PARAMETRIC EQUATIONS

Section 6.3

Parameter

A third variable “t” that is related to both x & y

Ex)The ant is LOCATED at a point (x,

y)

Its location changes based on TIME (t)

x(t) = the ant’s horizontal location at time “t”

y(t) = the ant’s vertical location at time “t”

x(t) = t2 – 2 , y(t) = 3tInterval: -3 ≤ t ≤ 1t x y

Rectangular Form vs. Parametric

Rectangular Form: an equation written in terms of only two variables (what you have used in math up to this point).

Parametric Form: an equation defined by a third variable “t”

Parameterization: changing from rectangular to parametric form

Eliminating the parameter: changing from parametric to rectangular form.

Parametric Rectangular

Step 1: solve one of the equations for t

Step 2: Substitute into the other

equation

Step 3: Simplify

*If the graph is a circle a different process is used

Ex 1) Change to the parametric equation below to rectangular form & identify the type of curve:

x = 1 – 2t , y = 2 – t

Ex 2) x = t2 , y = t + 1

Ex 3) x = t2 – 2 , y = 3t

Graph: x = t2 – 2 , y = 3t

Graphing Parametrics - Calculator

Change MODE to Par Your “X” will now become a “T”

Graph: x =2cosѲ, y = 4sinѲ

What type of graph is it?

What is the general equation for this type of graph?

Eliminate the paramter

Ex 4) x =2cosѲ, y = 4sinѲ

Ex 5) x = 5cosѲ, y = 5sinѲ

Rectangular Parametric

“Parametization” Let x = t (or whatever you want!)

Sub “t” (or whatever) in for “x” into y =

*Ellipse & circles – sub in “cos” & “sin”

Ex 6) Write a parametric equation representing: y=1– x2

Ex 7) Write a parametric equation representing 4x2 + 9y2 = 36

Ticket Out

A car is about to drive off a cliff. What are all the different aspects of the situation? What different measurements exist?

•Driving forward (horizontally)

•Falling downwards (vertically)

•Driving at a certain speed

(velocity)

•Time is passing

50 ft

10 ft

Velocity = 25 ft/s

x(t) = horizontal position @ time t

x(t) = 10 + 25t

Initial location

Rate

50 ft

10 ft

Velocity = 25 ft/s

y(t) = height @ time t

y(t) = 50 - 16t2

Initial location

“Free Fall”ft/s

x(t) = 25t + 10

y(t) = 16t2 + 501.Find the location of the car after 3 seconds

2. As a cargo plane ascends after takeoff, its altitude increases at a rate of 40 ft/s. while its horizontal distance from the airport increases at a rate of 240 ft/s.

Use the distance formula d = rt.

x = 240t

y = 40t

Describe the location of the cargo plane 20 seconds after take off.

x = 240t = 240(20) = 4800

y = 40t = 40(20) = 800Substitute t = 20.

At t = 20, the airplane has a ground distance of 4800 feet from the airport and an altitude of 800 feet.

3. A helicopter takes off with a horizontal speed of 5 ft/s and a vertical speed of 20 ft/s.

x = 5t

y = 20t

Describe the location of the helicopter at t = 10 seconds.

Substitute t = 10.x = 5t =5(10) = 50

y = 20t =20(10) = 200

At t = 10, the helicopter has a ground distance of 50 feet from its takeoff point and an altitude of 200 feet.