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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 31 "Tangent Planes and Normal Lines
" }}{PARA 257 "" 0 "" {TEXT -1 10 " Jim Herod" }}{PARA 258 "" 0 ""
{TEXT -1 20 "School of Mathemtics" }}{PARA 259 "" 0 "" {TEXT -1 12 "Ge
orgia Tech" }}{PARA 260 "" 0 "" {TEXT -1 21 "herod@math.gatech.edu" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA
0 "" 0 "" {TEXT -1 52 " In this worksheet, we consider surfaces S giv
en by" }}{PARA 0 "" 0 "" {TEXT -1 44 " R(x,y) = [x,
y, f(x, y)]" }}{PARA 0 "" 0 "" {TEXT -1 59 "where f is number valued \+
and [x, y] is a point in a region " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG
" }{TEXT -1 14 " in the plane." }}{PARA 0 "" 0 "" {TEXT -1 224 " T
his worksheet will provide an explanation for how a tangent plane for \+
the surface S is constructed in case f has continuous partial derivati
ves in x and y. If you understand this worksheet, you should know four
things:" }}{PARA 0 "" 0 "" {TEXT -1 81 "(1) how to compute the plane \+
tangent to the surface S at a point [a, b, f(a, b)]." }}{PARA 0 "" 0 "
" {TEXT -1 77 "(2) how to compute a normal line to the surface S at a \+
point [a, b, f(a, b)]." }}{PARA 0 "" 0 "" {TEXT -1 50 "(3) how to draw
pictures of these geometric ideas." }}{PARA 0 "" 0 "" {TEXT -1 54 "(4
) how to find a linear approximation for f at [a,b]." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 221 " Finally, at the e
nd of this worksheet, there is a pathological example. There is an exa
mple of function f which has partial derivatives with respect to x and
y, but the surface will not have a tangent plane at [0,0]." }}{PARA
0 "" 0 "" {TEXT -1 5 " " }}{PARA 0 "" 0 "" {TEXT -1 92 " As we
develop the steps leading to understanding, we carry with us the simp
le function " }{XPPEDIT 18 0 "f(x,y) = -(x^2+y^2)" "6#/-%\"fG6$%\"xG%
\"yG,$,&*$F'\"\"#\"\"\"*$F(F,F-!\"\"" }{TEXT -1 30 " with [x, y] in th
e rectangle " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG" }{TEXT -1 113 " = [-
2, 2] x [-2, 2]. For the purposes of pictures, we take the point of th
e construction to be [a, b] = [-1, 1]." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 44 "(1) Consider a surface R as described
above." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "f:=(x,y)->-(x^2+y
^2);\nplot3d(f(x,y),x=-2..2,y=-2..2);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 51 "(2) Consider the line L1 = [a + t, b
, f(a,b) + t " }{XPPEDIT 18 0 "diff(f,x)" "6#-%%diffG6$%\"fG%\"xG" }
{TEXT -1 8 "(a,b) ]." }}{PARA 0 "" 0 "" {TEXT -1 56 "(a) We draw the p
rojection of the line in the x-y plane." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 36 "a:=1; b:=1;\nplot([a+t,b,t=-3/2..1]);" }}}{PARA 0 ""
0 "" {TEXT -1 37 "(b) We draw the line in 3-dimensions." }}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 37 "dfdx:=subs(\{x=a,y=b\},diff(f(x,y),x));" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "L1:=t->[a+t,b,f(a,b)+t*dfdx]
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 184 "J1:=spacecurve(L1(t),
t=-3/2..1,axes=NORMAL,orientation=[10,80],color=RED):\nK:=plot3d(f(x,y
),x=-2..2,y=-2..2,axes=NORMAL,\n orientation=[10,80],color=
BLUE):\ndisplay3d(\{J1,K\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 50 "(3) Consider the line L2 = [a, b + t, f(a
,b) + t " }{XPPEDIT 18 0 "diff(f,y)" "6#-%%diffG6$%\"fG%\"yG" }{TEXT
-1 8 "(a,b) ]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 57 "(a) We draw the projection of the line in the x-y plane.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "a:=1; b:=1;\nplot([a,b+t
,t=-3/2..1]);" }}}{PARA 0 "" 0 "" {TEXT -1 37 "(b) We draw the line in
3-dimensions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "dfdy:=subs
(\{x=a,y=b\},diff(f(x,y),y));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 29 "L2:=t->[a,b+t,f(a,b)+t*dfdy];" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 187 "J2:=spacecurve(L2(t),t=-3/2..1,axes=NORMAL,orientati
on=[10,80],color=RED):\nK:=plot3d(f(x,y)-.1,x=-2..2,y=-2..2,axes=NORMA
L,\n orientation=[10,80],color=BLUE):\ndisplay3d(\{J2,K\});
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "(4) \+
We draw L1 and L2, It should be observed that these two lines are tang
ent to the surface." }}{PARA 0 "" 0 "" {TEXT -1 54 "(a) First, we draw
the projection of the two lines in " }{XPPEDIT 18 0 "Omega" "6#%&Omeg
aG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "plot(
\{[a+t,b,t=-3/2..1],[a,b+t,t=-3/2..1]\});" }}}{PARA 0 "" 0 "" {TEXT
-1 59 "(b) We draw the two lines superimposed with the graph of S." }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "display3d(\{J1,J2,K\});" }}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 119 "(5) A ve
ctor perpendicular to the surface can be found by taking the cross pro
duct of L1(1) - L1(0) with L2(1) - L2(0)." }}{PARA 0 "" 0 "" {TEXT -1
110 "(a) To see this vector, we draw the line through [a, b, f(a,b)] w
hich has the direction of the cross product.." }}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 13 "with(linalg):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 38 "N:=crossprod(L1(1)-L1(0),L2(1)-L2(0));" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "PL:=t->[a+t*N[1], b+t*N[2], f(a,b)+
t*N[3]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 "NL:=spacecurve
(PL(t),t=-3/2..1,axes=NORMAL,orientation=[10,80],\n color
=GREEN):\ndisplay(\{J1,J2,K,NL\});" }}}{PARA 0 "" 0 "" {TEXT -1 171 "(
b) Perhaps it is clear from the geometry that the normal line is perpe
ndicular to the two tangent lines. It was made up to have that propert
y! We check this algebracally." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1
0 67 "dotprod(L1(t)-L1(0),PL(t)-PL(0));\ndotprod(L2(t)-L2(0),PL(t)-PL(
0));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 161
"(6) We can now give an equation for the tangent plane. The tangent pl
ane is all [x, y, z] so that the dotproduct of [x-a, y-b, z-f(a,b)] wi
th the normal is zero." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "ta
ngenteqn:=dotprod([x-a,y-b,z-f(a,b)],N)=0;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 23 "Z:=solve(tangenteqn,z);" }}}{PARA 0 "" 0 "" {TEXT
-1 27 "We illustrate with a graph." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 114 "Plane:=plot3d(Z,x=a-1/2..a+1/2,y=b-1/2..b+1/2,axes=N
ORMAL,\n orientation=[10,80],color=RED,style=PATCH):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "display(\{Plane,K\});" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 143 "(7) This provides an understan
ding for what is the linear approximation for f at [a,b]. It is the va
lue for z obtained by solving the equation " }}{PARA 0 "" 0 "" {TEXT
-1 72 " 0 = dotprod([x - a, y - b, z - f(a
,b)], N)," }}{PARA 0 "" 0 "" {TEXT -1 21 "where N is the normal" }}
{PARA 0 "" 0 "" {TEXT -1 40 " [ - \+
" }{XPPEDIT 18 0 "diff(f,x)*(a,b)" "6#*&-%%diffG6$%\"fG%\"xG\"\"\"6$%
\"aG%\"bGF)" }{TEXT -1 4 ", - " }{XPPEDIT 18 0 "diff(f,y)*(a,b)" "6#*&
-%%diffG6$%\"fG%\"yG\"\"\"6$%\"aG%\"bGF)" }{TEXT -1 6 ", 1 ]." }}
{PARA 0 "" 0 "" {TEXT -1 5 "Thus," }}{PARA 0 "" 0 "" {TEXT -1 40 " \+
z = " }{XPPEDIT 18 0 "diff(f,x)*(a,b)
" "6#*&-%%diffG6$%\"fG%\"xG\"\"\"6$%\"aG%\"bGF)" }{TEXT -1 11 " * (x-a
) + " }{XPPEDIT 18 0 "diff(f,y)*(a,b)" "6#*&-%%diffG6$%\"fG%\"yG\"\"\"
6$%\"aG%\"bGF)" }{TEXT -1 18 " * (y-b) + f(a,b)." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 " In the context of th
e gradient, we could write this equation as" }}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 " \+
z = dotprod([ grad(f)(a,b), [x-a, y-b] ) + f(a,b)." }}{PARA 0 ""
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "We illustrate this w
ith formula for the tangent plane with a different example." }}{PARA
0 "" 0 "" {TEXT 256 9 "EXAMPLE. " }{TEXT -1 14 " Let f(x,y) = " }
{XPPEDIT 18 0 "x^2 + x*y - y^2" "6#,(*$%\"xG\"\"#\"\"\"*&F%F'%\"yGF'F'
*$F)F&!\"\"" }{TEXT -1 48 ". We find the tangent plane at [a,b] = [-1
, 1]." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "with(plots):with(li
nalg):\na:=-1; b:=1;\nf:=(x,y)->x^2+x*y-y^2;" }}}{EXCHG {PARA 0 "> "
0 "" {MPLTEXT 1 0 44 "fprime:=subs(\{x=-1,y=1\},grad(f(x,y),[x,y]));"
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Linear:=dotprod( fprime,[
x-a,y-b])+f(a,b);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 218 "Surfa
ce:=plot3d(f(x,y),x=-2..2,y=-2..2,axes=NORMAL,\n orientatio
n=[65,65],color=BLUE):\nTangent:=plot3d(Linear,x=a-1/2..a+1/2,y=b-1/2.
.b+1/2,axes=NORMAL,\n orientation=[65,65],color=RED,style=P
ATCH):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "display(\{Surface
,Tangent\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 21 "PATHOLOGICAL EXAMPLE\n"
}{TEXT -1 77 "Finally, here is a pathological example. By inspection, \+
one would agree that " }{XPPEDIT 18 0 "diff(f,x)*(0,0)" "6#*&-%%diffG6
$%\"fG%\"xG\"\"\"6$\"\"!F+F)" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "diff
(f,y)*(0,0)" "6#*&-%%diffG6$%\"fG%\"yG\"\"\"6$\"\"!F+F)" }{TEXT -1 48
" are zero but there is no tangent plane at [0,0]" }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 70 "plot3d(x*y/(x^2+y^2),x=0..1,y=0..1,orientatio
n=[-135,34],axes=NORMAL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0
"" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 "" {TEXT
-1 24 "Exercise for the student" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 13 "Let f(x,y) = " }{XPPEDIT 18 0 "x^3*y" "6#
*&%\"xG\"\"$%\"yG\"\"\"" }{TEXT -1 80 ". Find the plane tangent to the
graph of f at a = -1, b = 1. Draw both surfaces." }}}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}}{MARK "1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }