then we can rewrite by finding the ``magic factor’’sethian/math53_2020/... · this lecture is...

Math 53: Fall 2020, UC Berkeley Lecture 015 Copyright: J.A. Sethian

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Section 15.10: Change of Variable This lecture is one of those key moments in the course when all of sudden you really understand what is going on!

Let’s go back to 1D calculus

g(u) u x

Then we can rewrite by finding the ``magic factor’’

(1) Let x = g(u) (2) Then dx = g’(u) du

So we can write

Finally, we need to get the limits of integration correct, so if g(c)= a and g(d)=b, then

Example: Let f(x)=(2x)3, and suppose we want We can let x = g(u)= (1/2) u. Then dx = g’(u) du = ½ du, so

You have been doing this all along



Section 15.10: Change of Variable

The key to changing variables was the calculation of the magic factor: dx = g’(u) du

g(u) u x Which allowed us to go from

Okay, what if we had a function of two variables?

u

v

x

y

f(u,v)

g(u,v)

Could we find a magic factor?

that allows us to transform variables?

Example: We have already done this for polar coordinates!!!

r

q

x

y

So, for polar coordinate transformations, the magic factor is r

How do we find it in general?




Finding the magic factor general: (by the way, there’s nothing magic about it-it’s just really cool)

Step 1: Start with an arbitrary two-dimensional transformation

u

v

x

y S

(u1,v1) R

(x1,y1)

u

v

x=g(u,v)

y=h(u,v)

g(u,v)

h(u,v)

Step 2: Suppose we have We can write this in general as

T(u,v)= (x,y)




Example: Consider the transformation: x = u2 – v2, y = 2uv

What happens to the unit square 0≤u≤𝟏, 𝟎≤𝒗≤𝟏, under this transformation?

u

v

x

y R

(x1,y1) T (u1,v1)

What does the output region R look like?

Part 1: I am now going to do a funky example to show you that nice regions can be

transformed intro strange ones.



Example: x = u2 – v2, y = 2uv

u

v

x

y R

(x1,y1) T (u1,v1)

What does the output region R look like?

A

B

C

D

(1,0)

(0,1)

(0,0)

(1,1)

Let’s see what happens to the edges of the unit square

Edge A: For edge A, v=0 and 0≤u≤1, so the transformation is x=u2-v2 which = u2; y=2uv which = 0

So, x=u2 and y= 0 is a line from x=0 to x=1 u x

y

A (1,0) (0,0) (1,0) (0,0) A’

Edge B: For edge B, u=1 and 0≤v≤1, so the transformation is x=u2-v2 which = 1-v2; y=2uv which = 2v

So (y/2)=v, so x = 1-v2=1-(y/2)2=1-y2/4 which is a parabola

x

B

(1,0)

(1,1)

(1,0)

(0,2)

A’

B’

v

u

Edge C: For edge C, v=1 and 1≥u≥0, so the transformation is x=u2-v2 which = u2 -1; y=2uv which = 2u

So (y/2)=u, so x = u2 -1=(y/2)2-1=y2/4 -1 which is a parabola u x

C (1,1) (0,1)

(1,0) (0,0) A’

C’

x

Edge D: For edge D, u=0 and 1≥v≥0, so the transformation is x=u2-v2 which = -v2; y=2uv which = 0

So, x=-v2 and y= 0 is a line from x=-1 to x=0 u x

y

(0,0) (-1,0)

D’

v

D



Example: x = u2 – v2, y = 2uv So we just found that the uv unit square maps into:

u

v

x

y

(x1,y1)

T (u1,v1)

A

B

C

D

(1,0)

(0,1)

(0,0)

(1,1)

A’

B’ C’

D’

So, how do we find the magic factor such that ?

Here goes: Let’s agree that (x0,y0) = T(u0,v0) Standing at input (u0,v0), edges of the box DuDv get sent to different vectors in xy space

T(u,v)

u

v

The output (x,y) comes from an input (u,v). Then any point (x,y) in output space can be written as vector whose components depend on u and v

Our strategy will be to take the two vectors in the u,v plane making up a DuDv box, and see what they get transformed into. We will then take their cross-product to find the area of the transformed box—which is the magic stretch factor…

T(u,v)

u

v

T(bottom side)

T(left side) Area

[x coordinate of vector = g(u,v) y coordinate = h(u,v)]



So, how do we find the magic factor such that ?

Here goes: Let’s agree that (x0,y0) = T(u0,v0) Standing at input (u0,v0), edges of the box DuDv get sent to different vectors in xy space

T(u,v)

u

v

The output (x,y) comes from an input (u,v). Then any point (x,y) in output space can be written as vector whose components depend on u and v

Our strategy will be to take the two vectors in the u,v plane making up a DuDv box, and see what they get transformed into. We will then take their cross-product to find the area of the transformed box—which is the magic stretch factor…

T(u,v)

u

v

T(bottom side)

T(left side) Area

[x coordinate of vector = g(u,v) y coordinate = h(u,v)]

Part 2: I am now going to go back to the main question: and find magic factors in general!



Step 1: Tangent vector holding v fixed, at input (u,v) is

T(u,v)

u

v

T(bottom side)

T(left side) Area

Step 2: Remembering that

u

v

x=g(u,v)

y=h(u,v)

g(u,v)

h(u,v)

Step 3: So, now can find the vectors that make up the bottom and left sides:

bottom side vector =

left side vector =

Step 4: So we can take the cross-product to find the area

Called the Jacobian

Tangent vector holding u fixed, at input (u,v) is



u

v

x

y

f(u,v)

g(u,v)

So now have the result:

u

v

x

y S

(u1,v1) R

(x1,y1)

Let’s use it right away: Prove that the Jacobian for the polar transformation x=r cos q, y=r sin q is rdrdq

r

q

x=r cos q

y=r sin q

Jacobian is

[R plays the role of u, and q plays the role of v]

Do you remember that for polar coordinates, we had



u

v

x

y

f(u,v)

g(u,v)

Let’s do another example:

u

v

x

y S

(u1,v1) R

(x1,y1)

Using the change of variables x = u2 – v2, y = 2uv to evaluate where R is the region bounded by the x-axis

and the parabolas y2=4-4x,x≥0 and y2=4-4x, x≤0 x

y

Solution: The Jacobian for this mapping is

..I’ll let you finish it…..



And now you see how to find the magic factor!!!

x (u,v,w)

y (u,v,w)

z (u,v,w)

u

w z

x

y v

dx dy dz= [magic factor] du dv dw

Form the matrix of partials

Then the magic factor = | det A |

Let’s check it for things we know!

Cylindrical coordinates: x (r, q,z)

y (r, q,z)

z (r, q,z)

r

q

z z=z

x=r cos q

y=r sin q

det A= r cos q cos q – ( - r sin q sin q ) = r !!!

dx dy dz = r dr dq dz Cartesian element

Cylindrical element

Geometrically, we got



Spherical coordinates: x (r, q,z)

y (r, q,z)

z (r, q,z)

r

q

z z= r cos f

x=r sin f cos q

y=r sin f sin q

det A= r2 sin f!!!

dx dy dz = r2 sin f dr dq df And geometrically the book got

Here goes!

I leave it to you to check that:

And remembering that if then det A= a (ei-fh) – b (di-fg) + c (dh-eg)

then we can rewrite by finding the ``magic factor’’sethian/math53_2020/... · this lecture is...

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