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.