10.1
Curves Defined by
Parametric
Equations
Objectives and Learning Outcomes
By the end of this lesson, you will be able to:
Recognize curves defined by parametric
equations.
Recognize the direction of a particle moving on a
parametric curve.
Describe curves using parametric equations.
 A curve in the plane is usually described by an
equation (expression) in x & y.
 Another way to describe curves in the plane is
to express the x-coordinate and y-coordinate
as an arbitrary point (x,y) on the curve in
terms of a new third variable, say t (called a
parameter):
 If I is closed interval 𝑎 ≤ 𝑡 ≤ 𝑏, then 𝑓 𝑎 , 𝑔 𝑎
𝑓 𝑏 ,𝑔 𝑏
is the initial point and
is the terminal point of the curve.
 A parametrization of a curve is describing a curve by parametric
equations.
Example 1: For the following parametric equations
a) Identify the curve by finding a Cartesian equation for it.
b) Graph the curve and indicate the direction of motion.
i.
𝑥 = 3𝑡,
𝑦 = 18𝑡 2 , −∞ < 𝑡 < ∞
Solution:
𝑥
𝑥
𝑥 = 3𝑡 → 𝑡 =
⟹ 𝑦 = 18
3
3
ii. 𝑥 = 𝑐𝑜𝑠𝑡, 𝑦 = 𝑠𝑖𝑛𝑡,
2
⇒ 𝑦 = 2𝑥 2
0 ≤ 𝑡 ≤ 2𝜋
Solution:
𝑥 2 + 𝑦 2 = 𝑐𝑜𝑠 2 𝑡 + 𝑠𝑖𝑛2 𝑡 = 1
⇒ 𝑥2 + 𝑦2 = 1
t
Circle center (0,0), counter
clockwise
(x,y)
0
(1,0)
𝜋/2
(0,1)
𝜋
(-1,0)
2𝜋
(1,0)
iii.
𝑥 = 1 + 𝑠𝑖𝑛2𝑡, 𝑦 = 𝑐𝑜𝑠2𝑡,
0 ≤ 𝑡 ≤ 2𝜋
2
2
𝑥 − 1 = 𝑠𝑖𝑛2𝑡, 𝑦 = 𝑐𝑜𝑠2𝑡 ⇒ (𝑥 − 1)2 +𝑦 2 =𝑠𝑖𝑛 2𝑡 + 𝑐𝑜𝑠 2𝑡 = 1
⇒ 𝑥−1
2
+ 𝑦2 = 1
Two circles center
(1,0), Clockwise
iv.
⇒
𝑥 = 4𝑠𝑖𝑛𝑡, 𝑦 = 5𝑐𝑜𝑠𝑡,
𝑦
= 𝑐𝑜𝑠𝑡
𝑦 = 5𝑐𝑜𝑠𝑡 →
5
𝑥 2
4
+
𝑦 2
=
5
t
(x,y)
0
(1,1)
𝜋/4
(2,0)
𝜋/2
(1,-1)
𝜋
(1,1)
0 ≤ 𝑡 ≤ 2𝜋 𝑥 = 4𝑠𝑖𝑛𝑡 →
2
2
𝑥
𝑦
𝑐𝑜𝑠 2 𝑡 + 𝑠𝑖𝑛2 𝑡 = 1 ⇒ 2 + 2 = 1
4
5
t
Ellipse, Vertical ,center
(0,0), Vertices
(0,±5),clockwise
(x,y)
0
(0,5)
𝜋/2
(4,0)
𝜋
(0,-5)
𝑥
4
= 𝑠𝑖𝑛𝑡,
HW
v.
𝑥 = 1 + 𝑡 , 𝑦 = 𝑡 2 − 4𝑡 0 ≤ 𝑡 ≤ 5 (b only)
Example 2: Find a parametrization for The line passing through two points
(0,2) and (1,0)
Solution:
∆𝑦
0−2
Slope=
=
= −2
∆𝑥
1−0
a parametrization
The equation of the line is
i. 𝑥 = 𝑡, → 𝑦 = −2(𝑡 − 1),
𝑡 ∈ −∞, ∞
ii. 𝑥 − 1 = 𝑡 → 𝑦 = −2𝑡
⇒ 𝑥 = 𝑡 + 1,
𝑦 = −2𝑡,
𝑡 ∈ −∞, ∞
𝑦 − 0 = −2 𝑥 − 1
⇒ 𝑦 = −2(𝑥 − 1),
b. The line segment with endpoints (0,2) and (1,0)
Solution:
∆𝑦
0−2
Slope =
=
= −2
∆𝑥
1−0
a parametrization
𝑖) 𝑥 = 𝑡, 𝑦 = −2(𝑡 − 1),
The equation of the line is 𝑦 − 0 = −2 𝑥 − 1
When 𝑥 = 0 → 𝑡 = 0, 𝑥 = 1 → 𝑡 = 1 → 𝑡 ∈ [0,1]
𝑖𝑖) 𝑥 − 1 = 𝑡,
𝑦 = −2𝑡
When 𝑥 = 0 → 𝑡 = −1, 𝑥 = 1 → 𝑡 = 0 → 𝑡 ∈ [−1,0]
c.
i.
The lower half of the parabola 𝑥 − 1 = 𝑦 2
𝑦=𝑡⇒
𝑥 − 1 = 𝑡2
ii. 𝑥 − 1 = 𝑡 ⇒ 𝑡 = 𝑦 2 ,
⇒ 𝑥 = 𝑡 2 + 1 ⇒ 𝑦 = 𝑡, 𝑥 = 𝑡 2 + 1, 𝑡 ≤ 0
𝑡≥0
⇒ 𝑦 = −2(𝑥 − 1),