This document provides an overview of parametric equations and polar coordinates. It defines parametric equations as representing curves where the x- and y-coordinates are functions of a third variable called a parameter. Examples show how to graph curves defined by parametric equations by choosing values for the parameter and plotting the corresponding x- and y-coordinates. The document also discusses eliminating the parameter to obtain the Cartesian equation of the curve.

1. Chapter 10: Parametric Equations and Polar
Coordinates
Section 10.1: Curves Defined by Parametric Equations
Alea Wittig
SUNY Albany

2. Imagine a partical moving across a curve C, which does not pass
the Vertical Line Test.
The x and y coordinates of C are functions of time, t.

3. Parametric Equations
▶ Suppose x and y are both given as functions of third variable
t, called a parameter, by equations
x = f (t) y = g(t)
called parametric equations.
▶ As t varies, the point
(x, y) = (f (t), g(t))
varies and traces out a curve C called a parametric curve.
▶ Note: t does not always represent time.

4. Example 1
1/4
Sketch and identify the curve defined by the parametric equations
x = t2
− 2t y = t + 1

5. Example 1
2/4
▶ We start by making a table of points corresponding to
t = −2, −1, 0, 1, 2, 3, 4, . . .
▶ Choose both negative and positive values of t.
▶ Choosing integers for t also makes the computation of x and y
easier.
t x = t2 − 2t y = t + 1
−2 8 −1
−1 3 0
0 0 1
1 −1 2
2 0 3
3 3 4
4 8 5

6. Example 1
3/4
▶ Now we can plot points.
▶ Then connect the points
starting at t = 2 and
ending at t = 4
▶ We mark the curve with
arrows to show the
direction.
▶ One of the key differences
between a parametric curve
and a cartesian curve is
that parametric curves have
direction.

7. Example 1
4/4
▶ We notice that the curve looks like a transformation of the
curve y = ±
√
x
▶ Let’s check this by first solving for t in terms of y:
y = t + 1 =⇒ t = y − 1
▶ Then plugging this into the equation for x to eliminate the
parameter t:
x = t2
− 2t
= (y − 1)2
− 2(y − 1)
= y2
− 2y + 1 − 2y + 2
= y2
− 4y + 3
= (y − 2)2
− 1 =⇒
y = 2 ±
√
x + 1 ✓

8. Eliminating the Parameter
▶ The process used in example 1 to find a Cartesian equation in
x and y is called eliminating the parameter.
▶ In this process we are able to see the shape of the curve but
we lose information:
▶ where the particle is at any given time
▶ direction of motion

9. Restrictions on the Parameter
▶ Some parametric curves/equations may also have a restriction
on parameter t.
eg x = t2
− 2t, y = t + 1 for 0 ≤ t ≤ 2
is the parabola in Ex 1 that starts at (0, 1) and ends at (0, 3)
▶ In general, the curve with parametric equations
x = f (t) y = g(t) a ≤ t ≤ b
has initial point (f (a), g(a)) and terminal point
(f (b), g(b)).

10. Example 2
1/3
What curve is represented by the following parametric equations?
x = cos t y = sin t 0 ≤ t ≤ 2π

11. Example 2
2/3
t x = cos t y = sin t
0 1 0
π
6
√
3
2
1
2
π
4
√
2
2
√
2
2
π
3
1
2
√
3
2
π
2 0 1
.
.
.
.
.
.
.
.
.
π −1 0
.
.
.
.
.
.
.
.
.
3π
2 0 −1
.
.
.
.
.
.
.
.
.
2π 1 0
▶ We start by making a table
of points corresponding to
t = 0, π
6 , π
4 , π
3 , π
2 , . . . 2π
▶ We choose values of t that
will make computing x and
y easy.
▶ Initial point (1, 0)
▶ Terminal point (1, 0)

12. Example 2
3/3
▶ Plotting the points we have what appears to be the unit circle,
traversed counterclockwise one time.
▶ We can verify this by eliminating the parameter
▶ Using the trig identity we have:
cos2
θ + sin2
θ = 1
=⇒ x2
+ y2
= 1 ✓

13. Example 3
1/4
What curve is represented by the following parametric equations?
x = sin 2t y = cos 2t 0 ≤ t ≤ 2π

14. Example 3
2/4
▶ Again using the trig identity
cos2
θ + sin2
θ = 1 for all θ
=⇒ cos2
(2t) + sin2
(2t) = 1 for all t
=⇒ y2
+ x2
= 1
▶ So the curve represented is the unit circle but let’s investigate
this more:
▶ cos (2t) and sin (2t) each complete one full period from t = 0
to t = π.
▶ They complete two periods from t = 0 to t = 2π so the circle
is traversed twice.
▶ Initial point is
t = 0; x = sin 0 = 0, y = cos 0 = 1, → (0, 1)
▶ Terminal point is
t = 2π; x = sin 4π = 0, y = cos 4π = 1, → (0, 1)

15. Example 3
3/4
▶ Plotting a few values for t: t = 0, π
6 , π
4 , π
3 , π
2 , we see that
the circle is being traversed clockwise.
t 2t x = sin 2t y = cos 2t
0 0 0 1
π
6
π
3
√
3
2
1
2
π
4
π
2 1 0
π
3
2π
3
√
3
2 −1
2
π
2 π 0 −1

16. Example 3
4/4
▶ So the curve represented by parametric equations
x = sin 2t y = cos 2t 0 ≤ t ≤ 2π
is the unit circle traversed clockwise two times with initial and
terminal point (0, 1).

17. Example 4
1/5
Each of the following parametric equations has the Cartesian
equation x = y3. Graph the curve.
(a) x = t3, y = t
(b) x = −t3, y = −t
(c) x = t3/2, y =
√
t
(d) x = e−3t, y = e−t

18. Example 4
2/5
(a) x = t3, y = t
▶ Clearly x = y3.
▶ Plotting some points:
t x y
−2 −8 −2
−1 −1 −1
0 0 0
1 1 1
2 8 2
▶ The curve is traversed left
to right.

19. Example 4
3/5
(b) x = −t3, y = −t
▶ x = −t3 = (−t)3 = y3.
▶ Plotting some points:
t x y
−2 8 2
−1 1 1
0 0 0
1 −1 −1
2 −8 −2
▶ The curve is traversed from
right to left.

20. Example 4
4/5
(c) x = t3/2, y =
√
t
▶ Clearly x = y3.
▶ Plotting some points:
t x y
−2 undef undef
−1 undef undef
0 0 0
1 1 1
2
√
8
√
2
▶ traversed from left to right.
▶ not defined for t < 0.
▶ x ≥ 0 and y ≥ 0 for all
values of t.

21. Example 4
5/5
(d) x = e−3t, y = e−t
▶ Clearly x = y3.
▶ Plotting some points:
t x y
−2 e6 e2
−1 e3 e
0 1 1
1 1 1
2 1
e6
1
e2
▶ The curve is traversed from
right to left.
▶ x > 0 and y > 0 for all
values of t.