Download Module 10 lesson 6 Parametric Equations. When modeling the path

Module 10 lesson 6 Parametric Equations.
When modeling the path of an object, it is useful to use equations called
Parametric equations. Instead of using one equation with two
variables, we will use two equations and a third variable called a
parameter. This third variable allows us to determine not only where
the object has been, but it can tell us when the object was there.
A parametric equation in a plane consists of two equations
x = f(t) and y = g ( t )
Where x and y are ordered pairs and t is the parameter
( a constraint that you operate in , like time).
Here is an example of a parametric equation:
x = t and y = -2t when -2≤ 𝑡 ≤ 2
To sketch the graph we need to evaluate the parameter t within the given
interval to create our x and y values.
●
t
-2
-1
0
1
2
x
-2
-1
0
1
2
y
4
2
0
-2
-4
●
●
●
●
The arrows indicate the movement of the particle at the given parameter.
We can use our graphing calculator to graph parametric equations.
First, put your calculator into the parametric mode by hitting [MODE] and
choosing the [PAR] option. In the Mode menu, set your calculator to
[RADIAN] mode instead of [DEGREE] mode. They are used for trigonometric
Functions. After setting up your calculator for parametric mode, notice that
when you hit the Y= key, you no longer have a y1=. You now have a pair of
equations, an x and a y that are both functions of t. Simply enter the
parametric equations in for x and y. Notice that the key you have been
using for X is also marked T. In parametric mode, a T will automatically
appear instead of the X.
If we are given parametric equations we can write it as an equation in slope
intercept form.
Let’s say that we have parametric equations, x = -5 + 4t and y = 2-3t. We
can convert to a rectangular equation. This is called eliminating the
parameter.
First solve the equation x = -5 + 4t for the parameter, t.
So ( x + 5 ) / 4 = t.
Then substitute the expression into the other parametric equation for t.
So 𝑦 = 2 − 3 (
𝑥+5
4
)
then 4y = 8 -3(x+5 ) = -3x + 7
Or y = -3/4 x + 7/4.
This is a linear equation with slope of -3/4 and a y-intercept of 7/4.
Example Set 1:
Eliminate the parameter for the following parametric equations.
1. x = 3t-3
y = 2t +1
2. x = t + 2
y = t2
Going in the reverse, finding a parametric equation for a rectangular, is not
a unique solution. There is many interpretations depending on the
given parameter.
For example if given the equation y = 1 – 3x and the given parameter of t =
x, the parametric form then becomes x = t and y = 1-3t.
If the parameter is given as t = 1 – x then the parametric equations
become y = 1 – 3( 1-x ) = 1-3 + 3x or -2 +3x
So x = 1-t and y = -2+3x would be the final solution.
Exercise set 2
Find a set of parametric equations for y = 3x – 2 given the parameters:
1. t= x
2. t = 2-x
Parametric equations and Conic sections
Parametric equations can also represent the conic sections ( parabola, circle,
ellipse, hyperbola)
Let’s say that we have the parametric equations
X = t – 1 and y = 2t2 on [ 0, ∞)
Solving the x equation for t, we have t = x + 1. Substituting this into the y
equation we have:
Y = 2 ( x + 1 )2 . This is the equation for a parabola with vertex ( 0, 1 ).
By creating a table, we can then plot the graph using the parametric
equations.
t
0
1
2
3
x
-1
0
1
2
y
0
2
8
18
In the next example, let’s say that x = 3cos t and y = 5 sin t on [ 0, 2𝜋]
Solving both equations for sin t and cos t we have:
cos t = x/3 and sin t = y/5
We need a trigonometric relationship for sin and cos. Using
cos2t + sin2t = 1 and substituting in the above values, we have,
(x/3)2+ (y/5)2=1
Or
𝑥2
9
𝑦2
+ 25 = 1
This is the equation for the ellipse with center (0,0) with a horizontal major
axis.
Then next example is
x = cot t -3 and y = 5csc t
Solving for the trig function, we have
𝑥+3
𝑦+1
2
5
=cot t and
= csc 𝑡
The trig identity that we will use this time is
1 + cot2t = csc2t and substituting in the above equations, we have
1+(
𝑥+3 2
𝑦+1 2
)
=(
)
2
5
or
(𝑦+1)2
25
−
(𝑥−3)2
4
= 1 which is the equation of a hyperbola
with center ( 3, -1) and a vertical transverse axis.
Example set 3:
Transform the given parametric equations into rectangular form. Then
identify the conic .
X = 2 sec t and y = 7 tan t
Answers Example set 1
1. y = 2/3 x + 3
2. y = x2-4x+4
Answers Example set 2
1. x=t , y=3t-2
2. x=-t+2, y= -3t+4
Answers Example set 3
𝑥2
4
𝑦2
− 49 = 1 which is a hyperbola with center ( 0,0 ) and horizontal transverse
axis.