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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 21 "School of Mathematics" }}{PARA 259 "" 0 "" {TEXT -1 12 "G
eorgia 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 given 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 v
alued 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 "
This worksheet will provide an explanation for how a tangent plan
e for the surface S is constructed in case f has continuous partial de
rivatives in x and y. If you understand this worksheet, you should kno
w 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 t
o 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 -1 237 " Finally, at
the end of this worksheet, there is a pathological example. There is \+
an example of function f which has partial derivatives with respect to
x and y, f(0,0) = 0, and the surface will not have a tangent plane at
[0, 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 simple 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 the rectangle " }{XPPEDIT 18 0 "Omega" "6#%&OmegaG
" }{TEXT -1 113 " = [-2, 2] x [-2, 2]. For the purposes of pictures, w
e take the point of the 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) Consi
der the line L1 = [a + t, b, f(a,b) + t " }{XPPEDIT 18 0 "diff(f,x)
" "6#-%%diffG6$%\"fG%\"xG" }{TEXT -1 8 "(a,b) ]." }}{PARA 0 "" 0 ""
{TEXT -1 56 "(a) We draw the projection of the line in the x-y plane.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "a:=1; b:=1;\nplot([a+t,b
,t=-3/2..1]);" }}}{PARA 0 "" 0 "" {TEXT -1 52 "(b) We draw the line in
3-dimensions. The line will " }{TEXT 258 3 "not" }{TEXT -1 57 " lie i
n the x-y plane but will be parallel to the x-axis." }}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 37 "dfdx:=subs(\{x=a,y=b\},diff(f(x,y),x));" }}}{EXCHG
{PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "L1:=t->[a+t,b,f(a,b)+t*dfdx];" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 184 "J1:=spacecurve(L1(t),t=-3/2
..1,axes=NORMAL,orientation=[10,80],color=RED):\nK:=plot3d(f(x,y),x=-2
..2,y=-2..2,axes=NORMAL,\n orientation=[10,80],color=BLUE):
\ndisplay3d(\{J1,K\});" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 ""
0 "" {TEXT -1 50 "(3) Consider the line L2 = [a, b + t, f(a,b) + t "
}{XPPEDIT 18 0 "diff(f,y)" "6#-%%diffG6$%\"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 46 "(b) We draw the line in 3-dimensio
ns. It will " }{TEXT 259 3 "not" }{TEXT -1 58 " lie in the x-y plane, \+
but will be parallel to the y axis." }}{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 0 189 "J2:=spacecurve(L2(t),t=-3/2
..1,axes=NORMAL,orientation=[10,80],color=green):\nK:=plot3d(f(x,y)-.1
,x=-2..2,y=-2..2,axes=NORMAL,\n orientation=[10,80],color=B
LUE):\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 th
at these two lines are tangent 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#%&OmegaG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 ">
" 0 "" {MPLTEXT 1 0 62 "plot([[a+t,b,t=-3/2..1],[a,b+t,t=-3/2..1]],co
lor=[green,red]);" }}}{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 vector perpendicular to the
surface can be found by taking the cross product of L1(1) - L1(0) wit
h L2(1) - L2(0)." }}{PARA 0 "" 0 "" {TEXT -1 110 "(a) To see this vect
or, we draw the line through [a, b, f(a,b)] which 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(L
1(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 0 115 "NL:=spacecurve(PL(t),t=-3/2..1,axes=NORMAL,orie
ntation=[10,80],\n color=orange):\ndisplay(\{J1,J2,K,NL\}
);" }}}{PARA 0 "" 0 "" {TEXT -1 172 "(b) Perhaps it is clear from the \+
geometry that the normal line is perpendicular to the two tangent line
s. It was made up to have that property! We check this algebraically.
" }}{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 -1 162 "(6) We can now give an e
quation for the tangent plane. The tangent plane is all [x, y, z] so t
hat the dot product of [x-a, y-b, z-f(a,b)] with the normal is zero."
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "tangenteqn:=dotprod([x-a,y
-b,z-f(a,b)],N)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "Z:=so
lve(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=NORMAL,\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 understanding for what is the linear approxi
mation for f at [a,b]. It is the value for z obtained by solving the e
quation " }}{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 the gradient, we could write this equati
on 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 with formula for the tangent plane wit
h 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 th
e tangent plane at [a,b] = [-1, 1]." }}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 61 "with(plots):with(linalg):\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 "Surface:=plot3d(f(x,y),x=-2
..2,y=-2..2,axes=NORMAL,\n orientation=[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=PATCH):" }}}{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,orientation=[-135,34],axes=NORM
AL);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0
"> " 0 "" {MPLTEXT 1 0 0 "" }}}{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 "0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }
{PAGENUMBERS 0 1 2 33 1 1 }