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Neat Stuff from Vector Calculus
&
Related Subjects
Chris Hecker
[email protected]
definition six, inc. & Maxis
Related Subjects
scalar
calculus
linear
algebra
optimization &
constraints
vector
calculus
classical
mechanics,
dynamics
differential &
integral calculus
differential
geometry
Intro & Prerequisites
• this is a total hodge-podge, not a gentle introduction
– lack of rationale for caring about this stuff, trust me? :)
• stuff that I found non-intuitive or hard to figure out
• tour of fairly basic examples building on themselves to get
firm grounding and intuition in calculus concepts for the
kinds of math games do
– touch on lots of different areas during tour, many sidestreets
• comfortable with algebra, linear algebra
• need to understand scalar calc, at least at the plug’n’chug
level of differentiating functions
– f(x) = ax2+bx+c

f’(x) = 2ax + b
What is a Function?
• a function maps values from the domain to the
range uniquely
– can be multivalued in range, but not domain
f : Rn  Rm
y  f (x)
x
cos q
 f (q ) 
y
sin q
f ( x, y)  x 2  y 2  1  0
q
x
 f (u, v)
y
8
6
2
4
2
y  f (x)
1
0
-1
0
0
1
2
-1
z  f ( x, y)  x 2  y 2  1
What is a Derivative?
• derivative is another function that maps changes in
the domain to changes in the range, “normalized”
y  y  f ( x  x)
y f ( x  x)  f ( x)

x
x
y  f (x)
y
x
Derivatives are Linear
• the key insight of calculus: the change is so small
that you can ignore it anytime it’s multiplied by
itself...so, you can treat any continuous function as
linear if you’re zoomed in far enough (to 1st order)
– “continuity” keeps us from dividing by zero
– normalization makes the numbers finite
f ( x0  x)  f ( x0 )  f ( x0 )x  1 2 f ( x0 )x 2  
f ( x0  x)  f ( x0 )
 f ( x0 )
x
y  f (x)
y  f (x)
boom!
x0
x0
Random but Nifty Example of Ignoring Infinitessimals
(nothing up my sleeve!)
• prove infinitessimal rotations in 3D are vectors (add, commute, etc.)
– angular velocity is a vector, cross product differentiates rotating vectors
finite rotation: r '  Rr
RT R  1
infinitessimal rotation: r '  (1   )r  r  r
infinitessimal rotations add and commute:
(1  1 )(1   2 )  1  1   2  1 2  1  1   2
infinitessimal rotations are vectors:
(1   )(1   )T  1     T  1
3x3 skew symmetric matrices
   T
are isomorphic to vectors
angular velocity differentiates with cross product:
dr  r 'r   r  vec ( )  r
dr 1
v   dt  r  ~r    r
dt
Scalar Derivatives: a line
• how does y (or f(x)) change for a change in x?
– for lines, the change (derivative) is constant everywhere
– drawn as red vector, but actually a scalar, “slope”
y  f (x)
a
1
f ( x)  ax  b
f ' ( x)  a
df
a
dx
df  adx
f :R  R
df : R  R
Scalar Derivatives: a curve
• for curves, the derivative is position dependent
– derivative is a function itself, mapping change in
domain to change in range
– both direction and magnitude, but still a scalar
y  f (x)
f ( x)  ax 2  bx  c
f ' ( x)  2ax  b
df  (2ax  b)dx
f :R  R
df : R  R
The Shape of Matrix Operations
• vectors are columns of numbers in this talk, not rows
• we right-multiply matrices by column vectors: v’ = Mv
• matrix & vector ops “fit together” nicely
• keeping the shapes right is the key to sanity with vector calc
• “m by n * n by p = m by p”
b11
p
b11 b12 b13 b14
b21
Ab=c
b31
n b21 b22 b23 b24
AB=C
n
b31
b32
b33
b34
a11
a12
a13 c11
c12
c13
c14
m a21
a22
a23 c21
c22
c23
c24
a11
a21
a31
a12
a22
a32
a13 c11
a23 c21
a33 c31
a·b = c = aTb = c
a11
a12
• this is why “v’=vM” makes no sense for column vectors
– either you’re using row vectors, or you’re confused
– either way, you’re in for some pain when trying to do real math
• because all math books use columns for vectors and rows are special
• early computer graphics books got this backwards, and hosed everybody
a13
b11
b21
b31
c11
Shape of Derivatives
df needs to accept a domain linearly (ie. right-multiplied vector or scalar)
to produce a range
• scalar valued function of scalar
– y=f(x) dy =dfdx
df
d
• vector valued function of scalar
– v=f(t) dv = dfdt
• scalar valued function of vector
– z=f(x,y)
d
d
• vector valued function of vector
– p=f(u,v)
d
Shape of Derivatives (cont.)
• scalar valued function of vector: expands to row vector
– the resulting range value depends on all the domain values
– the differential needs a slot for a delta/change in each domain
dimension...so it must be a row vector, there’s no T or ·
– z=f(x,y)
f
f
f
f
f dx
dz

dx

dy

d
x

– dz=dfdx
x
y
x
x y dy
df = d
dz = dfdx = d
• vector valued function of vector: expands to matrix
– p=f(u,v)
df = d
dp = dfdx = d
Derivatives: a parametric function
“a vector function of a scalar”
• for a change in the parameter (domain), how does
the function (range) change?
– in this case, differential is a vector
x cos q

y sin q
q
dx  sin qdq  sin q


dq
dy
cos qdq
cos q
f : R  R2
df : R  R 2
Derivatives: a scalar function of a vector
“height field”
• differential is not in range of function
z  f ( x, y)  ax 2  bxy  cy 2  d
2
1.5
dz  df 
1
f
f
dx 
dy
x
y
0.5
 (2ax  by )dx  (bx  2cy )dy
0
0
0.5
1
1.5
2
 2ax  by
8
6
2
4
2
bx  2cy
dx
dy
column vector
1
0
-1
0
0
1
2
-1
row vector
f : R2  R
df : R 2  R
Derivatives: a scalar function of a vector
• view “height field” as implicit surface in 3d
g ( x , y , z )  z  f ( x, y )
– write g(x,y,z) > 0 above surface
– differential is surface normal (not unit)
dg 
g
g
g
dx 
dy 
dz
x
y
z
dx
8
6
  2ax  by
2
4
2
1
0
-1
 bx  2cy 1 dy
dz
0
0
1
 2ax  by
g
g 
  bx  2cy
x
1
T
2
-1
f : R3  R
df : R 3  R
Derivatives: a scalar function of a vector
• example plane
n p  d  0
1
n  1,
1
x
p y
z
x yzd 0
z  f ( x, y )   x  y  d
g ( x , y , z )  z  f ( x, y )
g ( x, y , z )  z  x  y  d
2
1
1
0
-1
0.5
-2
-1
0
-0.5
-0.5
0
0.5
1
-1
dg 
g
g
g
dx 
dy 
dz
x
y
z
dx
 1 1 1 dy
dz
f : R3  R
df : R 3  R
Derivatives: a scalar function of a vector
• example sphere
g ( x, y, z )  x 2  y 2  z 2  1
g
g
g
dg 
dx 
dy 
dz
x
y
z
dx
 2x
2y
2 z dy
dz
f : R3  R
df : R 3  R
A Surface Normal is Not a Vector!
this is why you need to keep shapes distinct
• vector is a difference between two points
• normal is “really” a mapping from a vector to a scalar
points &vectors transform like this:
a'  Ta
b'  Tb
v ba
v'  Tv  Tb  Ta  b'a'
v'  b'a'
n
normals transform like this:
v  n  vT n  0
v'T n'  0
n'  Sn, v'  Tv
(Tv)T Sn  0
vT T T Sn  0
TTS 1
v
a
b
S  T T
Normals and Vectors Example
• scaling an ellipse
g ( x, y)  ax 2  y 2  1
g
 2ax
x
x values for y=0
1
0.5
0.5
-0.5
-1
2y
1
1.5
2
a
1
x
1
4
 12
2
1
4
Derivative of Vector Mappings
• barycentric coordinates in 2D triangle
– vector function of a vector
– if square, can invert Jacobian to find du,dv given dp
– function is linear in this case, but works generally
– determinant of Jacobian is how areas distort under function
p2
e2
p  f (u, v)  p0  e1u  e2v
p
p1
e1
p0
f x
f
dp 
du  u
f y
u
u
f x
v du  e
1
f y dv
v
e2
du
dv
Jacobian matrix
f : R2  R2
df : R 2  R 2
Barycentric Coordinates in 3D Triangle
• now jacobian is 3x2
p  f (u, v)  p0  e1u  e2v
– can still find du,dv from dp with least squares
f x
u
f y
f
dp 
du 
u
u
f z
u
p2
e2
du
e2
dv
Ax  b
AT Ax  AT b
e1T
du
T
2
e
e1
e1T dp
T
2
e dp
p
e1
p0
f x
v
f y du
 e1
v dv
f z
v
p1
e2

dv
e1T e1
e2T e1

e1T
T
2
e
dp
e1T e2 du
e2T e2 dv
same as projecting dp down into triangle
f : R2  R3
df : R 2  R 3
Implicit Functions
• equalities (constraints) subtract off DOFs f(x,y) = 0
• explicit to implicit is easy: z = f(x,y)  g(x,y,z) = z – f(x,y) =0
• implicit to explicit is hard: f(x,y) = 0  y = f(x) y   1  x 2
• solving nonlinear equations, sometimes multiple or no solutions
• but, inverting it differentially is easy because of linearization:
f
f
dx 
dy  0
x
y
dy  
f
x
f
y
dx
f
dx  0
x
shape/rank of jacobian will tell you how constrained you are as well
f ( x, y)  x 2  y 2  1  0
dy   2 x 2 y dx
dy  dx
g ( x, y , z )  0
dy  0dx
f ( x, y , z )  0
Chain Rule for Vector Derivatives
• for scalar multiplication:
d (a  b)  da  b  a  db
• differential of a scalar function of a scalar is a scalar
Chain Rule for Vector Derivatives
• for scalar multiplication:
d (a  b)  da  b  a  db
• differential of a scalar function of a scalar is a scalar
• for dot product it’s a little wackier
• if vectors in dot are functions of scalars, it’s the same
d (a  b)  da  b  a  db
Chain Rule for Vector Derivatives
• for scalar multiplication:
d (a  b)  da  b  a  db
• differential of a scalar function of a scalar is a scalar
• for dot product it’s a little wackier
• if vectors in dot are functions of scalars, it’s the same
d (a  b)  da  b  a  db
• if vectors in dot are functions of vectors, need to watch shape!
d
– we know result must be row:
Chain Rule for Vector Derivatives
• for scalar multiplication:
d (a  b)  da  b  a  db
• differential of a scalar function of a scalar is a scalar
• for dot product it’s a little wackier
• if vectors in dot are functions of scalars, it’s the same
d (a  b)  da  b  a  db
• if vectors in dot are functions of vectors, need to watch shape!
d
– we know result must be row:
use transpose picture to reason about it:
d
d
???
d (aT b)  daT b  aT db
Chain Rule for Vector Derivatives
• for scalar multiplication:
d (a  b)  da  b  a  db
• differential of a scalar function of a scalar is a scalar
• for dot product it’s a little wackier
• if vectors in dot are functions of scalars, it’s the same
d (a  b)  da  b  a  db
• if vectors in dot are functions of vectors, need to watch shape!
d
– we know result must be row:
use transpose picture to reason about it:
d
d
d (aT b)  daT b  aT db
???
don’t want to use tensors, so we pull a fast one with the commutativity of dot (a·b=b·a)
d (aT b)  bT da  aT db
d
d
Dot Product Derivative Example
• derivative of squared length of vector
r  pq
r  r  r T r  r 2  ( p  q) 2

dr 2  d ( p  q)T ( p  q)

d (aT b)  bT da  aT db
dr 2  ( p  q)T d ( p  q)  ( p  q)T d ( p  q)
dr 2  2( p  q)T dp  2r T dp
q
dp
makes intuitive sense:
if dp is orthogonal, length
doesn’t change
r
p
Cross Product Works the Same Way
• if cross is a vector function of vectors...
d (a  b)  da  b  a  db
• result must be a matrix
• use the skew symmetric picture of cross product
~
~b  b
ab  a
a
d
d
???
same problem with tensors, so we pull the same (skew-)commutativity trick
~
~
~db
d ( a b)  b da  a
Cross Product Derivative Example
• differentiate a cross product of moving vector with constant vector
c  r n
dc  d (r  n)  d~
rn
~dr  ~
~dr
d~
r n  n
r dn  n
~
~b)  b
~db
d (a
da  a
dc  n  dr
n
dr
r
c
dc
again, makes intuitive sense:
dc changes orthogonally to dr changes
References
• Advanced Calculus of Several Variables
– Edwards, Dover
• Calculus and Analytic Geometry
– Thomas & Finney
• Classical Mechanics
– Goldstein