Precalculus by Richard Wright

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7-06 Parametric Equations

In physics, projectiles are objects that move only under the influence of gravity. Near the earth, the path of a projectile is a parabola that opens down. While it is possible to write a function to model the motion, it is more useful to have separate functions for the horizontal motion and the vertical motion. Both these functions are based on time.

$$\left\{\begin{align}x &= v_{x0} t\\y &= -9.8 t^2 + v_{y0} t + y_0\end{align}\right.$$

Parametric Equations

Parametric equations are separate equations for each of the dimensions. The variable in the equations is called the parameter and is often t or θ. To graph parametric equations, make a table of values where you choose values of the parameter and calculate x and y.

Graph Parametric Equations

Graph the curve given by

$$\left\{\begin{align}x &= t^2 - 1\\ y &= t - 2\end{align}\right.$$

Solution

Start by making a table by picking values of t and calculating x and y.

t	x	y
−2	3	−4
−1	0	−3
0	−1	−2
1	0	−1
2	3	0

Now graph the points formed from corresponding x and y coordinates.

Graph Parametric Equations

Graph \(\left\{\begin{align}x &= 3\sin θ\\ y &= \cos θ\end{align}\right.\).

Solution

Start by making a table by picking values of θ and calculating x and y.

θ	x	y
0	0	1
\(\frac{π}{4}\)	\(\frac{3\sqrt{2}}{2}\)	\(\frac{\sqrt{2}}{2}\)
\(\frac{π}{2}\)	3	0
\(\frac{3π}{4}\)	\(\frac{3\sqrt{2}}{2}\)	\(-\frac{\sqrt{2}}{2}\)
π	0	−1
\(\frac{5π}{4}\)	\(\frac{-3\sqrt{2}}{2}\)	\(\frac{-\sqrt{2}}{2}\)
\(\frac{3π}{2}\)	−3	0
\(\frac{7π}{4}\)	\(\frac{-3\sqrt{2}}{2}\)	\(-\frac{\sqrt{2}}{2}\)
2π	0	1

Now graph the points formed from corresponding x and y coordinates.

Graph \(\left\{\begin{align}x &= 2 \sec θ\\y &= \tan θ\end{align}\right.\).

Answer

Parametric Equations of Conics

Write Parametric Equations

Write a set of parametric equations for a vertical ellipse with center (2, 3), major axis length 10, and minor axis length of 8.

Solution

The equations for a vertical ellipse are \(\left\{\begin{align}x &= h + b \sin t\\y &= k + a \cos t\end{align}\right.\). The center is (h, k) = (2, 3) so h = 2 and k = 3. The variable a is half the major axis which is 10, so a = 5. b is half the minor axis length which is 8, so b = 4. Now fill in the equations.

$$\left\{\begin{align}x &= h + b \sin t\\y &= k + a \cos t\end{align}\right.$$

$$\left\{\begin{align}x &= 2 + 4 \sin t\\y &= 3 + 5 \cos t\end{align}\right.$$

Write Parametric Equations

Write a set of parametric equations for y² = 4x − 4.

Solution

Method 1:

This is a parabola, so use the equation for a horizontal parabola.

$$\left\{\begin{align}x &= pt^2 + h\\y &= 2pt + k\end{align}\right.$$

To use these p, h, and k need to be known, so put the equation in standard form.

$$y^2 = 4x - 4$$

$$y^2 = 4(x - 1)$$

When compared to (y − k)² = 4p(x − h), we find that p = 1, h = 1, and k = 0. Now fill in the parametric equations.

$$\left\{\begin{align}x &= pt^2 + h\\y &= 2pt + k\end{align}\right.$$

$$\left\{\begin{align}x &= t^2 + 1\\y &= 2t\end{align}\right.$$

Method 2:

Another way is to choose something to equal t. Since to solve y² = 4(x − 1) for y, the square root would have to be taken, it would be convenient to let t² = x − 1. Solve this for x to get

$$x = t^2 + 1$$

Substitute this for x in the equation to get

$$y^2 = 4(t^2 + 1 - 1)$$

Then solve for y.

$$y^2 = 4t^2$$

$$y = 2t$$

So the parametric equations are

$$\left\{\begin{align}x &= t^2 + 1\\y &= 2t\end{align}\right.$$

Eliminating the Parameter

It is possible to change parametric equations into regular functions by eliminating the parameter. To do this, solve one of the parametric equations for t and substitute the expression into the other equation. If the equations contain sine and cosine, or tangent and secant, it might be easier to solve each equation for the trigonometric function and use a Pythagorean Identity such as sin² θ + cos² θ = 1 or 1 + tan² θ = sec² θ.

Eliminate the Parameter

Eliminate the parameter in \(\left\{\begin{align}x &= t^2 - 1\\y &= 3t\end{align}\right.\).

Solution

Since these do not have trigonometric functions, start by solving one equation for t.

$$x = t^2 - 1$$

$$x + 1 = t^2$$

$$±\sqrt{x + 1} = t$$

Now substitute this into the other equation.

$$y = 3t$$

$$y = 3 \left(± \sqrt{x + 1}\right)$$

Eliminate the ± by squaring both sides.

$$y^2 = 9(x + 1)$$

This is a horizontal parabola.

Eliminate the Parameter

Eliminate the parameter in \(\left\{\begin{align}x &= 2 \tan θ\\y &= 4 \sec θ\end{align}\right.\).

Solution

Since these do have trigonometric functions, it will be easier to solve the equations for the trig functions.

$$\left\{\begin{align}x &= 2 \tan θ\\y &= 4 \sec θ\end{align}\right.$$

$$\left\{\begin{align}\frac{x}{2} &= \tan θ\\ \frac{y}{4} &= \sec θ\end{align}\right.$$

An identity that relates secant and tangent is

$$ 1 + \tan^2 θ = \sec^2 θ $$

$$1 + \left(\frac{x}{2}\right)^2 = \left(\frac{y}{4}\right)^2$$

$$1 + \frac{x^2}{4} = \frac{y^2}{16}$$

Rearrange the equation to put it in standard form.

$$\frac{y^2}{16} - \frac{x^2}{4} = 1$$

This is a vertical hyperbola.

Practice Exercises

1. What is a parameter?
2. Graph the parametric equations.
3. \(\left\{\begin{align}x &= 4t^2\\y &= 8t + 2\end{align}\right.\)
4. \(\left\{\begin{align}x &= 4t + 1\\y &= 2t^2\end{align}\right.\)
5. \(\left\{\begin{align}x &= 3 \sec θ\\y &= 2 \tan θ\end{align}\right.\)
6. \(\left\{\begin{align}x &= 2 \sin θ\\y &= 4 \cos θ\end{align}\right.\)
7. Write a set of parametric equations for the following conditions.
8. Parabola with vertex at (−2, −1) and focus (−5, −1).
9. Horizontal ellipse with center (0, 0), vertices (±5, 0), and foci (±3, 0).
10. y = 3x + 1
11. (y − 2)² = −16x
12. Eliminate the parameter.
13. \(\left\{\begin{align}x &= 4t^2\\y &= 8t + 2\end{align}\right.\)
14. \(\left\{\begin{align}x &= 3t\\y &= \frac{9}{t}\end{align}\right.\)
15. \(\left\{\begin{align}x &= 7 \sin θ\\y &= 4 \cos θ\end{align}\right.\)
16. \(\left\{\begin{align}x &= 6 \tan θ\\y &= 5 \sec θ\end{align}\right.\)
17. Problem Solving
18. On televised baseball games, sometimes the distance of a home run is given, but it was not measured. The distance was calculated. Let \(\left\{\begin{align}x &= 35t\\y &= -16t^2 + 30t\end{align}\right.\) model the path of the ball after it was hit. How far did the ball go? (Hint: Let y = 0 and find t.)

19. 

    A solar oven cooks food by reflecting the sunlight off a parabolic mirror. The food is placed at the focus where all the light is focused. Write a set of parametric equations to model the surface of the mirror if the vertex is at (0, 0) and the focus is 2 feet above the vertex.
20. Mixed Review
21. (7-05) Classify the conic and rewrite it in standard form by eliminating the Bxy term. x² − 2xy + y² − x − y = 0.
22. (7-04) Find the standard equation of the hyperbola with vertices (4, ±3) and foci (4, ±5).
23. (7-03) Find the center, vertices, covertices, and foci of the ellipse \(\frac{(x + 1)^2}{16} + \frac{y^2}{9} = 1\).
24. (7-02) Graph (x − 1)² = 4(y + 2).
25. (7-01) Find the distance from the point (2, 0) to the line 2x − y + 4 = 0.

Answers

1. The parameter is the value in the equations that is used to calculate x and y.

2. 

3. 

4. 

5. 

6. \(\left\{\begin{align}x &= -3t^2 - 2\\y &= -6t - 1\end{align}\right.\)
7. \(\left\{\begin{align}x &= 5 \cos t\\y &= 4 \sin t\end{align}\right.\)
8. \(\left\{\begin{align}x &= \frac{t}{3}\\y &= t + 1\end{align}\right.\)
9. \(\left\{\begin{align}x &= -4t^2\\y &= -8t + 2\end{align}\right.\)
10. \((y - 2)^2 = 16x\)
11. \(xy - 27 = 0\)
12. \(\frac{x^2}{49} + \frac{y^2}{16} = 1\)
13. \(\frac{y^2}{25} - \frac{x^2}{36} = 1\)
14. 65.625 feet
15. \(\left\{\begin{align}x &= 4t\\y &= 2t^2\end{align}\right.\)
16. Parabola; \(y′^2 = \frac{\sqrt{2}}{2}x′\)
17. \(\frac{y^2}{9}-\frac{(x - 4)^2}{16} = 1\)
18. C(−1, 0); V(−5, 0), (3, 0); CV(−1, −3), (−1, 3); F\((-1 - \sqrt{7}, 0)\), \((-1 + \sqrt{7}, 0)\)

19. 

20. \(\frac{8\sqrt{5}}{5}\)