Integration Along a Curve:

Kicking it up a notch

Presented by: Keith Ouellette University Of California, Los Angeles June 1, 2000

Motivation: Why do we want to integrate a function along a curve?

A Real Mathematician’s Answer:

Because we can. That’s why!

No, but really...

Motivation: A Massachusetts Dilemma In Boston, we freeze during the wintah.

School is often cancelled due to the hazids of snow and ice.

During a snowball fight, we notice the ice coating the telephone wiahs My buddy Maak the physics major says, “I bet you 10 bucks you can’t figure out the total mass of the ice on that wiah!”

You’re on.

But first I need to develop the theory of line integrals.

Setting up the Problem…...

The thickness of the ice varies as one moves along the wire r r = radius of wire R = radius of wire + ice coating R The confused person (an annoying recurring character)

Recall the Ole Physics formula: Mass = density * volume Density of ice = 0.92 g/cm 3 Area of a cross section of ice = p( R 2 - r 2 ) cm 2 Linear density f of ice on wire = 0.92 * p( R 2 - r 2 ) g/cm So...

The total mass of ice on wire = total accumulation of the linear density function along the wire

Parameterize the wire using a continuously differentiable (i.e. smooth) function a a :  [

a

,

b

] 

R

2 a (

t

)  (

x

(

t

),

y

(

t

)) x(t) y(t) Real-valued, continuously differentiable functions

Now we can at least approximate the total mass of the ice along the wire by these 4 easy steps: 1. Partition 2. Sample 3. Scale 4. Sum

1. Partition

Partition the arc a into n subarcs y How? a 1 By partitioning [a,b] into n subintervals, we induce a partition of a into n subarcs a n a( t n ) a( t )  ( x(t),y(t)) a( t 1 ) a( t 0 ) a( t i-1 ) a i a( t i ) x a = t 0 a = t 0 t 1 … t 1 … t i-1 t i-1 t i t i … … t n = b t n = b t

2. Sample

Recall f is the density function defined on a( [a,b]), our “frozen wire”.

On each subarc a i , choose a point a i * ( x i *,y i *) and sample f at those points.

a 0 * ( x 0 *,y 0 *) a i-1 * ( x i-1 *,y i-1 *) y a( t 0 ) a 1 a = t 0 a = t 0 a n a( t n ) a( t )  ( x(t),y(t)) a( t 1 ) t 1 … t 1 … a( t i-1 ) t i-1 a i t i a( t i ) … t i-1 t i … t n = b t n = b a n * ( x n *,y n *) x t

3. Scale

Now we scale those sampled values f(

a

i

*) 

f

(

x i *,y i *) by the length of the subarc denoted by

D

s i

D s i y a( t 0 ) a 0 a( t 1 ) a n-1 a = t 0 a( t i-1 ) t 1 … t i-1 a i-1 a( t i ) t i … a( t n ) a( t )  ( x(t),y(t)) t n = b x t

4. Sum

Now we sum those scaled sampled values to get what looks like a Riemann sum.

i n

  1

f

(

x i

*,

y i

* ) D

s i

But we want a way to actually calculate this mutha. So we need to bring it down to the case we know: the one variable case.

It seems so pointless.

Should I just give him the money right now?

Let’s do this.

(The Encouragement Slide)

No Way!!!!!

We gotta show him up!

Math majors, represent!

How, do you ask?

By relating everything to t.

f ( x i *,y i *) = f (a( t i *)) for some t i *in [t i-1 , t i ] D

s i

 a (

t i

)  a (

t i

 1 ) D y i a (t i-1 ) almost D s i Notice that for D s i small, a continuous curve looks locally linear a (t i ) D x i

Since a is continuously differentiable, a ' (

t

)  lim D

t

 0 a (

t

 D

t

)  a (

t

) D

t

So when D t is small, a (

t

 D

t

)  a (

t

)  a ' (

t

) D

t

 a ' (

t

) D

t

Since f is path integrable (continuity on a( [a,b]) is sufficient for this), we may sample f( a (t)) and partition however we choose: Let D t i = (b-a)/n = D t Let t i * = t i-1

So we have

i n

  1

f

(

x i

*,

y i

* ) D

s i



i n

  1

f

( a (

t i

*) ) a ' (

t i

*) D

t

which is a Riemann sum of the one variable real-valued function f( a (t))|| a ’(t)) || So letting L(P, f) and U(P, f) be our respective lower and upper Riemann sums...

Thus we define the path integral of f along the curve a  a

fds



a



b f

( a (

t

)) a ' (

t

)

dt

where

b a



f

( a (

t

)) a ' (

t

)

dt

 sup

P L

(

P

,

f

( a ) a ' )  inf

P U

(

P

,

f

( a ) a ' ) Note: If f (x,y)=1 on [a,b]  a

ds



b a

 a ' (

t

)

dt

= length of the curve a

So to answer Maak’s challenge… f(r,R) = linear density = 0.92* p (R 2 -r 2 ) R = Radius of wiah + ice r = Radius of wiah = 10cm a (t) = (t,t 2 ) parametrization of wire (parabola) the total mass of the ice is ………....

 a f(r, R)

ds

 

a



b

f( a (

t

)) a ' (

t

)

a



b

0.92

* p (t 4

dt

 

t

2 )

a



b

f((t, t 2 )) ( 1 , 2

t

) ( 1  4

t

2 )

dt dt

NOW SHOW ME THE MONEY!!!!

Words of Wisdom

Always Get A Prenuptual

-Dr. Jock Rader