chapter 8 sec 6

Vectors and Parametric Equations

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.


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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


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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


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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


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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


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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


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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


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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

Modeling Motion Using Parametric Equations


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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.


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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


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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.


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Objects in Motion Write the horizontal and vertical

components of the initial velocity.

v

vx

cosv

vy

sin

cosvvx

sinvvy


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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


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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


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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


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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


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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


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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.


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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


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Daily Assignment Chapter 8 Sections 6 & 7 Text Book

Pg 524○#13 – 31Odd;

Pg 531○#9 – 15 Odd;

chapter 8 sec 6

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