Download presentation

Presentation is loading. Please wait.

Published byArthur Paul Modified over 4 years ago

1
PARAMETRIC EQUATIONS Section 6.3

2
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)

3
x(t) = the ant’s horizontal location at time “t” y(t) = the ant’s vertical location at time “t” x(t) y(t) x(t) = t 2 – 2, y(t) = 3t Interval: -3 ≤ t ≤ 1 txy

4
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”

5
Parameterization: changing from rectangular to parametric form Eliminating the parameter: changing from parametric to rectangular form.

6
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

7
Ex 1) Change to the parametric equation below to rectangular form & identify the type of curve: x = 1 – 2t, y = 2 – t

8
Ex 2) x = t 2, y = t + 1

9
x = t 2 – 2, y = 3t Ex 3) x = t 2 – 2, y = 3t

10
Graph: x = t 2 – 2, y = 3t

11
Graphing Parametrics - Calculator Par Change MODE to Par Your “X” will now become a “T”

13
Graph: x =2cosѲ, y = 4sinѲ What type of graph is it? What is the general equation for this type of graph? Eliminate the paramter

14
Ex 4) x =2cosѲ, y = 4sinѲ

15
Ex 5) x = 5cosѲ, y = 5sinѲ

16
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”

17
Ex 6) Write a parametric equation representing: y=1– x 2

18
Ex 7) Write a parametric equation representing 4x 2 + 9y 2 = 36

19
Ticket Out

20
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

21
50 ft 10 ft Velocity = 25 ft/s x(t) = horizontal position @ time t x(t) = 10 + 25t Initial location Rate

22
50 ft 10 ft Velocity = 25 ft/s y(t) = height @ time t y(t) = 50 - 16t 2 Initial location “Free Fall” ft/s

23
x(t) = 25t + 10 y(t) = 16t 2 + 50 1.Find the location of the car after 3 seconds

24
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

25
Describe the location of the cargo plane 20 seconds after take off. x = 240t = 240(20) = 4800 y = 40t = 40(20) = 800 Substitute t = 20. At t = 20, the airplane has a ground distance of 4800 feet from the airport and an altitude of 800 feet.

26
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

27
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.

Similar presentations

© 2019 SlidePlayer.com Inc.

All rights reserved.

To make this website work, we log user data and share it with processors. To use this website, you must agree to our Privacy Policy, including cookie policy.

Ads by Google