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Workbook for Stewart’s "Calculus"

Cory Wilson

Section 10.2 Calculus with Parametric Curves

Subsection 10.2.1 Before Class

Figure 72. Pre-Class Video 1
Figure 73. Pre-Class Video 2

Subsubsection 10.2.1.1 Tangents

Derivative of a Parametric Curve.
If \(x=f(t)\) and \(y = g(t)\) are the parametric equations for a curve \(C\text{,}\) then the derivative \(\dfrac{dy}{dx}\) is given by
\begin{equation*} \dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt} \end{equation*}
provided that \(dx/dt \neq 0\)
From the chain rule,
\begin{equation*} \dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot \dfrac{dx}{dt} \end{equation*}
Rearranging gives the desired quantity.
Example 10.2.1.
For the circle \(x = \cos t\text{,}\) \(y = \sin t\text{,}\) what is the rate of change when \(\theta = \dfrac{\pi}{3}\text{?}\)
Solution.
\(\dfrac{dy}{dx} = -\dfrac{1}{\sqrt{3}}\)
Example 10.2.2.
What is the general formula for the rate of change of an ellipse, whose parametrization is given by \(x= a\cos t\text{,}\) \(y = b\sin t\) (\(0\leq t\leq 2\pi\))?
Solution.
\(\dfrac{dy}{dx} = -\dfrac{b}{a}\cot t\)
Example 10.2.3.
Find an equation for the tangent line to the curve \(x = t^3 + 1\text{,}\) \(y = t^4 + t\) at the point corresponding to the parameter value \(t=-1\text{.}\)
Solution.
\(y = -\dfrac{4}{3}x\)
Second Derivative of a Parametric Curve.
If \(x = f(t)\) and \(y = g(t)\) are the parametric equations for a curve \(C\) with derivative \(\dfrac{dy}{dx}\text{,}\) then the second derivative \(\dfrac{d^2y}{dx^2}\) is given by
\begin{equation*} \dfrac{d^2y}{dx^2} = \dfrac{\dfrac{d}{dt}(dy/dx)}{dx/dt} \end{equation*}
Provided that \(dx/dt \neq 0\)
\begin{align*} \dfrac{d^2y}{dx^2} \amp = \dfrac{d}{dx}\lrpar{\dfrac{dy}{dx}} \\ \amp = \dfrac{\dfrac{d}{dt}(dy/dx)}{dx/dt} \end{align*}
Example 10.2.4.
Find the value of the second derivative for the circle \(x = \cos t\text{,}\) \(y=\sin t\) when \(\theta = \dfrac{\pi}{3}\)
Solution.
\(\dfrac{d^2y}{dx^2}\bigg\rvert_{t = \pi/3} = \dfrac{8}{3\sqrt{3}}\)
Example 10.2.5.
Let \(C\) be a curve defined by the parametric equations \(x = 2t^2\text{,}\) \(y = t^3-t\text{.}\)
  1. Show that \(C\) has two tangents at the point \((2,0)\) and find their equations
  2. Find the points on \(C\) where the tangent is either horizontal or vertical
  3. Determine when the curve is concave up or concave down
  4. Sketch the curve using the information above
Solution.
  1. At \(t = \pm 1\text{,}\) the curve passes through \((2,0)\) so we know that the curve has up to two tangent lines at that point. Then, \(\dfrac{dy}{dx} = \dfrac{3t^2-1}{4t}\text{;}\) evaluating at \(t = 1\) gives \(\dfrac{dy}{dx} = \dfrac{1}{2}\) and at \(t = -1\) gives \(\dfrac{dy}{dx} = -\dfrac{1}{2}\text{.}\)
    This gives the tangent lines \(y = \dfrac{1}{2}(x-2)\) and \(y = -\dfrac{1}{2}(x-2)\)
  2. The tangent line is vertical when \(4t = 0\text{,}\) so when \(t = 0\text{.}\) This occurs at \((0,0)\text{.}\)
    The tangent line is horizontal when \(3t^2-1 = 0\text{,}\) or at \(t = \pm \dfrac{1}{\sqrt{3}}\text{.}\) This gives two points, \(\lrpar{\dfrac{2}{3}, - \dfrac{2}{3\sqrt{3}}}\) and \(\lrpar{-\dfrac{2}{3}, - \dfrac{2}{3\sqrt{3}}}\)
  3. Since \(\dfrac{d^2y}{dx^2} = \dfrac{6t}{4t} = \dfrac{3}{2}\text{,}\) the curve is always concave up.
  4. A sketch of the parametric curve on the interval \([-3,3]\)

Subsection 10.2.2 Pre-Class Activities

Example 10.2.6.

For the curve defined parametrically by \(x = 1 + \sqrt{t}\text{,}\) \(y = e^{t^2}\text{,}\) find an equation of the tangent line to the curve at the point \((2,e)\text{.}\) Then, eliminate the parameter to find a Cartesian expression for the curve.
Solution.
\(y - e = 4e(x-2)\) and \(y = e^{(x-1)^2}\)

Example 10.2.7.

For the following functions, find the first and second derivative.
  1. \(x = t^3 + 1\text{,}\) \(y = t^2-t\)
  2. \(x = t^2-1\text{,}\) \(y = e^t-1\)
  3. \(x = \cos 2t\text{,}\) \(y = \sin t\text{,}\) \(0\lt t \lt \pi\)
Solution.
  1. \(\dfrac{dy}{dx} = \dfrac{2t-1}{3t^2}\text{,}\) \(\dfrac{d^2y}{dx^2} = \dfrac{-6t^2 + 6t}{27t^6}\)
  2. \(\dfrac{dy}{dx} = \dfrac{e^t}{2t}\text{,}\) \(\dfrac{d^2y}{dx^2} = \dfrac{2te^t - 2e^t}{8t^3}\)
  3. \(\dfrac{dy}{dx} = -\dfrac{1}{4}\csc t\text{,}\) \(\dfrac{d^2y}{dx^2} = \dfrac{\dfrac{1}{4}\csc t\cot t}{-2\sin 2t}\)

Subsection 10.2.3 In Class

Example 10.2.8.

When a circle rolls on a flat surface, a fixed point on the circle will trace out a curve called a cycloid. The parametrization for a cycloid is given by \(x = r(\theta - \sin\theta)\text{,}\) \(y = r(1-\cos\theta)\text{,}\) where \(r\) is the radius of the circle.
  1. Does the value of the tangent depend on the radius of the circle?
  2. Compute the slope of the tangent line when \(\theta = \dfrac{\pi}{6}\text{.}\)
  3. At what points is the tangent horizontal? What about when it’s vertical?
Solution.
  1. No- \(\dfrac{dy}{dx} = \dfrac{\sin\theta}{1-\cos\theta}\text{,}\) which is independent of \(r\)
  2. Slope is \(dfrac{1/2}{1-\sqrt{3}/2}\)
  3. Tangent is vertical when \(\theta = (2k-1)\pi\) for \(k\in \Z\)and vertical for \(\theta = 2k\pi\) for \(k\in \Z\)

Example 10.2.9.

At what point(s) on the curve \(x = 3t^2 + 1\text{,}\) \(y = t^3-1\) does the tangent line have slope exactly \(\dfrac{1}{2}\text{?}\)
Solution.
\(t = 1\)

Subsubsection 10.2.3.1 Areas

Area Under a Parametric Curve.
Let \(C\) be a curve traved out exactly once by the parametric equation \(x = f(t)\) and \(y = g(t)\text{.}\) Then, the area under \(C\) between \(x = a\) and \(x = b\) is given by
\begin{equation*} A = \ds \int_a^b y\, dx = \int_\alpha^\beta g(t)f'(t)\, dt \end{equation*}
or
\begin{equation*} A = \ds \int_a^b x\, dy = \int_\alpha^\beta f(t)g'(t)\, dt \text{ as a function of }y \end{equation*}
where \(\alpha \leq t \leq \beta\) or \(\beta \leq t \leq \alpha\text{,}\) depending on direction of travel.
If \(y = F(x)\text{,}\) then the area is \(A = \int_a^b F(x)\, dx = \int_a^b y\, dx\text{.}\) In parametrics, \(F(x) = g(t)\) and \(\dfrac{dx}{dt} = f'(t)\iff dx = f'(t)\, dt\text{.}\)
If \(x = F(y)\text{,}\) then the area is \(A = \int_a^b F(y)\, dy = \int_a^b x\, dy\text{.}\) In parametrics, \(F(y) = f(t)\) and \(\dfrac{dy}{dt} = g'(t)\iff dy = g'(t)\, dt\)
Example 10.2.10.
Use the parametrization \(x = r\cos\theta\text{,}\) \(y = r\sin\theta\) (\(0\leq t\leq 2\pi\)) to show that the (unsigned) area of a circle is exactly \(\pi r^2\)
Solution.
If \(y = g(\theta) = r\sin\theta\) and \(x = f(\theta) = r\cos\theta\text{,}\) then \(f'(\theta) = -r\sin\theta\text{.}\) For the entire circle, we will use \(\alpha = 0\leq \theta \leq 2\pi = \beta\text{.}\) Then, we have
\begin{align*} \ds \int_\alpha^\beta y\, dx \amp = \ds \int_0^{2\pi} (r\sin\theta)(-r\sin\theta)\, d\theta\\ \amp = \ds -\int_0^{2\pi} r^2 \sin^2\theta\, d\theta \\ \amp = \ds -r^2\int_0^{2\pi} \sin^2\theta\, d\theta \\ \amp = -r^2\lrpar{\ds \dfrac{1}{2}x - \dfrac{1}{4}\sin 2x}\bigg\rvert_0^{2\pi} \\ \amp = -r^2\lrpar{\pi} \end{align*}
So we conclude the unsigned area is \(\pi r^2\text{.}\) Naturally, if we want the positive version, we could reverse the direction of the parametrization.
Example 10.2.11.
Show that the area under one arch of the cycloid \(x = r(\theta - \sin\theta)\text{,}\) \(y = r(1-\cos\theta)\) is exactly three times the area of the generating circle.
Solution.
If \(f(\theta) = r(\theta-\sin\theta)\) and \(g(\theta) = r(1-\cos\theta)\text{,}\) then we have \(f'(\theta) = r(1-\cos\theta)\text{.}\) Our parametrization runs \(0\leq \theta \leq 2\pi \text{,}\) as each arch can be shown to be distance \(2\pi r\) from each other.
\begin{align*} \ds \int_\alpha^\beta y\, dx \amp \ds \int_0^{2\pi} r^2(1-\cos\theta)^2\, d\theta \\ \amp = r^2\lrpar{\dfrac{1}{4}(6x-8\sin x + \sin 2x)}\bigg\rvert_0^{2\pi}\\ \amp = 3\pi r^2 \end{align*}
Example 10.2.12.
Find the area enclosed by the curve \(x = t^2-2t\text{,}\) \(y = \sqrt{t}\text{,}\) and the \(y-\)axis.
Solution.
\(x = 0 \iff t = 0,2\text{.}\) Then, we integrate with respect to \(y\) and note that \(f(t) = t^2-2t\text{,}\) \(g(t) = \sqrt{t}\) and \(g'(t) = \dfrac{1}{2}t^{-1/2}\text{.}\) So,
\begin{align*} \ds \int_\alpha^\beta x\, dy \amp = \ds \int_0^2 (t^2-2t)\lrpar{\dfrac{1}{2}t^{-1/2}}\, dt \\ \amp = \ds \int_0^2 \dfrac{1}{2}t^{3/2} - t^{1/2}\, dt\\ \amp = \ds \lrpar{\dfrac{1}{5}t^{5/2} - \dfrac{2}{3}t^{3/2}}\bigg\rvert_0^2\\ \amp = \dfrac{4\sqrt{2}}{5} - \dfrac{4\sqrt{2}}{3} \end{align*}
Example 10.2.13.
Use the parametric equations \(x = a\sin\theta\text{,}\) \(y = b\cos\theta\text{,}\) \(0\leq \theta \leq 2\pi\text{,}\) to show that the area contained in an ellipse is \(\pi ab\)
Solution.
The work is virtually identical to the unsigned area problem above; the difference is now that we have parameters \(a,b\) showing up.

Subsubsection 10.2.3.2 Arc Length

Arc Length of a Parametric Curve.
If a curve \(C\) is described by the parametric equations \(x = f(t)\text{,}\) \(y = g(t)\text{,}\) for \(\alpha \leq t\leq \beta\text{,}\) where \(f'\) and \(g'\) are continuous on \([\alpha,\beta]\) and \(C\) is traversed exactly once as \(t\) ranges from \(\alpha\) to \(\beta\text{,}\) then the length of \(C\) is given by
\begin{equation*} \ds L = \int_\alpha^\beta \sqrt{(f')^2 + (g')^2}\, dt \end{equation*}
Adapt the proof for the arc length formula from Section 8.1; the Mean Value Theorem gives that \(\Delta x_i = f'(t_i)\Delta t\) and \(\Delta y_i = g'(t_i)\Delta t\) so that
\begin{equation*} |P_{i-1}P_i| = \sqrt{(\Delta x_i)^2 + (\Delta y_i)^2} = \sqrt{(f'(t_i))^2 + (g'(t_i))^2}\Delta t \end{equation*}
Pushing to the integral gives the result.
Example 10.2.14.
Find the length of one arch of the cycloid \(x = r(\theta - \sin\theta)\text{,}\) \(y = r(1-\cos\theta)\)
Solution.
\(f'(\theta) = r(1-\cos\theta)\) and \(g'(\theta) = r(\sin\theta)\text{,}\) so
\begin{align*} \ds \int_0^{2\pi} \sqrt{(r(1-\cos\theta))^2 + (r(\sin\theta))^2}\, d\theta \amp = \ds \int_0^{2\pi} r\sqrt{(1-\cos\theta)^2 + (\sin\theta)^2}\, d\theta \\ \amp = \ds \int_0^{2\pi} r\sqrt{2-2\cos\theta}\, d\theta\\ \amp = \ds \int_0^{2\pi} r\sqrt{4\sin^2\lrpar{\dfrac{\theta}{2}}}\, d\theta \\ \amp = \ds 2r \int_0^{2\pi} \sin\lrpar{\dfrac{\theta}{2}}\, d\theta\\ \amp = \ds 2r \lrpar{-2\cos\lrpar{\dfrac{\theta}{2}}}\bigg\rvert_0^{2\pi}\\ \amp = 4r(1+1)\\ \amp = 8r \end{align*}
Example 10.2.15.
Prove that the circumference of a circle of radius \(r\) is \(2\pi r\)
Solution.
Chose parametrization \(x = r\cos 2\pi \theta\text{,}\) \(y = r\sin 2\pi\theta\text{.}\) Then, \(x' = -2\pi r\sin 2\pi\theta\) and \(y' = 2\pi r\cos 2\pi \theta\) and
\begin{equation*} \ds \int_0^1 \sqrt{(-2\pi r\sin 2\pi\theta)^2 + (2\pi r\cos 2\pi \theta)^2}\, d\theta = \int_0^1 2\pi r\, d\theta = 2\pi r \end{equation*}
Example 10.2.16.
Find the exact length of the curve \(x = 1+3t^2\text{,}\) \(y = 4+2t^3\text{,}\) \(0\leq t\leq 1\)
Solution.
\(\ds \int_0^1 6\sqrt{t^2 + t^4}\, dt = 4\sqrt{2} - 2\)
Example 10.2.17.
Find the exact length of the curve \(x = e^t\cos t\) and \(y = e^t\sin t\text{,}\) \(0\leq t\leq \pi\)
Solution.
\(\int_0^1 \sqrt{(e^t\cos t -e^t\sin t)^2 + (e^t\cos t + e^t\sin t)^2}\, dt = \int_0^\pi \sqrt{2}e^t = \sqrt{2}(e^\pi - 1)\)

Subsection 10.2.4 After Class Activities

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