APEX Calculus

We start by making a table of \(t\text{,}\) \(x\) and \(y\) values as shown in Figure 11.1.4.(a). Plotting these points gives an indication of what the graph looks like. In Figure 11.1.4.(b), we indicate these points and sketch the full graph. We also highlight \(\vec r(-1)\) and \(\vec r(2)\) on the graph.

\(t\)	\(t^3-t\)	\(\ds \frac{1}{t^2+1}\)
\(-2\)	\(-6\)	1/5
\(-1\)	\(0\)	1/2
\(0\)	\(0\)	\(1\)
\(1\)	\(0\)	1/2
\(2\)	\(6\)	1/5

We can again plot points, but careful consideration of this function is very revealing. Momentarily ignoring the third component, we see the \(x\) and \(y\) components trace out a circle of radius 1 centered at the origin. Noticing that the \(z\) component is \(t\text{,}\) we see that as the graph winds around the \(z\)-axis, it is also increasing at a constant rate in the positive \(z\) direction, forming a spiral. This is graphed in Figure 11.1.6. In the graph \(\vec r(7\pi/4)\approx (0.707,-0.707,5.498)\) is highlighted to help us understand the graph.

We can graph \(\vec r_1\) and \(\vec r_2\) easily by plotting points (or just using technology). Let's think about each for a moment to better understand how vector-valued functions work.

We can rewrite \(\vec r_1(t) = \la\, 0.2t,0.3t\,\ra\) as \(\vec r_1(t) = t\la 0.2,0.3\ra\text{.}\) That is, the function \(\vec r_1\) scales the vector \(\la 0.2,0.3\ra\) by \(t\text{.}\) This scaling of a vector produces a line in the direction of \(\la 0.2,0.3\ra\text{.}\)

We are familiar with \(\vec r_2(t) = \la\, \cos(t) ,\sin(t) \,\ra\text{;}\) it traces out a circle, centered at the origin, of radius 1. Figure 11.1.9.(a) graphs \(\vec r_1(t)\) and \(\vec r_2(t)\text{.}\)

Adding \(\vec r_1(t)\) to \(\vec r_2(t)\) produces \(\vec r(t) = \la\,\cos(t) + 0.2t,\sin(t) +0.3t\,\ra\text{,}\) graphed in Figure 11.1.9.(b). The linear movement of the line combines with the circle to create loops that move in the direction of \(\la 0.2,0.3\ra\text{.}\) (We encourage the reader to experiment by changing \(\vec r_1(t)\) to \(\la 2t,3t\ra\text{,}\) etc., and observe the effects on the loops.)

Multiplying \(\vec r(t)\) by 5 scales the function by 5, producing \(5\vec r(t) = \langle 5\cos(t) +1,5\sin(t) +1.5\rangle\text{,}\) which is graphed in Figure 11.1.9.(c) along with \(\vec r(t)\text{.}\) The new function is “5 times bigger” than \(\vec r(t)\text{.}\) Note how the graph of \(5\vec r(t)\) in Figure 11.1.9.(c) looks identical to the graph of \(\vec r(t)\) in Figure 11.1.9.(b). This is due to the fact that the \(x\) and \(y\) bounds of the plot in Figure 11.1.9.(c) are exactly 5 times larger than the bounds in Figure 11.1.9.(b).

This problem is not very difficult if we approach it in a clever way. We start by letting \(\vec p(t)\) describe the position of the point \(p\) on the circle, where the circle is centered at the origin and only rotates clockwise (i.e., it does not roll). This is relatively simple given our previous experiences with parametric equations; \(\vec p(t) = \la \cos(t) , -\sin(t) \ra\text{.}\)

We now want the circle to roll. We represent this by letting \(\vec c(t)\) represent the location of the center of the circle. It should be clear that the \(y\) component of \(\vec c(t)\) should be 1; the center of the circle is always going to be 1 if it rolls on a horizontal surface.

The \(x\) component of \(\vec c(t)\) is a linear function of \(t\text{:}\) \(f(t) = mt\) for some scalar \(m\text{.}\) When \(t=0\text{,}\) \(f(t) = 0\) (the circle starts centered on the \(y\)-axis). When \(t=2\pi\text{,}\) the circle has made one complete revolution, traveling a distance equal to its circumference, which is also \(2\pi\text{.}\) This gives us a point on our line \(f(t) = mt\text{,}\) the point \((2\pi, 2\pi)\text{.}\) It should be clear that \(m=1\) and \(f(t) = t\text{.}\) So \(\vec c(t) = \la t, 1\ra\text{.}\)

We now combine \(\vec p\) and \(\vec c\) together to form the equation of the cycloid: \(\vec r(t) = \vec p(t) + \vec c(t) = \la \cos(t) + t,-\sin(t) +1\ra\text{,}\) which is graphed in Figure 11.1.12.

The function \(\vec r(t)\) traces out the unit circle, though at a different rate than the “usual” \(\la \cos(t) ,\sin(t) \ra\) parametrization. At \(t_0=-1\text{,}\) we have \(\vec r(t_0) = \la 0,-1\ra\text{;}\) at \(t_1=1\text{,}\) we have \(\vec r(t_1) = \la 0,1\ra\text{.}\) The displacement of \(\vec r(t)\) on \([-1,1]\) is thus \(\vec d = \la 0,1\ra - \la 0,-1\ra = \la 0,2\ra\text{.}\)

A graph of \(\vec r(t)\) on \([-1,1]\) is given in Figure 11.1.15, along with the displacement vector \(\vec d\) on this interval.

We computed in Example 11.1.14 that the displacement of \(\vec r(t)\) on \([-1,1]\) was \(\vec d = \la 0,2\ra\text{.}\) Thus the average rate of change of \(\vec r(t)\) on \([-1,1]\) is:

We interpret this as follows: the object followed a semi-circular path, meaning it moved towards the right then moved back to the left, while climbing slowly, then quickly, then slowly again. On average, however, it progressed straight up at a constant rate of \(\la 0,1\ra\) per unit of time.

We can quickly see that the displacement on \([-1,5]\) is the same as on \([-1,1]\text{,}\) so \(\vec d = \la 0,2\ra\text{.}\) The average rate of change is different, though:

As it took “3 times as long” to arrive at the same place, this average rate of change on \([-1,5]\) is \(1/3\) the average rate of change on \([-1,1]\text{.}\)