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Lessons for Multivariable Calculus

Mark Fitch

Section 1.6 Parameterizing Curves

Every curve can be parameterized many ways. Here we continue to look at how that effects our calculations. We will also work on choosing parameterizations for convenience.

Subsection 1.6.1 Effect of Parameterization

Example 1.6.1.

Here we will consider two parameterizations of the same line and see what effect the parameterization has on calculations.
Consider the line given by both
\begin{equation*} \ell_1(t)=(2,3,7)+(3,-4,12)t \end{equation*}
\begin{equation*} \ell_2(t)=(2,3,7)+\left(\frac{3}{13},\frac{-4}{13},\frac{12}{13}\right)t \end{equation*}
We want to find a value \(t\) that gives us the point \((8,-5,31)\text{.}\)
\begin{align*} \ell_1(t) = & (8,-5,31).\\ (2,3,7)+(3,-4,12)t = & (8,-5,31).\\ (3,-4,12)t = & (8,-5,31)-(2,3,7).\\ (3,-4,12)t = & (6,-8,24).\\ 3t = & 6.\\ t = & 2.\\ (3,-4,12)2 = & (6,-8,24). \end{align*}
\begin{align*} \ell_2(t) = & (8,-5,31).\\ (2,3,7)+\left(\frac{3}{13},\frac{-4}{13},\frac{12}{13}\right)t = & (8,-5,31).\\ \left(\frac{3}{13},\frac{-4}{13},\frac{12}{13}\right)t = & (8,-5,31)-(2,3,7).\\ \left(\frac{3}{13},\frac{-4}{13},\frac{12}{13}\right)t = & (6,-8,24).\\ \frac{3}{13}t = & 6.\\ t = & 26.\\ \left(\frac{3}{13},\frac{-4}{13},\frac{12}{13}\right)26 = & (6,-8,24). \end{align*}
Notice if we interpret \(t\) as time, that \(\ell_1\) reaches the point much earlier than \(\ell_2\) does (\(2 \le 26\)). We can think of \(\ell_1\) as being a pamaterization that represents a faster transit of this line.

Activity 23.

Using the idea of parameterizations encoding different speeds from Example 1.6.1, we will consider the effect on calculations.
Consider the line given by both
\begin{equation*} \ell_1(t)=(2,3,7)+(3,-4,12)t \end{equation*}
\begin{equation*} \ell_2(t)=(2,3,7)+\left(\frac{3}{13},\frac{-4}{13},\frac{12}{13}\right)t \end{equation*}
(a)
First we consider the effect of parameterization on limits.
(i)
Calculate \(\lim_{t \to 2} \ell_1(t)\)
(ii)
Calculate \(\lim_{t \to 26} \ell_2(t)\)
(iii)
Are the limits the same or different? does this make sense with respect to the different parameterizations?
(b)
Next we consider the effect of parameterization on derivatives.
(i)
Calculate \(\ell_1^\prime(2)\text{.}\)
(ii)
Calculate \(\|\ell_1^\prime(2)\|\text{.}\)
(iii)
Calculate \(\ell_2^\prime(26)\text{.}\)
(iv)
Calculate \(\|\ell_2^\prime(26)\|\text{.}\)
(v)
Compare the derivatives.
(c)
Finally, we consider the effect of parameterization on arclength.
(i)
Calculate the arclength of \(\ell_1\) for \(t \in [0,2]\text{.}\)
(ii)
Calculate the arclength of \(\ell_2\) for \(t \in [0,26]\text{.}\)
(iii)
Compare the lengths.

Subsection 1.6.2 Parameterizing by Arclength

In Example 1.6.1 the parmaterization \(\ell_2\) has a ratio of increase of one unit distance for each increase of one unit input. This is considered to be parametrized by arclength, because the rate is measured in unit length. This is different from a ratio of increase of one unit time for each increase of one unit input. The results are then in speed rather than distance.
Parmetrizing the line in terms of distance was relatively easy: we changed the slope to be a unit vector. While we can use a similar method for non-linear curves, the calculations become much more complicated. We will rely on technology to help us with the calculations.

Example 1.6.2.

We will parameterize the circle
\begin{equation*} C=(5\cos \theta, 5\sin \theta, 1) \end{equation*}
with respect to arclength. With a line the arclength is given by the slope vector. Here we must calculate the arclength.
\begin{align*} s = & \int_0^\theta \sqrt{(-5\sin\alpha)^2+(5\cos\alpha)^2+0^2} \; d\alpha\\ = & \int_0^\theta \sqrt{25(\cos^2\alpha+\sin^2\alpha)} \; d\alpha\\ = & \int_0^\theta 5 \; d\alpha\\ = & \left. 5\theta \right|_0^\theta\\ = & 5\theta. \end{align*}
Now we can solve for arclength (\(s\)) with respect to the parameter. \(\theta=s/5\text{.}\) Our new parameterization is
\begin{equation*} C_u=(5\cos(s/5),5\sin(s/5),1)\text{.} \end{equation*}
Note the unit circle given by
\begin{equation*} C = (\cos \theta, \sin \theta, 1) \end{equation*}
is already parameterized with respect to arclength because a radian is a measure of arclength.
Arclength integrals can be quite complicated. In practice we may rely on software to evaluate the necessary integral for us.
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