its normal \(B(t)\text{,}\) twisting motion involves the change in \(B(t)\text{.}\)
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Definition 2.2.30.

Let \(\gamma(s): [0,L] \rightarrow \RR^3\) be a parametric curve parametrized by arclength. Its torsion is written \(\tau(s)\) and defined by the formula
\begin{equation*} \tau(s) = \left| \frac{d B(s)}{ds} \right|\text{.} \end{equation*}
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Torsion is a scalar that measures the change in \(B\text{,}\) the change in the normal of the osculating plane. It measures the rate of twisting, the tendency of curving paths to change their plane of motion.
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As with curvature, I want to calculate torsion in an arbitrary parametrization.
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Proof.

As in the proof for torsion, I can setup an arbitrary parametrization using the arclength function \(s = s(t)\text{.}\) Then I can write \(B(s(t))\) for the binormal in the arbitrary parametrization. The chain rule calculates the derivative of the binormal vector.
\begin{equation*} \frac{dB(t)}{dt} = \frac{dB(t(s))}{ds} \frac{ds}{dt} \end{equation*}
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I can rearrange this expression.
\begin{equation*} \frac{dB(s)}{ds} = \frac{\frac{dB(t)}{dt}}{\frac{dt}{ds}} = \frac{B^\prime(t)}{|\gamma^\prime(t)|} \end{equation*}
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Omiting the proof here, it is true that \(B^\prime(t)\) points in the direction of the negative of the normal, \(-N(t)\text{.}\) To get the torsion, which is the length of the vector in the previous formula, I can take the dot product with the unit vector in this direction, which is simply \(-N(t)\text{.}\)
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Now every curve has three associated scalars (speed, curvature, torsion) and three matching vectors (tangent, normal and binormal). Together, this information completely describes motion in \(\RR^3\text{.}\) For convenience, the next list collects the definitions of all six of these mathematical objects.
\begin{align*} \gamma(t) \amp = (x(t), y(t), z(y))\\ \gamma^\prime(t) \amp = (x^\prime(t), y^\prime(t), z^\prime(t))\\ v(t) = |\gamma^\prime(t)| \amp = \sqrt{x^\prime(t)^2 + y^\prime(t)^2 + z^\prime(t)^2 }\\ T(t) \amp = \frac{\gamma^\prime(t)}{|\gamma^\prime(t)|}\\ \kappa(t) \amp = \frac{|T^\prime(t)|}{|\gamma^\prime(t)|}\\ N(t) \amp = \frac{ T^\prime(t)}{|T^\prime(t)|}\\ B(t) \amp = T(t) \times N(t)\\ \tau(t) \amp = - \frac{B^\prime(t)}{|\gamma^\prime(t)|} \cdot N(t) \end{align*}
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Example 2.2.32.

Let \(r\) and \(h\) be two positive scalars. Consider the helix \(\gamma(t) = (r\cos t, r\sin t, ht)\text{.}\) In the helix, \(r\) is the radius of the circular movement of the helix, and \(h\) is the rate of linear movement along the axis of the helix. I’ll calculate all the information about the motion along the helix: the speed (\(v\)), curvature (\(\kappa\)), torsion (\(\tau\)),the tangent, the normal and finally the binormal.
\begin{align*} \gamma^\prime(t) \amp = (-r \sin t, r \cos t, h)\\ v = |\gamma^\prime(t)| \amp = \sqrt{r^2 \sin^2 t + r^2 \cos^2 t + h^2} = \sqrt{r^2 + h^2}\\ T(t) \amp = \frac{\gamma^\prime(t)}{|\gamma^\prime(t)|} = \frac{1}{\sqrt{r^2+h^2}} (-r \sin t, r \cos t, h)\\ T^\prime(t) \amp = \frac{1}{\sqrt{r^2 + h^2}} ( -r \cos t, -r \sin t, 0)\\ |T^\prime(t)| \amp = \frac{1}{\sqrt{r^2 + h^2}} \sqrt{r^2 \cos^2 t + r^2 \sin^2 t} = \frac{r}{\sqrt{r^2 + h^2}}\\ \kappa(t) \amp = \frac{|T^\prime(t)|}{|\gamma^\prime(t)|} = \frac{\frac{r}{\sqrt{r^2 + h^2}}}{\sqrt{r^2 + h^2}} = \frac{r}{r^2 + h^2}\\ N(t) \amp = \frac{ T^\prime(t)}{|T^\prime(t)|} = \frac{1}{\sqrt{r^2+h^2}}{r}{\sqrt{r^2 + h^2}} (-r\cos t, -r \sin t, 0)\\ \amp = \frac{1}{r} (-r \cos t, -r \sin t, 0) = (-\cos t, -\sin t, 0)\\ B(t) \amp = T(t) \times N(t) = \frac{1}{\sqrt{r^2 + h^2}} (-r \sin t, r \cos t, h) \times (-\cos t, - \sin t, 0)\\ \amp = \frac{1}{\sqrt{r^2 + h^2}} (h \sin t, -h \cos t, r)\\ B^\prime(t) \amp = \frac{1}{\sqrt{r^2 + h^2}} (h \cos t, h \sin t, 0)\\ \tau(t) \amp = - \frac{B^\prime(t)}{|\gamma^\prime(t)|} \cdot N(t) \\ \amp = \frac{1}{r^2 + h^2} (h \cos t, h \sin t , 0 ) \cdot (-\cos t, - \sin t, 0) = \frac{h}{r^2 + h^2} \end{align*}
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Look at the three scalars.
\begin{align*} v \amp = \sqrt{r^2 + h^2} \\ \kappa \amp = \frac{r}{r^2 + h^2} \\ \tau \amp = \frac{h}{r^2 + h^2} \end{align*}
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All three, speed, curvature and torsion, are constant here. This helix has constant motion, curvature and torsion
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Following the observation at the end of the previous example, I can summarize what kind of shapes of curves exist for constant and zero values of the three scalars describing the motion of parametric curves.
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\(v = 0\) \(\kappa = 0\) \(\tau = 0\) \(\implies\) no movement at all
\(v = c\) \(\kappa = 0\) \(\tau = 0\) \(\implies\) straight line, no curving or twisting
\(v = c_1\) \(\kappa = c_2\) \(\tau = 0\) \(\implies\) circle, constant speed and curvature, no twisting
\(v = c_1\) \(\kappa = c_2\) \(\tau = c_3\) \(\implies\) helix, constantly moving, curving and twisting
Like the straight line and the circle, the helix is the unique curve which has constant non-zero speed, torsion and curvature.
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Subsection 2.2.8 Acceleration and Movement in Space

Now that I understand how parametric curves describe motion in \(\RR^3\text{,}\) I can do a little physics. This section tries to undertstand what acceleration means in the language of parametric curves. The starting point is simple: acceleration should be the second derivative of the position and position is just the output of the parametric curve.
\begin{equation*} a(t) = \gamma^{\prime \prime}(t) = \frac{d}{dt} \gamma^\prime(t) \end{equation*}
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This means acceleration is a vector. I will try to work out its direction and magnitude, using \(T\text{,}\) \(N\) and \(B\) as reference for directions. Recall the definition of the unit tangent.
\begin{equation*} T(t) = \frac{\gamma^\prime(t)}{|\gamma^\prime(t)|} \end{equation*}
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I solve for \(\gamma^\prime(t)\text{.}\)
\begin{equation*} \gamma^\prime(t) = \underbrace{|\gamma^\prime|}_{\text{ scalar } } \cdot \underbrace{T(t)}_{\text{ direction } } \end{equation*}
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I differentiate this, using the product rule.
\begin{equation} \frac{d}{dt} \gamma^\prime(t) = \left( \frac{d}{dt} |\gamma^\prime(t)| \right) T(t) + |\gamma^\prime(t)| \frac{d}{dt} T(t)\tag{2.2.1} \end{equation}
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Recall the definition of the normal.
\begin{equation*} N(t) = \frac{T^\prime(t)}{|T^\prime(t)|} \end{equation*}
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I isolate \(T^\prime(t)\text{.}\)
\begin{equation} T^\prime(t) = |T^\prime(t)| N(t)\tag{2.2.2} \end{equation}
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Finally, recall the definition of curvature.
\begin{equation*} \kappa(t) = \frac{|T^\prime(t)|}{|\gamma^\prime(t)|} \implies |T^\prime(t)| = \kappa(t) |\gamma^\prime(t)| \end{equation*}
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I use this expression to replace \(|T^\prime(t)|\) in Equation (2.2.2).
\begin{equation*} T^\prime(t) = \kappa(t) |\gamma^\prime(t)| N(t) \end{equation*}
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Finally, I put this expression for \(T^\prime(t)\) into the second term of Equation (2.2.1), then group and label the terms.
\begin{equation*} a(t) = \underbrace{\frac{d}{dt} |\gamma^\prime(t)|}_{\text{ linear acceleration } } T(t) + \underbrace{\kappa(t) |\gamma^\prime(t)|^2}_{\text{ angular acceleration } } N(t) \end{equation*}
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Definition 2.2.33.

Let \(\gamma(t): [a,b] \rightarrow \RR^3\) be a parametric curve. Its linear acceleration is the vector \(\frac{d}{dt} |\gamma^\prime(t)| T(t)\) and its angular acceleration is the vector \(\kappa(t) |\gamma^\prime(t)|^2 N(t)\text{.}\) Linear acceleration described how quickly the speed changes. Angular acceleration describes how quickly the direction changes.
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Example 2.2.34.

For the helix, \(|\gamma^\prime(t)|\) is constant, so the acceleration is entirely angular.
\begin{equation*} a(t) = 0 T(t) + r N(t) = r N(t) \end{equation*}
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This makes sense, since the linear speed doesn’t every change; only the direction changes as the helix curves and twists.
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You might wonder why this little bit of physics didn’t include torsion. There is one derivative to get to the tangent and a second derivative to get to the normal and the binormal. To calculate torsion requires the derivative of the binormal: this is three derivatives away from the original curve. Torsion only shows up as a third derivative. The above discussion of acceleration only used second derivatives; indeed, acceleration can only describe motion in a plane. To get torsion and motion that diverges from its osculating plan requires a third derivative.
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