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Notebook[{
Cell[TextData[{
StyleBox["Lab 3B - Tangent Planes",
FontSize->24,
FontWeight->"Bold"],
"\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n \
questions to: drake@math.umn.edu"
}], "Text",
CellFrame->True,
TextAlignment->Center,
FontColor->GrayLevel[1],
Background->RGBColor[0, 0, 1]],
Cell[CellGroupData[{
Cell["Introduction", "Section"],
Cell[TextData[{
"The last lab investigated properties of continuity, differentiability, \
partial derivatives, and the relationships between them. In this lab, all the \
functions we consider will be differentiable, which as we learned means that \
there's a linear transformation that approximates the function very well at \
each point. If the function goes from ",
Cell[BoxData[
\(TraditionalForm\`\[DoubleStruckCapitalR]\^2\)]],
"to \[DoubleStruckCapitalR], then the linear transformation must do the \
same\[LongDash]and in this simple and common case, we call the transformation \
a tangent plane. Today we'll use several different methods to find the \
specific plane tangent to a surface at a given point."
}], "Text"],
Cell[TextData[{
"We'll be doing some parametric plot in this lab. If you don't like the \
little error messages ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" gives you when doing those, execute the line below. It doesn't really \
change how anything works, it just makes things look a bit nicer."
}], "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell[BoxData[
\(Off[ParametricPlot3D::ppcom]\)], "Input"]
}, Closed]],
Cell[CellGroupData[{
Cell["Plotting Planes", "Section"],
Cell["\<\
Before we begin considering issues of tangency, let's review how to \
plot planes.
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, \(-1\), 2}, {s, \(-1\), 1}]\)], "Input"],
Cell["\<\
You can keep the parametric equation of the plane in \
``un-multiplied out'' form and use ParametricPlot3D, but it may complain. \
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, \(-1\), 2}, {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\n",
StyleBox["Example",
FontWeight->"Bold"],
"\n\nPlot 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",
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]]
}, Closed]],
Cell[CellGroupData[{
Cell["Tangent Planes Via Tangent Vectors", "Section"],
Cell["\<\
Let's begin by plotting curves in a surface that are parallel to \
the coordinate planes. We're going to find the plane tangent to the surface\
\
\>", "Text"],
Cell[BoxData[
\(z\ = \ x\^2\ + \ 3 xy\ + \ y\^2\)], "DisplayFormula"],
Cell["\<\
at the point (1,-2,-1). If we look at the cross section of this \
surface where x = 1, then we will get a path that lies in the surface and is \
parallel to the yz plane. It's easier to see this if we plot it out; first, \
the surface itself.\
\>", "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["\<\
We'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; the y-value can be \
arbitrary\[LongDash]we'll use t for that\[LongDash]and the z-value is \
described by 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["\<\
The parametric description of the restriction to y = -2 is very \
similar:\
\>", "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. I used \"Mesh \
-> False\" back in the Plot3D command to make it easier to see the \
curves.\
\>", "Text"],
Cell[BoxData[
\(\(\(Show[surf, \ one, two]\)\(\[IndentingNewLine]\)\)\)], "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[{
\(\(d\/dt\) c1 \((t)\)\ = \ \((0, \ 1, \
2 t + 3)\)\), "\[IndentingNewLine]",
\(\(d\/ds\) c2 \((s)\)\ = \ \((1, \ 0, \
2 s - 6)\)\)}], "DisplayFormula"],
Cell["\<\
The point of intersection occurs when t = -2 for c1, so the tangent \
vector there is (0,1,-1). The intersection is at s = 1 for c2, yielding a \
tangent vector of (1,0,-4). There's nothing magical about s and t being the \
same or being equal to one; we just need the values of s and t that place us \
at the point of intersection.
Now we know everything we need to describe the plane: we know a point it \
passes through (namely 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:", "Text"],
Cell[BoxData[{
\(tanplane\ =
ParametricPlot3D[
p[s, t], \ {s, \(-1\), 1}, \ {t, \(-1\),
1}]\), "\[IndentingNewLine]",
\(Show[surf, \ tanplane]\)}], "Input"],
Cell[TextData[{
"The two pictures of the plane may look quite different, but that's just \
because ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is using different-sized boxes to plot the plane in. Use the ViewPoint \
command to look at the above picture from several different viewpoints and \
convince yourself that this plane really is tangent to the surface. 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.\n\nIn the last lab, we \
were very careful about differentiability, existence of partial derivatives, \
and the like. In this case, f(x,y) is continuous and differentiable and \
everything you could want. You should calculate its partial derivatives at \
(x,y) = (1,-2), make sure they are continuous, and compare them to the \
tangent vectors you found above.\n\n",
StyleBox["Curves Not Parallel To Coordinate Planes",
FontSize->16,
FontWeight->"Bold"],
StyleBox["\n",
FontSize->16,
FontVariations->{"CompatibilityType"->0}],
StyleBox["\nAnother thing we discovered in the last lab is that the \
existence of partial derivatives does not even guarantee continuity. One \
rather fuzzy way to explain why is that the partials do not ``look in enough \
directions''. Conversely, if a function is differentiable at a point, we can \
``look in any direction'' to find tangent vectors. To make this somewhat more \
precise, let's redo the above example, but instead of restricting f(x,y) to \
the planes x=1 and y=-2, let's restrict to the planes y = 2x -4 and y = -x \
-1.\n\nLet's define c3 and c4 to represent the restrictions to y = 2x - 4 and \
y = -x-1 respectively.",
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["\<\
Now let's look at the original surface with these two curves:\
\>", \
"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",
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, \(-1\), 1}, {s, \(-1\), 1}]\), "\[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 better be, and in this case they \
are. You'll get a chance to verify that these two different methods produce \
identical planes.\
\>", "Text"],
Cell[TextData[{
StyleBox["Exercise 2",
FontSize->16,
FontWeight->"Bold"],
"\n\nFind the plane tangent to the function ",
StyleBox["g(x,y) = ",
FontSlant->"Italic"],
"sin",
StyleBox["(x+y)",
FontSlant->"Italic"],
" at the point (x,y) = (\[Pi]/3, \[Pi]/3) using the above method. First, \
use two paths parallel to the xz and yz planes. Include in your answer:\n\t\
(i) the parametric descriptions of the paths in simplest form (i.e., write \
``2t - 7''instead of ``g(t,-3) + 1'' or whatever);\n\t\n\t(ii) the tangent \
vectors at (x,y) = (\[Pi]/3, \[Pi]/3);\n\t\n\t(iii) the parametric \
description and cartesian equation of the tangent plane.\n\nIt will probably \
be helpful to plot graphs as we did above; you don't need to include those in \
your report, but you may if you like.\n\nNow use the paths along the planes",
StyleBox[" ", "InlineFormula"],
StyleBox["y = ", "InlineFormula",
FontSlant->"Italic"],
StyleBox["-", "InlineFormula"],
StyleBox[Cell[BoxData[
\(TraditionalForm\`1\/2\)], "InlineFormula"], "InlineFormula"],
StyleBox["x + ", "InlineFormula",
FontSlant->"Italic"],
StyleBox[Cell[BoxData[
\(TraditionalForm\`\[Pi]\/2\)], "InlineFormula",
FontSlant->"Italic"], "InlineFormula"],
" and ",
StyleBox["y",
FontSlant->"Italic"],
" =",
StyleBox[" x",
FontSlant->"Italic"],
" and do (i), (ii), and (iii) above. Did you get the same tangent plane in \
each case?\n\n(Hint for finding cartesian equations: find a normal vector.)"
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]]
}, Closed]],
Cell[CellGroupData[{
Cell["Tangent Planes Via Gradients", "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[
\(\(\((\[PartialD]f\/\[PartialD]x, \ \[PartialD]f\/\[PartialD]y)\)\(.\)\)\
\)], "DisplayFormula"],
Cell["\<\
Hmmm. This is a two-dimensional vector, and to define a plane in \
3-space, we'll definitely need vectors with three components. Sometimes we \
can ``convert'' two-dimensional vectors into three-dimensional ones by \
tacking on a zero in the third component\[LongDash]but that won't work here; \
every tangent plane would then be perpendicular to the xy plane, and that's \
certainly not the case! (why would every tangent plane be perpendicular to \
the xy plane?) We can fix this problem with some minor tweaking: given \
f(x,y), we'll find the gradient of a new function:\
\>", "Text"],
Cell[BoxData[
\(g \((x, y, z)\)\ = \ z\ - \ f \((x, y)\)\)], "DisplayFormula"],
Cell[TextData[{
"This is now a function of three variables, and its gradient ",
Cell[BoxData[
\(TraditionalForm\`is\)]]
}], "Text"],
Cell[BoxData[
\(\(\((\(-\(\[PartialD]f\/\[PartialD]x\)\), \ \(-\(\[PartialD]f\/\
\[PartialD]y\)\), \ 1)\)\(.\)\)\)], "DisplayFormula"],
Cell[TextData[{
"The crucial feature about this vector is that ",
StyleBox["this vector is perpendicular to the surface z = f(x,y)",
FontSlant->"Italic"],
". To explain why, we'll need to think about level sets. The zero level set \
of g(x,y,z) is the set of points in 3-space where g(x,y,z) = 0, which means"
}], "Text"],
Cell[BoxData[{
\(0\ = \ z\ - \ f \((x, y)\)\), "\[IndentingNewLine]",
\(z\ = \ f \(\((x, y)\)\(.\)\)\)}], "DisplayFormula"],
Cell[TextData[{
"So the zero level set is just ",
"a fancy way of referring to our original surface. g(x,y,z) is identically \
zero at all those points\[LongDash]in other words, the value of g is not \
changing at all. The gradient vector represents the direction of fastest \
change, and as you saw in lecture, the direction of fastest change is ",
StyleBox["perpendicular ",
FontSlant->"Italic"],
"to the directions in which the function doesn't change at all (the \
explanation is also in your text on pages 216-218). The gradient of g(x,y,z) \
is perpendicular to the zero level set of g(x,y,z), which is the same thing \
as our original surface. That gradient can therefore be used as a normal \
vector for the tangent plane. How cool is ",
StyleBox["that",
FontSlant->"Italic"],
"?"
}], "Text"],
Cell[TextData[{
StyleBox["Exercise 3",
FontSize->16,
FontWeight->"Bold"],
"\n\nUse this gradient method to find the cartesian equation of the plane \
tangent to\n\n\t",
Cell[BoxData[
\(h \((x, y)\)\ = \
Exp[\(-\((x\^2 + y\^2)\)\)]\ - \
Exp[\(-\((x\^2 + \((y - 1)\)\^2)\)\)]\)]],
"\n\nat the point (0,1, h(0,1)). Show the graph of h(x,y) and this tangent \
plane together on the same plot. Be sure to show your work and explain your \
reasoning."
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]]
}, Closed]],
Cell[CellGroupData[{
Cell["Tangent Plane Conspiracy Theory", "Section"],
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"],
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"],
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 4",
FontSize->16,
FontWeight->"Bold"],
"\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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