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Calculus 3 (Math 221) Notes and Study Guide

Brenton LeMesurier

Section 1.2 Calculus of Parametric Curves

References.

Derivatives of Parametric Equations.

Often we are interested in the slope of a parameteric curve \((x, y) = (F(t), G(t))\text{,}\) meaning intuitively \(m = dy/dx\text{,}\) but we do not have an explicit formula \(y = f(x)\text{.}\) Fortunately, rather than having to solve for \(f\) by eliminating th parameter \(t\text{,}\) the needed derivative and slope can be computed by implicit differentiation: if \(y = G(t) = f(x) = f(F(t))\) then
\begin{equation*} \frac{dy}{dt} = \frac{dG}{dt} = f'(F(t))F'(t) = f'(x)F'(t) = \frac{dy}{dx}\frac{dx}{dt}; \end{equation*}
which is the familiar intuitive pattern of the Chain Rule. This can than be solved to get
\begin{equation} \frac{dy}{dx} = \frac{{dy}/{dt}}{{dx}/{dt}},\tag{1.2.1} \end{equation}
so long as the denominator \(dx/dt\) is non-zero.
Note that this will be a formulas in terms of \(t\text{,}\) not \(x\text{!}\)

Example 1.2.1. Slope of the spiral \((x,y) = (t\cos(t), t \sin(t))\).

\(\)
First,
\begin{equation*} dx/dt = \cos(t) - t \sin(t) \text{ and } dy/dt = \sin(t) + t \cos(t), \end{equation*}
so the slope at the point given by any value of the parameter \(t\) is given by
\begin{equation*} \frac{dy}{dx} = \frac{\sin(t) + t \cos(t)}{\cos(t) - t \sin(t)} \end{equation*}

Example 1.2.2. Slope of the cycloid \((x,y) = (t - \sin(t), 1 - \cos(t))\).

\(\)
\(dx/dt = 1 - \cos(t)\) and \(dy/dt = \sin(t)\text{,}\) so the slope at the point given by any value of the parameter \(t\) is given by
\begin{equation*} \frac{dy}{dx} = \frac{\sin(t)}{1-\cos(t)}, \end{equation*}
except where the denominator is zero; that happens when \(\cos(t) = 1\text{,}\) which is for \(t\) an integer multiple of \(2\pi\text{;}\) \(t = 2 n \pi\text{,}\) where also \(\sin(t) = 0\text{.}\)
This is the points \((x,y) = (2 n \pi, 0)\) where the cycloid “touches down”, and where the graphs done in Section 1.1 suggested that something strange was happening.

Second-Order Derivatives.

Once one has a formula for the first derivative \(dy/dx\) (albeit in terms of \(t\)), computing second and higher derivatives is relatively strightforward; no furhter implicit differentition is needed. First,
\begin{equation*} \frac{d^2y}{dx^2} = \frac{d}{dx}\left[ \frac{dy}{dx} \right] \end{equation*}
Next, use the above Equation (1.2.1) with \(y\) replaced by \(dy/dx\text{:}\)
\begin{equation*} \frac{d^2y}{dx^2} = \frac{d(dy/dx)/dt}{dx/dt} \end{equation*}
Since the method above gives \(dy/dx\) as a function of \(t\text{,}\) we can evaluate the two derivatives here.

Exercise 1.2.3. The concavity of the cycloid.

Compute the second derivative \(\ds \frac{d^2y}{dx^2}\) of the above cycloid.
Graphs suggest that this curve is always concave down, so check that.

Integrals Involving Parametric Equations: Area Under a Curve.

If a curve \(y=f(x)\text{,}\) \(a \leq x \leq b\) also has a parametric form \(x=F(t)\text{,}\) \(y=G(t)\text{,}\) \(\alpha \leq t \leq \beta\text{,}\) with \(f\) an increasing function and \(y=g(t) \geq 0\text{,}\) then it lies over a region \(a \leq x \leq b\) with \(x(\alpha)=a\text{,}\) \(x(\beta)=b\text{,}\) and it makes sense to talk of the area between this curve and the \(x\)-axis.
If we could eliminate the parameter and describe the curve as \(y=F(x)\text{,}\) this area would be \(A=\int_a^b F(x) \, dx\text{,}\) but in fact, we do not need to get an explicit formulas for \(F(x)\text{!}\) Instead, use the (inverse) substitution \(x=f(t)\) to get
\begin{equation} A = \int_{x=a}^{b} y \, dx = \int_{x=a}^{b} f(x) \, dx = \int_{t=\alpha}^{\beta} f(x(t)) \frac{dx}{dt} dt = \int_{t=\alpha}^{\beta} f(x(t)) F'(t) dt = \int_{t=\alpha}^{\beta} y \frac{dx}{dt} dt.\tag{1.2.2} \end{equation}
where we use the fact that \(y=F(x(t))\) and also \(y=g(t)\text{.}\)
That is, we get the intuitive change of variables
\begin{equation} A = \int_{x=a}^{b} y \, dx = \int_{t=\alpha}^{\beta} y \frac{dx}{dt} dt.\tag{1.2.3} \end{equation}
As always, note how the limits of integration change when the dummy variable is changed by substitution!

Exercise 1.2.4.

Compute the area under one arch of the cycloid \((x, y) = (a(t-\sin(t)), a(1 - \cos(t)))\)

Arc Length of a Parametric Curve.

The formula for the arc length of a curve \(y=f(x)\) can be converted to parametric form when \(x=F(t)\text{,}\) as was done for areas:
\begin{align} L \amp= \int_{x=a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx\notag\\ \amp= \int_{t=\alpha}^{\beta} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \frac{dx}{dt}\ dt\notag\\ \amp= \int_{t=\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dx}\frac{dx}{dt}\right)^2}\ dt\notag\\ \amp= \int_{t=\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2}\ dt\tag{1.2.4} \end{align}
A good intuitive way to see this is that each “infinitesimally” small piece of the curve has length
\begin{equation} ds = \sqrt{(dx)^2 + (dy)^2} = \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\tag{1.2.5} \end{equation}
sometimes called the arc length differential; the arc length is then the “sum” or integral of these infinitesimal lengths: \(\ds L= \int ds.\)
This idea can be used to show that in fact for any parametric curve with \(f'(t)\) and \(g'(t)\) continuous for \(\alpha \leq t \leq \beta\text{,}\) the arc length is, as above,
\begin{equation} L = \int ds = \int_{t=\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\tag{1.2.6} \end{equation}
The curve does not have to be in the form of the graph of a function, and in particular, \(x\) need not be an increasing function of the parameter.

Exercise 1.2.5.

Compute the circumference of a circle or radius \(R\text{,}\) parameterized as \((x, y) = (R \cos(t), R\sin(t))\)

Exercise 1.2.6.

Compute the length of one arch of the cycloid \((x, y) = (a(t-\sin(t)), a(1 - \cos(t)))\)

Surface Area Generated by a Parametric Curve (Omitted).

This topic is not covered in this course, but I include this brief introduction; it is discussed further in Section 7.2 of the OpenStax Calculus text.
 2 
openstax.org/books/calculus-volume-2/pages/7-2-calculus-of-parametric-curves
The area of the surface produced by rotating a parametric curve about the \(x\)-axis can be computed, and the most intuitive way to see the result is to work with a surface area differential \(dS\text{,}\) much as the arc length differential \(ds\) was used above.
When an infinitesimal part of the parametric curve \(x=F(t),\, y=G(t)\) of arc length \(ds\) is rotated about the \(x\)-axis, it produces an angled strip of width \(ds\text{,}\) radius \(y\text{,}\) circumference \(2 \pi y\text{,}\) and thus with infinitesimal area given by the surface area differential \(dS = 2 \pi y \, ds\text{.}\)
Thus, with appropriate limits of integration, the surface area is
\begin{equation} S = \int dS = \int 2 \pi y \, ds = \int_{t=\alpha}^b 2\pi y \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt\tag{1.2.7} \end{equation}

Exercise 1.2.7.

Compute the area of the “football shaped” surface produced by rotating one arch of the cycloid \((x, y) = (a(t-\sin(t)), a(1 - \cos(t)))\) about the \(x\)-axis.

Study Guide.

Study Calculus Volume 3, Section 1.2
 3 
openstax.org/books/calculus-volume-3/pages/1-2-calculus-of-parametric-curves
; in particular
  • Theorems 1 2, 3
  • Examples 4, 5, 6, 7, 8
  • Checkpoints 4, 5, 6, 7, 8
  • and one or several exercises from each of the following groupls: 62–65, 66–70, 71–74, 75–77, 88–90, 104–107 (areas under curves), 108–112 (arclengths: no need to evaluate for 112, just setup the integral).
Note that we omit the final topic of Surface Area Generated by a Parametric Curve
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