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Course Notes for Calculus II

Remkes Kooistra

Section 6.2 Calculus on Parametric Curves

Subsection 6.2.1 Slopes

In the study of algebraic plane curves in Section 2.1, implicit derivatives were used to calculate the slope of loci and classify singularities where the slope was undefined. The calculus of loci was restricted to properties of the shape, such a slopes and singularities. In contrast, there is a rich calculus for parametric curves which calculates slopes as well as local directions of movement, speed of movement along the curve, and total distance travelled. Since there is a time variable, time derivatives will measure the movement of a point along the curve.
First, as with loci, I want to calculate \(\frac{dy}{dx}\text{:}\) the slope of the curve.

Definition 6.2.1.

If \(\gamma(t) = (x(t), y(t))\) is a parametric curve, then the slope of the parametric curve is given by the expression
\begin{equation*} \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \end{equation*}
If either of the derivatives are undefined, or if \(x^\prime(t) = 0\) at some value of \(t\text{,}\) the slope is undefined.
The situation for undefined slopes is similar but not exactly the same as singularities for algebraic plane curve. When \(x^\prime(t)\) approaches \(0\) but \(y^\prime(t)\) does not, the slope will approach infinity. In the limit, I can conclude that there will be a vertical tangent. This is very much the same as for loci, where an infinite limit in the implicit derivaitve indicates a vertical tangent. However, self-intersection of a parametric curve is no longer a problem, since the intersection will occur for different values of \(t\text{.}\) Each path through the self intersection will have its own slope. Finally, when the derivatives are not defined, the curve may have a sharp corner like the cusps for algebraic plane curves.

Example 6.2.2.

Figure 6.2.3. The Folium of Descartes
The folium of Descartes has the following parametric description.
\begin{align*} \gamma(t) = \left( \frac{3t}{1+t^3},\frac{3t^2}{1+t^3} \right) \amp \amp t \in (-\infty,-1) \cup (-1, \infty) \end{align*}
The domain for this curve comes in two disconnected parts, so I could think of this as two individual curves. When \(t \in (-\infty, -1)\text{,}\) I get the part of the curve below the \(x\) axis, and when \(t \in (-1, \infty)\text{,}\) I get the part of the curve above the \(x\) axis. Though it looks like the two parts connect in Figure 6.2.3, the lower part of the curve only approaches \((0,0)\) from below but never actually reaches it.
The folium is historically interesting because it was (as the name suggests) studied by Descartes. He was concerned, specifically, with the slope of the folium; this was a question he couldn’t answer with the mathematics of his day. This question was answered by Newton in the following decades, using the techniques of calculus. Let me follow in these steps and calculate the slope of the folium.
\begin{align*} \frac{dx}{dt} \amp = \frac{3(1+t^3) - 3t3t^2}{(1+t^3)^2}\\ \frac{dy}{dt} \amp = \frac{6t(1+t^3) - 3t^23t^2}{(1+t^3)^2}\\ \frac{dy}{dx} \amp = \frac{6t + 6t^4 - 9t^4}{3 + 3t^3 - 9t^3} = \frac{6t-3t^4}{3-6t^3} = \frac{2t-t^4}{1-2t^3} \end{align*}
This slope is undefined when \(t = \sqrt[3]{\frac{1}{2}}\text{.}\) I expect a vertical tangent at this point. In Figure 6.2.3, I can see the vertical tangent at the point on the loop with the largest \(x\) coordinate.

Example 6.2.4.

I will calculate the slope of the logarithmic spiral.
\begin{align*} \gamma(t) \amp = (e^{t} \cos t, e^{t} \sin t)\\ \frac{dx}{dt} \amp = e^{t} \cos t - e^{t} \sin t\\ \frac{dy}{dt} \amp = e^{t} \sin t + e^{t} \cos t\\ \frac{dy}{dx} \amp = \frac{e^{t} \sin t + e^{t} \cos t} {e^{t} \cos t - e^{t} \sin t} = \frac{\sin t + \cos t}{ \cos t - \sin t} \end{align*}
This slope undefined when \(\cos t = \sin t\) or \(1 = \tan t\text{.}\) This is a regular occurence, but that makes sense, since there are infinitely many locations on the curve where there is a vertical tangent. The slope is \(0\) similarly when \(1 = -\cot t\text{,}\) showing infinitely many places where there is a horizontal tangent.

Subsection 6.2.2 Velocity and Distance on a Parametric Curve

Slopes are common to both parametric curves and loci. Movement, however, is unique to parametric curve. I know what to describe the calculus of this movement, starting with velocity and distance. The variable \(s\) is conventionally used to represent distance along a curve. Velocity is the rate of change of distance, so it should be \(\frac{ds}{dt}\text{.}\) The following argument tries to construct a way to calculate this velocity.
The functions \(x(t)\) and \(y(t)\) represent movement of the two coordinates of the points on a curve. If there is a change in coordinates \(\Delta x\) and \(\Delta y\text{,}\) then the change in total distance is given by the pythagoreaon theorem: \(\Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2}\text{.}\) If these change in \(x\) and \(y\) happen over some fixed interval \(\Delta t\text{,}\) then I can divide the pythagorean combination by this \(\Delta t\text{.}\) On the right side, I can bring the \(\Delta t\) into the square root.
\begin{equation*} \frac{\Delta s}{\Delta t} = \sqrt{ \left( \frac{\Delta x}{\Delta t} \right)^2 + \left( \frac{\Delta y}{\Delta t} \right)^2} \end{equation*}
Then I can take the limit of this expression as \(\Delta t \rightarrow 0\text{.}\) Under the square root, the limit will produce the derivative of the two coordinate variables. On the left, it will produce the derivative of the new distance variable \(s\text{.}\)

Definition 6.2.5.

Let \(\gamma(t) = (x(t), y(t))\) be a parametric curve. The speed of the parametric curve is the derivative \(\frac{ds}{dt}\) and is calculated as follows.
\begin{equation*} \frac{ds}{dt} = \sqrt{\left( \frac{dx}{dt} \right)^2 + \left( \frac{dy}{dt} \right)^2} \end{equation*}
If I just differentiated \(\gamma(t)\) component-wise, I would get \(\gamma\prime(t) = (x^\prime(t), y^\prime(t))\text{.}\) Using this convention, the speed is the length of this derivative: \(\frac{ds}{dt} = |\gamma^\prime(t)|\text{.}\) This is a very natural language for describing derivatives of parametric curve, but it realies on a bit more vector algebra than I’ve established in this course. In Multivariable Calculus, the calculus of parametric curves (particular in three dimensions) is extended using this kind of vector algebra. For now, let me just calculate some examples of velocity of cuves using the expression in the definition above.

Example 6.2.6.

The curve \(\gamma(t) = (t,t)\) has derivatives \(x^\prime(t) = 1\) and \(y^\prime(t) = 1\text{.}\) Therefore, \(\frac{ds}{dt} = \sqrt{1+1} = \sqrt{2}\text{;}\) the curve has a constant speeed of \(\sqrt{2}\) units of distance per unit of time. In each unit of time, the curve increases from \((0,0)\) to \((1,1)\text{,}\) then to \((2,2)\text{,}\) then to \((3,3)\) and so ong. The distance between any two consecutive points is \(\sqrt{2}\text{,}\) so the curve does cover \(\sqrt{2}\) units of distance per unit of time. It makes sense that the speed is constant.

Example 6.2.7.

The curve \(\gamma(t) = (t^2,t^2)\) traces the same line. It has derivatives \(x^\prime(t) = 2t\) and \(y^\prime(t) = 2t\text{,}\) so the speed is \(\frac{ds}{dt} = \sqrt{4t^2 + 4t^2} = 2\sqrt{2} t\text{.}\) Though it travels the same line as the previous example, it accelerates as it travels along that line.

Example 6.2.8.

The curve \(\gamma(t) = \left(\frac{1}{t}, \frac{1}{t} \right)\) again traces the same line, though in the opposite direction. It’s derivatives are \(x^\prime(t) = \frac{-1}{t^2}\) and \(y^\prime(t) = \frac{-1}{t^2}\text{.}\) The speed is \(\frac{ds}{dt} = \sqrt{\frac{1}{t^4} + \frac{1}{t^4}} = \frac{\sqrt{2}}{t^2}\text{,}\) which gets slower and slower. This makes sense, since it only approaches the origin as \(t \rightarrow \infty\text{.}\) In that approach, it gets slower and slower so that it never actually reaches the origin.

Example 6.2.9.

Recall the cycloid: \(\gamma(t) = (a(t-\sin t), a(1-\cos t))\text{.}\) It has derivatives \(\frac{dx}{dt} = a - a \cos t\) and \(\frac{dy}{dt} = a \sin t\text{.}\) Here is the velocity calculation.
\begin{align*} \frac{ds}{dt} \amp = \sqrt{(a-a\cos t)^2 + a^2 \sin^2 t}\\ \amp = \sqrt{a^2 - 2a^2\cos t + a^2 \cos^2 t + a ^2 \sin^2 t}\\ \amp = \sqrt{a^2 - 2a^2 \cos t + a^2} = a\sqrt{2(1-\cos t)} \end{align*}
Notice that \(1-\cos t\) is always non-negative, so the square root is well defined. Curiously, at \(t=0\text{,}\) \(t=2\pi\) and any other multiple of \(2\pi\text{,}\) the velocity is momentarily zero.
The cycloid is the path of a point on the edge of a wheel as the wheel rolls. When \(t\) is a multiple of \(2\pi\text{,}\) the point on the wheel is momentarily in contact with the ground and its speed is momentarily zero. That means that on a rolling wheel, the point touching the ground at any instant is momentarily stationary. This is something to ponder the next time you watch a train or other fast moving wheeled vehicle: at every instant in time, there is at least one point on each wheel which isn’t moving at all.

Subsection 6.2.3 Arclength

In the previous section, I defined \(\frac{ds}{dt}\) to be the speed of the curve. To get distance travelled, I need to integrate speed.

Definition 6.2.10.

The distance travelled along a parametric curve, as the parameter goes from \(a\) to \(b\text{,}\) is the integral of speed.
\begin{equation*} L = \int_a^b \frac{ds}{dt} dt = \int_a^b \sqrt{\frac{dx}{dt}^2 +\frac{dy}{dt}^2}dt \end{equation*}
The distance travelled along a parametric curve is typically called the arclength.
I’ll go directly into examples of arclength calculations.

Example 6.2.11.

I’ll start with the circle: \(\gamma(t) = (\cos t, \sin t)\) for \(t \in [0,2\pi]\text{.}\) Its length is calculated by the arclength integral. The derivatives are \(x^\prime(t) = -\sin t\) and \(y^\prime(t) = \cos t\text{.}\)
\begin{equation*} L = \int_0^{2\pi} \sqrt{ \sin^2 t+ \cos^2 t}dt = \int_0^{2\pi} dt = 2\pi \end{equation*}
Unsurprisingly, the arclength of the circle is its circumference: \(2\pi\text{.}\) Let me check another parameterization to make sure it also recovers this arclength. Take \(\gamma(t) = (\cos 3t, \sin 3t)\) for \(t \in [0, 2\pi/3]\text{.}\) The derivatives are \(x^\prime(t) = -3\cos 3t\) and \(y^\prime = 3 \cos 3\text{.}\)
\begin{equation*} L = \int_0^{\frac{2\pi}{3}} \sqrt{ 9\sin^2 3t+ 9\cos^2 3t}dt = \int_0^{\frac{2\pi}{3}} 3 dt = \frac{2\pi}{3} 3 = 2\pi \end{equation*}
The alternate parametrization still calculates the correct circumference.

Example 6.2.12.

I’ll return to the cycloid \(\gamma(t) = (t-\sin t, 1-\cos t)\) for \(t \in [0,2\pi]\text{.}\) The derivatives are \(x^\prime = 1-\cos t\) and \(y^\prime = \sin t\text{.}\) Here is the arclength calculation.
\begin{align*} L \amp = \int_0^{2\pi} \sqrt{(x^\prime)^2 + (y^\prime)^2 } dt\\ \amp = \int_0^{2\pi} \sqrt{1 - 2\cos t + \cos^2 t + \sin^2 t} dt\\ \amp = \int_0^{2\pi} \sqrt{2 - 2\cos t} dt = \sqrt{2} \int_0^{2\pi} \sqrt{1 - \cos t} dt \end{align*}
I’ll use a half-angle identity to deal with \(\sqrt{1 - \cos t}\text{.}\) Since \(\sin^2 t = \frac{1 - \cos 2t}{2}\text{,}\) I multiply by two to scale the angle, I can get \(2 \sin^2 \left(\frac{t}{2} \right) = 1 - \cos t\text{.}\) I’ll replace the inside of the square root with this expression.
\begin{align*} \amp = \sqrt{2} \int_0^{2\pi} \sqrt{2} \sqrt{\sin^2 \frac{t}{2}} dt = 2 \int_0^{2\pi} \left| \sin \frac{t}{2} \right| dt \end{align*}
Since \(\sin \frac{t}{2}\) has period \(4\pi\text{,}\) over the integration range \([0, 2\pi]\) this sine function only covers the first half of a period. Therefore, it is positive and I can drop the absolute value.
\begin{align*} \amp = 2 \int_0^{2\pi} \sin \frac{t}{2} dt = (2)(2) \left. \left( - \cos \frac{t}{2} \right) \right|_0^{2\pi} = 4( \cos 0 - \cos \pi) = 4 \cdot 2 = 8 \end{align*}

Example 6.2.13.

An interesting example is calculating the perimeter of an ellipse. It’s an impotant calculation historically, since it can determine the length of elliptical orbits (among other applications). Assume that the ellipse is centred at the origin and positioned so its largest semi-axis is along the \(x\) axis. Let \(a\) be the semi-major axis and \(b\) be the semi-minor. Then \(\gamma(t) = (a \cos t, b \sin t)\) for \(t \in [0,2\pi]\) describes the ellipse as a parametric curve.
Before calculating, it is convenient to also define a new quantity: the eccentricity of the ellipse. This is defined as \(e = \frac{\sqrt{a^2-b^2}}{a}\) (assuming that \(a\geq b\)) and it always a number in \([0,1)\text{.}\) Eccentricity is a nice way of measuring how close an ellipse is to a circle. If \(e = 0\) then \(a=b\) and the ellipse is exactly a circle. As \(e \rightarrow 1\) the ellipse becomes less circular and more elongated. With the eccentricity defined, I’ll try to calculate the arclength of the parametric curve the circumference of the ellipse. (This \(e\) is not to be confused with the notation for the exponential function. Sorry to re-use a letter that already has an important purpose, but the use of \(e\) for eccentricity here is conventional and, in this integral at least, I can avoid any exponential functions that would confuse the issue.)
\begin{align*} L \amp = \int_0^{2\pi} \sqrt{ (x^\prime)^2 + (y^\prime)^2} dt\\ \amp = \int_0^{2\pi} \sqrt{ a^2 \sin^2 t + b^2 \cos^2 t} dt\\ \amp = \int_0^{2\pi} \sqrt{a^2 \sin^2 t + (b^2 - a^2) \cos^2 t + a^2 \cos^2 t} dt\\ \amp = \int_0^{2\pi} a\sqrt{1 + \frac{b^2-a^2}{a^2} \cos^2 t} dt\\ \amp = a \int_0^{2\pi} \sqrt{1 - \frac{a^2 - b^2}{a^2} \cos^2 t} dt \end{align*}
The expression before the sine term here is exactly the square of eccentricity, so I’ll replace it with \(e\text{.}\)
\begin{equation*} = a \int_0^{2\pi} \sqrt{ 1 - e^2 \cos^2 t}dt \end{equation*}
If \(e=0\text{,}\) then the cosine disappears and I have an easy integral. In this form, however, the integration is very difficult. This is so difficult, in fact, that it has a special name: this is an elliptic integral of the second kind. These integrals have been studied for three hundred years, and with good cause, since they have no elementary anti-derivatives. Even without an elementary anti-derivative, however, the behaviour can be investigated. This has led to many insights in geometry; elliptic curves (a type of algebraic plane curve) come from the long process of trying to understand this integral.

Subsection 6.2.4 A Substitution Rule Lemma

Before moving on with the discussion of parametric curves, I’m going to interupt for a short section to prove a lemma about the substitution rule for integration. (A ‘lemma’ in mathematic is a short statement, usually used a part of a larger proof). I’m going to need this lemma in the next section, but I don’t want to interupt the narrative flow of that section, so I’m going to prove the lemma in advance.

Proof.

To prove this, I’ll calculate the derviative of the composition \(f(g(u))\text{.}\)
\begin{equation*} \frac{d}{du} f(g(u)) = f^\prime(g(u)) \frac{dg}{du} = f^\prime(t) \frac{dg}{du} \end{equation*}
I replaced \(g(u)\) with \(t\) in the first term on the right, since the substitution is, indeed, \(t = g(u)\text{.}\) Then I’ll solve for the first term on the right.
\begin{equation*} f^\prime(t) = \frac{d}{du} f(g(u)) \frac{1}{\frac{dg}{du}} \end{equation*}
This is the desired transformation, completing the proof.

Subsection 6.2.5 Independence of Parametrization

In Example 6.2.11, I looked at a couple different parametrizations of the circle and checked the calculation to make sure that both parametrizations produced the correct circumference. This is a special case of a much more general issue. Parametric curves come with parametrizations and different parametrization can describe the same shape. Some notions, like speed, should naturally depend on the parametrizations. However, other notions like total distance should be intrinsic to the shape, not the rate of movement along the shape. With parametric objects, I have to distinguish between that which depends on the parameter (like speed) and that which is intrinsic to the shape (like arclength). For the latter, I have to make sure that if I use a parametrization to calculate an intrinsic quantity, the result is independent of the parametrization. For now, I’m concerned just with arclength.

Proof.

Let me start by writing the arclength integral in the parameter \(t\)
\begin{equation*} L = \int_{a}^{b} \sqrt{ \frac{dx}{dt}^2 + \frac{dy}{dt}^2} dt \end{equation*}
Now I will apply the substitution \(t = g(u)\text{.}\) The differential changes as \(dt = g^\prime(u) du \text{.}\) For the two derivatives inside the square root, I can apply Lemma 6.2.14 to both. This lets me do the entire substitution.
\begin{equation*} = \int_{g^{-1}(a)}^{g^{-1}(b)} \sqrt{ \left( \frac{dx(g(u))}{du} \frac{1}{\frac{dg}{du}} \right)^2 +\left( \frac{dy(g(u))}{du} \frac{1}{\frac{dg}{du}} \right)^2} \frac{dg}{du} du \end{equation*}
The denominator in both of the terms in the square root is the same, so I can make this a commmon denominator. Then since the denominator is squared, I can take it out of the square root.
\begin{align*} \amp= \int_{g^{-1}(a)}^{g^{-1}(b)} \sqrt{ \frac{ \left( \frac{dx(g(u))}{du} \right)^2 + \left( \frac{dy(g(u))}{du} \right)^2 }{\frac{dg}{du}}} \frac{dg}{du} du \\ \amp = \int_{g^{-1}(a)}^{g^{-1}(b)} \sqrt{ \left( \frac{dx(g(u))}{du} \right)^2 + \left( \frac{dy(g(u))}{du} \right)^2} \frac{1}{{\frac{dg}{du}}} \frac{dg}{du} du \end{align*}
Then the two derivatives outside the square root simply cancel.
\begin{equation*} = \int_{g^{-1}(a)}^{g^{-1}(b)} \sqrt{ \left( \frac{dx(g(u))}{du} \right)^2 + \left( \frac{dy(g(u))}{du} \right)^2} du \end{equation*}
This is precisely the expression for arclength calculation in the parameter \(u\text{.}\) I started with the expression for arclength in \(t\) and ended with the expression for arclength in \(u\text{;}\) I conclude that the arclength is the same regardless of which parameter is used to calculate it.
I have proved that arclength is independent of parametrization. There are other properties of the shape of a parametric curve (such as curvature) that will be introduced in Multivariable Calculus; all of these geometric properties will also be independent of paramtrization.
Even though these properties are independent of parametrization, there is no way to calculate them without using a parametrization. This is a surprisingly common situation in mathematics. Many objects are defined in a way that seems to involve an arbitrary choice, such as a parametrization. (As I said earlier, each shape has infinitely many parametrization.) To work with the object, a choice must be made. But mathematicians want the properties to be independent of the choice, which requires careful and often difficult proofs.
Ideally, there would be a special or privileged choice, to make it easier to access the properties of the object. For parametric curves, there is such a thing. It will be defined in the next section.

Subsection 6.2.6 Parametrization by Arclength

Definition 6.2.16.

The parametrization by arclength is the parametrization in the special variable \(s\) where \(\frac{ds}{dt} = 1\text{,}\) i.e., the unique parametrization where movement along the curve always has a speed of one unit of distance per unit of time. The starting point of the curve is at \(s = 0\) and the range of \(s\) is \(s \in [0, L]\) where \(L\) is the length of the curve. The use of the variable \(s\) for parametrization by arclength is conventional; I will only use \(s\) as a parameter when I want to indicate the parametrization by arclength (as oppsed to the use of \(t\) and \(u\) as paramaters in various calculations so far).
The parameter \(s\) is simply the distance travelled along the curve. If the movement always moves one unit of distance per unit of time, then after \(s\) units of time, the movement has covered \(s\) units of distance, for all choices of the parameter \(s\text{.}\) Therefore, it is approprate to treat the parameter \(s\) as the distance covered.
Now that it’s defined, I would like an algorithm to construct the parametrization by arclength. In order to do this, I need a new function: the arclength function.

Definition 6.2.17.

Let \(\gamma(t)\) be a parametric curve for \(t \in [a,b]\text{.}\) The the function defined by the following integral is called the arclength function of the parametric curve.
\begin{equation*} s(t) = \int_a^t \sqrt{(x^\prime(u))^2 + (y^\prime(u))^2} du \end{equation*}
The domain of \(s(t)\) is \(t \in [a,b]\text{,}\) the parameter domain for the curve. The range of \(s\) is \([0,L]\text{,}\) since when \(t=a\text{,}\) the integral has no interval to integrate over, and when \(t = b\text{,}\) the integral just gives the entire length of the curve. Since the parameter variable \(t\) is outside the integral, I needed to defined the temporary variable \(u\) to do the integration. (The same variable should never show up as a variable of integration and a variable existing outside the intergral.) The arclength function measures how much distance the curve has covered as the parameter goes from the start \(a\) to \(t\text{.}\)
The arclength function is guaranteed to be an increasing function, since it is an integral of a positive function. Therefore it is invertible. Typically, the inverse is written \(t(s)\) where \(s \in [0, L]\text{.}\) Then I can substitute \(t(s)\) for \(t\) to get \(\gamma(t(s))\text{,}\) a reparametrization.
What has this reparametrization accomplished? I’ve turned the arclength function into the parameter. Therefore, this is the parametrization by arclength. It is the unique parametrization with speed one, where the parameter and length along the curve are always the same. Let me summarize the algorithm.
  1. Calculate the arclength function \(s(t)\text{.}\)
  2. Invert the arclenth function to get \(t(s)\text{.}\)
  3. Replace \(t\) in the definition of the curve with \(t(s)\) to produce the parametrization by arclength.

Example 6.2.18.

Consider the circle of radius \(4\) which is described by this parametric curve: \(\gamma(t) = (4 \cos t, 4 \sin t)\) for \(t \in [0, 2\pi]\text{.}\) I’d like to parametrize this curve by by arc-length, so I step through the the algorithm. First I calcualte the arclength function.
\begin{align*} s(t) \amp = \int_0^t \sqrt{(x^\prime(u))^2 + (y^\prime(u))^2} du\\ \amp = \int_0^t \sqrt{16 \sin^2 u + 16 \cos^2 u} du\\ \amp = \int_0^t 4 du\\ s(t) \amp = 4t \end{align*}
This is a reasonable function to invert. The inverse is \(t(s) = \frac{s}{4}\text{.}\) The starting value for the parameter \(s\) is \(0\text{,}\) as always. The ending value is the length of the whole curve. That is \(s(2\pi) = 4(2\pi) = 8\pi\text{,}\) which is the circumference of the circle of radius \(4\text{.}\) Then I can replace \(t\) in the original curve to write the parametrization by arclength.
\begin{align*} \gamma(s) \amp = \left(4 \cos \frac{s}{4}, 4 \sin \frac{s}{4} \right) \amp \amp t \in [0,8\pi] \end{align*}
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