in chapter 1, we talked about parametric equations

In chapter 1, we talked about parametric equations.Parametric equations can be used to describe motion that is not a function.

x f t y g t

If f and g have derivatives at t, then the parametrized curve also has a derivative at t.

The formula for finding the slope of a parametrized curve is:

dy

dy dtdxdxdt

This makes sense if we think about canceling dt.

The formula for finding the slope of a parametrized curve is:

dy

dy dtdxdxdt

We assume that the denominator is not zero.

To find the second derivative of a parametrized curve, we find the derivative of the first derivative:

dydtdxdt

2

2

d y

dx dy

dx

1. Find the first derivative (dy/dx).2. Find the derivative of dy/dx with respect to t.

3. Divide by dx/dt.

22 3

2Find as a function of if and .

d yt x t t y t t

dx

22 3

2Find as a function of if and .

d yt x t t y t t

dx

1. Find the first derivative (dy/dx).

dy

dy dtydxdxdt

21 3

1 2

t

t

2. Find the derivative of dy/dx with respect to t.

21 3

1 2

dy d t

dt dt t

2

2

2 6 6

1 2

t t

t

Quotient Rule

3. Divide by dx/dt.

2

2

d y

dx

dxdt

dydt

2

2

2 6 6

1 2

1 2

t t

t

t

2

3

2 6 6

1 2

t t

t

Ldx

dt

dy

dtdt

a

b F

HGIKJ FHG

IKJz 2 2

The equation for the length of a parametrized curve is similar to our previous “length of curve” equation:

1. Revolution about the -axis 0x y 2 2

2b

a

dx dyS y dt

dt dt

2. Revolution about the -axis 0y x

2 2

2b

a

dx dyS x dt

dt dt

This curve is:

sin 2

2cos 5

x t t t

y t t t

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”

This could be substituted in the formula for the angle between vectors to give:

1cos

u v

u v

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