The Beginning of

Parametric Equations

(Sec. 6.3a)(Sec. 6.3a)

Consider a rock dropped from the top of a 420-foot tower…

The rock’s height y above the ground t seconds latercan be modeled with the following equation:

y = –16t + 4202

Since the horizontal position of the rock never changes,it can be modeled by an equation such as:

x = 2.5

These are examples of These are examples of parametric equationsparametric equations, with, witha a parameter parameter of t (often, t represents time)of t (often, t represents time)

Definition: Parametric Curve, Parametric Equations

The graph of the ordered pairs (x, y) where

x = f(t), y = g(t)

are functions defined on an interval I of t-values is aparametric curve. The equations are parametricequations for the curve, the variable t is a parameter,and I is the parameter interval.

Look back at our first equations:

y = –16t + 4202

x = 2.5

Let’s graph these in ourLet’s graph these in our calculator, and see if wecalculator, and see if we can interpret this graph…can interpret this graph…

More examples

2 2x t 3y t

3 1t 2 3t

For each of the given parameter intervals, use yourcalculator to graph the parametric equations

3 3t

How do the graphs differ from each other???How do the graphs differ from each other???

(A) (B) (C)

More examples

Complete the table for the parametric equations and thenplot each of the points.

t

x

y

11

00

cosx t siny t0 2 3 2 2

00

11

––11

00

00

––11

11

00

Where’s the graph???Where’s the graph???

More examples

Now, let’s explore these functions on the calculator…Now, let’s explore these functions on the calculator…What happens when we manipulate the

range for t, or the “t-step” value???

t

x

y

11

00

cosx t siny t0 2 3 2 2

00

11

––11

00

00

––11

11

00

More examples

3cosx t 3siny t

Sketch a graph of the following parametric equationsby hand, then verify your work using a calculator.

3cosx t 3siny tt x y

0 1 0

2 0 1

1 0

3 2 10

4 2

42

4

3 4 2

42

4 Verify with a calculator!!!Verify with a calculator!!!

Eliminating the Parameter,Parametrization of Curves

In some situations, we can In some situations, we can eliminate the parametereliminate the parameterfrom parametric equations, obtaining a rectangularfrom parametric equations, obtaining a rectangularequation that represents the curve…equation that represents the curve…

Ex: x = 1 – 2t, y = 2 – t, – < t <

8 8Solve the first equation for t:

Substitute for t in the second equation:

y = .5x + 1.5y = .5x + 1.5 Identify the curve!!!

2 1t x 11

2t x

2y t

12 12

y x

Eliminate the parameter and identify the graph:Eliminate the parameter and identify the graph:

x = t – 2, y = 3t

y = + 3 x + 2y = + 3 x + 2 Identify the curve!!!

2

––Can we graph the curve in both forms???Can we graph the curve in both forms???

Eliminate the parameter and identify the graph:Eliminate the parameter and identify the graph:

x + y = 4 x + y = 4

2cosx t 2siny t 0 2πt First, check the graph…First, check the graph…

22 22

Using Using vectorsvectors, we can also find parametric equations, we can also find parametric equationsfor a line or a line segment:for a line or a line segment:

Find a parametrization of the line through the pointsA(–2, 3) and B(3, 6).

x

y

O

A(–2, 3)

B(3, 6)

P(x, y)OA + AP = OP

AP = OP – OA

OA + AB = OB

AB = OB – OA

AP must be a scalarmultiple of AB

(let the scalar be “t”)

OA 2,3 ,OB 3,6 ,OP ,x y BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

Using Using vectorsvectors, we can also find parametric equations, we can also find parametric equationsfor a line or a line segment:for a line or a line segment:

Find a parametrization of the line through the pointsA(–2, 3) and B(3, 6).

x

y

O

A(–2, 3)

B(3, 6)

P(x, y)OP – OA = t(OB – OA)

OA 2,3 ,OB 3,6 ,OP ,x y BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB

AP = tAB

x + 2, y – 3 = t 5, 3

x + 2 = 5t y – 3 = 3t

x = –2 + 5t y = 3 + 3t

Try graphing theseTry graphing these parametrics!!!parametrics!!!

Now, how do we find the equation of the segmentthrough the same two points???

x

y

O

A(–2, 3)

B(3, 6)

x = –2 + 5t y = 3 + 3t

What happens when weplug in t = 0 and 1???

So, use the same equations, but restrict trestrict t!!!

x = –2 + 5t y = 3 + 3t 0 < t < 1

t = 0 produces point A,t = 1 produces point B

Find a parametrization for the line segment withendpoints (5,2) and (–2,–4).

5 7x t One possibility:

2 6y t 0 1t

Find a parametrization for the circle with center (–2,–4)and radius 2.

2 2cosx t One possibility:

4 2siny t

0 2t