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Coordinated Calculus

Nathan Wakefield, Christine Kelley, Marla Williams, Michelle Haver, Lawrence Seminario-Romero, Robert Huben, Aurora Marks, Stephanie Prahl, Based upon Active Calculus by Matthew Boelkins

Section 3.7 Parametric Equations

Supplemental Videos.

The main topics of this section are also presented in the following videos:
Up to this point, we have been representing a graph by a single equation involving two variables such as \(x\) and \(y\text{.}\) However, some items are not really well described by such equations. In this section, you will study situations in which it is useful to introduce a third variable to represent a curve in the plane. This process is commonly called parameterization and is the basis for our study of parametric curves.

Subsection 3.7.1 Parametric Equations

Consider the path of an object that is propelled into the air at an angle of \(45^{\circ}\text{.}\) In a physics class, you might learn that this object would follow a parabolic path. That is,
\begin{equation*} y=-\frac{x^2}{72}+x \end{equation*}
However, this equation does not tell the whole story. While the equation does tell you where the object has been, it does not tell you when the object was at a given point \((x, y)\) on the path. To determine this time, you can introduce a third variable \(t\text{,}\) called a parameter. It is possible to write both \(x\) and \(y\) as functions of \(t\) to obtain the parametric equations
\begin{align*} x(t) \amp= 24 \sqrt{2}t\\ y(t) \amp =-16t^2+24\sqrt{2}t \end{align*}
The parametric equations are graphed in Figure 3.69 below. Using the parametric equations, we can state properties such as: at time \(t = 0\text{,}\) the object is at the point \((0, 0)\) and at time \(t = 1\text{,}\) the object is at the point \((24 \sqrt{2},24\sqrt{2}-16)\text{.}\)
Figure 3.69. A plot of the parametric equations \(x(t)= 24 \sqrt{2}t\text{,}\) \(y(t) =-16t^2+24\sqrt{2}t\text{.}\) Notice that when \(t=\frac{3\sqrt{2}}{4}\text{,}\) \(x=36\) and \(y=18\) (you can get these values by plugging \(t=\frac{3\sqrt{2}}{4}\) into the equations for \(x\) and \(y\)).

Example 3.70.

Suppose the position of an object is given by the parametric equations
\begin{align*} x(t) \amp= 24 \sqrt{2}t\\ y(t) \amp =-16t^2+24\sqrt{2}t. \end{align*}
After \(t=3\) seconds, what is the position of the object if the origin represents the initial position?
Hint.
Try plugging \(t=3\) into each equation.
Answer.
After \(3\) seconds the object is located at \((24 \sqrt{2}3,144-72 \sqrt{2}).\)
Solution.
By plugging \(t=3\) into each of the equations for \(x(t)\) and \(y(t)\) we see that
\begin{align*} x(3) \amp= 24 \sqrt{2}(3)\\ y(3) \amp =-16(3)^2+24\sqrt{2}(3) \end{align*}
Therefore, after \(3\) seconds the object is located at \((24 \sqrt{2}3,144-72 \sqrt{2}).\)
With these examples in hand we are now in a position to formally define parametric equations.

Parametric Equations.

Parametric equations are equations that specify the values of \(x\) and \(y\) in terms of a third variable \(t\) called a parameter. We often represent parametric curves in the form
\begin{align*} x(t) \amp= f(t)\\ y(t) \amp =g(t). \end{align*}
where \(f\) and \(g\) are functions and the parameter \(t\) varies over some interval \(a \lt t \lt b\text{.}\) When no interval is specified we assume that \(t\) accepts all real values.

Subsection 3.7.2 Graphing Parametric Equations

Graphs of curves sketched from parametric equations can have very interesting shapes, as exemplified in Figure 3.71. In this section we will cover some methods to sketch parametric curves.
Figure 3.71. Parametric equations can give some very interesting graphs.
One of the most effective ways to sketch a parametric curve is to create a table of values by choosing various values of \(t\) and computing both \(x(t)\) and \(y(t)\text{.}\) The following example demonstrates this procedure.

Example 3.72.

Sketch the curve given by the parametric equations
\begin{align*} x(t) \amp= t^2-4\\ y(t) \amp = \frac{t}{2}. \end{align*}
Hint.
A reasonable first step is to cerate a table of values for \(t=-2,-1,0,1,2,3\text{.}\) Why these values of \(t\text{?}\) Choosing "good" values of \(t\) is not always a straight forward task and in general you must choose values that give you a "reasonable" picture of the curve. You also might wonder how many values of \(t\) are appropriate. Again, the answer is enough values that you can be reasonably confident you have captured the shape of the curve. So what does this mean? Often it helps to choose a few positive and negative values close to zero, calculate these values, plot the points, and then determine if you need more values to give your curve a clear shape.
Answer.
We get the following graph by starting at \(t=3\)
Figure 3.73. Plot of the parametric equations \(x(t)=t^2-4\) and \(y(t)=\frac{t}{2}\) for \(-3 \lt t \lt 3\text{.}\)
Solution.
We begin by creating a table of values (For more details on how we chose these value see the hint).
Table 3.74. Table of values for the parametric equations \(x(t)=t^2-4\) and \(y(t)=\frac{t}{2}\)
t x(t) y(t)
-2 0 -1
-1 -3 -\frac{1}{2}
0 -4 0
1 -3 \frac{1}{2}
2 0 1
3 5 \frac{3}{2}
We plot these values on a coordinate plane and then connect the values with appropriate curves as in Figure 3.75. It is often helpful to place arrows on the curve to indicate how the curve changes as \(t\) increases.
Figure 3.75. Plot of the parametric equations \(x(t)=t^2-4\) and \(y(t)=\frac{t}{2}\) for \(-2 \lt t \lt 3\text{.}\)
You may notice in Figure 3.75 that the curve seems to start abruptly at \(t=-2\text{.}\) In fact, this is just an artifact of our starting our table of values at \(t=-2\text{.}\) If we instead started our table at \(t=-3\) then we would get a slightly different image, Figure 3.73.
Figure 3.76. Plot of the parametric equations \(x(t)=t^2-4\) and \(y(t)=\frac{t}{2}\) for \(-3 \lt t \lt 3\text{.}\)
In addition to the method of plotting points, many graphing software programs can effectively plot parametric curves. For convenience, one usch utility based on GeoGebra is included in [cross-reference to target(s) "parametricapp" missing or not unique].
Figure 3.77. GeoGebra app for plotting parametric equations.

Example 3.78.

Describe the motion of the particle whose coordinates at time t are expressed with the parametric equations: \(x = \cos(t)\) and \(y = \sin(t)\text{.}\)
Solution.
Since \((\cos(t))^2 + (\sin(t))^2 = 1\text{,}\) we have \(x^2 + y^2 = 1\text{.}\) That is, at any time \(t\) the particle is at a point \((x, y)\) on the unit circle \(x^2 + y^2 = 1\text{.}\) We plot points at different times to see how the particle moves on the circle. The particle completes one full trip counterclockwise around the circle every \(2\pi\) units of time. Notice how the \(x\)-coordinate goes back and forth from \(-1\) to \(1\) while the \(y\)-coordinate goes up and down from \(-1\) to \(1\text{.}\) The two motions combine to trace out a circle.
Table 3.79. Table of values for the parametric equations \(x = \cos(t)\) and \(y = \sin(t)\text{.}\)
t x y
0 1 0
\(\frac{\pi}{2}\) 0 1
\(\pi\) -1 0
\(\frac{3\pi}{2}\) 0 -1
\(2\pi\) 1 0
Figure 3.80. Plot of the parametric equations \(x = \cos(t)\) and \(y = \sin(t)\text{.}\)
The first thing we note is that \((\cos(3t))^2 + (\sin(3t))^2 = 1\) and so we still have the fact that, at any time \(t\) the particle is at a point \((x, y)\) on the unit circle \(x^2 + y^2 = 1\text{.}\) So, what effect is the \(3\) in \(x = \cos(3t)\) and \(y = \sin(3t)\) having? Consider the following table of values:
Table 3.81. Table of values for the parametric equations \(x = \cos(3t)\) and \(y = \sin(3t)\text{.}\)
t x y
0 1 0
\(\frac{\pi}{6}\) 0 1
\(\frac{\pi}{3}\) -1 0
\(\frac{\pi}{2}\) 0 -1
\(\frac{2\pi}{3}\) 1 0
In other words, the motion of the particle described by \(x = \cos(3t)\) and \(y = \sin(3t)\) is nearly identical to the motion of a particle described by \(x = \cos(t)\) and \(y = \sin(t)\) except that the particle described by \(x = \cos(3t)\) and \(y = \sin(3t)\) is moving three times as fast. The motion of the particule is displayed by the graph in Figure 3.82.
Figure 3.82. Plot of the parametric equations \(x = \cos(3t)\) and \(y = \sin(3t)\text{.}\)

Subsection 3.7.3 Motion in a Straight Line and Derivatives

Suppose an object moves with constant speed along a straight line through the point \((x_0, y_0)\text{.}\) Recall that they key to motion in a straight line is that the rate of change is constant. In other words, both the \(x\) and \(y\)-coordinates have a constant rate of change. For convenience, let \(a = \frac{dx}{dt}\) and \(b = \frac{dy}{dt}\) be the rate of change in the \(x\) and \(y\) directions, respectively. Then at time \(t\) the object has coordinates \(x = x_0 + at\text{,}\) \(y = y_0 + bt\text{.}\) Notice that \(a\) represents the change in \(x\) in one unit of time, and \(b\) represents the change in \(y\) in one unit of time (see Figure 3.83). Thus the line has slope \(m = \frac{b}{a}\text{.}\) Therefore, the parametric equations for motion in a straight line are given by \(x = x_0 + at\text{,}\) \(y = y_0 + bt\text{.}\)
Figure 3.83. Plot of the parametric equations \(x = \cos(3t)\) and \(y = \sin(3t)\text{.}\)
Now that we have developed the notion of motion in a straight line for parametric equations we can discuss tangent lines for parametric equations.

Parametric Equations for the Tangent Line.

An object moving along a line through the point \((x_0, y_0)\text{,}\) with \(\frac{dx}{dt} = a\) and \(\frac{dy}{dt} = b\text{,}\) has parametric equations
\begin{equation*} x = x_0 + at \mbox{, } y = y_0 + bt. \end{equation*}
The slope of the line is \(m = \frac{b}{a}\text{.}\)

Example 3.84.

Find the tangent line at the point \((1, 2)\) to the curve defined by the parametric equations \(x = t^3\text{,}\) \(y = 2t\text{.}\)
Hint.
Try finding the tangent line at the point for each equation separately.
Answer.
the tangent line has parametric equations
\begin{equation*} x = 1 + 3t\mbox{, } y = 2 + 2t. \end{equation*}
Solution.
At time \(t = 1\) the particle is at the point \((1, 2)\text{.}\) The velocity in the \(x\)-direction at time \(t\) is \(v_x = \frac{dx}{dt} = 3t^2\text{,}\) and the velocity in the \(y\)-direction is \(v_y = \frac{dy}{dt} = 2\text{.}\) So at \(t = 1\) the velocity in the \(x\)-direction is 3 and the velocity in the \(y\)-direction is \(2\text{.}\) Thus the tangent line has parametric equations
\begin{equation*} x = 1 + 3t\mbox{, } y = 2 + 2t. \end{equation*}
With the notion of a tangent line in hand we are in the position to be able to talk about the speed and velocity of an object whose position is given by parametric equations. In general, given an object traveling in the \(xy\)-plane we can break its movement into two pieces. Movement in the "\(x\)-direction" and movement in the "\(y\)-direction." We can also talk about rate-of-change in the "\(x\)-direction" (\(\frac{dx}{dt}\)) and rate-of-change in the "\(y\)-direction" (\(\frac{dy}{dt}\)). If we think of these rates of change as vectors then we can use the Pythagorean theorem to talk about the instantaneous speed \(v\) , see Figure 3.85.
Figure 3.85. Calculation of instantaneous speed based on individual directional component rates of change using Pythagorean theorem.
Using this idea we have the following definition.

Instantaneous Speed.

The instantaneous speed of a moving object is defined to be
\begin{equation*} v=\sqrt{\left(\dfrac{dx}{dt} \right)^2+\left(\dfrac{dy}{dt} \right)^2} \end{equation*}
The quantity \(v_x = \frac{dx}{dt}\) is the instantaneous velocity in the \(x\)-direction; \(v_y = \frac{dy}{dt}\) is the instantaneous velocity in the \(y\)-direction. The velocity vector \(\mathbf{v}\) is written
\begin{equation*} \mathbf{v}=v_x \mathbf{i}+v_y\mathbf{j}. \end{equation*}

Example 3.86.

A particle moves along a curve in the \(xy\)-plane with \(x(t) = 2t + e^t\) and \(y(t) = 3t -4\text{,}\) where \(t\) is time. Find the velocity vector and speed of the particle when \(t = 1\text{.}\)
Hint.
Use the formula above after finding the derivative of each equation.
Answer.
The speed at \(t = 1\) is
\begin{equation*} v=\sqrt{(2+e)^2+3^2} \approx 5.591\text{.} \end{equation*}
Solution.
Differentiating gives \(\frac{dx}{dt} = 2 + e^t\text{,}\) \(\frac{dy}{dt} = 3\text{.}\) When \(t = 1\) we have \(v_x = 2 + e\text{,}\) \(v_y = 3\text{.}\) So the velocity vector is
\begin{equation*} \mathbf{v}=(2+e)\mathbf{i}+3\mathbf{j} \end{equation*}
The speed at \(t = 1\) is
\begin{equation*} v=\sqrt{(2+e)^2+3^2} \approx 5.591\text{.} \end{equation*}
We now formally define the slope of the tangent line to a parametric curve and concavity of a parametric curve.

Slope of Tangent Line to Parametric Curves.

\begin{equation*} \dfrac{dy}{dx}=\dfrac{\dfrac{dy}{dt}}{\dfrac{dx}{dt}} \end{equation*}

Concavity of Parametric Curves.

\begin{equation*} \dfrac{d^2y}{dx^2}=\dfrac{\dfrac{d}{dt} \left(\dfrac{dy}{dx}\right)}{\dfrac{dx}{dt}} \end{equation*}

Example 3.87.

A particle moves in the \(xy\)-plane so that its position at time \(t\) is given by \(x = \sin(2t)\text{,}\) \(y = -\cos(2t)\) for \(0 \leq t \lt 2\pi\text{.}\)
  1. At what time does the particle first touch the \(x\)-axis? What is the speed of the particle at that time?
  2. Is the particle ever at rest?
Hint.
Recall the \(x\)-axis is where \(y=0\text{.}\) Use the above equation for the speed. If the particle is at rest, think about what the value of the velocity is.
Answer.
  1. The particle touches the \(x\)-axis when \(t=\frac{\pi}{4}\text{.}\) At this point, the speed of the particle is
    \begin{equation*} 2\text{.} \end{equation*}
  2. The particle is never at rest.
Solution.
  1. The particle touches the \(x\)-axis when \(y=0\text{,}\) that is when \(\cos(2t)=0\text{.}\) This occurs when \(t=\frac{\pi}{4}\text{.}\) At this point, the speed of the particle is given by
    \begin{equation*} \sqrt{4\cos^2(\frac{\pi}{2})+4\sin^2(\frac{\pi}{2})}=2\text{.} \end{equation*}
  2. Since \(\sqrt{4\cos^2(2t)+4\sin^2(2t)}=2\text{,}\) the particle is never at rest.

Exercises 3.7.4 Exercises

1. Evaluating Parametric Equations.

Find the coordinates at times t = 0, 4, 7 of a particle following the path \(x = 5\text{,}\) \(y = 7+7t^{3}\text{.}\)
t = 0:
t = 4:
t = 7:

2. Sketching Parametric Equations.

Assume \(t\) is defined for all time. Enter the letter of the graph below which corresponds to the curve traced by the parametric equations. Think about the range of \(x\) and \(y\text{,}\) and whether there is periodicity and or symmetry.
  1. \(\displaystyle x=t^3-5t+2;\quad y=3-t^2\)
  2. \(\displaystyle x=t+\cos(10t);\quad y=t^2+\sin(t)\)
  3. \(\displaystyle x=\sin(t);\quad y=\cos(t)-2\cos(2t)\)
  4. \(\displaystyle x= t+\sin(5t);\quad y=t+\cos(5t)\)
  5. \(\displaystyle x=\sin(t+\sin(7t));\quad y=\cos(t)\)

3. Parametric Lines.

A line is parameterized by \(x= 8+7t\) and \(y = 4+4t\text{.}\)
(a) Which of the following points are on the section of the line obtained by restricting \(t\) to nonnegative numbers (for each, enter Y if the point is on the section, and N if not)?
\(\left(-6,-4\right)\) :
\(\left(22,12\right)\) :
\(\left(43,24\right)\) :
Then, give one more point that is on the section of the line obtained by this restriction:
(b) What are the endpoints of the line segment obtained by restricting \(t\) to \(-4\le t\le -3\text{?}\)
left endpoint :
right endpoint :
(c) How should \(t\) be restricted to give the part of the line to the left of the \(y\)-axis (give your answer as an interval for \(t\text{,}\) for example, (3,8) or [-2,Inf))?
\(t\) must be in :

4. Slope in Parametric Equations.

The functions \(f(t)\) and \(g(t)\) are shown below.
\(f(t)\) \(g(t)\)
If the motion of a particle whose position at time \(t\) is given by \(x=f(t)\text{,}\) \(y=g(t)\text{,}\) sketch a graph of the resulting motion and use your graph to answer the following questions:
(a) The slope of the graph at \(\left(0.25,0.5\right)\) is
(enter undef if the slope is not defined)
(b) At this point the particle is moving
  • neither left nor right
  • to the left
  • to the right
and
  • neither up nor down
  • up
  • down
.
(c) The slope of the graph at \(\left(1.75,0.5\right)\) is
(enter undef if the slope is not defined)
(d) At this point the particle is moving
  • neither left nor right
  • to the left
  • to the right
and
  • neither up nor down
  • up
  • down
.

5. Derivatives of Parametric Equations.

A particle moves with its position given by \(x = \cos\!\left(t\right)\) and \(y = \sin\!\left(\frac{t}{2}\right)\text{,}\) where positions are given in feet from the origin and time \(t\) is in seconds.
Find the speed of the particle.
Speed =
(include units
 54 
/webwork2_files/helpFiles/Units.html
)
Find the first positive time when the particle comes to a stop.
\(t =\)
(include units
 55 
/webwork2_files/helpFiles/Units.html
)
If \(n\) is any odd integer, write a formula (in terms of \(n\)) for all positive times \(t\) at which the particle comes to a stop.
\(t =\)
(include units
 56 
/webwork2_files/helpFiles/Units.html
)

6. Tangent Lines and Parametric Equations.

Find parametric equations for the tangent line at \(t = 2\) for the motion of a particle given by \(x(t) = 3t^{2}+4\text{,}\) \(y(t) = 6t^{3}\text{.}\)
For the line,
\(x(t) =\)
\(y(t) =\)
(Note that because the correctness of a parametrically described line depends on both \(x(t)\) and \(y(t)\text{,}\) both of your answers may be marked incorrect if there is an error in one of them.)

7. Horizontal Tangent Lines and Parametric Equations.

Suppose a curve is traced by the parametric equations
\begin{equation*} x = 3 \big(\sin(t)+\cos(t)\big) \end{equation*}
\begin{equation*} y = 38 - 15\cos^2(t) - 30 \sin(t) \end{equation*}
as \(t\) runs from \(0\) to \(\pi\text{.}\) At what point \((x,y)\) on this curve is the tangent line horizontal?
\(x =\)
\(y =\)

8. Speed and Parametric Equations.

Determine the speed \(s(t)\) of a particle with a given trajectory at a time \(t_0\) (in units of meters and seconds).
\begin{equation*} c(t) = (\ln(t^2 + 1), t^3), \, t_0 = 8 \end{equation*}

9. Calculating Speed Using Parametric Equations.

Find the speed of the cycloid \(c(t) = (6 t - 6 \sin t, \, 6 - 6 \cos t)\) at points where the tangent line is horizontal.

10. Intersecting Parametric Equations.

Two particles move in the \(xy\)-plane. At time \(t\text{,}\) the position of particle \(A\) is given by \(x(t) = 5 t - 5\) and \(y(t) = 3 t - k\text{,}\) and the position of particle \(B\) is given by \(x(t) = 4 t\) and \(y(t) = t^2 - 2t - 1\text{.}\)
(a) If \(k = -1\text{,}\) do the particles ever collide?
  • yes
  • no
  • it is not possible to determine for certain
(Be sure that you are able to explain your answer!)
(b) Find \(k\) so that the two particles are certain to collide.
\(k =\)
(c) At the time the particle collide in (b), which is moving faster?
  • particle A
  • particle B
  • neither particle (they are moving at the same speed)
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