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CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 2.1 Definitions and First Examples

In the last chapter, we studied vector valued functions of a single variable, like, for example, the velocity \(\vv(t)\) of a particle at time \(t\text{.}\) Suppose however that we are interested in a fluid. There is a, possibly different, velocity at each point in the fluid. So the velocity of a fluid is really a vector valued function of several variables. Such a function is called a vector field.

Definition 2.1.1.

  1. A vector field in the plane is a rule which assigns to each point \((x,y)\) in a subset, \(D\text{,}\) of the \(xy\)-plane, a two component vector \(\vv(x,y)\text{.}\)
  2. A vector field in space is a rule which assigns to each point \((x,y,z)\) in a subset of \(\bbbr^3\text{,}\) a three component vector \(\vv(x,y,z)\text{.}\)
Here are two typical applications that naturally involve vector fields.
  • If \(\vv(x,y,z)\) is the velocity of a moving fluid at position \((x,y,z)\text{,}\) then \(\vv\) is called a velocity field.
  • If \(\vF(x,y,z)\) is the force at position \((x,y,z)\text{,}\) then \(\vF\) is called 1  a force field.
Imagine
  • The whole world is filled with an incompressible fluid. Call it water.
  • Somehow you find a way to produce still more water at the origin. Say you create \(4\pi m\) litres per second.
  • This forces the water to flow outward. Let's suppose that it flows symmetrically outward from the origin.
Let's find the resulting vector field \(\vv(x,y,z)\text{.}\) As the flow is to be symmetric, the velocity of the water at the point \((x,y,z)\)
  • has to be pointing radially outward from the origin. That is, the direction of the velocity vector \(\vv(x,y,z)\) has to be the unit radial vector
    \begin{equation*} \hat\vr(x,y,z) = \frac{x\hi + y\hj + z\hk}{\sqrt{x^2+y^2+z^2}} \end{equation*}
  • The magnitude of the velocity, i.e. the speed \(|\vv(x,y,z)|\) of the water, has to depend only on the distance from the origin. That is, the speed can only be some function of
    \begin{equation*} r(x,y,z) = \sqrt{x^2+y^2+z^2} \end{equation*}
Thus the velocity field is of the form
\begin{equation*} \vv(x,y,z) = v\big(r(x,y,z)\big)\,\hat\vr(x,y,z) \end{equation*}
We just have to determine the function \(v(r)\text{.}\) Fix any \(r \gt 0\) and concentrate on the sphere \(x^2+y^2+z^2=r^2\text{.}\) It is sketched in red in the figure below.
During a very short time interval \(dt\) seconds, \(4\pi m\,dt\) litres of water is created at the origin (which is the red dot). As the water is incompressible, \(4\pi m\,dt\) litres of water must exit through the sphere during the same time interval to make room for the newly created water.
But, at the surface of the sphere the water is flowing radially outward with speed \(v(r)\text{.}\) So during the time interval in question the water near the surface of the sphere moves outward a distance \(v(r)\,dt\text{,}\) and in particular the water that was in the thin spherical shell \(\ \ r-v(r)\,dt \le \sqrt{x^2+y^2+z^2} \le r\ \ \) at the beginning of the time interval exits through the sphere \(\sqrt{x^2+y^2+z^2} = r\) during the time interval. The shell is sketched in gray in the figure above. The volume of water in the gray shell is essentially the surface area of the shell, which is \(4\pi r^2\text{,}\) times the thickness of the shell, which is \(v(r)\,dt\text{.}\) So, equating the volume of water created inside the sphere with the volume of water that exited the sphere,
\begin{equation*} 4\pi m\,dt = (4\pi r^2) \big(v(r)\,dt\big) \implies v(r) = \frac{4\pi m}{4\pi r^2} =\frac{m}{r^2} \end{equation*}
Thus our vector field is
\begin{equation*} \vv(x,y,z) = \frac{m}{r(x,y,z)^2}\,\hat\vr(x,y,z) \end{equation*}
If the world were two, rather than three dimensional 2 , and the source created \(2\pi m\) litres per second, the same argument leads to
\begin{equation*} 2\pi m\,dt = (2\pi r) \big(v(r)\,dt\big) \implies v(r) = \frac{2\pi m}{2\pi r} =\frac{m}{r} \end{equation*}
and to the vector field
\begin{equation*} \vv(x,y) = \frac{m}{r(x,y)}\,\hat\vr(x,y)\qquad r(x,y) = \sqrt{x^2+y^2}\qquad \hat\vr(x,y) = \frac{x\hi + y\hj}{\sqrt{x^2+y^2}} \end{equation*}
To get a mental image of what this field looks like, imagine sketching, for each point \((x,y)\text{,}\) the vector \(\frac{m}{r(x,y)}\,\hat\vr(x,y)\) with its tail at \((x,y)\text{.}\) Note that the vector \(\frac{m}{r(x,y)}\,\hat\vr(x,y)\)
  • points radially outward and
  • has length \(\frac{m}{r(x,y)}\) which
    • depends only on \(r=|(x,y)|\) and
    • is very long when \((x,y)\) is near the origin and
    • decreases in length like \(\frac{1}{r}\) as \(r=|(x,y)|\) increases.
Here is a sketch of a bunch of such vectors.
Figure 2.1.3.
Note that as \(|(x,y)|\rightarrow 0\text{,}\) the magnitude of the velocity \(|\vv(x,y)|\rightarrow\infty\text{.}\) This is a consequence of our idealized assumption that we are producing water at a single point (the origin).
In this example, we sketch the vector field
\begin{equation*} \vv(x,y) = \Om\big(-y\hi +x\hj\big) \end{equation*}
where \(\Om\) is just a strictly positive constant. We give an efficient procedure for getting a rough sketch, which still provides a pretty realistic picture of the vector field, and which also generalises to other vector fields. First concentrate on the horizontal component \(\hi\cdot\vv(x,y)\) of the vector field and determine in which part of the \(xy\)-plane it is zero, in which part it is positive and in which part it is negative.
\begin{equation*} \hi\cdot\vv(x,y) =-\Om y\ \ \begin{cases} =0 &\text{if } y=0\\ \lt 0 &\text{if } y \gt 0\\ \gt 0 &\text{if } y \lt 0 \end{cases} \end{equation*}
Next repeat with the vertical component.
\begin{equation*} \hj\cdot\vv(x,y) = \Om x\ \ \begin{cases} =0 &\text{if } x=0 \\ \lt 0 &\text{if } x \lt 0 \\ \gt 0 &\text{if } x \gt 0 \end{cases} \end{equation*}
This naturally divides the \(xy\)-plane into nine parts according to whether each of the components is positive, \(0\) or negative —
  • \(\hi\cdot\vv \gt 0\) and \(\hj\cdot\vv \gt 0\) in \(\big\{\ (x,y)\in\bbbr^2\ \big|\ y \lt 0,\ x \gt 0\ \big\}\)
  • \(\hi\cdot\vv \gt 0\) and \(\hj\cdot\vv=0\) in \(\big\{\ (x,y)\in\bbbr^2\ \big|\ y \lt 0,\ x=0\ \big\}\)
  • \(\hi\cdot\vv \gt 0\) and \(\hj\cdot\vv \lt 0\) in \(\big\{\ (x,y)\in\bbbr^2\ \big|\ y \lt 0,\ x \lt 0\ \big\}\)
  • \(\hi\cdot\vv=0\) and \(\hj\cdot\vv \gt 0\) in \(\big\{\ (x,y)\in\bbbr^2\ \big|\ y=0,\ x \gt 0\ \big\}\)
  • and so on
Now think of \(\vv(x,y)\) as being the velocity at \((x,y)\) of a flowing fluid.
  • Look at the first bullet point above. It says that in the first of the nine parts, namely \(\big\{\ (x,y)\in\bbbr^2\ \big|\ y \lt 0,\ x \gt 0\ \big\}\text{,}\) which is the fourth quadrant, the horizontal component \(\hi\cdot\vv \gt 0\) signifying that the fluid is flowing rightwards. Indicate this in the sketch by drawing a rightward pointing horizontal arrow at some generic point in the middle of the fourth quadrant. (It's the blue arrow in the figure below.) The vertical component \(\hj\cdot\vv \gt 0\) signifying that the fluid is also moving upwards. Indicate this in the sketch by drawing an upward pointing vertical arrow at the same generic point in the fourth quadrant. (It's the red arrow in the figure below.)
  • Next, look at the second bullet point above. It says that on the second of the nine parts, namely \(\big\{\ (x,y)\in\bbbr^2\ \big|\ y \lt 0,\ x=0\ \big\}\text{,}\) which is the bottom half of the \(y\)-axis, the horizontal component \(\hi\cdot\vv \gt 0\text{,}\) signifying that the fluid is moving rightwards. Indicate this in the sketch by drawing a rightward pointing horizontal arrow at some generic point in the middle of the bottom half of the \(y\)-axis. (It's the second blue arrow in the figure below.) The vertical component \(\hj\cdot\vv=0\) signifying that the fluid has no vertical motion at all. Indicate this in the sketch by not drawing any vertical arrow on the bottom half of the \(y\)-axis.
  • and so on
By the time we have looked at all nine regions we will have built up the following sketch.
Figure 2.1.5.
From this sketch we see that, for example, in the first quadrant,
  • the fluid is moving upwards and to the left and
  • the fluid crosses the \(x\)-axis vertically (so that close to the \(x\)-axis, the arrows will be almost vertical) and
  • the fluid crosses the \(y\)-axis horizontally (so that close to the \(y\)-axis, the arrows will be almost horizontal) and
  • there is one point, namely \((0,0)\text{,}\) where the vector field is exactly zero. It's the black dot in the centre of the figure above. Furthermore \(\vv(x,y)=\Omega(-y\hi+x\hj)\) is smaller when \((x,y)\) is closer to \((0,0)\) and \(\vv(x,y)\) is larger when \((x,y)\) is farther from \((0,0)\text{,}\)
Putting all of this accumulated wisdom together, we come up with this better sketch of the vector field.
Figure 2.1.6.
This shows the field swirling around the origin in a counterclockwise direction. Hence the name “vortex”.
In this example, we illustrate another way in which vector fields arise. Model a pendulum by a mass \(m\) that is connected to a hinge by an idealized rod that is massless 3  and of fixed length \(\ell\text{.}\) Denote by \(\theta\) the angle
between the rod and vertical. The forces acting on the mass are
  • gravity and
  • the tension in the rod, whose magnitude, \(\tau\text{,}\) automatically adjusts itself so that the distance between the mass and the hinge is fixed at \(\ell\text{.}\)
In the optional 4  Section 2.5, we show that the angle \(\theta(t)\) obeys the second order nonlinear 5  differential equation
\begin{equation*} \difftwo{\theta}{t} +\frac{g}{\ell}\sin\theta=0 \end{equation*}
It is often much more convenient to deal with first order, rather than second order, differential equations. The second order pendulum equation above may be reformulated 6  as a system of first order ordinary differential equations, by the simple expedient of defining
\begin{equation*} x(t)=\theta(t)\qquad y(t)=\theta'(t) \end{equation*}
So \(x(t)\) is the angle at time \(t\) and \(y(t)\) is the angular velocity at time \(t\text{.}\) Then,
\begin{alignat*}{4} x'(t)&=&\theta'(t)&=y(t)\cr y'(t)&=&\ \theta''(t)&=-\frac{g}{\ell}\sin x(t) \end{alignat*}
Usually, one does not write in the \((t)\) dependence explicitly.
\begin{align*} x'&=y\\ y'&=-\frac{g}{\ell}\sin x \end{align*}
The right hand sides form the vector field
\begin{equation*} \vv\big((x,y)\big)=\Big(y\,,\,-\frac{g}{\ell}\sin x\Big) \end{equation*}
We can sketch this vector field, just as we sketched the vector field of Example 2.1.4. Noting that the horizontal component
\begin{equation*} \hi\cdot\vv(x,y) = y\ \ \begin{cases} =0 &\text{if } y=0 \\ \gt 0 &\text{if } y \gt 0 \\ \lt 0 &\text{if } y \lt 0 \end{cases} \end{equation*}
and the vertical component.
\begin{equation*} \hj\cdot\vv(x,y) = -\frac{g}{\ell}\sin x\ \ \begin{cases} =0 &\text{if } x=0,\ \pm\pi, \pm2\pi, \cdots \\ \gt 0 &\text{if } -\pi \lt x \lt 0,\ \ \pi \lt x \lt 2\pi,\text{ etc.}\\ \lt 0 &\text{if } 0 \lt x \lt \pi, \ \ 2\pi \lt x \lt 3\pi,\text{ etc.} \end{cases} \end{equation*}
we have
  • rightward motion 7  when \(y \gt 0\)
  • leftward motion when \(y \lt 0\)
  • downward motion when \(0 \lt x \lt \pi\text{,}\) \(2\pi \lt x \lt 3\pi\text{,}\) \(\cdots\) and
  • upward motion when \(-\pi \lt x \lt 0\text{,}\) \(\pi \lt x \lt 2\pi\text{,}\) \(\cdots\text{.}\)
This gives us the collection of arrows in the figure
Our full sketch will be less cluttered if we make all arrows the same length. This gives
which is a sketch of what is called the direction field of our vector field (see below).
In the next section, we'll learn how to use vector field sketches to sketch solution trajectories.

Definition 2.1.8.

The direction field of a vector field \(\vv(x,y,z)\) is the vector field
\begin{equation*} \vV(x,y,z) =\begin{cases} \frac{\vv(x,y,z)}{|\vv(x,y,z)|} &\text{if } \vv(x,y,z)\ne\vZero \\ \vZero & \text{if } \vv(x,y,z) = \vZero \end{cases} \end{equation*}

Exercises Exercises

Exercise Group.

Exercises — Stage 1
1.
Below is a sketch of the vector field \(\vv(x,y)\text{.}\)
Find the regions where the \(x\)-coordinates and \(y\)-coordinates are positive, negative, and zero:
\begin{equation*} \vv(x,y)\cdot \hi \begin{cases} \gt 0 & \mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \\ =0 &\mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \\ \lt 0&\mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \end{cases} \end{equation*}
\begin{equation*} \vv(x,y) \cdot \hj \begin{cases} \gt 0 & \text{ when } \fbox{$\vphantom{L}\qquad\qquad $} \\ =0 & \text{ when } \fbox{$\vphantom{L}\qquad\qquad $} \\ \lt 0&\text{ when } \fbox{$\vphantom{L}\qquad\qquad $} \end{cases} \end{equation*}
You may assume that \(\vv(x,y)\) behaves as expected at the points you don't see. That is, the samples are representative of a smooth, continuous vector-valued function. You may also assume the tick marks on the axes correspond to unit distances.
2.
Below is a sketch of the vector field \(\vv(x,y)\text{.}\)
Find the regions where the \(x\)-coordinates and \(y\)-coordinates are positive, negative, and zero:
\begin{equation*} \vv(x,y)\cdot \hi \begin{cases} \gt 0 & \mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $}\\ =0 &\mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \\ \lt 0&\mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \end{cases} \end{equation*}
\begin{equation*} \vv(x,y)\cdot \hj \begin{cases} \gt 0 & \mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \\ =0 &\mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \\ \lt 0&\mbox{ when } \fbox{$\vphantom{L}\qquad\qquad $} \end{cases} \end{equation*}
You may assume that the samples shown are representative of the general behaviour of \(\vv(x,y)\text{.}\) You may also assume the tick marks on the axes correspond to unit distances.
3.
A platform with many small conveyor belts is aligned on a coordinate plane. Every conveyor belt moves an object on top of it in the direction of the origin, and a conveyor belt at position \((x,y)\) causes an object on top of it to move with speed \(|y|\text{.}\) Assume the objects do not interfere with one another.
Give a vector-valued formula for the velocity of an object at position \((x,y)\text{.}\)
4.
Let \(\vF = P\,\hi + Q\,\hj\) be the two-dimensional vector field sketched below.
Determine the signs of \(P\text{,}\) \(Q\text{,}\) \(\pdiff{Q}{x}\) and \(\pdiff{Q}{y}\) at the point \(A\text{.}\)
5.
Imagine that the vector field \(\vv(x,y) = x\,\hi+y\,\hj\)
is the velocity field of a moving fluid.
  1. At time \(0\) you drop a twig into the fluid at the point \((1,1)\text{.}\) What is the approximate position of the twig at time \(t=0.01\text{?}\)
  2. At time \(0\) you drop a twig into the fluid at the point \((0,0)\text{.}\) What is the position of the twig at time \(t=0.01\text{?}\)
  3. At time \(0\) you drop a twig into the fluid at the point \((0,0)\text{.}\) What is the position of the twig at time \(t=10\text{?}\)
6.
Imagine that the vector field \(\vv(x,y) = 2x\,\hi -\hj\)
is the velocity field of a moving fluid. At time \(0\) you drop a twig into the fluid at the point \((0,0)\text{.}\) What is the position of the twig at time \(t=10\text{?}\)

Exercise Group.

Exercises — Stage 2
7.
Friendly (?) bees fly towards your face from all directions. The speed of each bee is inversely proportional to its distance from your face. Find a vector field for the velocity of the swarm.
8.
Sketch the vector field \(\vv(x,y)=(x^2,y)\text{.}\)
9.
Sketch the direction field of \(\vv(x,y) = \left( \sqrt{x^2+y^2} , \sqrt{(x-1)^2+(y-1)^2}\right)\text{.}\)
10.
Sketch the direction field of \(\vv(x,y)=(x^2+xy,y^2-xy)\text{.}\)
11.
Sketch the vector field \(\displaystyle \vv(x,y)=\left[\frac{1/3}{\sqrt{x^2+y^2}}(x,y)+\frac{1/3}{\sqrt{(x-1)^2+y^2}}(x-1,y)\right]\text{.}\)
12.
Sketch each of the following vector fields, by drawing a figure like Figure 2.1.3.
  1. \(\vv(x,y) = x\,\hi+y\,\hj\text{.}\)
  2. \(\vv(x,y) = 2x\,\hi -\hj\text{.}\)
  3. \(\vv(x,y) = \frac{y\,\hi -x\,\hj}{\sqrt{x^2+y^2}}\text{.}\)
13.
A body of mass \(M\) exerts a force of magnitude \(\frac{GM}{D^2}\) on a particle of unit mass distance \(D\) away from itself, where \(G\) is a physical constant. The force acts in the direction from the particle to the body.
Suppose a mass of 5 kg sits at position \((0,0)\text{,}\) a mass of 3 kg sits at position \((2,3)\text{,}\) and a mass of 7 kg sits at position \((4,0)\) on a coordinate plane. Give the vector field \(\vf(x,y)\) of the net gravitational force exerted on a unit mass at position \((x,y)\text{.}\)

Exercise Group.

Exercises — Stage 3
14.
  1. A pole leans against a vertical wall. The pole has length 2, and it touches the wall at height \(H=1\text{.}\) The pole slides down, still touching the wall, with its height decreasing at a rate of \(\diff{H}{t}=0.5\text{.}\)
    Find a vector function \(\vv:\mathbb [0,2] \to \mathbb R^2\) for the velocity, when \(H=1\text{,}\) of a point on the pole that is \(p\) units from the lower end, using the coordinate system from the sketch above.
  2. The frame of an umbrella is constructed by attaching straight, rigid poles to a common centre. The poles are all the same length, so they form radii of a circle.
    The frame is lifted from the centre of the circle. The edges of the frame drag on the ground, keeping the frame in the shape of a right circular cone that is becoming taller and thinner.
    Suppose the length of each pole is 2 metres, and the centre of the frame is being lifted at a rate of 50 cm/s. Give a vector field for the velocity \(\vV(x,y,z)\) of a point \((x,y,z)\) on the frame when its centre is 1 metre above the ground.
    Let the ground have height \(z=0\text{,}\) and let the centre of the frame sit directly above the origin.
No, force fields are not only a sci-fi trope. Gravity is an example of a force field.
You might want to think about what happens in \(d\) dimensions for general \(d\text{.}\)
While we are idealizing, let's put everything in a vacuum.
In the optional Section 2.5 we also include frictional forces. In this example, we do not, so set the \(\beta\) of Section 2.5 to zero here.
It is common, when considering only small amplitude oscillations, to approximate \(\sin\theta\) by \(\theta\text{.}\) This converts our nonlinear differential equation into a linear differential equation.
This “hack” generalizes easily and is commonly used when generating, by computer, approximate solutions to higher order ordinary differential equations.
Note that this is rightward motion of the point \((x,y)\text{,}\) not of the pendulum itself.
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