PIVOTTRANSFORMS

ANNEX1

TANGENTPLANESTOSURFACES

Ifasurfaceisdefinedbytwoparametersu,vi.e.

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

and(x,y,z)isthetangentpointofaplanewhile(,,)arethecoordinatesofavariablepointinthatplane
thentheplaneisgivenbytheequation

y , z z , x x , y
x y z =0
u , v u , v u ,v

wheree.g.

y y
y , z u v

u , v z z
u v

(c.f.forexamplePartialDifferentiationbyR.P.Gillespie).

If(x,y,z)=(0,0,0)then

y , z z , x x , y
=0
u , v u , v u , v

which givesaplanethroughtheoriginparallel tothetangentplanewiththedeterminants asits plane

Itfollowsthatthedirectioncosinesofthenormaltothesurfaceat(x,y,z)areproportionalto

{ y , z z , x x , y
, ,
u , v u , v u , v }
Weselecttheparametersu,vasfollows

u=theverticalheightofGaboveO(seemaintext)
v=theheighthofthecontourplane.

Thenfrom(13)inthemaintextwehave
 kv y 3 mx 3
y u , v =
1m2
x u , v =my u , v
z u , v =v

notingthata+b=e+c=1(c.f.Figure9).

Sinceb=handa=1b,andrecalling(11)and(12),wehave

k1 v
b=
k 1 vk 2 1v
k 2 1v
a=
k 1 vk 2 1v
a [ ex 2u x 1 x 2 ]
x 3=
e [ aeu ]
a [ ey 2 u y 1 y 2 ]
y 3=
e [ aeu ]

Thefollowingpartialderivativesarethenobtained:

y3 a [ ae y1 y 2 ey 2 ]
=
u e aeu 2
x3 a [ ae x 1 x 2 ex 2 ]
=
u e aeu 2
y3 k 2 b2 ue [ ey 2 u y 1 y 2 ]
=
v k 1 ev 2 aeu 2
x3 k 2 b 2 ue [ ex 2u x 1 x 2 ]
=
v k 1 ev 2 aeu 2

Sincez=vwehave
z
=0
u
z
=1
v

Wenowrequire m/uand m/vwhichrequiresustofindasuitableexpressionform. Referringto

Itintersectsthecirclex2+y22xx12yy1+(x12+y12R2)=0inthepointsgivenby

(mx+q)2+x22xx12y1(mx+q)+(x12+y12R2)=0
whichmayberearrangedasaquadraticequationinx:

x2(m2+1)+2x(mqy1mx1)+(q22y1q+x12+y12R2)=0

Settingthedescriminanttozerotogiveusequalroots(foratangent)wefind

m2(R2x12)+2mx1(y1q)+(R2+2qy1y12q2)=0

Substitutingq=y3mx3andsimplifyinggives

m2[R2(x1x3)2]+2m(x1x3)(y1y3)+R2(y1y3)2=0

which is a quadratic equation inm forthetwotangents tothe circle in Figure9,interms of known

m
=
[
m x 1 x 3
y3
u
y 1 y 3
x3
u ] [
m 2 x 1 x 3
x3
u
R
R
u ]
y 1y 3
y3
u
R
R
u
u [ ]
m R2 x 1 x 3 2 x 1 x 3 y 1 y3

Similarlyweget

m
=
[
m x 1 x 3
y3
v
y 1 y 3
x3
v ] [
m2 x 1 x 3
x3
v ]
y1 y 3
y3
v
v [ ]
m R2 x 1 x 3 2 x 1 x 3 y 1 y3

Allthesubsidiarypartialderivativesaregivenaboveexcept R/u,whichneedstobederivedfroman
expressionforRwhichisindependentofm,anddeterminedbythevortex.

InthemaintextwesawthatRisgivenby

R=W
eu u
u e
soclearly
R
=0
v
anddifferentiatingwrtugives

R W u
=
u u2 e [ eu e ]

Nowwecanfindthepartialderivativesofx,yandzwrtuandv:
 =
u u
y kv y3 mx 3
1m 2
=
{
kv y 3
1m2 u
m
x3
u
x 3
m 2m y 3 mx 3 m
u

1m2 u }
y
=
kv y 3
v 1m2 v
m
{
x3
v
x 3
v

1m2 v

1m 2 }
m 2m y 3 mx 3 m k y 3 mx 3

x my y m
= =m y
u u u u
x y m
=m y
v v v

Thisgivesuswhatweneedtocalculate

y , z z , x x ,y
, and
u , v u , v u , v