Lesson 14 a - parametric equations

DEFINITION: PARAMETRIC EQUATIONS
If there are functions f and g with a common domain
T, the equations x = f(t) and y = g(t), for t in T, are
parametric equations of the curve consisting of all
points ( f(t), g(t) ), for t in T. The variable t is the
parameter.
The equations x = t + 2 and y = 3t – 1
for example are parametric equations and t is the
parameter. The equations define a graph. If t is assigned
a value, corresponding values are determined for x and
y. The pair of values for x and y constitute the
coordinates of a point of the graph. The complete graph
consists of the set of all points determined in this way

as t varies through all its chosen values. We can
eliminate t between the equations and obtain an
equation involving x and y. Thus, solving either equation
for t and substituting in the other, we get
3x – y = 7
The graph of this equation, which also the graph of
the parametric equations, is a straight line.
Example 1: Sketch the graph of the parametric
equations x = 2 + t and y = 3 – t2 .

t -3 -2 -1 0 1 2 3
x -1 0 1 2 3 4 5
y -6 -1 2 3 2 -1 -6

y

t=0
●
t=-1● ● t=1

x
t=-2 ● ● t=2

t=-3 ● ● t=3

Example 2: Eliminate the parameter between x = t + 1
and y = t2 + 3t + 2 and sketch the graph.
Solution:
Solving x = t + 1 for t, we have t = x – 1.
Substitute into y = t2 + 3t + 2, then
y = (x – 1)2 + 3(x – 1) + 2
y = x2 – 2x + 1 + 3x – 3 + 2
y = x2 + x
Reducing to the standard form,
y + ¼ = x2 + x + ¼
y + ¼ = (x + ½)2 , a parabola with V(-½,-¼)
opening upward

y

2

1

0 1 2
x
-2 -1
-1

-2

Example 3: Eliminate the parameter between x = sin t
and y = cos t and sketch the graph.
Solution:
Squaring both sides of the parametric equations, we
have
x2 = sin2 t and y2 = cos2 t
And adding the two equations will give us
x2 + y2 = sin2 t + cos2 t
But
sin2 t + cos2 t = 1
Therefore
x2 + y2 = 1 , a circle with C(0, 0) and r = 1

y

2

1

0 1 2
x
-2 -1

-1

-2

Example 4: Find the parametric representation for the
line through (1, 5) and (-2, 3).
Solution:
Letting (1, 5) and (-2, 3) be the first and second points,
respectively, of
x = x1 + r(x2 – x1)
and
y = y1 + r(y2 – y1)
We then have
x = 1 + r(-2 – 1) and y = 5 + r(3
– 5)
x = 1 – 3r y = 5 – 2r

Example 5: Eliminate the parameter between
x = sin t + cos t and y = sin t.
Solution:
Solving sin2 t + cos2 t = 1 for cos t, we have
cos t = 1 − sin2 t
Substitute into x = sin t + cos t , then
x = sin t + 1 − sin2 t
But y = sin t and y2 = sin2 t
Therefore x=y+ 1 − y2
x–y= 1 − y2
Squaring both sides
(x – y)2 = 1 – y2

Exercises:
Eliminate the parameter and sketch the curve.
• x = t2 + 1, y = t + 1
• x = t2 + t – 2 , y = t + 2
• x = cos θ , y = cos2 θ + 8 cos θ
• x = 4 cos θ , y = 7 sin θ
• x = cos θ , y = sin 2θ
• x = 1 + cos 2θ , y = 1 – sin θ

Lesson 14 a - parametric equations

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Lesson 14 a - parametric equations