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# Lecture Notes

Below are the lecture notes from my course on multivariable calculus, as well as some handouts I give to students for some of the experiments in the introductory physics lab courses. Note that I update my notes each term, but this website may not link to the latest editions.

### Notes on MTH3015: Calculus III

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Week 3: Vector-valued Functions; Curves

Sujeet Akula

Vector-valued Functions

We discuss here the class of functions $f : \R\to\R^m$, where $m=2,3$. In the case that $m=2$, we mean to say that $f$ maps a real number (a scalar) to a two-dimensional vector. Similarly, for $m=3$, $f$ would map a scalar to a three-dimensional vector. This can be conveniently expressed algebraically by considering the components of a vector-valued functions to be independent scalar functions as in: \begin{equation} \f(t) = f_1(t)\vi + f_2(t)\vj = (f_1(t),f_2(t))_O .\end{equation} For three-dimensions this would be: \begin{equation} \f(t) = f_1(t)\vi + f_2(t)\vj + f_3(t)\vk = (f_1(t),f_2(t),f_3(t))_O .\end{equation} Note that in this course, we will always assume that vector-valued functions are based at the origin $O$, and so writing the subscript each time may be omitted. The functions $f_1$, $f_2$, and $f_3$ are the usual scalar functions, i.e., of the sort $f_{1,2,3} : \R \to \R$. To be clear, this means that they are functions of a single real variable which map to another single, real variable.

Basic Operations

We have already discussed at length operations that may be performed on vectors, but now that we have introduced the vector-valued functions, it may not be clear if the same operations and properties hold. In fact, they do: if $\f(t)$ is a vector-valued function, then $\f(t)$ for any specific value of $t$ is a constant vector; the sort that we had been studying in the previous sections. Thus, we establish the natural definitions of sums and products of vector-valued functions: \begin{align} (\f + \g)(t) &:= \f(t) + \g(t) \\ (\alpha\f)(t) &:= \alpha\f(t) \\ (\f\cdot\g)(t) &:= \f(t)\cdot\g(t) \\ (\f\times\g)(t) &:= \f(t)\times\g(t) . \end{align} These statements just serve to declare that vector-valued functions specified at any given time behave like normal vectors. The only difference is that since they are functions, we have additional properties, which we will introduce later. We must also note that since all of our vector-valued functions are based at the origin, the operations that yield another vector will also be based at the origin. Of course, the dot-product of two vector functions is not another vector function but a scalar function.

Example Let $\f(t) = (t,t^2,t^3)_O$ and $\g(t) = (\cos t, \sin t, 1)$. We will compute a few simple operations on these vector functions. \begin{align} (4\f - 2\g)(t) &= (4t - 2\cos t, 4t^2 - 2\sin t, 4t^3 - 2)_O ,\\ (\f\cdot3\g)(t) &= 3t\cos t + 3t^2\sin t + 3t^3 ,\\ (\f\times\g)(t) &= (t^2 - t^3\sin t, t^3\cos t -t, t\sin t - t^2\cos t)_O . \end{align}

The Calculus of Vector-valued Functions

We will discuss now the limits of vector functions, which naturally brings us to differentiation. Mostly, the trick will be to apply the calculus operation on the individual components, since they behave like normal scalar functions, of which we already know the calculus. Suppose that the following limits exist \begin{equation} \lim\limits_{t\to t_0} f_1(t) = L_1 \text{ , } \lim\limits_{t\to t_0} f_2(t) = L_2 \text{ , and }\lim\limits_{t\to t_0} f_3(t) = L_3 . \end{equation} Then, for $\f(t) = (f_1(t),f_2(t),f_3(t))_O$, we define \begin{equation} \lim\limits_{t\to t_0}\f(t) := \left(\lim\limits_{t\to t_0} f_1(t), \lim\limits_{t\to t_0} f_2(t), \lim\limits_{t\to t_0} f_3(t) \right)_O = \left(L_1,L_2,L_3\right)_O . \end{equation} Similarly, if $f_1$, $f_2$, and $f_3$ are continuous and differentiable at $t$, then $\f$ is also continuous and differentiable at $t$, and we define \begin{equation} \f'(t) := \left(f_1'(t), f_2'(t), f_3'(t)\right)_O . \end{equation} Equivalently, we can define the derivative of a vector function in a way analogous to the definition of a derivative for scalar functions: \begin{equation} \f'(t) := \lim\limits_{\Delta t \to 0}\frac{\f(t+\Delta t) - \f(t)}{\Delta t} . \end{equation} Of course, the 2D versions of these definitions work the same way with only two components. Also, as we have already stated, vector functions will always be based at the origin $O$, so the limit or the derivative of a vector function will also be be based at $O$.

Example Let $\f(t) = (t,t^2,t^3)_O$ and $\g(t) = (\cos t, \sin t, 1)$. Then \begin{align} \f'(t) &= (1,2t,3t^2)_O \text{ , and} \\ \g'(t) &= (-\sin t, \cos t, 0)_O . \end{align}

Example Let $\f(t) = (t,t^2,t^3)_O$ and $\g(t) = (\cos t, \sin t, 1)$. Then \begin{align} \frac{d}{dt}(\f\cdot\g) &= \frac{d}{dt}(t\cos t + t^2\sin t + t^3) \\ &= (1+t^2)\cos t + t\sin t + 3t^2 . \end{align} Since $(\f\cdot\g)(t)$ is a scalar function, the process of differentiation is quite trivial. However, one might observe the result can also be expressed as $(1,2t,3t^2)_O\cdot(\cos t,\sin t,1)_O + (t,t^2,t^3)_O\cdot(-\sin t,\cos t, 0)_O = \f'(t)\cdot\g(t) + \f(t)\cdot\g'(t) .$ This is a simple illustration of the product rule discussed in the theorem below.

Theorem Let \f and \g be vector functions which are differentiable at $t$. Let $u$ be a scalar function also differentiable at $t$. And, let $\alpha$ be a constant scalar. Then

• $\alpha\f$ and $\f + \g$ are differentiable at $t$, and \begin{gather} (\alpha\f)'(t) = \alpha\f'(t) , \\ (\f + \g)'(t) = \f'(t) + \g'(t) . \end{gather}
• $u\f$, $\f\cdot\g$, and $\f\times\g$ (in 3D) are all differentiable at $t$, and \begin{gather} (u\f)'(t) = u'(t)\f(t) + u(t)\f'(t) , \\ (\f\cdot\g)'(t) = \f'(t)\cdot\g(t) + \f(t)\cdot\g'(t) \text{ , and} \\ (\f\times\g)'(t) = \f'(t)\times\g(t) + \f(t)\times\g'(t) . \end{gather}

The elements of this theorem as easily proven by writing $\f$ and $\g$ in terms of arbitrary components and performing each calculation. We now discuss the chain rule for vector functions. It is easily understood, but there is one important quality to consider. Vector functions are asymmetric in their mapping, since it is of the form $f:\R\to\Rthree$ (or $f:\R\to\Rtwo$). This means that some compositions are not possible. For example, we cannot write $u\circ\f$ since $\f$ gives a vector in $\Rthree$ (or $\Rtwo$), and since $u:\R\to\R$, i.e., $u$ is a scalar function, it cannot map vectors. Similarly, we cannot create compositions like $\f\circ\g$. We can, however, study composite functions of the form $\f\circ u$. In case you are not familiar with this notation, $(\f\circ u)(t)$ is read as "$f$ composed of $u$ at $t$" and it is equivalently written as $(\f\circ u)(t) = f(u(t))$.

Theorem Let $u$ be a scalar function, differentiable at $t$, and let $\f$ be a vector function differentiable at $u(t)$. Then the composite function $\f\circ u$ is differentiable at $t$, and \begin{equation} (\f\circ u)'(t) = u'(t)\f'(u(t)) . \end{equation}

Proof This is a simple calculation. As an outline: we assume arbitrary functions as the components of $\f$; apply the compositions to each component; perform the regular chain rule; factor the result. \begin{align} \text{Let } \f(t) &= (f_1(t),f_2(t),f_3(t))_O . \\ (\f\circ u)'(t) &= \frac{d}{dt}[\f(u(t))] \\ &= \frac{d}{dt}[(f_1(u(t)),f_2(u(t)),f_3(u(t)))_O] \\ &= (f_1'(u(t))u'(t),f_2'(u(t))u'(t),f_3'(u(t))u'(t))_O \\ &= u'(t)(f_1'(u(t)),f_2'(u(t)),f_3'(u(t)))_O \\ &= u'(t)\f'(u(t)) \\ &= \left(u'(\f'\circ u)\right)(t) . \end{align}

Curves Parametrized by Vector Functions

Now that we have studied vector functions, we will discuss their geometric interpretation, curves. We have already studied a special case of curves, lines, also parametrized by vector functions, where the form was $\vect{r}(t) = \vect{r_0} + t\vect{d}$. But now, we consider the more general case of a vector function $\f(t)$, which need not be a line at all. One can think of the independent variable as representing time, and then imagine $\f$ as an arrow that changes with time, pointing to parts of the curve, essentially tracing it out. The curve is labeled as $\C$, and we say that $\C$ is parametrized by $\f$. Let's begin by attempting to parametrize the unit circle centered at the origin with the following vector function: \begin{equation} \f(t) = (\cos t, \sin t)_O , t\in[0,2\pi] . \end{equation} Of course, we could also write this as \begin{equation} \f(t) = (\cos t, \sin t)_O , t\in[-\pi,\pi] . \end{equation} One way to see that this traces out the unit circle is by examining the norm of $\f$: \begin{equation} ||\f||(t) = \sqrt{\cos^2t+\sin^2t} = 1 . \end{equation} So, $\f$ always points exactly one unit away from the origin. Then, examining the trigonometry of the parametrization indicates that it traces out the entire unit circle. It's important to realize that the parametrization of any given curve is by no means unique: another parametrization of the unit circle might be: \begin{equation} \g(t) = (\cos 2t, \sin 2t)_O , t\in[0,\pi] . \end{equation}

Example Let's examine a 3D curve. Consider the curve $\C$ parametrized by \begin{equation} \f(t) = (\cos 2\pi t, \sin 2\pi t, t) , t\in[0,1] . \end{equation} If we neglect the third component, this would of course be again the unit circle. The third component sketches the circle linearly increasing in $z$, so $\C$ is actually like a winding coil. A plot of this curve is given in Fig.(\ref{3dcoil}). \begin{figure}[h] \begin{center} \includegraphics[scale=0.6]{3d_coil_plot.pdf} \caption{\label{3dcoil} 3D parametric plot from the previous example.} \end{center} \end{figure}

Tangent Line, Tangent Vector

So, we now understand that geometrically, a vector function $\f$ represents a curve. We will now consider the geometric meaning of $\f'$. In fact, the meaning of $\f'$ is very analogous to the meaning of the derivative of a scalar function. For a scalar function $g$, $g'(t)$ is the slope of the line tangent to $g$ at $t$. $\f'$ is also understood in terms of a tangent line, except now the tangent line would be something of the form $\vect{r}(t) = \vect{r_0} + t\vect{d}$, and the 'slope' is not a scalar; instead, the slope is the direction vector $\vect{d}$.

Definition Let $\C$ be a curve parametrized by the differentiable function $\f$, and let $P$ be the point on the curve with position vector $\f(t_0)$. Suppose that $\f'(t_0) \ne \vz$. The tangent line to $\C$ at $P$ is the line through $P$ with direction vector $\f'(t_0)$. So, this tangent line is parametrized by \begin{equation} \vect{r}(u) = \f(t_0) + u\f'(t_0) . \end{equation} Note that it absolutely essential that $\f'(t_0) \ne \vz$ because $\vz$ has no direction. We can now also define the tangent vector to a curve at some point, $P$, which will be parallel to the direction vector of the tangent line but is a constant vector based at $P$. Of course, even for a given $P$ and curve $\C$, the tangent vector would not be a unique vector.

Definition Let $\C$ be a curve parametrized by the differentiable function $\f$, and let $P$ be the point on the curve with position vector $\f(t_0)$. Suppose that $\f'(t_0) \ne \vz$. A tangent vector to $\C$ at $P$ is any vector based at $P$ which is parallel to $\f'(t_0)$.

Example Find the tangent line and a tangent vector to the twisted twisted cubic $\f(t) = (t,t^2,t^3)_O$ at the point $P=(1,1,1)$ on the curve. \\ First, we must find what value of $t$ gives the position vector of $P$. Clearly, the position vector of $P$ is $\f(t)$. Now, we compute the direction of the line tangent to $\C$ at $P$, which is given by the vector $\f'(1)$, so long as it is not $\vz$. We see that $\f'(t) = (1,2t,3t^2)_O$, so clearly $\f'(1)=(1,2,3)_O \ne \vz$. Therefore, the tangent line to $\C$ at $P$ is $\vect{r}(t) = (1,1,1)_O + t(1,2,3)_O .$ Since the tangent vector is any vector parallel to the direction vector of the tangent line but based at $P$, both $\vect{v} = (1,2,3)_P \text{ , and } \vect{w} = (-6,-12,-18)_P$ are tangent vectors to $\C$ at $P$. The algebraic interpretation of $\f'$ is familiar. If one takes $\f$ to be the position of a particle, this would describe a position traveling along the curve parametrized by $\f$ then $\f'$ is the velocity, $\f''$ is the acceleration, and $||\f'||$ is the speed.

Arc-length

Let's consider a finite portion $\C$ of a some curve, where the endpoints have position vectors $\f(a)$ and $\f(b)$. We can partition the interval for $t \in [a,b]$ into the sequence $\{t_0, t_1, \dots, t_n\}$ where $t_0 = a$ and $t_n = b$. Then, the sequence of points $\{P_0, P_1, \dots, P_n\}$ will lie on the curve $\C$ and have position vectors $\{\f(t_0), \f(t_1), \dots, \f(t_n)\}$. The distance between two successive points on $\C$, $P_{i-1}$ and $P_i$, is approximately given by $||\f'(t_i)||(t_i-t_{i-1})$, which is obtained by linearizing the displacement along the curve. Then, we can approximate the length of $\C$ by summing up these distances: \begin{equation} \text{Length of $\C$ } = \sum_{i=1}^n ||\f'(t_i)||(t_i-t_{i-1}) . \end{equation} You should recognize this type of summation because it is crucial: it is a Riemann sum. Thus, we can convert it to an integral by making the familiar substitution for $\displaystyle\sum \Delta t\ldots \to \int dt\dots$, where $\Delta t = (t_i-t_{i-1})$, which gives \begin{equation} \text{Length of $\C$ } = \int_a^b dt ||\f'(t)|| . \end{equation}

Definition Let $\C$ be a curve parametrized by the differentiable function $\f$, and suppose that $||\f'||$ is integrable overt $[a,b]$. Then, the arc-length of the portion of $\C$ between $\f(a)$ and $\f(b)$ is \begin{equation} \int_a^b dt ||\f'(t)|| . \end{equation} Note that since $||\f'(t)||$ is a scalar function, the calculus necessary here is that of ordinary scalar functions. The only vector operation necessary here is to compute the norm of the vector function, to get the scalar function.

Polar Curves

In polar co-ordinates, we have access to a special class of parametrized curves, which are of the form \begin{equation} r = f(\theta) , \label{polar_curve} \end{equation} where $f$ is some real-valued scalar function. This means that the polar co-ordinate $r$ which you should recall is the distance from the origin to the point is given by a function of the polar angle to the point. This is a special class of curves because it is a special case of the more general class of curves that we studied in the previous section. This is to say that we can construct a vector function that is equivalent to any polar curve of this form. We can accomplish this by considering that vector functions are specified in Cartesian co-ordinates. If we make the substitutions \begin{equation} x = r\cos\theta \text{, and } y=r\sin\theta , \end{equation} then the polar curves of the form Eq. \eqref{polar_curve} will be parametrized by the vector function \begin{equation} \g(\theta) = (f(\theta)\cos\theta, f(\theta)\sin\theta)_O . \label{polar_vfn} \end{equation} The unit circle parametrization that we studied in the previous section would be given by setting $f(\theta) = 1$.

Arc-length

Since we already know how to write a polar curve as a vector function, it is easy to compute the arc-length. For the curve $\C$ which is parametrized by $\g$ given in Eq. \eqref{polar_vfn}, the arc-length integrand will require us to compute $||\g'(\theta)||$: \begin{align} ||\g'(\theta)|| &= ||(f'(\theta)\cos\theta -f(\theta)\sin\theta, f'(\theta)\sin\theta + f(\theta)\cos\theta)|| \\ ||\g'(\theta)||^2 &= (f'(\theta)\cos\theta -f(\theta)\sin\theta)^2 + (f'(\theta)\sin\theta + f(\theta)\cos\theta)^2 \\ &= f'(\theta)^2\cos^2\theta - 2f(\theta)f'(\theta)\cos\theta\sin\theta + f(\theta)^2\sin^2\theta \\ & + f'(\theta)^2\sin^2\theta + 2f(\theta)f'(\theta)\cos\theta\sin\theta + f(\theta)^2\cos^2\theta \\ &= f'(\theta)^2(\cos^2\theta + \sin^2\theta) + f(\theta)^2(\cos^2\theta + \sin^2\theta) \\ &= f(\theta)^2 + f'(\theta)^2 . \end{align} This is most of the proof necessary for the theorem which follows.

Theorem Let $\C$ be the polar curve given by $r = f(\theta)$, for $\theta \in [a,b]$. Suppose that $f$ is differentiable and that $f'$ is integrable over $[a,b]$. Then, the length of $\C$ is \begin{equation} L(\C) = \int_a^b\sqrt{f(\theta)^2 + f'(\theta)^2} d\theta \end{equation}

Example The curve $\C$ given by $r=\cos\theta$, $\theta\in[-\pi/2,\pi/2]$ has length $L(\C) = \int_{-\pi/2}^{\pi/2}\sqrt{(-\sin\theta)^2+(\cos\theta)^2} d\theta = \int_{-\pi/2}^{\pi/2} 1 d\theta = \pi .$

Area Enclosed by a Polar Curve

To compute the area enclosed by the polar curve $\C$ given by \begin{equation} r=f(\theta) , \theta\in[a,b] , \end{equation} we begin as usual by partitioning the interval of $\theta$ into $\{\theta_0, \theta_1, \dots, \theta_n\}$. The area between two successive points (by approximating the curve to be locally circular) is approximately \begin{equation} \frac{\theta_k - \theta_{k-1}}{2\pi} \pi f(\theta_k)^2 = \frac{1}{2}f(\theta_k)^2(\theta_k - \theta_{k-1}) . \end{equation} Then, if we add up all of the partial areas going around $\C$, we arrive at the Riemann sum \begin{equation} \sum_{k=1}^n \frac{1}{2}f(\theta_k)^2(\theta_k - \theta_{k-1}) , \end{equation} which as usual will converge to an integral.

Definition Suppose that $f$ is continuous on $[a,b]$. Then, the area enclosed by the polar curve $r=f(\theta)$, for $\theta\in[a,b]$, and the polar lines $\theta=a$ and $\theta=b$ is \begin{equation} \int_a^b \frac{1}{2}f(\theta)^2 d\theta . \end{equation}

Example The area enclosed by the polar curve $r=\cos\theta$, for $\theta\in[-\pi/2,\pi/2]$ is \begin{align} \int_{-\pi/2}^{\pi/2}\frac{1}{2}\cos^2\theta d\theta &= \frac{1}{4}\int_{-\pi/2}^{\pi/2}(1+\cos2\theta) d\theta \\ &= \frac{1}{4}\left[\theta + \frac{1}{2}\sin2\theta\right]_{-\pi/2}^{\pi/2} \\ &= \frac{\pi}{4} . \end{align}