chapter 8 sec 6
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Vectors and Parametric Equations. Chapter 8 SEC 6. Vector Equations. A vector equation and equations known as parametric equations gives us a way to track the position of a moving object for given a moment of time. Tracking an airplanes movement. Parametric Equations. - PowerPoint PPT PresentationTRANSCRIPT
Pre-calculus Chapter 8 Sections 6 & 7
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Vector Equations A vector equation and equations known as
parametric equations gives us a way to track the
position of a moving object for given a moment of
time.
Tracking an airplanes movement.
Pre-calculus Chapter 8 Sections 6 & 7
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If a line passes through the point P1 and P2 and is parallel to the vector
, the vector is also parallel to
Thus must be a scalar multiple of
Using the scalar t, we get Notice both are vectors. This is called the vector equation of a line.
Since is parallel to the line, it is called a directional vector. The scalar t is called a parameter.
21,aaa
Parametric Equations
.a 21PP
21PP .a
.21 atPP
a
Pre-calculus Chapter 8 Sections 6 & 7
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Write a vector equation describing a line passing through P1(1, 4) and parallel to
Let the line l through P1 (1, 4) be parallel to For any
point P (x, y) on l,
Since P1 P is on l and is parallel to for some
value t. By substitution we have
A vector equation can be used to describe the coordinate for a point on a line for any value of the parameter t.
Example 1
.a
.2 ,3 a
.4,11 yxPP
atPPa 1,
2,34,1 tyx
Pre-calculus Chapter 8 Sections 6 & 7
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• For when t = 4 we get
Then x – 1 = 12 and y – 4 = –8 to find ordered pair (13, –4).
When t = 0 we get (1, 4) , t often represents time (Hence the t)
An object moving along will be at (1, 4) at time t = 0 and point (13, –4) at time t = 4.
Parametric Vectors2,34,1 tyx
.8,122,344,1 yx
2,34,1 tyx
Pre-calculus Chapter 8 Sections 6 & 7
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As we saw, the vector equation can be written as two equations relating to the horizontal and vertical components.
x – x1 = ta1 and y – y1 = ta2
x = x1 + ta1 y = y1 + ta2
Parametric Vectors
2111 ,, aatyyxx
Pre-calculus Chapter 8 Sections 6 & 7
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Find the parametric equation for a line parallel to and passing through the point (–2, –4). Then make a table of values and graph the line.
Use the general form of the parametric equations of a line with
x = x1 + ta1 y = y1 + ta2
x = – 2 + 6t y = – 4 – 3t
Example 23,6 q
.4,2, and 3,6, 1121 yxaa
t x y– 1
0
1
2
t x y– 1 – 8 – 1
0
1
2
t x y– 1 – 8 – 1
0 – 2 – 4
1
2
t x y– 1 – 8 – 1
0 – 2 – 4
1 4 – 7
2 10 – 10
Pre-calculus Chapter 8 Sections 6 & 7
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Write a parametric equation of y = – 4x + 7.
In the equation x is the independent variable and y is the dependant variable.
In parametric equations t is the independent variable and x and y are dependant.
So, we set the independent variables x and t equal, then we can write two parametric equations in terms of t.
x = t and y = – 4t + 7If we make a table of values and plot both the parametric and linear equations we will see they describe the same line.
Example 3
Pre-calculus Chapter 8 Sections 6 & 7
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Example 4Write an equation in slope-intercept form of the line whose parametric equations are x = – 2 + t and y = 4 – 3t.
1st - Solve for t.
x = – 2 + t
x + 2 = t ty
3
4y = 4 – 3t
y – 4 = – 3t
3
42
yx
463 yx
23 xy
Pre-calculus Chapter 8 Sections 6 & 7
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Objects in Motion
Object that are launched, like a football, are called projectiles.
The path of a projectile is called its trajectory.
The horizontal distance is its range.Physicists describe the motion in terms of its
position, velocity and acceleration.
All can be represented by vectors.
Pre-calculus Chapter 8 Sections 6 & 7
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Objects in Motion A punted kicks a ball and the
initial trajectory is described by: The magnitude and direction θ will
describe a vector with components . As the ball moves gravity acts on the vertical
direction, while horizontal is unaffected . If we discount air friction the horizontal
speed is constant throughout flight.
v
. and yx vv
Pre-calculus Chapter 8 Sections 6 & 7
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Objects in Motion Note the vertical component
is large and positive in the beginning.
Decreasing to zero at the top At the end the vertical speed is the same
magnitude but the opposite direction. Parametric equations can represent the position
of the ball relative to the starting point in terms of time.
Pre-calculus Chapter 8 Sections 6 & 7
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Objects in Motion Write the horizontal and vertical
components of the initial velocity.
v
vx
cosv
vy
sin
cosvvx
sinvvy
Pre-calculus Chapter 8 Sections 6 & 7
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Example 1Find the initial horizontal velocity and vertical velocity of a ball kicked with an initial velocity of 18 feet per second at an angle of 37°with the ground.
cosvvx
37sin18yv 37cos18xv
sinvvy
ft/sec 14xv ft/sec 11yv
Pre-calculus Chapter 8 Sections 6 & 7
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It’s about time Vertical velocity is affected by gravity, so we must
adjust the vertical component by subtracting the vertical displacement of gravity we can determine the vertical position after t seconds .
The equation for gravity of a free falling object is
Vertical distance
timeVertical Velocity
2
2
1sin gtvty
seconds.in is,/32or /8.9 where2
1 222 t sftsmggth
Displacement due to gravity
Pre-calculus Chapter 8 Sections 6 & 7
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It’s about time Because the horizontal velocity is unaffected by
gravity, the horizontal position can be found after t seconds by the following.
cosvtx
Horizontal distance
timeHorizontal
velocity
Pre-calculus Chapter 8 Sections 6 & 7
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Example 2Sammy Baugh of the Washington Redskins has the record for the highest average punting record for a lifetime average of 45.16 yards. Suppose that he kicked the ball with an initial velocity of 26 yards per second at an angle of 72°.
A.How far has the ball traveled horizontally and what is its vertical height at the end of 3 seconds.
cosvtx
23322
172sin783 y
72cos263x
2
2
1sin gtvty
yards 1.24x
yards 1.26or ft 5.78y
Because g = 32 ft/sec2
26yds = 78 ft
Pre-calculus Chapter 8 Sections 6 & 7
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Example 2Sammy Baugh of the Washington Redskins has the record for the highest average punting record for a lifetime average of 45.16 yards. Suppose that he kicked the ball with an initial velocity of 26 yards per second at an angle of 72°.
A.How far has the ball traveled horizontally and what is its vertical height at the end of 3 seconds.
B.Suppose that the kick returner lets the ball hit the ground instead of catching it. What is the hang time?
The height as the ball hits the ground is 0. So we need to find the time, t when the height or y = 0.
2
2
1sin0 gtvt
2322
172sin780 tt
21618.740 tt tt 1618.740
t1618.74
seconds 4.5about or 63.4t
Pre-calculus Chapter 8 Sections 6 & 7
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Initial height
Parametric equations describe objects launched from the ground. If objects are launched from above ground you must add in the initial height to the vertical component, y.
Pre-calculus Chapter 8 Sections 6 & 7
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Example 3A baseball is thrown at an angle of 5.1° with the horizontal at a speed of 85 mile per hour. The distance from the pitcher to home is 60.5 feet. If the ball is released 2.9 feet above the ground, how far above the ground is the ball when it crosses home plate?
Remember to convert 85 mph to ft/sec…
cosvtx
hgtvty 2
2
1sin
ft/sec 66.124 1.5cos66.1245.60 t
483.t
9.2483.322
11.5sin66.124483. 2 y
ft 4.5about or 519.4y