Skip to main content

Workbook for Stewart’s "Calculus"

Cory Wilson

Section 10.1 Curves Defined by Parametric Equations

Subsection 10.1.1 Before Class

Figure 70. Pre-Class Video 1
Figure 71. Pre-Class Video 2

Subsubsection 10.1.1.1 Parametric Equations

Question 10.1.1.
The picture below does not describe a function? Why?
A graph of the parametric function given by \(x(t) = t^3-3t, y(t) = 3t^2-9\)
Solution.
It fails the vertical line test on \([-2,2]\)(i.e. there exist multiple outputs for individual inputs).
Definition 10.1.2. Parameter/Parametric Equation.
Let \(C\) be a curve, and let the functions \(x =f(t)\) and \(y = g(t)\) describe the \(x-\) and \(y-\)coordinate of \(C\text{.}\) The variable \(t\) is called the parameter, and the function \(x = f(t)\) and \(y = g(t)\) are called parametric equations.
Example 10.1.3.
Consider the parametric equations \(x(t) = t-2\) and \(y(t) = t^2 - 2t\text{.}\)
  1. Fill out the table below:
    \(t\) \(x\) \(y\) \(t\) \(x\) \(y\)
    \(-2\) \(\) \(\) \(2\) \(\) \(\)
    \(-1\) \(\) \(\) \(3\) \(\) \(\)
    \(0\) \(\) \(\) \(4\) \(\) \(\)
    \(1\) \(\) \(\)
  2. Plot the points from the table on the graph below. Connect the plotted points to see the parametric curve, and draw arrows to indicate its direction of travel. What kind of function do you see?
    An empty plot on \([-4,4]\times [-1,8]\)
Solution.
  1. \(t\) \(x\) \(y\) \(t\) \(x\) \(y\)
    \(-2\) \(-4\) \(8\) \(2\) \(0\) \(0\)
    \(-1\) \(-3\) \(3\) \(3\) \(1\) \(3\)
    \(0\) \(-2\) \(0\) \(4\) \(2\) \(8\)
    \(1\) \(-1\) \(-1\)
  2. The function is a parabola
    A plot of the parametric equation given in the problem on \([-4,4]\)
Example 10.1.4.
Rather than plotting points, we could have directly seen the function. We can do this by a process called eliminating the parameter.
  1. Given the parametric equations \(x= t-2\) and \(y= t^2-2t\text{,}\) eliminate the parameter to show that the parametric curve is given by the function \(y =x^2-2x\)
  2. If we wanted to only include the right portion of the graph shown, what sort of restriction would we need to put on \(t\text{?}\)
Solution.
  1. Note that if \(x=t-2\text{,}\) then \(t=x+2\text{.}\) We can replace this in the equation for \(y\text{:}\) \(y(x) = (x+2)^2 - 2(x+2) = x^2-2x\)
  2. We need the right portion, which starts at the vertex. According to the table in the previous example, this happens when \(t=1\text{.}\) So, the restriction should be \(t\geq 1\)
Example 10.1.5.
Eliminate the parameter to find the curve represented by the parametric equations \(x = \sqrt{t}\text{,}\) \(y = t^2+1\text{.}\) There is a natural restriction on the time interval; what is it?
Solution.
Eliminating the parameter gives \(y=x^4+1\text{.}\) Since \(x=\sqrt{t}\text{,}\) the natural restriction on the time parameter is \(t\geq 0\)
Example 10.1.6.
  1. What curve is represented by the parametric equations \(x = \cos t\text{,}\) \(y = \sin t\text{,}\) \(0\leq t\leq 2\pi\text{?}\)
  2. The parametric equations \(x = \cos 2t\text{,}\) \(y = \sin 2t\text{,}\) \(0\leq t\leq 2\pi\) represent the same curve as in part (a). What is the difference between the parametrizations in (a) and (b)?
  3. What if in (b) we instead took the time interval \(0\leq t\leq \pi\text{?}\) What would be the difference between (a) and (b) in that case?
Solution.
  1. A circle of radius 1: \((\cos t)^2 = \cos^2t = x^2\) and \((\sin t)^2 = \sin^2t = y^2\text{,}\) so by the Pythagorean theorem, \(cos^2t + \sin^2t = x^2+y^2=1\)
  2. The parametrization in (b) traces the circle twice rather than once
  3. Changing the time interval results in the same graph, but only traces it once rather than twice

Subsection 10.1.2 Pre-Class Activities

Example 10.1.7.

Use this space to write any questions you might have from the videos
Solution.
Answers vary

Example 10.1.8.

For each of the following parametric curves, sketch the curve and indicate the direction in which the curve is traced as \(t\) increases. Then, eliminate the parameter to find a Cartesian equation for the curve.
  1. \(x = 2t -1 \text{,}\) \(y = \dfrac{1}{2}t + 1\)
  2. \(x = t^2 -3\text{,}\) \(y = t+2\text{,}\) \(-3\leq t\leq 3\)
  3. \(x = \sin t\text{,}\) \(y = 1-\cos t\text{,}\) \(0\leq t\leq 2\pi\)
Solution.
  1. The graph of the parametric equation \(x = 2t -1 \text{,}\) \(y = \dfrac{1}{2}t + 1\)
    Eliminating the parameter gives \(y = \dfrac{1}{4}x+\dfrac{5}{4}\)
  2. The graph of the parametric equation \(x = t^2 -3\text{,}\) \(y = t+2\text{,}\) \(-3\leq t\leq 3\)
    Eliminating the parameter gives \(y = 2\pm \sqrt{x+3}\)
  3. The graph of the parametric equation \(x = \sin t\text{,}\) \(y = 1-\cos t\text{,}\) \(0\leq t\leq 2\pi\)
    Eliminating the parameter gives \(x^2 + (y-1)^2 = 1\)

Subsection 10.1.3 In Class

Subsubsection 10.1.3.1 Examples

Example 10.1.9.
Sketch the curve \(x = \sin \lrpar{\dfrac{1}{2}\theta}\text{,}\) \(y = \cos \lrpar{\dfrac{1}{2}\theta}\text{,}\) \(-\pi\leq\theta\leq\pi\text{.}\) Then, eliminate the parameter to find a Cartesian expression for the curve.
Solution.
Sketch of the curve \(x = \sin \lrpar{\dfrac{1}{2}\theta}\text{,}\) \(y = \cos \lrpar{\dfrac{1}{2}\theta}\text{,}\) \(-\pi\leq\theta\leq\pi\)
Eliminating the parameter gives \(x^2 + y^2 = 1\) on the interval \([-1,1]\)
Example 10.1.10.
Do the same thing for the parametric equations \(x = e^t\text{,}\) \(y = e^{-2t}\text{.}\)
Solution.
Sketch of the curve \(x = e^t\text{,}\) \(y = e^{-2t}\text{,}\) \(-2\leq t\leq 4.75\)
Eliminating the parameter gives \(y=x^{-2}\)
Example 10.1.11.
Describe the motion of a particle with position \(x = 5+2\cos (\pi t)\text{,}\) \(y =3 + 2\sin (\pi t)\) as \(t\) varies in the interval \([1,2]\text{.}\)
Solution.
Make a table of values to help see the path of the curve. The curve begins at \((3,3)\) and moves counterclockwise, creating a quarter circle of radius 2.
Example 10.1.12.
Match the graphs of the parametric equations \(x = f(t)\) and \(y = g(t)\) in (a)-(d) with the parametric curves labeled I - IV. Justify your response.
Description
Solution.
Some of these can be tricky.
  1. Notice that the x-coordinate should begin at 1, move to 2, then back to 1. At the same time, the y-coordinate should start at 0, move to 1, back to 0, down to -1, and back to 0.
    The only curve which keeps the x-coordinate positive is III, so this must be the answer.
  2. The x-coordinate oscillates between 0, 2, and -2; the y coordinate does the same, but faster. Specifically, we’re looking for 4 distinct positive and negative "strands" for the horizontal component, and 6 distinct strands for the vertical component.
    The graph which fits the description is I.
  3. Notice here that the x-coordinate switches between positive and negative a few times, but the y-coordinate is non-negative, moving between 0 and 2.
    This makes IV the appropriate graph.
  4. By elimination, d matches with II. But we can also see this by matching the behavior of the y-coordinate.
Example 10.1.13.
Develop a parametrization which traces out a circle three times in the time interval \(0\leq t \leq \pi\text{.}\)
Solution.
Answers vary, but the most straightforward answer is likely \(x(t) = \cos(6t), y(t) = \sin(6t)\)
Example 10.1.14.
Compare the curves represented by the parametric equations given. How do they differ?
  1. \(x = t^3\text{,}\) \(y = t^2\)
  2. \(x = e^{-3t}\text{,}\) \(y =e^{-2t}\)
  3. \(x = t^6\text{,}\) \(y = t^4\)
Solution.
The graphs of (b) and (c) appear to be parametrized in such a way that the left branch of (a) is naturally excluded. The distinction between (b) and (c) is that it appears that (c) is traced out faster in time than (b).

Subsection 10.1.4 After Class Activities

Example 10.1.15.

Sketch the parametric curve \(x = t^2\text{,}\) \(y = t^3\text{,}\) and eliminate the parameter to find a Cartesian equation for the curve.
Solution.
The curve \(x = t^2\text{,}\) \(y = t^3\) on the interval \([-2,2]\)
Eliminating the parameter gives \(y=x^{3/2}\)

Example 10.1.16.

Do the same thing for the curve \(x = t^2\text{,}\) \(y = \ln t\)
Solution.
A sketch of the curve \(x = t^2\text{,}\) \(y = \ln t\) on \([-2,2]\)
Eliminating the parameter gives \(y= \dfrac{1}{2}\ln x\)

Example 10.1.17.

Describe the motion of a particle with position \(x = 2 + \sin t\text{,}\) \(y = 1 + 3\cos t\) on the interval \(\pi/2\leq t\leq 2\pi\text{.}\)
Solution.
The particle traces an ellipse beginning at \((3,1)\text{,}\) going clockwise, and ending at \((2,4)\)

Example 10.1.18.

Compare the curves represented by the parametric equations below. How do they differ?
  1. \(x = t\text{,}\) \(y = t^{-2}\)
  2. \(x = \cos t\text{,}\) \(y = \sec t\)
  3. \(x = e^t\text{,}\) \(y = e^{-2t}\)
Solution.
Part (c) traces out only the right hand side of the graphs that (a) and (b) trace out, because of the natural domain restriction from the parametrization. (b) traces out the same curve as (a), but does so more slowly.

Example 10.1.19.

The graphs of \(x = f(t)\) and \(y = g(t)\) are given below. Use the graphs to sketch the corresponding parametric curve; use arrows to indicate the direction in which the curve travels.
\(x(t) = t^2\) on the left and \(y(t)=-t\)
Solution.
A sketch of the curve \(x(t)=t^2,y(t) = -t\) on \([-2,2]\)
<Prev^TopNext>