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Notebook[{
Cell[TextData[{
StyleBox["Lab 3C - Tangent Planes",
FontSize->24,
FontWeight->"Bold",
FontVariations->{"Underline"->True}],
"\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n \
questions to: rogness@math.umn.edu, drake@math.umn.edu"
}], "Text",
CellFrame->True,
TextAlignment->Center,
FontColor->GrayLevel[1],
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Cell[CellGroupData[{
Cell[TextData[StyleBox["Introduction",
FontSize->16]], "Section"],
Cell[TextData[{
"In the last few labs, we've investigated properties of continuity, \
differentiability, partial derivatives, directional derivatives, and \
gradients. In this lab, all of the functions we consider will be \
differentiable, which as we learned means that we can approximate the \
function at any given point with a linear approximation, or \"tangent \
plane.\"\n\nThere's an important point here: earlier we talked about whether \
a function is differentiable at the point ",
Cell[BoxData[
\(TraditionalForm\`x\&\[RightVector] = a\&\[RightVector]\)]],
"; if we say that a function is \"differentiable\" without any modifier, we \
mean that it is differentiable at ",
StyleBox["every",
FontSlant->"Italic"],
" point, or at least every point where it is defined. Go to the following \
web page:\n\nhttp://www.math.umn.edu/~rogness/multivar/tanplane.shtml\n\n(Or, \
alternatively, expand the subsection below labeled \"Tangent Plane Example\" \
and evaluate the cell there.) This demo shows you part of a paraboloid with \
a tangent plane at a certain point; you can click and drag the point to move \
it around, and LiveGraphics3D will automatically show you the tangent plane \
at the new point.\n\nThis is intended to reinforce the point that a \
\"differentiable\" function doesn't have one single tangent plane (or linear \
approximation, if you prefer that language). Rather, it has a different \
linear approximation at each point!\n\nToday we'll learn various methods of \
finding the equation of the tangent plane / linear approximation of a \
differentiable function. In some sense this is one of the most important \
things you can learn in this course, and all of the material for the first \
month of the class has led to this point. In particular, we'll use a lot of \
the ideas from the previous two labs."
}], "Text"],
Cell[CellGroupData[{
Cell["Tangent Plane Example", "Subsection"],
Cell["\<\
As with the code in the previous example, you don't have to worry \
about understanding the commands here. It's just intended for people who \
would like to view the example without opening up a web browser.
This is a slightly modified version of an example on the LiveGraphics3D \
website, so I make no claim of originality.\
\>", "Text",
CellFrame->True,
Background->GrayLevel[0.833326]],
Cell[BoxData[
\(\(\( (*\ plane, \ paraboloid, \ point, \
vector\ *) \)\(\[IndentingNewLine]\)\(\(independentVariables = {x \
\[Rule] .5, y \[Rule] .5};\)\n
\(dependentVariables = {x \[Rule]
If[x < \(-1\), \(-1\), If[x > 1, 1, x]],
y \[Rule] If[y < \(-1\), \(-1\), If[y > 1, 1, y]],
z \[Rule] x*x + y*y};\)\n
minx = \(-1\); miny = \(-1\); maxx = 1; maxy = 1;\[IndentingNewLine]
n = 10; dx = \((maxx - minx)\)/n;
dy = \((maxy - miny)\)/n;\[IndentingNewLine]
\(point = {PointSize[0.04], RGBColor[1, 0, 0], Point[{x, y, z}],
Point[{1. *x, 1. *y, z + .2}],
Point[{1. *x, 1. *y, z - .2}]};\)\[IndentingNewLine]
\(paraboloid =
Table[Polygon[{{i, j, i*i + j*j}, {i + dx,
j, \((i + dx)\)*\((i + dx)\) + j*j}, {i + dx,
j + dy, \((i + dx)\)*\((i + dx)\) + \((j + dy)\)*\((j +
dy)\)}, {i, j + dy,
i*i + \((j + dy)\)*\((j + dy)\)}}], {i, minx, maxx - dx/2,
dx}, {j, miny, maxy - dy/2, dy}];\)\n
\(plane =
Table[Polygon[{{i,
j, \((i - x)\)*2*x + \((j - y)\)*2*y + z}, {i + dx,
j, \((i + dx - x)\)*2*x + \((j - y)\)*2*y + z}, {i + dx,
j + dy, \((i + dx - x)\)*2*x + \((j + dy - y)\)*2*y + z}, {i,
j + dy, \((i - x)\)*2*x + \((j + dy - y)\)*2*y + z}}], {i,
minx, maxx - dx/2, dx}, {j, miny, maxy - dy/2,
dy}];\)\[IndentingNewLine]\[IndentingNewLine]
\(g =
Graphics3D[{paraboloid, point, plane}, BoxRatios \[Rule] {1, 1, 1},
Axes \[Rule] True,
AxesLabel \[Rule] {"\", "\", "\"}];\)\n\
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JLink\ loaded\ and\ Java\ installed\ in\ math2374 .
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y>1,1,y]],z\[Rule]x*x+y*y}\>", "\", \
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}, Closed]]
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Cell[CellGroupData[{
Cell[TextData[StyleBox["Plotting Planes",
FontSize->16]], "Section"],
Cell["\<\
Before we begin considering issues of tangency, let's review how to \
plot planes. It might help you to look in your textbook and review Cartesian \
and parametric equations for planes; you can also ask your TA for help.
The nicest situation is when we can describe the plane as a function of x and \
y; in that case, we can write the equation of the plane as ``z = ax + by + \
c'' where a, b, and c are real numbers. We can just use Plot3D in those \
cases; for instance, if we had z = 3x - y, we could just use\
\>", "Text"],
Cell[BoxData[
\(Plot3D[3 x\ - \ y, \ {x, \(-2\), 2}, {y, \(-2\), 2}]\)], "Input"],
Cell["\<\
to plot the plane, adjusting the range of x and y if necessary.
Even if we can describe a plane with a nice, easy equation, it is sometimes \
more convenient or enlightening to describe the plane with parametric \
equations. If we are given a point and two non-parallel vectors, there is a \
unique plane containing that point parallel to each of the vectors. Let's say \
our vectors are (2,3,5) and (-7,-11,13), and we wish to find the plane \
parallel to those two vectors passing through the point (17, 29, 0). In \
parametric equations, we could write\
\>", "Text"],
Cell[BoxData[
\(\((17, 29, 0)\)\ + \ t\ \((2, 3, 5)\)\ + \
s \((\(-7\), \(-11\), 13)\)\)], "DisplayFormula"],
Cell["\<\
where s and t range over all real numbers. To plot this plane, we \
use the properties of scalar multiplication and vector addition, and use \
ParametricPlot3D:\
\>", "Text"],
Cell[BoxData[
\(ParametricPlot3D[{17 + 2*t - 7*s, \ 29 + 3*t - 11*\ s, \
5*t + 13*s}, \ {t, \(-1\), 1}, {s, \(-1\), 1}]\)], "Input"],
Cell[TextData[{
"You can keep the parametric equation of the plane in ``un-multiplied out'' \
form and use ParametricPlot3D, although earlier versions of ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" might complain a bit before showing you the graph. Either way works; \
mathematically they are exactly equivalent. The following form may be easier \
to write."
}], "Text"],
Cell[BoxData[
\(ParametricPlot3D[{17, \ 29, 0}\ + \ t*{2, 3, 5} +
s*{\(-7\), \(-11\), 13}, \ {t, \(-1\), 1}, {s, \(-1\), 1}]\)], "Input"],
Cell[TextData[{
"One important thing to note is that we've only plotted a small portion of \
the plane. Above we chose values of t and s between -1 and 1, but we can \
choose any range we like. The ranges on s and t don't have to be the same. \
You don't have to use s and t if you don't like; ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" will let you use pretty much any variable name you like. The world is \
your oyster.\n\nThere's one other point you should be aware of. In this \
class you've seen how difficult it can be sometimes to represent \
three-dimensional pictures on a two-dimensional computer screen or piece of \
paper. A picture of a plane can be particularly hard to decipher; it's \
sometimes difficult to tell which part is sloping uphill, which part is going \
downhill, or even whether a plane is horizontal or vertical. As an example, \
look at the previous picture and try to decide how the plane is situated in \
space. Then evaluate the following command, which will make the same picture \
pop up in an interactive window. Rotate the plane around and see how \
accurately you interpreted the two-dimensional picture."
}], "Text"],
Cell[BoxData[
\(ParametricPlot3D[{17, \ 29, 0}\ + \ t*{2, 3, 5} +
s*{\(-7\), \(-11\), 13}, \ {t, \(-1\), 1}, {s, \(-1\), 1}]\)], "Input"],
Cell[CellGroupData[{
Cell["Example", "Subsection"],
Cell["Plot the plane that's described by", "Text"],
Cell[BoxData[
\(s \((0, 0, 1)\)\ + \ t \((0.2, \ 0.2, \ 2)\)\)], "DisplayFormula"],
Cell["\<\
when s and t range from 2 to 3. It won't look very \
good\[LongDash]it's ``too skinny''. Experiment with some different ranges on \
s and t until you get a nicer picture\[LongDash]one that's a little more \
``square''.\
\>", "Text"],
Cell[TextData[{
StyleBox["Exercise 1 - A Thought Experiment",
FontSize->16,
FontWeight->"Bold"],
"\n\nAre there planes that are not the graph of a function of x and y? If \
so, which planes are they? If there are no such planes, explain why.\n\nAre \
there planes that ",
StyleBox["cannot",
FontSlant->"Italic"],
" be described by a parametrization? If so, which planes are they? If there \
are no such planes (i.e., any plane whatsoever can be described by a \
parametrization), explain why.\n\nExplain your reasoning!"
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]]
}, Open ]]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Tangent Planes Via Tangent Vectors",
FontSize->16]], "Section"],
Cell[TextData[{
"We're going to start out this section with a crucial observation. Go back \
and think about the interpretation of a directional derivative -- it might \
help to reopen Lab 3B and run the example there, or go to:\n\n\
http://www.math.umn.edu/~rogness/multivar/dirderiv.shtml\n\nWe sliced through \
a surface with a vertical plane, and the intersection of the plane and \
surface gave us a ",
StyleBox["cross-section",
FontSlant->"Italic"],
" of the surface. (This is represented by the curvy part of the blue line; \
the straight blue lines are just there to help you visualize the vertical \
plane that you're using as a meat cleaver.) The red arrow is part of the \
tangent line of the cross-section.\n\nHere's the crucial observation: because \
the arrow is tangent to a cross-section through that point, then the arrow \
must be tangent to the ",
StyleBox["surface",
FontSlant->"Italic"],
" at that point! In other words, that arrow is actually part of the \
tangent plane at that point. As you rotate the cross-section, the arrow \
moves around a lot, so this might be hard to believe. There's another online \
example you can look at which might help convince you:\n\n\
http://www.math.umn.edu/~rogness/multivar/tanplane_withvectors.shtml\n\n\
(Alternatively, you can expand the subsection titled \"Interactive Example of \
Tangent Plane with Vectors\" down below and evaluate that cell to start the \
example from within ",
StyleBox["Mathematica",
FontSlant->"Italic"],
". In either case you should read on so that you know what you're looking \
at.)\n\nThe picture shows you the same surface as the previous directional \
derivative example, but now we've got ",
StyleBox["two",
FontSlant->"Italic"],
" cross-sections of the surface, both of which go through the point ",
Cell[BoxData[
\(TraditionalForm\`\((1, 0, 0)\)\)]],
". The red and purple vectors are tangent to the cross-sections at the \
point. The green vector is the cross product of the red and purple vectors.\n\
\nNote that we've also drawn in a portion of the plane tangent to the surface \
at the point (1,0,0). Rotate the cross-sections (using the slider on the \
bottom) and the entire picture (by clicking and dragging on the picture) \
until you're convinced that the tangent vectors are ",
StyleBox["always",
FontSlant->"Italic"],
" contained in the tangent plane. (This isn't always immediately obvious \
because LiveGraphics3D sometimes has trouble drawing lines and planes that \
overlap.)"
}], "Text"],
Cell[CellGroupData[{
Cell["Interactive Example of Tangent Plane with Vectors", "Subsection"],
Cell["\<\
As always with these special examples, you don't have to worry \
about understanding the code here. Just evaluate the next cell to start the \
example. Then collapse this subsection (so you don't have to look at the \
messy commands) and continue reading the lab.\
\>", "Text",
CellFrame->True,
Background->GrayLevel[0.833326]],
Cell[BoxData[{
\(\(f[x_, y_] = 2 y*Exp[\(-x^2\) - y^2];\)\), "\[IndentingNewLine]",
\(\(gf[x_, y_] = {D[f[x, y], x], D[f[x, y], y]};\)\ (*\
Gradient\ of\ f\ *) \), "\n",
\(\(\(Lf[x_, y_] =
f[1, 0] + gf[1, 0] . {x - 1, y};\)\(\[IndentingNewLine]\)
\)\), "\n",
\(\(\(SliderX[xmin_, xmax_, yval_, zval_,
ptlabel_] := \[IndentingNewLine]{{RGBColor[1, 0, 0],
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Text[ptlabel, {x, yval, zval* .8}]};\)\(\n\)
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\(\(n = 16;\)\), "\n",
\(xmin = \(-1\); xmax = 3.15; ymin = \(-2\); ymax = 2;\), "\n",
\(dx = \((xmax - xmin)\)/n; dy = \((ymax - ymin)\)/n;\), "\n",
\(\(mesh = {GrayLevel[0.7],
Table[Line[{{i, j, f[i, j]}, {i + dx, j, f[i + dx, j]}, {i + dx,
j + dy, f[i + dx, j + dy]}, {i, j + dy, f[i, j + dy]}, {i,
j, f[i, j]}}], {i, xmin, xmax, dx}, {j, ymin, ymax,
dy}]};\)\), "\n",
\(\(\(point = {RGBColor[0, 1, 0], PointSize[0.03],
Point[{1, 0, 0}]};\)\(\[IndentingNewLine]\)
\)\), "\n",
\(\(n = 22;\)\), "\n",
\(rmin = \(-2\); rmax = 2;\), "\n",
\(\(dr = \((rmax - rmin)\)/n;\)\), "\n",
\(\(dt = .9*Pi/2;\)\), "\[IndentingNewLine]",
\(\(section1 = {RGBColor[0, 0, 1], Thickness[0.005],
Table[Line[{{1 + \((r)\) Cos[t], \((r)\) Sin[t],
2 \((r)\) Sin[t]
Exp[\(-1\) - r^2 - 2 r*Cos[t]]}, {1 + \((r + dr)\)
Cos[t], \((r + dr)\) Sin[t],
2 \((r + dr)\) Sin[t]
Exp[\(-1\) - \((r + dr)\)^2 -
2 \((r + dr)\)*Cos[t]]}}], {r, rmin, rmax - dr,
dr}]};\)\), "\[IndentingNewLine]",
\(\(section2 = {RGBColor[0, 0, 1], Thickness[0.005],
Table[Line[{{1 + \((r)\) Cos[t + dt], \((r)\) Sin[t + dt],
2 \((r)\) Sin[t + dt]
Exp[\(-1\) - r^2 - 2 r*Cos[t + dt]]}, {1 + \((r + dr)\)
Cos[t + dt], \((r + dr)\) Sin[t + dt],
2 \((r + dr)\) Sin[t + dt]
Exp[\(-1\) - \((r + dr)\)^2 -
2 \((r + dr)\)*Cos[t + dt]]}}], {r, rmin, rmax - dr,
dr}]};\)\[IndentingNewLine]\[IndentingNewLine]\), "\
\[IndentingNewLine]",
\(dx = .1; dy = .1;\), "\[IndentingNewLine]",
\(\(\(tanplane = {RGBColor[ .9, .8, .8],
Table[Polygon[{{i, j, Lf[i, j]}, {i + dx, j,
Lf[i + dx, j]}, {i + dx, j + dy, Lf[i + dx, j + dy]}, {i,
j + dy, Lf[i, j + dy]}}], {i, .5, 1.4,
dx}, {j, \(- .5\), .4, dy}]};\)\(\n\)
\)\), "\[IndentingNewLine]",
\(\(vector1 = {RGBColor[1, 0, 0], Thickness[0.01],
Arrow3D[{1, 0, 0}, {1 + 1 Cos[t], 1 Sin[t],
2 Sin[t]/E}]};\)\), "\n",
\(\(vector2 = {RGBColor[1, 0, 1], Thickness[0.01],
Arrow3D[{1, 0, 0}, {1 + 1 Cos[t + dt], 1 Sin[t + dt],
2 Sin[t + dt]/E}]};\)\), "\[IndentingNewLine]",
\(\(normalvector = {RGBColor[0, 1, 0], Thickness[0.01],
Arrow3D[{1, 0, 0}, {1, \(-2\)/E, 1}]};\)\), "\[IndentingNewLine]",
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Graphics3D[{mesh, point, vector1, vector2, tanplane, normalvector,
section1, \ section2, SliderX[0, Pi, \(-3\), \(-1\), "\<\>"],
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Boxed \[Rule] False,
PlotRange \[Rule] {{\(-1\), 3.15}, {\(-3\), 2}, {\(-1.5\), .8}},
ViewPoint \[Rule] {2, \(-3\), 1.5},
Lighting \[Rule] False];\)\n\[IndentingNewLine] (*\
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Cell[CellGroupData[{
Cell["\<\
Using Tangent Vectors to find Parametric Equations of Tangent \
Planes\
\>", "Subsection"],
Cell[TextData[{
"So now you've seen that tangent vectors of cross sections through a point \
are in the tangent plane at that point. But wait -- all we need to write \
down the parametric equation of a plane is a point on the plane and two \
vectors in the plane! So, in this situation, we've got all the information \
we need. Given two cross sections of a surface which intersect at a point ",
Cell[BoxData[
\(TraditionalForm\`x\&\[RightVector] = a\&\[RightVector]\)]],
", the equation of the tangent plane is:"
}], "Text"],
Cell[BoxData[
\(TraditionalForm\`p(s, t) =
a\&\[RightVector] +
s\[CenterDot]\((tan . \
vector\ of\ first\ cross\ section\ at\ the\ point\ a\&\
\[RightVector])\)\ + \
t\[CenterDot]\(\((tan . \
vector\ of\ second\ cross\ section\ at\ the\ point\ a\&\
\[RightVector])\)\(.\)\)\)], "DisplayFormula",
TextAlignment->Center],
Cell["\<\
Let's do a real-life example now, by finding the plane tangent to \
the surface\
\>", "Text"],
Cell[BoxData[
\(z\ = \ x\^2\ + \ 3 xy\ + \ y\^2\)], "DisplayFormula",
TextAlignment->Center,
FontSize->14],
Cell[TextData[{
"at the point (1,-2,-1). First let's plot the surface; we'll use the ",
StyleBox["Mesh",
FontWeight->"Bold"],
" option to turn off the grid, so that later on we can see our cross \
sections a little better."
}], "Text"],
Cell[BoxData[{
\(f[x_, y_]\ = \ x^2\ + \ 3\ x*y\ + \ y^2\), "\[IndentingNewLine]",
\(surf\ = \
Plot3D[f[x, y], \ {x, 0, 2}, {y, \(-3\), \(-1\)}, \
Mesh \[Rule] False]\)}], "Input"],
Cell[TextData[{
"For our first cross section, let's just slice the surface with the plane \
x=1. [Note that this ",
StyleBox["will",
FontSlant->"Italic"],
" give us a cross section through the point (1,-2,-1). If you're not sure \
why this is true, as your TA.]\n\nWe'll use a parametric description for the \
curve that represents the restriction of f(x,y) to x=1. The x-value is always \
1; we'll let y vary -- let's say y=t in our parametric equation -- and the \
z-value is described by ",
Cell[BoxData[
\(TraditionalForm\`f(1, t)\)]],
":"
}], "Text"],
Cell[BoxData[{
\(c1[t_]\ = \ {1, \ t, \ f[1, t]}\), "\[IndentingNewLine]",
\(one\ =
ParametricPlot3D[
Evaluate[c1[t]], \ {t, \(-3.1\), \(-0.9\)}]\)}], "Input"],
Cell[TextData[{
"For our second cross section, let's use the vertical plane y=-2. The \
parametric description of this cross section is very similar: we fix y=-2, \
and let x vary -- let's say x=s in our parametric equation, and the z-value \
is just ",
Cell[BoxData[
\(TraditionalForm\`f(s, \(-2\))\)]],
"."
}], "Text"],
Cell[BoxData[{
\(c2[s_]\ = \ {s, \ \(-2\), \ f[s, \ \(-2\)]}\), "\[IndentingNewLine]",
\(two\ =
ParametricPlot3D[Evaluate[c2[s]], \ {s, \(-0.1\), 2.1}]\)}], "Input"],
Cell["\<\
Now we'll plot the surface and these curves together. Remember, we \
used \"Mesh -> False\" back in the Plot3D command to make it easier to see \
the curves.\
\>", "Text"],
Cell[BoxData[
\(Show[surf, \ one, two]\)], "Input"],
Cell[TextData[{
"Note how the paths run right through the surface. The nice thing about \
going to the trouble of using these paths is that",
StyleBox[" the tangent vectors of the paths are also tangent to the \
surface",
FontSlant->"Italic"],
". We can use those two tangent vectors to describe the plane that is \
tangent to the surface at the point where the curves intersect. It's easy to \
find the tangent vectors for the paths; we differentiate with respect to t \
(or s) and plug in the appropriate values of t or s. We get"
}], "Text"],
Cell[BoxData[{
\(dc1[t_] = D[c1[t], t]\), "\[IndentingNewLine]",
\(dc2[s_] = D[c2[s], s]\)}], "Input"],
Cell[TextData[{
"We're interested in the tangent vectors at the point (1,-2,-1), so we have \
to do a little bit of work here. For what value of ",
Cell[BoxData[
\(TraditionalForm\`t\)]],
" does ",
Cell[BoxData[
\(TraditionalForm\`c1(t) = \((1, \(-2\), \(-1\))\)\)]],
"? Well, we can solve that easily enough:"
}], "Text"],
Cell[TextData[{
Cell[BoxData[
\(TraditionalForm\`c1(t) = \((1, \(-2\), \(-1\))\)\)]],
"\n\n",
Cell[BoxData[
\(TraditionalForm\`\((1, t,
1 + 3 t + t\^2)\) = \((1, \(-2\), \(-1\))\)\)]]
}], "Text",
TextAlignment->Center,
FontSize->14],
Cell["\<\
Look at the y-component; it's pretty clear that c1 hits the \
intersection point (1,-2,-1) when t=-2. You can check that t=-2 gives you \
the right z-value, too. That means our first tangent vector is:\
\>", "Text"],
Cell[BoxData[
\(dc1[\(-2\)]\)], "Input"],
Cell["\<\
Similarly, the point of intersection occurs at s = 1 for c2 (check \
this!), yielding a tangent vector of (1,0,-4):\
\>", "Text"],
Cell[BoxData[
\(dc2[1]\)], "Input"],
Cell[TextData[{
"Now we know everything we need to describe the plane: we know a point it \
passes through, namely ",
Cell[BoxData[
\(TraditionalForm\`\((1, 2, f(1, \(-2\))\)\)]],
"), and two vectors parallel to the plane. In parametric terms, the plane \
is"
}], "Text"],
Cell[BoxData[
\(p[s_, t_]\ = \ {1, \(-2\), f[1, \(-2\)]}\ + \ t*{0, 1, \(-1\)}\ + \
s*{1, 0, \(-4\)}\)], "Input"],
Cell["\<\
Let's plot the plane and the surface together in a LiveGraphics3D \
pop-up window:\
\>", "Text"],
Cell[BoxData[{
\(\(tanplane\ =
ParametricPlot3D[
p[s, t], \ {s, \(-1\), 1}, \ {t, \(-1\),
1}];\)\), "\[IndentingNewLine]",
\(\(Show[surf, \ tanplane];\)\), "\[IndentingNewLine]",
\(\)}], "Input"],
Cell[TextData[{
"Rotate the picture and see if you can convince yourself that this plane \
really is tangent to the surface. (We've talked about how appearances can be \
deceiving, but in this case your eyes aren't lying to you.) Note how the \
plane is above the surface at some points and below it at others; that's \
perfectly fine. It's analogous to the line tangent to the curve y = ",
Cell[BoxData[
\(TraditionalForm\`sin\ x\)]],
" at x = 0, which goes above and beneath the curve:"
}], "Text"],
Cell[BoxData[
\(Plot[{Sin[x], x}, {x, \(-Pi\)/2, Pi/2},
AspectRatio \[Rule] Automatic]\)], "Input"],
Cell["\<\
Experience has taught us that the most confusing part of this \
process is where we set t=-2 and s=1. Before you go on, make sure you \
understand what we did there.
There's nothing magical going on; we're just trying to find the values of s \
and t so that c1(t) and c2(s) equal the point we're interested in. In \
particular, s and t don't have to be equal; they also don't have to be equal \
to 1, or 0, or any other \"nice\" value.\
\>", "Text",
CellFrame->True,
Background->GrayLevel[0.833326]]
}, Open ]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Curves Not Parallel to Coordinate Planes",
FontVariations->{"CompatibilityType"->0}]], "Subsection"],
Cell[TextData[{
StyleBox["In the previous example, we used the cross sections determined by \
x=1 and y=-2, but it's not necessary for us to choose such nice equations -- \
in fact, it's not even necessary that the values for x and y follow a \
straight line; we just need to make sure the x- and y-values follow some \
curve which goes through our point -- (1,-2), in this case. For example we \
could have used a cross section where the x- and y-values are on the parabola \
",
FontVariations->{"CompatibilityType"->0}],
Cell[BoxData[
\(TraditionalForm\`y = \(-x\^2\) - 1\)]],
". [Check that this parabola ",
StyleBox["does",
FontSlant->"Italic"],
" in fact go through (1,-2).] ",
"It would look like this:"
}], "Text"],
Cell[BoxData[{
\(\(f[x_, y_]\ = \
x^2\ + \ 3\ x*y\ + \ y^2;\)\), "\[IndentingNewLine]",
\(\(cs[t_] = {t, \(-t^2\) - 1,
f[t, \(-t^2\) - 1]};\)\), "\[IndentingNewLine]",
\(\(parabcs =
ParametricPlot3D[cs[t], {t, 0, 1.4}];\)\), "\[IndentingNewLine]",
\(\(Show[surf, parabcs,
AxesLabel \[Rule] {"\", "\", "\"}];\)\)}], "Input"],
Cell[TextData[{
StyleBox["You should look at the above commands until you're sure you \
understand the definition for ",
FontVariations->{"CompatibilityType"->0}],
Cell[BoxData[
\(TraditionalForm\`cs1[t]\)]],
". Ask your TA if you need help. However, in practice we usually don't \
use complicated cross sections like this because it's simply easier to work \
with \"straight ones.\"",
StyleBox["\n\nIt's certainly not necessary for the vertical planes to be as \
simple as x=1 or y=-2, however. Let's work through the previous example \
again, but this time, instead of restricting f(x,y) to the planes x=1 and \
y=-2, let's restrict to the planes ",
FontVariations->{"CompatibilityType"->0}],
Cell[BoxData[
\(TraditionalForm\`y = 2 x - 4\)]],
".",
StyleBox[" and ",
FontVariations->{"CompatibilityType"->0}],
Cell[BoxData[
\(TraditionalForm\`y = \(-x\) - 1\)]],
".",
StyleBox[" [You should check that these do in fact slice through the \
surface at the point ",
FontVariations->{"CompatibilityType"->0}],
Cell[BoxData[
\(TraditionalForm\`\((x, y)\) = \((1, \(-2\))\)\)]],
".]",
StyleBox["\n\nLet's define c3 and c4 to represent the restrictions to y = \
2x - 4 and y = -x-1 respectively. Again, make sure you understand the \
following definitions; ask for help if you're not sure about them.",
FontVariations->{"CompatibilityType"->0}]
}], "Text"],
Cell[BoxData[{
\(c3[s_]\ = \ {s, \ 2 s\ - 4, \
f[s, 2 s - 4]}\), "\[IndentingNewLine]",
\(c4[t_]\ = \ {t, \ \(-t\) - 1, \ f[t, \(-t\) - 1]}\)}], "Input"],
Cell[TextData[{
"Now let's look at the original surface with these two curves; you could \
replace the ",
StyleBox["Show",
FontWeight->"Bold"],
" command with ",
StyleBox["Show",
FontWeight->"Bold"],
" if you prefer."
}], "Text"],
Cell[BoxData[{
\(three\ =
ParametricPlot3D[
Evaluate[c3[s]], \ {s, 0.5, 1.5}]\), "\[IndentingNewLine]",
\(four\ = \
ParametricPlot3D[
c4[t], \ {t, \(-0.1\), 2.1}]\), "\[IndentingNewLine]",
\(Show[surf, three, four]\)}], "Input"],
Cell[TextData[{
"The tangent vectors are again found by differentiation: the tangent \
vectors are (1, 2, 22s-28) and (1, -1, -2t-1) and we are interested in the \
point when s and t both equal one, so we get (1,2,-6) and (1,-1,-3). Just \
like above, ",
StyleBox["the vectors tangent to the paths at the point of intersection are \
both tangent to the surface at that point",
FontSlant->"Italic"],
". The resulting plane is then"
}], "Text"],
Cell[BoxData[{
\(p2[s_, t_]\ = \ {1, \(-2\), f[1, \(-2\)]}\ + \ t*{1, 2, \(-6\)}\ + \
s*{1, \(-1\), \(-3\)}\), "\[IndentingNewLine]",
\(tanplane2\ = \
ParametricPlot3D[
p2[s, t], {t, \(- .5\), .5}, {s, \(- .5\), .5}]\), "\
\[IndentingNewLine]",
\(Show[surf, \ tanplane2]\)}], "Input"],
Cell["For comparison, here's the first tangent plane we found:", "Text"],
Cell[BoxData[
\(Show[surf, tanplane]\)], "Input"],
Cell["\<\
Are these planes the same? They had better be, and in this case \
they are. The pictures don't look exactly the same, but that's because we \
used different cross sections, and therefore we have different vectors in our \
parametric equations for the plane. You'll get a chance to verify that these \
two different methods produce identical planes.\
\>", "Text"]
}, Open ]],
Cell[CellGroupData[{
Cell["\<\
Using Tangent Vectors to find Cartesian Equations of Tangent Planes\
\
\>", "Subsection"],
Cell["\<\
Sometimes we need to find a Cartesian equation for a tangent plane. \
Recall that the Cartesian equation of a plane looks like\
\>", "Text"],
Cell[BoxData[
\(A \((x - x\_0)\) + B \((y - y\_0)\) + C \((z - z\_0)\) =
0\)], "DisplayFormula",
TextAlignment->Center,
FontSize->14],
Cell[TextData[{
"where ",
Cell[BoxData[
\(TraditionalForm\`\((x\_0, y\_0, z\_0)\)\)]],
" is a point on the plane, and the vector ",
Cell[BoxData[
\(TraditionalForm\`\((A, B, C)\)\)]],
" is normal (i.e. perpendicular) to the plane. Now think back to what \
we've been doing so far: given a surface, we choose a point; then we find two \
curves through that point; then we find the tangent vectors at that point.\n\n\
So we've got a point, and now we need a normal vector. Why not just use the \
cross product of the two tangent vectors? They're both in the tangent plane, \
so their cross product is perpendicular to the plane.\n\nHere's the \
parametric equation of the tangent plane we just found, which touches the \
surface at ",
Cell[BoxData[
\(TraditionalForm\`\((1, \(-2\), \(-1\))\)\)]],
". "
}], "Text"],
Cell[BoxData[
\(p2[s_, t_]\ = \ {1, \(-2\), f[1, \(-2\)]}\ + \ t*{1, 2, \(-6\)}\ + \
s*{1, \(-1\), \(-3\)}\)], "Input"],
Cell[TextData[{
"The two tangent vectors are ",
Cell[BoxData[
\(TraditionalForm\`\((1, \(-1\), \(-3\))\)\)]],
" and ",
Cell[BoxData[
\(TraditionalForm\`\((1, 2, \(-6\))\)\)]],
", so our normal vector is their cross product. ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" can compute cross products with a command named, not surprisingly, ",
StyleBox["Cross",
FontWeight->"Bold"],
":"
}], "Text"],
Cell[BoxData[
\(Cross[{1, \(-1\), \(-3\)}, {1, 2, \(-6\)}]\)], "Input"],
Cell["Therefore our Cartesian equation for this same plane is", "Text"],
Cell[BoxData[{
\(\(\(12 \((x - 1)\) + 3 \((y + 2)\) + 3 \((z + 1)\) =
0\)\(\[IndentingNewLine]\)
\)\), "\[IndentingNewLine]",
\(z = \(-4\) x - y + 1\)}], "DisplayFormula",
TextAlignment->Center],
Cell[TextData[{
"Let's plot this using ",
StyleBox["Plot3D",
FontWeight->"Bold"],
" and show it with the original surface:"
}], "Text"],
Cell[BoxData[{
\(\(tanplane3 =
Plot3D[\(-4\) x - y + 1, {x, 0,
2}, {y, \(-1\), \(-3\)}];\)\), "\[IndentingNewLine]",
\(\(Show[surf, tanplane3];\)\)}], "Input"],
Cell["\<\
Are you convinced that this is the same tangent plane? It \
certainly looks that way, and in fact it is!\
\>", "Text"],
Cell[TextData[{
StyleBox["Exercise 2",
FontSize->16,
FontWeight->"Bold"],
"\n\nIn this exercise you're going to find the linear approximation (or \
tangent plane, if you prefer) of the function ",
Cell[BoxData[
\(TraditionalForm\`g(x, y) = x\^2 - y\^2\)]],
"at the point ",
Cell[BoxData[
\(TraditionalForm\`\((x, y)\) = \((1, 2)\)\)]],
". Your work should include the following steps:\n\n(a) Find two curves on \
the surface which go through the point ",
Cell[BoxData[
\(TraditionalForm\`\((1, 2)\)\)]],
"; one of them can be the cross section given by ",
Cell[BoxData[
\(TraditionalForm\`x = 1\)]],
", but the other should be something different than the cross section given \
by ",
Cell[BoxData[
\(TraditionalForm\`y = 2\)]],
".\n\n(b) Find the derivatives of each of these curves, and find their \
tangent vectors at the point ",
Cell[BoxData[
\(TraditionalForm\`\((1, 2)\)\)]],
". \n\n(c) Give simplified parametric and cartesian equations for the \
plane.\n\nYour writeup should include a good picture of both the surface and \
the tangent plane. You should put some thought into the ranges of the x- and \
y-values in your picture; if you're too close, you won't be able to tell the \
difference between the plane and the surface, but if you're too far away, the \
resulting picture may not be very useful. In particular, if your picture \
clearly shows that your plane is ",
StyleBox["not",
FontSlant->"Italic"],
" tangent, or if it's impossible to tell, you will most likely lose \
points.",
"\n\n",
StyleBox["Exercise 3",
FontSize->16,
FontWeight->"Bold"],
"\n\nIn this exercise you're going to find the linear approximation (or \
tangent plane, if you prefer) of the function ",
Cell[BoxData[
\(TraditionalForm\`g(x, y) = \(sin(x)\) \(cos(y)\)\)]],
"at the point ",
Cell[BoxData[
\(TraditionalForm\`\((x, y)\) = \((\[Pi]/4, \(-\[Pi]\)/4)\)\)]],
". Your work should include the following steps:\n\n(a) Find two curves on \
the surface which go through the point (\[Pi]/4,-\[Pi]/4); one of them can be \
the cross section given by ",
Cell[BoxData[
\(TraditionalForm\`y = \(-\[Pi]\)/4\)]],
", but the other should be something different than the cross section given \
by ",
Cell[BoxData[
\(TraditionalForm\`x = \[Pi]/4\)]],
".\n\n(b) Find the derivatives of each of these curves, and find their \
tangent vectors at the point ",
Cell[BoxData[
\(TraditionalForm\`\((\[Pi]/4, \(-\[Pi]\)/4)\)\)]],
". \n\n(c) Give simplified parametric and cartesian equations for the \
plane.\n\nYour writeup should include a good picture of both the surface and \
the tangent plane. You should put some thought into the ranges of the x- and \
y-values in your picture; if you're too close, you won't be able to tell the \
difference between the plane and the surface, but if you're too far away, the \
resulting picture may not be very useful. In particular, if your picture \
clearly shows that your plane is ",
StyleBox["not",
FontSlant->"Italic"],
" tangent, or if it's impossible to tell, you will most likely lose \
points.",
"\n"
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]]
}, Open ]]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Tangent Planes Via Gradients",
FontSize->16]], "Section"],
Cell["\<\
Another way to find the tangent plane to a surface is to use the \
gradient vector. Recall that the gradient vector of a function f(x,y) is \
defined to be\
\>", "Text"],
Cell[BoxData[
\(\[Del]\&\[RightVector] f \((x,
y)\) = \(\((\[PartialD]f\/\[PartialD]x, \ \
\[PartialD]f\/\[PartialD]y)\)\(.\)\)\)], "DisplayFormula",
TextAlignment->Center],
Cell[TextData[{
"Hmmm. This is a two-dimensional vector, and to define a plane in 3-space, \
we'll definitely need vectors with three components. We get around this by \
doing some minor rearranging. Our surface is defined by ",
Cell[BoxData[
\(TraditionalForm\`z = f(x, y)\)]],
". We could move everything over to the same side and say that our surface \
is defined by the equation"
}], "Text"],
Cell[BoxData[
\(TraditionalForm\`z - f(x, y) = 0\)], "DisplayFormula",
TextAlignment->Center,
FontSize->14],
Cell[TextData[{
"You might be thinking that this is an odd step; if anything, it's a more \
complicated way to write it. That's true, and it's going to get a little \
worse. Rather than saying our surface is defined by this equation, let's \
defined a ",
StyleBox["new",
FontSlant->"Italic"],
" function:"
}], "Text"],
Cell[BoxData[
\(g \((x, y, z)\)\ = \ z\ - \ f \((x, y)\)\)], "DisplayFormula",
TextAlignment->Center],
Cell[TextData[{
"And now we say that our function is the ",
StyleBox["level set",
FontSlant->"Italic"],
" defined by ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = 0\)]],
". So now instead of the nice, simple statement ",
Cell[BoxData[
\(TraditionalForm\`z = f(x, y)\)]],
", we're suddenly talking about level sets. Yikes! There's a reason for \
this, however...\n\nNotice that ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z)\)]],
" is a function of three variables, and its gradient ",
Cell[BoxData[
\(TraditionalForm\`is\)]]
}], "Text"],
Cell[BoxData[
\(\[Del]\&\[RightVector] g \((x, y,
z)\) = \(\((\(-\(\[PartialD]f\/\[PartialD]x\)\), \ \
\(-\(\[PartialD]f\/\[PartialD]y\)\), \ 1)\)\(.\)\)\)], "DisplayFormula",
TextAlignment->Center],
Cell[TextData[{
"Now recall from lecture (or the previous lab) that ",
StyleBox["the gradient of a function g is perpendicular to the level sets \
of g",
FontSlant->"Italic"],
". In other words:\n\n-- We can start with a point (x,y,z) which is on the \
level set ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z) = 0\)]],
".\n-- Because of the way we defined ",
Cell[BoxData[
\(TraditionalForm\`g(x, y, z)\)]],
", this is the same as saying the point is on the surface ",
Cell[BoxData[
\(TraditionalForm\`z = f(x, y)\)]],
".\n-- Furthermore, ",
Cell[BoxData[
\(\[Del]\&\[RightVector] g \((x, y, z)\)\)],
TextAlignment->Center],
"will be perpendicular (or normal) to our surface at that point.\n-- In \
particular, we can use ",
Cell[BoxData[
\(\[Del]\&\[RightVector] g \((x, y, z)\)\)],
TextAlignment->Center],
"as a normal vector to define the equation for our tangent plane!\n\nLet's \
use this method quickly to find the same tangent plane we've been working \
with this whole time. Remember that the function is given by:"
}], "Text"],
Cell[BoxData[
\(f[x_, y_]\ = \ x^2\ + \ 3\ x*y\ + \ y^2\)], "Input"],
Cell["So we can do our magic \"new\" function like this:", "Text"],
Cell[BoxData[
\(g[x_, y_, z_] = z - f[x, y]\)], "Input"],
Cell[TextData[{
"We're interested in the gradient at the point ",
Cell[BoxData[
\(TraditionalForm\`\((1, \(-2\), \(-1\))\)\)]],
":"
}], "Text"],
Cell[BoxData[{
\(gradg[x_, y_, z_] = Grad[g[x, y, z]]\), "\[IndentingNewLine]",
\(gradg[1, \(-2\), \(-1\)]\)}], "Input"],
Cell["Thus the Cartesian equation of the tangent plane is:", "Text"],
Cell[BoxData[{
\(\(\(4 \((x - 1)\) + \((y + 2)\) + \((z + 1)\) =
0\)\(\[IndentingNewLine]\)
\)\), "\[IndentingNewLine]",
\(z = \(-4\) x - y + 1\)}], "DisplayFormula",
TextAlignment->Center],
Cell["\<\
Which is exactly what we found before. Good!
Although the \"gradient method\" of finding a tangent plane might seem more \
complicated -- particularly if you don't like level sets -- it's very, very \
useful. We didn't have to find any cross sections, find tangent vectors, \
solve for values of s and t, and so on. We just rewrote the original \
equation, computed one gradient, and we were basically done!\
\>", "Text"],
Cell[TextData[{
StyleBox["Exercise 4",
FontSize->16,
FontWeight->"Bold"],
"\n\nUse the gradient method to find the cartesian equation of the plane \
tangent to\n\n",
Cell[BoxData[
\(TraditionalForm\`f(x, y) = \(Exp(\(-x\^2\) - y\^2)\)*x\)]],
"\n\nat the point ",
Cell[BoxData[
\(TraditionalForm\`\(\((0, 0, f(0, 0))\)\(.\)\)\)]],
" Show the graph of f(x,y) and this tangent plane together on the same \
plot. Be sure to show your work and explain your reasoning.\n\n",
StyleBox["Exercise 5",
FontSize->16,
FontWeight->"Bold"],
"\n\nUse the gradient method to find the cartesian equation of the plane \
tangent to\n\n",
Cell[BoxData[
\(TraditionalForm\`f(x, y) = y*\(Cos(x)\)\)]],
"\n\nat the point ",
Cell[BoxData[
\(TraditionalForm\`\(\((0, \[Pi]/4, f(0, \[Pi]/4))\)\(.\)\)\)]],
" Show the graph of f(x,y) and this tangent plane together on the same \
plot; use the ranges -\[Pi] to \[Pi] for both x and y. Be sure to show your \
work and explain your reasoning. In particular, with the suggested ranges \
the \"true\" tangent plane looks like it sticks through part of the surface. \
You may want to explain why that's the case."
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Tangent Plane Conspiracy Theory",
FontSize->16]], "Section"],
Cell["\<\
You are not responsible for the material in this section, but we've \
included it for those people who are interested in the mathematics behind all \
of this; it turns out that all of the different methods of finding tangent \
planes aren't really all that different!\
\>", "Text",
CellFrame->True,
Background->GrayLevel[0.833326]],
Cell[TextData[{
"In this section, we'll learn that, despite all appearances to the \
contrary, the ``tangent vectors'' and ``gradient vector'' methods are \
surreptitiously working together and are engaged in a ",
StyleBox["secret government conspiracy!",
FontSlant->"Italic"],
"\n\nOkay, there's no government conspiracy (well...none that ",
StyleBox["I'm",
FontSlant->"Italic"],
" aware of), but it is true that the two methods are working together\
\[LongDash]in fact, they are just different ways of looking at the same \
thing. Let's talk about directional derivatives for a bit before explaining \
why they're the same.\n\nAbove we were talking about partial derivatives and \
directions, which should remind you of directional derivatives. In the \
Tangent Vectors section, when specifying ``the planes y = 2x - 4 and y = -x - \
1'', we were really talking about direction vectors. In the above cases, the \
corresponding direction vectors would be"
}], "Text"],
Cell[BoxData[
\(\(\(\(1\/\@5\) \((1,
2)\)\ \ \ \ and\ \ \ \ \ \(1\/\@2\) \((1, \(-1\))\)\)\(,\)\)\)], \
"DisplayFormula",
TextAlignment->Center],
Cell[TextData[{
"respectively. We are basically using the slope of a line: y = 2x - 4 is a \
line (in two dimensions) with slope 2. The vector (1,2) is parallel to that \
line, and above we just normalized it. The other direction vector was \
obtained in a similar way. Once we know direction vectors, we just need to \
take the dot product with the gradient vector to find the directional \
derivative. The conspiracy is already unravelling.\n\nFor the function f(x,y) \
= ",
Cell[BoxData[
\(TraditionalForm\`x\^2\ + \ 3 xy + \ y\^2\)]],
" at the point (1,-2), and the first direction vector\[LongDash]call it u\
\[LongDash]we have"
}], "Text"],
Cell[BoxData[
\(Df\_u = \ \(\((\[PartialD]f\/\[PartialD]x, \ \
\[PartialD]f\/\[PartialD]y)\)\[CenterDot]\((1\/\@5,
2\/\@5)\)\ = \ \(\((\(-4\), \ \(-1\))\)\[CenterDot]\((1\/\@5,
2\/\@5)\)\ = \ \(-\(6\/\@5\)\)\)\)\)], "DisplayFormula",
TextAlignment->Center],
Cell[TextData[{
"We interpret this by saying ``in the direction of u, f is increasing at a \
rate of ",
Cell[BoxData[
\(TraditionalForm\`\(-6\)/\@5\)]],
".'' The upshot of this is that (",
Cell[BoxData[
\(TraditionalForm\`1/\@5\)]],
", ",
Cell[BoxData[
\(TraditionalForm\`2/\@5\)]],
", ",
Cell[BoxData[
\(TraditionalForm\`\(-6\)/\@5\)]],
") is a tangent vector to f(x,y), and it's parallel to \
(1,2,-6)\[LongDash]the tangent vector we found by using the path we described \
in the first section. We get this vector by taking the direction vector\
\[LongDash]a two-dimensional vector\[LongDash]and adding a third component, \
which is given by the directional derivative of f(x,y). In general, this \
means that"
}], "Text"],
Cell[BoxData[
\(\((u1, \ u2, \ \(D\_u\) f)\)\)], "DisplayFormula"],
Cell[TextData[{
"is a vector tangent to the surface z = f(x,y). You should do this same \
process with the other direction vector above and make sure that you end up \
with a tangent vector parallel to (1,-1,-3).\n\nOf course, there's nothing \
special about those two directional vectors above. We can use ",
StyleBox["any",
FontSlant->"Italic"],
" two non-parallel direction vectors and we'll get the same \
plane\[LongDash]which is exactly what you would hope for, since a function \
can have only one tangent plane (or no tangent plane, if the function isn't \
differentiable). The following exercise will help you expose the conspiracy\
\[LongDash]er, show that the tangent planes from the ``tangent vectors \
method'' and from the ``gradient vectors method'' are in fact the same. We'll \
do that by comparing the normal vectors of the resulting planes."
}], "Text"],
Cell[TextData[{
StyleBox["Exercise ",
FontSize->16,
FontWeight->"Bold"],
"\n\n(Notice that this isn't in a red box; you don't have to hand this \
exercise in. You might find it interesting, however.)\n\nAs discussed in the \
Gradient Vector section, the gradient of z - f(x,y) represents a normal \
vector to the tangent plane, so our main objective is to find the normal \
vector of the plane we get from the tangent vector method. We'll use an \
arbitrary differentiable function f(x,y) and two non-parallel direction \
vectors u = (u1, u2) and v = (v1, v2).\n\n\t(i) Find the vectors tangent to \
f(x,y) in the directions of u and v. Describe them in terms of u1, u2, v1, \
v2, and the partials of f (\[PartialD]f / \[PartialD]x and \[PartialD]f / \
\[PartialD]y).\n\n\t(ii) Find the normal vector of the plane spanned by those \
two vectors.\n\t\n\t(iii) Show that the normal vector you found in (ii) is \
parallel to the gradient vector of the function g(x,y,z) = z - f(x,y).\n\t\n\t\
(iii) What does this imply about the tangent planes found using the two \
different methods?\n "
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"This lab is a descendent of an earlier lab on Tangent Planes written by \
Cindy Kaus. Dan Drake did a major rewrite in 2002, and then I overhauled it \
again in February 2004. We changed the focus a little bit, and added a few \
things here and there; having taught this lab for a few years now, we've \
learned what the sticking points are for students, so I've tried to add extra \
explanations in these parts. I also changed the exercises, except for #1, \
which Dan wrote, and the pseudo-exercise in the \"Conspiracy Theory\" \
section, which is also Dan's. I changed the rest of them so we'd have a \
little variety; the previous exercises have been used for the last 4 years or \
so.\n\nIt's a little difficult to completely describe who did what, but \
here's an attempt:\n\n-- I don't believe any of the writing is Cindy's, but \
the particular surface and cross sections used in the \"Tangent Vectors\" \
sections (and exercises 2 and 3) are hers.\n\n-- The \"Plotting Planes\" and \
\"Tangent Plane Conspiracy Theory\" sections are all Dan's; at most I changed \
a few words here or there, or added a ",
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" command.\n\n-- The other sections are fairly major rewrites of Dan's \
original sections. A lot of his original text remains. Some of it remains \
but has been modified a bit, because of reorganization, etc. Some of it is \
new text added by me. To make things more complicated, it's all intertwined; \
many paragraphs combine Dan's writing with mine.\n\nThis would be a major \
mess, but fortunately we've all agreed to use the same license, and it works \
out. The current version of the lab is copyright 2004 by Jonathan Rogness \
(rogness@math.umn.edu) and is protected by the Creative Commons \
Attribution-NonCommercial-ShareAlike License. You can find more information \
on this license at http://creativecommons.org/licenses/by-nc-sa/1.0/. (As \
mentioned, parts of this are copyright 2000 by Cindy Kaus and 2002 by Dan \
Drake and are protected under the same license.)\n\nAlthough it's not \
specifically required by the license, I'd appreciate it if you let me know if \
you use parts of our labs, just so I can keep track of it. Please send me \
any questions or comments!"
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