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CLP-4 Vector Calculus

Joel Feldman, Andrew Rechnitzer, Elyse Yeager

Section 1.2 Reparametrization

There are invariably many ways to parametrize a given curve. Kind of trivially, one can always replace \(t\) by, for example, \(3u\text{.}\) But there are also more substantial ways to reparametrize curves. It often pays to tailor the parametrization used to the application of interest. For example, we shall see in the next couple of sections that many curve formulae simplify a lot when arc length is used as the parameter.
Here are three different parametrizations of the semi-circle \(x^2+y^2=r^2\text{,}\) \(y\ge 0\text{.}\)
  • The first uses the polar angle \(\theta\) as the parameter. We have already seen, in Example 1.0.1, the parametrization
    \begin{align*} &\vr_1(\theta) = \big(r\cos\theta\,,\,r\sin\theta\big)\\ &0\le \theta\le \pi \end{align*}
  • The second uses \(x\) as the parameter. Just solving \(x^2+y^2=r^2\text{,}\) \(y\ge 0\) for \(y\) as a function of \(x\text{,}\) gives \(y(x) = \sqrt{r^2-x^2}\) and so gives the parametrization
    \begin{align*} &\vr_2(x) = \big(x\,,\,\sqrt{r^2-x^2}\,\big)\\ &-r\le x\le r \end{align*}
  • The third uses arc length from \((r,0)\) as the parameter. We have seen, in Example 1.1.6, that the arc length from \((r,0)\) to \(\vr_1(\theta)\) is just \(s=r\theta\text{.}\) So the point on the semicircle that is arc length \(s\) away from \((r,0)\) is
    \begin{align*} &\vr_3(s) = \vr_1\Big(\frac{s}{r}\Big)\\ &= \Big(r\cos\frac{s}{r}\,,\,r\sin\frac{s}{r}\Big)\\ &0\le s\le \pi r \end{align*}
We shall see that, for some purposes, it is convenient to use parametrization by arc length. Here is a messier example in which we reparametrize a curve so as to use the arc length as the parameter.
We saw in Example 1.1.9, that, as \(t\) runs from \(0\) to \(\frac{\pi}{2}\text{,}\) \(\vr(t) = a \cos^3 t\,\hi+a\sin^3t\,\hj\) runs from \((a,0)\) to \((0,a)\) along the astroid \(x^{2/3}+y^{2/3}=a^{2/3}\text{.}\) Suppose that we want a new parametrization \(\vR(s)\) chosen so that, as \(s\) runs from \(0\) to some appropriate value, \(\vR(s)\) runs from \((a,0)\) to \((0,a)\) along \(x^{2/3}+y^{2/3}=a^{2/3}\text{,}\) with \(s\) being the arc length from \((a,0)\) to \(\vR(s)\) along \(x^{2/3}+y^{2/3}=a^{2/3}\text{.}\)
We saw, in Example 1.1.9, that, for \(0\le t\le \frac{\pi}{2}\text{,}\) \(\diff{s}{t}=\frac{3a}{2}\sin(2t)\) so that the arclength from \((a,0)=\vr(0)\) to \(\vr(t)\) is
\begin{equation*} s(t) = \int_0^t\frac{3a}{2}\sin(2t')\,\dee{t'} =\frac{3a}{4}\big[1-\cos(2t)\big] \end{equation*}
which runs from \(0\text{,}\) at \(t=0\text{,}\) to \(\frac{3a}{2}\text{,}\) at \(t=\frac{\pi}{2}\text{.}\) This is relatively clean and we can invert \(s(t)\) to find \(t\) as a function of \(s\text{.}\) The value, \(T(s)\text{,}\) of \(t\) that corresponds to any given \(0\le s\le\frac{3a}{2}\) is determined by
\begin{equation*} s=\frac{3a}{4}\big[1-\cos\big(2T(s)\big)\big]\qquad \iff\qquad T(s)=\frac{1}{2}\arccos\Big(1-\frac{4s}{3a}\Big) \end{equation*}
and
\begin{equation*} \vR(s) = \vr\big(T(s)\big) = a\cos^3\big(T(s)\big)\hi + a\sin^3 \big(T(s)\big)\hj \end{equation*}
We can simplify \(\cos^3\big(T(s)\big)\) and \(\sin^3\big(T(s)\big)\) by just using trig identities to convert the \(\cos\big(2T(s)\big)\) in \(s=\frac{3a}{4}\big[1-\cos\big(2T(s)\big)\big]\) into \(\cos\big(T(s)\big)\)'s and \(\sin\big(T(s)\big)\)'s.
\begin{align*} s=\frac{3a}{4}\big[1-\cos\big(2T(s)\big)\big] &=\frac{3a}{4}\big[1-\big\{2\cos^2\big(T(s)\big)-1\big\}\big]\\ &\iff \cos^2\big(T(s)\big)=1-\frac{2s}{3a}\\ s=\frac{3a}{4}\big[1-\cos\big(2T(s)\big)\big] &=\frac{3a}{4}\big[1-\big\{1-2\sin^2\big(T(s)\big)\big\}\big]\\ &\iff \sin^2\big(T(s)\big)=\frac{2s}{3a} \end{align*}
Consequently the desired parametrization is
\begin{equation*} \vR(s) = a\left[1-\frac{2s}{3a}\right]^{3/2}\hi + a\left[\frac{2s}{3a}\right]^{3/2}\hj \qquad 0\le s \le \frac{3a}{2} \end{equation*}
which is remarkably simple.

Exercises Exercises

Exercise Group.

Exercises — Stage 1
1.
A curve \(\vr(s)\) is parametrized in terms of arclength. What is \(\displaystyle\int_1^t |\vr'(s)|\,\dee{s}\) when \(t \ge 1\text{?}\)
2.
The function
\begin{equation*} \vr(s)=\sin\left(\frac{s+1}{2}\right)\hi+\cos\left(\frac{s+1}{2}\right)\hj+\frac{\sqrt3}{2}(s+1)\hk \end{equation*}
is parametrized in terms of arclength, starting from the point \(P\text{.}\) What is \(P\text{?}\)
3.
A curve \(\vR=\va(t)\) is reparametrized in terms of arclength as \(\vR=\vb(s)=\va(t(s))\text{.}\) Of the following options, which best describes the relationship between the vectors \(\va'(t_0)\) and \(\vb'(s_0)\text{,}\) where \(t(s_0)=t_0\text{?}\)
You may assume \(\va'(t)\) and \(\vb'(s)\) exist and are nonzero for all \(t,s\ge0\text{.}\)
  1. they are parallel and point in the same direction
  2. they are parallel and point in opposite directions
  3. they are perpendicular
  4. they have the same magnitude
  5. they are equal

Exercise Group.

Exercises — Stage 2
4. (✳).
  1. Let
    \begin{equation*} \vr(t) = (2 \sin^3 t , 2\cos^3 t, 3 \sin t \cos t) \end{equation*}
    Find the unit tangent vector to this parametrized curve at \(t = \pi/3\text{,}\) pointing in the direction of increasing \(t\text{.}\)
  2. Reparametrize the vector function \(\vr(t)\) from (a) with respect to arc length measured from the point \(t = 0\) in the direction of increasing \(t\text{.}\)
5. (✳).
This problem is about the logarithmic spiral in the plane
\begin{equation*} \vr(t) = e^t (\cos t, \sin t),\qquad t \in \bbbr \end{equation*}
  1. Find the arc length of the piece of this spiral which is contained in the unit circle.
  2. Reparametrize the logarithmic spiral with respect to arc length, measured from \(t = -\infty\text{.}\)

Exercise Group.

Exercises — Stage 3
6.
Define
\begin{equation*} \vr(t)=\left(\frac{1}{\sqrt{1+t^2}}, \frac{\arctan t}{\sqrt{1+t^{-2}}}, \arctan t\right) \end{equation*}
for \(0 \le t\text{.}\) Reparametrize the function using \(z=\arctan t\text{,}\) and describe the curve it defines. What is the geometric interpretation of the new parameter \(z\text{?}\)
7.
Reparametrize the function \(\vr(t)=(\tfrac12 t^2 , \tfrac13 t^3)\) in terms of arclength from \(t=-1\text{.}\)
Since we specified the derivatives are nonzero, there's no messiness about vectors being parallel to a zero vector.
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