The Beginning of Parametric Equations (Sec. 6.3a).
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Transcript of The Beginning of Parametric Equations (Sec. 6.3a).
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The Beginning of
Parametric Equations
(Sec. 6.3a)(Sec. 6.3a)
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Consider a rock dropped from the top of a 420-foot tower…
The rock’s height y above the ground t seconds latercan be modeled with the following equation:
y = –16t + 4202
Since the horizontal position of the rock never changes,it can be modeled by an equation such as:
x = 2.5
These are examples of These are examples of parametric equationsparametric equations, with, witha a parameter parameter of t (often, t represents time)of t (often, t represents time)
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Definition: Parametric Curve, Parametric Equations
The graph of the ordered pairs (x, y) where
x = f(t), y = g(t)
are functions defined on an interval I of t-values is aparametric curve. The equations are parametricequations for the curve, the variable t is a parameter,and I is the parameter interval.
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Look back at our first equations:
y = –16t + 4202
x = 2.5
Let’s graph these in ourLet’s graph these in our calculator, and see if wecalculator, and see if we can interpret this graph…can interpret this graph…
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More examples
2 2x t 3y t
3 1t 2 3t
For each of the given parameter intervals, use yourcalculator to graph the parametric equations
3 3t
How do the graphs differ from each other???How do the graphs differ from each other???
(A) (B) (C)
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More examples
Complete the table for the parametric equations and thenplot each of the points.
t
x
y
11
00
cosx t siny t0 2 3 2 2
00
11
––11
00
00
––11
11
00
Where’s the graph???Where’s the graph???
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More examples
Now, let’s explore these functions on the calculator…Now, let’s explore these functions on the calculator…What happens when we manipulate the
range for t, or the “t-step” value???
t
x
y
11
00
cosx t siny t0 2 3 2 2
00
11
––11
00
00
––11
11
00
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More examples
3cosx t 3siny t
Sketch a graph of the following parametric equationsby hand, then verify your work using a calculator.
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3cosx t 3siny tt x y
0 1 0
2 0 1
1 0
3 2 10
4 2
42
4
3 4 2
42
4 Verify with a calculator!!!Verify with a calculator!!!
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Eliminating the Parameter,Parametrization of Curves
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In some situations, we can In some situations, we can eliminate the parametereliminate the parameterfrom parametric equations, obtaining a rectangularfrom parametric equations, obtaining a rectangularequation that represents the curve…equation that represents the curve…
Ex: x = 1 – 2t, y = 2 – t, – < t <
8 8Solve the first equation for t:
Substitute for t in the second equation:
y = .5x + 1.5y = .5x + 1.5 Identify the curve!!!
2 1t x 11
2t x
2y t
12 12
y x
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Eliminate the parameter and identify the graph:Eliminate the parameter and identify the graph:
x = t – 2, y = 3t
y = + 3 x + 2y = + 3 x + 2 Identify the curve!!!
2
––Can we graph the curve in both forms???Can we graph the curve in both forms???
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Eliminate the parameter and identify the graph:Eliminate the parameter and identify the graph:
x + y = 4 x + y = 4
2cosx t 2siny t 0 2πt First, check the graph…First, check the graph…
22 22
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Using Using vectorsvectors, we can also find parametric equations, we can also find parametric equationsfor a line or a line segment:for a line or a line segment:
Find a parametrization of the line through the pointsA(–2, 3) and B(3, 6).
x
y
O
A(–2, 3)
B(3, 6)
P(x, y)OA + AP = OP
AP = OP – OA
OA + AB = OB
AB = OB – OA
AP must be a scalarmultiple of AB
(let the scalar be “t”)
OA 2,3 ,OB 3,6 ,OP ,x y BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
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Using Using vectorsvectors, we can also find parametric equations, we can also find parametric equationsfor a line or a line segment:for a line or a line segment:
Find a parametrization of the line through the pointsA(–2, 3) and B(3, 6).
x
y
O
A(–2, 3)
B(3, 6)
P(x, y)OP – OA = t(OB – OA)
OA 2,3 ,OB 3,6 ,OP ,x y BBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBBB
AP = tAB
x + 2, y – 3 = t 5, 3
x + 2 = 5t y – 3 = 3t
x = –2 + 5t y = 3 + 3t
Try graphing theseTry graphing these parametrics!!!parametrics!!!
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Now, how do we find the equation of the segmentthrough the same two points???
x
y
O
A(–2, 3)
B(3, 6)
x = –2 + 5t y = 3 + 3t
What happens when weplug in t = 0 and 1???
So, use the same equations, but restrict trestrict t!!!
x = –2 + 5t y = 3 + 3t 0 < t < 1
t = 0 produces point A,t = 1 produces point B
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Find a parametrization for the line segment withendpoints (5,2) and (–2,–4).
5 7x t One possibility:
2 6y t 0 1t
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Find a parametrization for the circle with center (–2,–4)and radius 2.
2 2cosx t One possibility:
4 2siny t
0 2t