Section 7.2 Calculus of Parametric Curves
References.
- Calculus, Early Transcendentals by Stewart, Section 10.2.
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{7.2.1}
\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 \(x\text{!}\)
\(\)
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 7.2.2. 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 7.1 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*}
\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.
Checkpoint 7.2.3. 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{7.2.2}
\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{7.2.3}
\end{equation}
As always, note how the limits of integration change when the dummy variable is changed by substitution!
Checkpoint 7.2.4.
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 arc-length in Section 2.4 can also be converted when \(y=f(x)\) also has a parametric form, using the same substitution \(x=F(t)\) as for areas:
\begin{equation*}
L = \int_{x=a}^{b} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} dx
= \int_{t=\alpha}^{\beta} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \frac{dx}{dt} \, dt
= \int_{t=\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dx}\frac{dx}{dt}\right)^2} \, dt
= \int_{t=\alpha}^{\beta} \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} \, dt
\end{equation*}
As in Chapter 2, 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{7.2.4}
\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 as in Section 2.4 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{7.2.5}
\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.
Note that when \(x=t\)—so the curve is just the graph of \(y=g(x)\)— this gives the same formula as in Equation (2.4.1) of Section 2.4. So if you only learn one arc length formula, it should be the one here!
Checkpoint 7.2.5.
Compute the circumference of a circle or radius \(R\text{,}\) parameterized as \((x, y) = (R \cos(t), R\sin(t))\)
Checkpoint 7.2.6.
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
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{7.2.6}
\end{equation}
Checkpoint 7.2.7.
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 2, Section 7.2 3 ; 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
openstax.org/books/calculus-volume-2/pages/7-2-calculus-of-parametric-curvesopenstax.org/books/calculus-volume-2/pages/7-2-calculus-of-parametric-curvesopenstax.org/books/calculus-volume-2/pages/7-2-calculus-of-parametric-curves
