Section 16:3 The Fundamental Theorem of Calculus for Line …sethian/math53_2020/lectures/... · 2020. 10. 29. · Step 2: Pick (a,y) at same y level as the endpoint, giving a straight

Math 53: Fall 2020, UC Berkeley Lecture 017 Copyright: J.A. Sethian

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Section 16:3 The Fundamental Theorem of Calculus for Line Integrals

This is the chapter when you learn that the fundamental theory of calculus applied in multi-dimensions, over far more complex input regions.

x=a x=b

y=f(x)

1D Fundamental Theorem Calculus

z=f(x,y)

“integral of a derivative of f(x) over a region” =

“difference of function over boundary”

Curve C(t) = (a(t),b(t), c(t) )

Multi-Dimensional Fundamental Theorem Calculus

We’re now going to figure this out



Remember last time that

Line integral over a curve C of a function f(x,y) from R2 to R

Curve C=(x(t),y(t))

t=a

t=b

x

y f(x,y) defined in all of space

Example: f(x,y)=x2+y2

Want to evaluate

C(t) = (t,t2), 1 ≤t ≤2

t=1 (x=1,y=1)

t=2 (x=2,y=4)

C(t)

Dt=.1 t=1.1: (x=1.1,y=(1.1)2)

DC= (12+[(2(1)]2)1/2 (.1)

Contribution to integral = [f(1,1) DC]= (12+[(2(1)]2)1/2 (.1)

Dt=.1

t=1.2: (x=1.2,y=(1.2)2) Dt=.1

DC= ((1.)2+[(2(1.1)]2)1/2 (.1)

Contribution to integral = [f(1,(1.1)2) DC]= (12+[(2(1.1)]2)1/2 (.1)



Line integral over a curve C of a vector field from R2 to R2

Curve C=(x(t),y(t))

t=a

t=b

x

y defined in all of space

Example:

Want to evaluate

C(t) = (t,t2), 1 ≤t ≤2

t=1 (x=1,y=1)

t=2 (x=2,y=4)

C(t)

Contribution to integral:

u1= (0,1)

u2= (0,1)



Section 16:2 Integration of a vector field over a curve C

Def: The line integral of along C is

To summarize:

We now show how to do line integrals in “pieces”

Step 1: Let So P and Q are the two components of the vector field

Step 2: We also have that

Step 3: So that means



Again:

Line integrals over a vector field from R2 to R2

F(x,y)=(P(x,y),Q(x,y))

Curve C=(x(t),y(t))

t=a

t=b

Collect all the tangential components of along C

x

y

t=a

t=b

3D: Collect all the tangential components of along C

x

y



Section 16:3 The Fundamental Theorem of Calculus for Line Integrals

For 1D calculus, we had that

x

y

Theorem: In more dimensions, we have that

Wow! The integral of the derivative of a function is just the difference between its values at the two endpoints!

[The integral of the derivative of a function is just the difference between its values at the two endpoints]

Example: Suppose

Consider the vector field given by

Find where C is the curve:

x

z

p Answer:

y




Let’s prove it

Step 1: Recall that

Step 2: Letting this becomes

Step 3: Remember that

Step 4: Substituting this into ***, we have

***

thus

Using 1D Fundamental Theom. Calculus

Done!



On to a new idea…. all results are true in 2D, 3D, 4D, …..

t=a

t=b

C1

C2 Def.: We say that a vector field is path-independent if

is the same no matter what path you take

Claim: Let C be any closed loop. Then

if and only if (iff) is path-independent

Proof: Step 1: Suppose We want to show that from t=a

to t=b along two different paths, we get the same thing

So, we want to show that

Proof: Observe that C1 – C2 is a closed path. So

Proof: Step 2: The same proof in reverse. Suppose that is path-independent. We want to show that

Proof: Let C be a closed curve. Then pick any two points t=a and t=b on it. Then

C



If C is any closed loop. Then if and only if (iff) is path-independent

C

Now, you are starting to wonder: what is the relationship between the above and the “fundamental theorem” ?


Obviously, if the vector field comes from a gradient (that is ) then is path-independent, since

The last expression is the same regardless of which curve C you pick

What do we have so far?

Conservative: There exists a scalar f(x,y,z) such that

The vector field is path-independent

An obvious question: Can this arrow be made both ways?



The question: if is path-independent, must there exist a scalar f(x,y,z) s.t. ?

The answer is YES –but the proof is a bit complicated. Here goes:

Goal: Given = (P,Q) and a point (a,b) in the domain, we will show how to build a scalar function f such that

(a,b)

(x,y)

Step 1: Define the function

Since is path-independent it doesn’t matter what path we take from (a,b) to (x,y)

Step 2: Pick (a,y) at same y level as the endpoint, giving a straight line C2 from (a,y) to (x,y)

(a,y)

Then

C1

C2

Step 3: The tricky part: Notice that the integral over the curve C1 does not depend on x

So:

Step 4: Remember that So:

Fund. Theorem of Calculus So fx = P…The exact same argument shows that fy = Q



So now we’ve got:



Given a vector field F(x,y,z), how can we break into this mess and figure out if it is conservative?

A stunning claim: if = (P(x,y), Q(x,y)), then it is conservative iff Py = Qx

Proof: If is conservative, then f(x,y) such that

E

So, Py = fxy and Qx = fyx and by Clairaut’s theorem, these are the same.

It’s even true in the other direction (but too hard to prove right now).

(requires that the region be simply connected (no holes), and a bunch of other stuff)



So now we’ve got: Given a vector field (P,Q)



Example: show that the vector field is path-independent

Py = 2xy Qx = 2xy so yes!!!

Over any closed curve C



Example: Suppose (P,Q) = (3+2xy, x2-3y2). (1) Is there a function f(x,y) such that

(2) Find that function f(x,y)

(3) Evaluate over the curve

Answer to #1

We check to see if Answer to #2

We try to build f(x,y) through integration. We require that fx=P and fy = Q

fx = P

fy = Q which must satisfy which =x2-3y2 so dConst(y)/dy=-3y2

so Const(y) = -y3 so f(x,y) = 3x + x2y – y3 Answer to #3

Since the vector field comes from a gradient, then we need only evaluate at endpoints

So

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Section 16:3 The Fundamental Theorem of Calculus for Line …sethian/math53_2020/lectures/... · 2020. 10. 29. · Step 2: Pick (a,y) at same y level as the endpoint, giving a straight