Transcript Parametric Equations - Pleasanton Unified School District

Parametric Equations
• ordered pairs (x , y) are based upon a third
variable, t, called the parameter
• t usually represents time
• there are two equations, one for x and one
for y, each in terms of t.
• An interval, I, is provided to define the
values for t [ tmin , tmax ]
Ex.
x  t 1
y t2
I  [3 , 3]
2
t
-3
-2
-1
0
1
2
3
x
y
Ex.
x  t 1
y t2
I  [3 , 3]
2
t
-3
-2
-1
0
1
2
3
x
8
3
0
-1
0
3
8
y
-1
0
1
2
3
4
5
Graphing Parametric Eq.
• Graphs can be made without a calculator by
plotting the points found in the previous
table
Graphing Parametric Eq.
• Graphs can be made without a calculator by
plotting the points found in the previous
table
• Graphs can be made with a calculator by
changing the mode
MODE  PAR (instead of FUNC)
Graphing Parametric Eq.
• Graphs can be made without a calculator by
plotting the points found in the previous
table
• Graphs can be made with a calculator by
changing the mode
MODE  PAR (instead of FUNC)
go to y = to see how that screen has changed
The graph of the parametric curve from the example
above looks like:
Eliminating the Parameter
• Parametric equations can be changed into a
rectangular equation with x and y by
“eliminating the parameter”
• Solve one equation for t and then substitute
it into the other equation and simplify.
Ex. #1
x = 2t + 1
y=t–1
solution: solve the 2nd equation for t and then
substitute the expression into the 1st equation
t=y+1
x = 2(y + 1) + 1
x = 2y + 3
x – 3 = 2y
.5x – 1.5 = y or y = .5x – 1.5
( the equation of a line)
Ex #2
x t 2
y  3t
2
y = 3t  t = y/3
(solve 2nd equation for t)
x = (y/3)2 – 2
(substitute into the 1st equation)
x = y2/9 – 2
(simplify)
y2 = 9(x + 2)
(the equation of a parabola)
Ex #3 (special case) x = 2 cos t
y = 2 sin t
I = [0 , 2π]
If you graph this it looks like a circle, let’s see why.
x2 + y2 = 4cos2t + 4sin2t
= 4(cos2t + sin2t)
= 4(1)
=4
so x2 + y2 = 4 (the
equation of circle)
Practice Problem #1
Eliminate the parameter:
and describe the graph
x  t 1
y  2t
2
Solution:
t=x+1
y = 2(x + 1)2 or
y = 2x2 + 4x + 2 (parabola which opens up)
Finding the Parametrization of a Line
• Given two points, finding parametric equations to
describe the line or segment containing them
• Use the following formula (not found in the book)
given points ( x1 , y1 ) and ( x2 , y2 )
x  x1  ( x2  x1 )t
y  y1  ( y2  y1 )t
Notes:
• it doesn’t matter which point is #1 or #2
• for a segment, set I = [0 , 1]
• for a line, set I = (-  , )
Practice Problem #2
Find the parametrization of the line segment
between (– 4 , 5) to (3 , – 2).
Solution:
x = – 4 + (3 – (– 4))t
y = 5 + (– 2 – 5)t
x = – 4 + 7t
y = 5 – 7t
I = [0 , 1]
or
x = 3 – 7t
y = –2 + 7t
I = [0 , 1]