Lecture 3

Lecture 3

Author: Kemal Ahmed Instructor: Dr. French Course: Math 2ZZ3 Date: 2013-07-03

Math objects made using MathType.

Chapter 12.4 – Complex Fourier Series

Recall: the Fourier Series of a function defines a periodic function with a fundamental period, 2

T = p. Thus, you can re-write the series in the following way:

( )

in x

n n

f x c e ω ∞

−

=−∞

=

∑

,where 2

T

π

ω = , where ω is the fundamental angular frequency.

The plot points,

(

n_ω,c_n

)

, are called the frequency spectrum of f. e.g. 1)

Find the frequency spectrum of the periodic wave that is the periodic extension of

( )

1, 2 0

1, 0 2

x f x

x

− − < <



=  _{< <}



First, you need to find the Fourier coefficients

( )

( )

( )

2

2 2

2 2

2 2

0 2

2 0

0 2

2 0

2 1

d 4

1 1

1 d 1 d

4 4

1 2 1 2

4 4

x

x n

n n

in n

in in

x x

in in

x x

p

c f x c x

e x e x

e e

in in

π

π π

π π

π π

− −

− −

−

= =

− −

=− =

=

=

= − +

−

= ⋅ + ⋅

− −

∫

∫

∫

Specify n≠ 0.

n = 0:

( )

( )

( )

( )

2

0 0

2 2

2

0 2

2 0

1

d 4

1

d 4

1 1

1 d 1 d

4 4

0

c f x e x

f x x

x x

−

−

−

=

=

= − +

=

∫

∫

∫

∫

You know this because it’s odd.

( )

1 n 1

i nπ

 

= _ − − _

( )

( )

_{1 1} 2_{, 0}

x n

n in

n i

f x e n

n

π

π

=∞

=−∞

 

∴ =

∑

_ − − _ ≠ ← 2

x in

e π is from the cos(x).

2 4

T = p=

( )

2 2

0, 0

1 1

, 0

n n

T

n c

n n

π π ω

π

= =

= 

 =  − −

≠ 



3

0, , , ,...

2 2

nω= ±π ± ±π π

(

)

( )

( )

(

)

(

)

1

,

0, 0, 0

1 1 2

1, , ,

2 2

2

1, ,

2

2, , 0

2, , 0 n

n c

n

n

n

n

n

ω

π π

π π

π π π

π

=

 _{− −}  _{ } = _ _⇒_{ }  

 

  = − _− _   =

= − −

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Chapter 9.1 – Vector Functions

Higher derivatives: done component-wise

( )

( ) ( ) ( )

( )

( ) ( ) ( )

, ,

, ,

r t f t g t h t r t f t g t h t

=

′′ = ′′ ′′ ′′

Chain rule: If r is a differentiable vector function ands=u t

( )

is a differentiable scalar function, then

( ) ( )

d d d

d d d

s

t s t

s u t

= ⋅

′ ′

=

r r

r

Note:r=r

Rules of differentiation

Letr r₁, ₂be differentiable vector functions andu t

( )

is a differentiable vector function. The following properties are all derived from properties of scalar functions.

i) 1

( )

2

( )

1

( )

2

( )

d

dt t t t t

′ ′

+ = +

 

r r  r r

ii)

( ) ( )

1

( ) ( )

1

( ) ( )

1

d

dt u t t u t t u t t

′ ′

= +

 

 r  r r

ii) 1

( ) ( )

2 1

( ) ( )

2 1

( ) ( )

2

d

dt t t t t t t

′ ′

⋅ = ⋅ + ⋅

 

r r  r r r r

iv) 1

( )

2

( )

1

( )

2

( )

1

( )

2

( )

d

dr t t t t t t

′ ′

× = × + ×

 

r r  r r r r

e.g. 2) Differentiate

( )

(

( )

)

( )

(

( )

)

( )

( )

( )

( ) ( ) ( )

( )

( ) ( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( )

d d d

d d d

d d

1 2

t t t t t t t t t

t t t

t t t t t t

t

t t t t t t t

t t t t t

 ⋅ = _ _⋅ + ⋅  

   

′

= ⋅ + ⋅ _ _

′ ′

= ⋅ + ⋅_ + _

′

= ⋅ + ⋅

r r r r r r

r r r r

r r r r r

r r r r

f,g,h integrable, and r

( )

t = f t

( ) ( ) ( )

,g t ,h t , then have

( )

t dt= f t

( )

d ,t g t

( )

d ,t h t

( )

dt

∫

r

∫

∫

∫

and

( )

d

( )

d ,

( )

d ,

( )

d

b b b b

a t t= a f t t a g t t ah t t

∫

r

∫

∫

∫

The indefinite integral of r is another vector function,R c+ , such that R′

( ) ( )

t =r t and c is a

vector constant.

e.g. 3)

Find the vector function of r satisfying:

( )

( )

( )

1 2

12 , 3 , 2

1 0,1, 0

1 2, 0, 1

t t t−

′′ = − ′ =

= −

r

r

r

( )

( )

( )

1 2 1 2

2

d

12 , 3 , 2 d

6 , 6 , 2

1 6, 6, 2

0,1, 0

6, 7, 2

t r t t

t t t

t t t

−

′ = ′′ = −

= − + ′ = − +

=

⇔ = − −

∫

∫

r

c

r c

c

Do this on your own, but the answer is as follows:

( )

3

2

3 2

2 6 6, 4 7 3, 2

t = t − + −t t + −t t − t

r

Arc Length

Letr

( )

t = f t

( ) ( ) ( )

,g t ,h t be smooth, then the length of the curve traced byr

( )

t is given by:

( )

( )

( )

( )

2 2 2

d

d b

a b a

s f t g t h t t

t t

′ ′ ′

= _ _ +_ _ +_ _ ′

=

∫

∫

r

e.g. 4)

Line segment 2

 from (0,0) to (0,2)

( )

0,

0 2

t t

t

= ≤

r

Line segment between two points, p p₁, ₂inn:

( ) (

t = −1 t

)(

1− 2

)

+t 2

r p p p

0≤ ≤t 1

This expression is linear since you only have t in powers of 1 or 0 Apply this to the example:

( ) (

1

)

0, 0 0, 2

0 1

0, 2 , 0 1

t t t

t

t t

= − + ≤ ≤

= ≤ ≤

r

Arc length functional:

( )

t

( )

d

a

s t =

∫

r′ u u

u is like a “dummy variable” which gets “integrated away” This gives s in terms of t.

Now, try to invert the relationship; find t in terms of s. i.e. t (s). Plug t (s) back intor

( )

t ⇒r

( )

t s

( )

e.g. 5) Helix

( )

t = 2 cos

( )

t , 2 sin 2 ,

( )

t t, ≥0

r

Reparametrize curve with respect to arc length, in direction of increasing t:

( )

( )

( )

( )

(

( )

)

2

(

( )

)

2 ₂

2 sin , 2 cos ,1

2 sin 2 cos 1

5

t t t

t t t

′ = −

′ = − + +

=

r

r

( )

( )

( )

0

0

d 5d 5

t t

s t u u

t s t t

′ =

=

=

∫

∫

r

( )

5

s t s

⇒ =

Plug t back in:

( )

( )

2 cos , 2 sin ,

5 5 5

s s s

t s = _ _ _ _

   

r , s≥ 0

Chapter 9.2 – Motion on a Curve

( )

t = f t

( ) ( ) ( )

,g t ,h t

r is the position vector of a moving particle at time, t. Thusr′

( )

t =v

( )

t is the velocity vector of the particle at time, t.

( ) ( )

t t

( )

t

′′ = = ′

r a v is acceleration of the particle at time, t.

Speed of the particle is v

( )

t = r′

( )

t

( )

(

)

(

( )

)

(

( )

)

2 2 2

2 2 2

d d d

d d d

x y z

t t t

f t g t h t

     

= _{ } +_{ } +_{ }

     

′ ′ ′

= + +

Notice: relate r to arc length:

( )

( )

( )

( )

( )

0

d

' speed

t t

s t u u

s t t t

′ =

′

⇒ = = =

∫

r

r v

e.g. 6)

Circular motion in a plane can be described by:

( )

( )

( )

( )

( )

( )

( )

( )

( )

( )

0 0

0

0 0

2 2

0 0

2

cos , sin ,

sin , cos

cos , sin

t t t

t t t

t t t

t

ω ω

ω

ω ω ω ω

ω ω ω ω

ω

= ∈

′ = −

′′ = − −

= −

r r r

r

r r r

r r r

r



( )

t

a points opposite the position vector; centripetal acceleration

Claim:v⊥r

Length of r is constant

( )

( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

2 ₂

2

2

d d 0

0 2 0

t c

t c

t t c

t t c

t

t t t t

t t t t

=

⇔ =

⇔ ⋅ =

 

⇔ _ ⋅ = _

′ ′

= ⋅ + ⋅

′

⇔ = ⋅

′

⇔ = ⋅

r

r

r r

r r

r r r r

r r

r r

Q.E.D. (quod erat demonstrandum) – Latin for “as proven” e.g. 7)

Shell fired from ground level with initial speed, 48ft

s , angle of elevation, 6 π_.

Find:

a)

Vector function of shell’s trajectory:

( )

( )

( )

2 2

ft m

s s

0,1

9.8 32

d

0, 32

t g

j

g

t t t

t

= − =

≈ ≈ =

= − +

∫

a j

v a

c

Find v(0). The magnitude is known to be 48.

( )

0 = 24 3, 24

v

Use to solve for c:

( )

( )

( )

2

24 3, 24

24 3, 32 24

d 24 3 , 16 24

t t

t t t t t t

⇒ =

= − +

=

∫

= − + +

c v

r v D

Haver

( )

0 = 0, 0 because fired from the ground Use it to solve for D; obtainD= 0, 0

( )

2

24 3 , 16 24

t = t − t + t

r

b)

Find the maximum altitude: maximize y.

( )

( )

3 2 3 4

32 24 0

9ft

y t t

t y

′ = − + =

⇒ = =

c)

Range of shell

x-coordinate when shell hits the ground

( )

2

3 2

0 16 24

0,

y t t t

t

= = − +

⇒ =

( )

3

2 ?

36 3ft

x =

=

d)

Speed at impact

( )

3 ft

2 = 24 3, 24− =48s

v

9.3 − Curvature and Components of Acceleration

( )

t

′

r is tangent tor

( )

t

( )

( )

_{( )}

t

T t

t

′ =

′

r

r is known as the unit tangent vector.

Note:d

( )

d

s

t

t = r′ , so if the curve defined by r is reparametrized with respect to arc length, then

by the chain rule:

d d d

d d d

s t = ⋅ t

r r

r

And

( )

( )

( )

d

d _d

d d

d

t

t _{T t}

s

s t

t

′

= = =

′

r r r

r

Length of T we traverse c does not change. Direction of T does change. Amount of change in direction of T tells us how “curvy” c is.

( )

t

r , a vector function, defining a smooth curve, c. If s is arc-length parameter and d d

T t

= r is the unit tangent, then we define the curvature of c to be a scalar quantity, κ, where

d d

T s

κ =

Problem: c is not always in terms of s. Use the chain rule!

( )

_{( )}

( )

d

d d d d _d

d

d d d d

d

T

T T s T _t

s

t s t s

t T t

t

t

κ

= ⋅ ⇔ =

′

⇔ =

′

r

e.g. 8)

Find the curvature of a circle of radius, a.

( )

t = acos

( )

t , sina

( )

t r

Compute r′

( )

t =?

( )

( )

( )

( )

2 2

( )

2 2

( )

sin , cos

sin cos

t a t a t

t a t a t a

′ = −

′ = + =

r r

( )

( )

_{( )}

( )

( )

( )

( )

1

sin , cos sin , cos

t T t

t

a t a t

a

t t

′ =

′

= −

= −

r r

( )

( )

( )

( )

( )

_{( )}

( )

cos , sin

1

1

T t t t

T t

T t t

a t

κ

′ = − − ′ =

′ = =

′

r

Principle/unit normal,

( )

( )

( )

T t N t

T t

′ =

′

We call this the normal vector because: Claim:N ⊥T

Proof:T ⊥T′

T has constant length

2 2

1

1 1

T T

= = =

[

]

d

1 d

2 0

T T t T T T T

⋅ = ′ ⋅ =

′ ⊥

Note: this is a potential proof that will probably be in the exam.

( )

t = ′

( )

t

v r , set speed v

( )

t = ←ν greek letter, nu Then, plug it in to the formula for T:

( )

( )

t

T t

v T T ν

=

= =

v v

v v

( )

( )

( )

d

( )

d

t t T t T t

t

ν ν

′ ′

⇒a =v = +

Acceleration is made of two pieces: one parallel to T, the other parallel toT′. Have κ

( )

t T t

( )

ν

′

= , and

( )

T t

( )

_{( )}

N t

T t

′ =

′

Combine to get: T t′

( )

=νκN t

( )

( )

( )

( )

( )

( )

2 d

d

N T

t N t T t

t N t T t

ν ν κ

= +

= +

a

a a

, where:

2

n =ν κ

a : normal component of acceleration

d d

T t

ν =

a : tangential component of acceleration

Binormal

( )

( )

( )

B t =T t ×N t is the binormal.

( )

B t is unit length

( )

( )

( )

( )

( )

( )

( )

(

)

(

( )

sin

)

B t T t N t

B t T t N t

T t N t θ

= × = × =

( )( )

2 1 1 sin

2 1

,

T N

B N T

π θ

π

⊥ ⇒ =

 

= _{ }

  =

⊥

Call ,T N B, the Frenet frame.

Plane spanned by T and N is the osculating plane (important). Plane spanned by N and B is the normal plane.

Plane spanned by T and B is the rectifying plane. Identities:

B T N N B T T N B

= × = × = ×

e.g. 9)

( )

1 2 1 3

2 3

, ,

t = t t t r

Find T, N, B, κ at point t = 1

( )

( )

( )

( )

( )

_{( )}

( )

2

1 1 1

3 3 3

2 2 4

1, , 1 1,1,1

1 3

1

1 , ,

1 1

t t t

T

t t t

′ =

′ =

′ =

′

= =

′

′ = + +

r r r

r r r

( )

( )

_{( )}

2

2 4

1

1, , 1

t

T t t t

t _{t t}

= =

′ _{+ +}

r r

Use the Leibnez rule.

( )

(

) (

)

( )

( ) ( )

3 2

3 2

2 4 3 2

1

2 _{2 4}

1

1 2 4 1, , 0,1, 2

1

1 1

1 3 6 1,1,1 0,1, 2

2 3

1 1

, 0,

3 3

T t t t t t t t t

t t T

− −

−

′ = + + + +

+ + −

′ = +

− =

( )

1 2 1 2 2

1

3

3 3

T′ = _ − _ +_ _ =

   

( )

1

( )

_{( )}

1 1

3 1 1

, 0,

2 3 3

1 1

, 0,

2 2

T N

T

′ =

′

= −

= −

( )

1

( )

1

( )

1

1 1 1 1 1

, , , 0,

3 3 3 2 2

1 1 1

3 3 3

1 1

0

2 2

1 1 1 1 1 1 1 1 1 1

0 0

2 2 2 3 2 3 2 3 2 3

1 3 1

, ,

6 6 6

B =T ×N

−

= ×

=

−

 

− −

   

   

= _ ⋅ − ⋅ _− _ ⋅ − ⋅ _+ _ ⋅ −_{  }

       

− =

i j k

i j k

+ − + − + − + − +

( )

_{( )}

( )

1 23 2

1

3

1 3

T

κ = ′ = =

′

r

More Dot Product Stuff

( )

(

)

( )

(

)

T N

T N

T N

T N

t T N

T N

T N

T T T N

ν ν

= +

⋅ = ⋅ +

= ⋅ + ⋅

= ⋅ + ⋅

a a a

v a v a a

a v a v

a a

Tν

=a

T

ν

= v = v

v

d d

T t

ν

ν =

⋅ =

⋅ =

′ ′′⋅ =

′

a

v a

v a v r r

r

(

T N

)

T N

T

T N

T N

T T ν

× = × +

= × + ×

= ×

v a v a a

a v a v

a _N

N

T N T N

ν ν

+ ×

= ×

a a

Take lengths⇒ × =v a a_Nν B

1

N ν

×

⇔ =

′ ′′

× ×

= =

′

v a a

v a r r

v r

Solve for κ:

2

2 κν

κ ν

′ ′′× =

′ ′ ′′× ⇔ =

′

r r r r r

r

( )

t 3

κ = ′ ′′× ′

r r r