10.3 Vector Valued Functions

Greg Kelly, Hanford High School, Richland, Washington

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

8

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.

Note: The magnitude of the velocity is by definition identical to the speed which is a scalar (not a vector) and never negative; however, velocity is a vector because it has direction and magnitude

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

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

velocity: 3sin 3cos4 4 4

v i j

3 3

2 2 i j

acceleration: 3cos 3sin4 4 4

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

3 3

4 2 2

v i j

3 3

4 2 2

a i j

speed:4

v

2 23 3

2 2

9 9

2 2 3

direction:

/ 4

/ 4

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 line where .1t

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

position: 5,11 Slope=9

12

To write equation:

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