10.4 MINIMAL PATH PROBLEMS

10.5 MAXIMUM AND MINIMUM PROBLEMS IN MOTION AND ELSEWHERE

10.6 VECTOR FUNCTIONS FOR MOTION IN A PLANE

Warning:

Only some of this is review.

Quantities that we measure that have magnitude but not direction are called scalars.

Quantities such as force, displacement or velocity that have direction as well as magnitude are represented by directed line segments.

A

B

initialpoint

terminalpoint

AB��������������

The length is AB��������������

A

B

initialpoint

terminalpoint

AB��������������

A vector is represented by a directed line segment.

Vectors are equal if they have the same length and direction (same slope).

A vector is in standard position if the initial point is at the origin.

x

y

1 2,v v

The component form of this vector is: 1 2,v vv

A vector is in standard position if the initial point is at the origin.

x

y

1 2,v v

The component form of this vector is: 1 2,v vv

The magnitude (length) of 1 2,v vv is:2 2

1 2v v v

P

Q

(-3,4)

(-5,2)

The component form of

PQ��������������

is: 2, 2 v

v(-2,-2) 2 2

2 2 v

8

2 2

If 1v Then v is a unit vector.

0,0 is the zero vector and has no direction.

Vector Operations:

1 2 1 2Let , , , , a scalar (real number).u u v v k u v

1 2 1 2 1 1 2 2, , ,u u v v u v u v u v

(Add the components.)

1 2 1 2 1 1 2 2, , ,u u v v u v u v u v

(Subtract the components.)

Vector Operations:

Scalar Multiplication:1 2,k ku kuu

Negative (opposite): 1 21 ,u u u u

v

vu

u

u+vu + v is the resultant vector.

(Parallelogram law of addition)

The angle between two vectors is given by:

1 1 1 2 2cosu v u v

u v

This comes from the law of cosines.See page 524 for the proof if you are interested.

The dot product (also called inner product) is defined as:

1 1 2 2cos u v u v u v u v

Read “u dot v”

Example:

3,4 5,2

3 5 4 2 23

The dot product (also called inner product) is defined as:

1 1 2 2cos u v u v u v u v

This could be substituted in the formula for the angle between vectors (or solved for theta) to give:

1cos

u v

u v

Find the angle between vectors u and v:

2,3 , 2,5 u v

1cos

u v

u v

Example:

1 2,3 2,5cos

2,3 2,5

1 11cos

13 29

55.5

Application: Example 7

A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?

N

E

Application: Example 7

A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?

N

Eu

Application: Example 7

A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?

N

E

v

u

60o

Application: Example 7

A Boeing 727 airplane, flying due east at 500mph in still air, encounters a 70-mph tail wind acting in the direction of 60o north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?

N

E

v

u

We need to find the magnitude and direction of the resultant vector u + v.

u+v

N

E

v

u

The component forms of u and v are:

u+v

500,0u

70cos60 ,70sin 60v

500

70

35,35 3v

Therefore: 535,35 3 u v

538.4 22535 35 3 u v

and: 1 35 3tan

535 6.5

N

E

The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5o north of east.

538.4

6.5o

p

Any vector can be written as a linear

combination of two standard unit vectors.

,a bv

1,0i 0,1j

,a bv

,0 0,a b

1,0 0,1a b

a b i j

The vector v is a linear combination

of the vectors i and j.

The scalar a is the horizontal

component of v and the scalar b is

the vertical component of v.

We can describe the position of a moving particle by a vector, r(t).

tr

If we separate r(t) into horizontal and vertical components,

we can express r(t) as a linear combination of standard unit vectors i and j.

t f t g t r i j f t i

g t j

In three dimensions the component form becomes:

f g ht t t t r i j k

Graph on the TI-89 using the parametric mode.

MODE Graph……. 2 ENTER

Y= ENTERxt1 t cos t

yt1 t sin t

WINDOW

GRAPH

cos sin 0t t t t t t r i j

8p

Graph on the TI-89 using the parametric mode.

cos sin 0t t t t t t r i j

MODE Graph……. 2 ENTER

Y= ENTERxt1 t cos t

yt1 t sin t

WINDOW

GRAPH

Most of the rules for the calculus of vectors are the same as we have used, except:

Speed v t

velocity vectorDirection

speed

t

tv

v

“Absolute value” means “distance from the origin” so we must use the Pythagorean theorem.

Example 5: 3cos 3sint t t r i j

a) Find the velocity and acceleration vectors.

3sin 3cosd

t tdt

r

v i j

3cos 3sind

t tdt

v

a i j

b) Find the velocity, acceleration, speed and direction of motion at ./ 4t p

Example 5: 3cos 3sint t t r i j

3sin 3cosd

t tdt

r

v i j 3cos 3sind

t tdt

v

a i j

b) Find the velocity, acceleration, speed and direction of motion at ./ 4t p

velocity: 3sin 3cos4 4 4

p p p v i j

3 3

2 2 i j

acceleration: 3cos 3sin4 4 4

p p p a i j

3 3

2 2 i j

Example 5: 3cos 3sint t t r i j

3sin 3cosd

t tdt

r

v i j 3cos 3sind

t tdt

v

a i j

b) Find the velocity, acceleration, speed and direction of motion at ./ 4t p

3 3

4 2 2

p v i j

3 3

4 2 2

p a i j

speed:4

p v

2 23 3

2 2

9 9

2 2 3

direction:

/ 4

/ 4

pp

v

v3/ 2 3/ 2

3 3

i j

1 1

2 2 i j

Example 6: 3 2 32 3 12t t t t t r i j

2 26 6 3 12d

t t t tdt

r

v i j

a) Write the equation of the tangent where .1t

At :1t 1 5 11 r i j 1 12 9 v i j

position: 5,11 slope:9

12

tangent: 1 1y y m x x

311 5

4y x

3 29

4 4y x

3

4

The horizontal component of the velocity is .26 6t t

Example 6: 3 2 32 3 12t t t t t r i j

2 26 6 3 12d

t t t tdt

r

v i j

b) Find the coordinates of each point on the path where the horizontal component of the velocity is 0.

26 6 0t t 2 0t t

1 0t t 0, 1t

0 0 0 r i j

1 2 3 1 12 r i j

1 1 11 r i j

0,0

1, 11

p

…interesting how they posed this question. They are not asking for the acceleration at t = 1 they are asking how much of the acceleration helped the velocity! This is not as easy to compute.

This is not the acceleration at t = 1. This is how much of the acceleration helped the velocity!