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Notebook[{
Cell[TextData[{
StyleBox["Lab 6B - Surface Integrals of Vector Fields",
FontSize->24,
FontWeight->"Bold",
FontVariations->{"Underline"->True}],
"\nMath 2374 - University of Minnesota\nhttp://www.math.umn.edu/math2374\n\
questions to: drake@math.umn.edu or rogness@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[{
"\tBatman:\"There's an eclipse of the sun due.\"\n\tRobin:\"But that's only \
for half a minute!\"\n\tBatman:\"That's all we'll need, if my calculus is \
correct...\"\n\nBy now you're familiar with line integrals of vector fields, \
which in a certain sense measure how much the vector field is \"pushing \
particles\" along the path -- over a closed path, we called that ",
StyleBox["circulation",
FontSlant->"Italic"],
". What about surface integrals? That is, what about integrating a vector \
field over some parametrized ",
StyleBox["surface",
FontSlant->"Italic"],
" in three-dimensional space?\n\nFor surfaces, it no longer makes much \
sense to ask how much the vector field is pushing particles along the \
surface. Instead, the question we'll want to answer is: how much is the \
vector field pushing particles ",
StyleBox["across",
FontSlant->"Italic"],
" the surface?\n\nLet's make that a little bit more precise. Say the vector \
field represents velocity. Consider a very small bit of the surface (a small \
parallelogram, for example), so small that we can consider the vector field \
to be constant over that parallelogram. In that case, the sides of the \
parallelogram and the vector representing the vector field form a \
parallelepiped. There's a picture and some more descriptions of this on pages \
344-345 of your text.\n\nTo find the volume of that parallelepiped, we use \
the scalar triple product: we take the cross product of the sides of the \
parallelogram, and then find the dot product of that with the vector from the \
vector field. We interpret the result as \"amount of fluid (or particles) \
moving across the surface per unit time\".\n\nThe precise definition of a \
surface integral (or flux integral; the two terms are interchangeable) of a \
vector field is in the next section."
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Definition",
FontSize->16]], "Section"],
Cell[TextData[{
"Just for your reference, we'll repeat the definition of a surface integral \
of a scalar function.\n\nLet ",
Cell[BoxData[
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StyleBox["F",
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StyleBox[\(U \[Subset] \[DoubleStruckCapitalR]\^3\[LongRightArrow]\
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FontSize->14]}], TraditionalForm]]],
"be a continuous vector field, and let M be a smooth surface lying in U \
that is parametrized by ",
StyleBox["f",
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"(s,t), where ",
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" (or ",
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")",
StyleBox[" of F over M",
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" is:\n"
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RowBox[{"(",
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RowBox[{"(",
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\[PartialD]t\),
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Cell[TextData[{
"Note that this definition is very similar to the one for surface integrals \
of scalar functions. For scalar functions we multiplied the scalar function \
by the length of the normal vector. In this case we'll take the dot product \
of our vector field with the normal vector. Remember, the dot product of two \
vectors is a number, so our final integrand will actually still be a scalar \
function.\n\nThere are at least two ways to convince yourself that this \
definition makes sense. As we said above, ",
Cell[BoxData[
FormBox[
StyleBox[\(F\[CenterDot]n\),
FontWeight->"Bold"], TraditionalForm]]],
" represents the scalar triple product of the partial derivatives and ",
Cell[BoxData[
FormBox[
StyleBox["F",
FontWeight->"Bold"], TraditionalForm]]],
", which we know is equal to the volume of the parallelepiped spanned by \
those vectors (up to a possible minus sign).\n\nAnother way to see this is to \
rewrite the dot product: ",
Cell[BoxData[
FormBox[
RowBox[{
StyleBox[\(F\[CenterDot]n\),
FontWeight->"Bold"], " ", "=", " ",
RowBox[{"||",
StyleBox["F",
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StyleBox["n",
FontWeight->"Bold"], "||", " ", \(cos\ \[Theta]\)}]}],
TraditionalForm]]],
". In this case, we see that the length of the normal vector is the area of \
the spanned parallelogram, which is multiplied by the length of ",
StyleBox["F",
FontWeight->"Bold",
FontSlant->"Italic"],
" at that point. The cosine takes care of the ``tilt'': if the vector field \
is moving in the same direction as the surface, it isn't pushing any \
particles at all across the surface; in that case, ",
Cell[BoxData[
\(TraditionalForm\`cos\ \[Theta]\ = \ \(cos\ \((\[Pi]/2)\)\ = \
0\)\)]],
" (remember that the normal vector will be perpendicular to ",
StyleBox["F",
FontWeight->"Bold",
FontSlant->"Italic"],
" in this case). Similarly, if ",
StyleBox["F",
FontWeight->"Bold",
FontSlant->"Italic"],
" is moving perpendicular to the surface, ",
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FormBox[
RowBox[{"||",
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StyleBox["n",
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" will be multiplied by ",
Cell[BoxData[
\(TraditionalForm\`cos\ 0\ = \ 1\)]],
", so ",
StyleBox["F",
FontWeight->"Bold",
FontSlant->"Italic"],
" has as much effect as it can.\n\nIt's important to get the orientation of \
the surface right. The integral represents the flux from the \"inside\" of \
the surface to the \"outside\" -- the direction in which the normal vector \
points is considered \"outside\". Every time you do a surface integral of a \
vector field, you should make sure you're using a normal vector that points \
in the correct direction (which should be specified in the problem)."
}], "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Driving in the Rain",
FontSize->16]], "Section"],
Cell[TextData[{
"Suppose there's a nice spring shower outside and you're out driving a car. \
You're sitting at a stoplight, and the wipers are at just the right speed to \
keep the windshield clean. The light turns green, and you start driving at \
30mph down the road. All of a sudden you can't see a thing -- there's too \
much water on the windshield, and you have to turn your wipers to a higher \
setting. Why is that? We can actually explain it using flux integrals.\n\n\
Suppose our windshield is a rectangular piece of glass, represented by the \
plane ",
Cell[BoxData[
\(TraditionalForm\`z = 2 - 2 x\)]],
" over the rectangle 0\[LessEqual]x\[LessEqual]0.5, -1.25\[LessEqual]y\
\[LessEqual]1.25. Here's a parametrization and picture of it; as you can \
guess by the slanting of the plane, we're assuming that the car is driving \
along the positive x-axis."
}], "Text"],
Cell[BoxData[{
\(\(f[x_, y_] = {x, y, 2 - 2 x};\)\), "\[IndentingNewLine]",
\(windshield =
ParametricPlot3D[f[x, y], {x, 0, 0.5}, {y, \(-1.25\), 1.25},
AxesLabel \[Rule] {"\", "\", "\"}]\)}], "Input"],
Cell[TextData[{
"Rain is really just a bunch of water falling from the sky. If we're \
sitting at a stoplight, then the rain is falling straight down, say, at 10 \
m/s. So the velocity of the raindrops is described by the vector field ",
StyleBox["G",
FontWeight->"Bold"],
"(x,y,z)=(0,0,-10). ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" can plot this vector field for us, as long as we load the correct package \
first:"
}], "Text"],
Cell[BoxData[{
\(G[x_, y_, z_]\ = \ {0, 0, \(-10\)}\), "\[IndentingNewLine]",
\(rain\ = \
PlotVectorField3D[
G[x, y, z], \ {x, \(- .5\), \ 1}, \ {y, \(-2\), 2}, {z, .5,
2.5}]\)}], "Input"],
Cell[TextData[{
StyleBox["Mathematica",
FontSlant->"Italic"],
" can also show us a picture of the windshield in the rain. You can choose \
which of the following two cells to evaluate, depending on whether you want \
an interactive picture or not."
}], "Text"],
Cell[BoxData[
\(Show[windshield, rain]\)], "Input"],
Cell[BoxData[
\(ShowLive[windshield, rain]\)], "Input"],
Cell[TextData[{
"Now a flux integral measures the amount of fluid crossing a surface. If \
we compute the flux integral of this vector field over the windshield, this \
will actually tell us how much rain is hitting the windshield. (Because it's \
made of glass, it can't actually flow through, so instead it just splatters \
all over, and forces the driver to turn on the wipers.)\n\nThe steps we need \
to take to compute this integral are very similar to the ones we used last \
week. We need to find ",
StyleBox["G",
FontWeight->"Bold"],
"(",
StyleBox["f",
FontWeight->"Bold"],
"(x,y)), but since ",
StyleBox["G",
FontWeight->"Bold"],
" is a constant vector field (it never changes), we know that ",
StyleBox["G",
FontWeight->"Bold"],
"(",
StyleBox["f",
FontWeight->"Bold"],
"(x,y)) = (0,0,-10) at all points. Let's run the following command anyway \
so that ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" knows about it, and so that you know how to use the command in the \
future:"
}], "Text"],
Cell[BoxData[
\(Apply[G, f[x, y]]\)], "Input"],
Cell["\<\
We need to compute the normal vector, which for a plane is easy \
(for this plane, it's (2,0,1)), but for more complicated situations, you'll \
have to use something like the following command to compute the cross product \
of the partial derivatives:\
\>", "Text"],
Cell[BoxData[
\(n = Cross[D[f[x, y], x], D[f[x, y], y]]\)], "Input"],
Cell[TextData[{
"And now we're ready to integrate. The integrand is ",
StyleBox["G",
FontWeight->"Bold"],
"(",
StyleBox["f",
FontWeight->"Bold"],
"(x,y))\[CenterDot]",
Cell[BoxData[
\(TraditionalForm\`n\&\[RightVector]\)]],
"."
}], "Text"],
Cell[BoxData[{
\(\(integrand = Apply[G, f[x, y]] . n;\)\), "\[IndentingNewLine]",
\(Integrate[integrand, {x, 0, .5}, {y, \(-1.25\), 1.25}]\)}], "Input"],
Cell[TextData[{
"Why is the answer negative? The answer has to do with the normal vector. \
Our normal vector is pointing in the direction that we're travelling -- look \
at ",
Cell[BoxData[
\(TraditionalForm\`n\&\[RightVector]\)]],
" and look at the picture of the windshield until you believe this. In \
some sense the normal vector defines which direction is \"outward.\" Flux \
integrals measure the net flow from inside the surface to outside the \
surface, so in this case a positive answer would mean water was coming from \
the inside of the car; water hitting the windshield from the outside \
represents negative flow."
}], "Text"],
Cell["\<\
Here's a question for you to think about: what are the correct \
units for our answer?\
\>", "Text",
CellFrame->True,
Background->GrayLevel[0.849989]],
Cell[TextData[{
"Remember, this first part of the example was for when the car is sitting \
still. Now suppose we're driving down the street at 50 km/h, or just over \
31mph. Since we expressed the speed of the rain in meters per second, we \
should do the same with the car; you can check that 50km/h is about 13.89 \
m/s.\n\nIn our earlier example, the velocity of the rain was described by the \
vector field (0,0,-10). In this case, relative to the car, the velocity of \
the rain is described by the vector field, ",
Cell[BoxData[
\(TraditionalForm\`H(x, y, z)\ = \ \((\(-13.89\), 0, \(-10\))\)\)]],
". Before going on, convince yourself that this is the correct formula! \
(Think about the negative signs in particular.)\n\nHere's a picture of the \
new situation. Again, you can choose whether you want an interactive picture \
or not:"
}], "Text"],
Cell[BoxData[{
\(\(H[x_, y_, z_] = {\(-13.89\),
0, \(-10\)};\)\), "\[IndentingNewLine]",
\(newrain =
PlotVectorField3D[
H[x, y, z], {x, \(-1\), 1}, {y, \(-2\), 2}, {z, 0,
3}]\), "\[IndentingNewLine]",
\(Show[windshield, newrain]\)}], "Input"],
Cell[BoxData[{
\(\(H[x_, y_, z_] = {\(-13.89\),
0, \(-10\)};\)\), "\[IndentingNewLine]",
\(newrain =
PlotVectorField3D[
H[x, y, z], {x, \(-0.5\), 1}, {y, \(-2\), 2}, {z, 0.5,
2.5}]\), "\[IndentingNewLine]",
\(ShowLive[windshield, newrain]\)}], "Input"],
Cell["\<\
How much rain is hitting the windshield this time? Let's compute \
the integral:\
\>", "Text"],
Cell[BoxData[
\(\(\( (*\ New\ vector\ field\ is\ H\ *) \)\(\[IndentingNewLine]\)\( (*\
Parametrization\ f[x,
y]\ already\ defined\ *) \)\(\[IndentingNewLine]\)\(\
\[IndentingNewLine]\)\(\(integrand =
Apply[H, f[x, y]] .
Cross[D[f[x, y], x], \ D[f[x, y], y]];\)\[IndentingNewLine]
Integrate[integrand, {x, 0, .5}, {y, \(-1.25\), 1.25}]\)\)\)], "Input"],
Cell["\<\
So there is nearly four times as much water hitting the windshield \
when we're driving at 30 mph! Now do you understand why you have to turn \
your wipers up?
The same principle applies, to a lesser extent, to other modes of transport. \
If you've ever biked during a light shower, you know that you get a lot more \
wet than if you were walking. On a slightly different note, when people are \
walking and it starts to rain, often they start running instead of walking. \
If you run, you'll be out in the rain for a shorter time, but because you're \
going faster, you'll be hit by more water during the time that you're still \
outside. It would be an interesting project to determine if it's really \
worth running or not!\
\>", "Text"]
}, Closed]],
Cell[CellGroupData[{
Cell[TextData[StyleBox["Exercises",
FontSize->16]], "Section"],
Cell[TextData[{
StyleBox["Exercise 1",
FontWeight->"Bold"],
"\n\nLet's re-do the above example, but instead of driving a boring, \
everyday car, let's imagine you are driving the state-of-the-art, \
gadget-laden vehicle of the Dark Knight of Gotham City, the Batmobile. We'll \
assume that this Batmobile is somewhat similar to the version used in the \
campy 60's television series and has a hemispherical windshield. (For a \
picture of the original campy batmobile, and more information than you'd ever \
need to know, go to http://www.javelinamx.com/Batmobile/ in your browser.)\n\n\
(i) What is a parametrization of this hemisphere? Assume the radius of the \
hemisphere is 1 and that the center is at the origin. You can think of this \
shape as a sphere, but we're only interested in the part from the \"equator\" \
to the \"north pole\". \n\n(ii) While driving, rain will only be hitting the \
front half of the windshield. Describe a parametrization of this portion of \
the windshield. You should be able to just restrict the domain of your \
parameters a bit from your parametrization in (i).\n\n(iii) Let's say Batman \
is inside a building, foiling a nefarious plot by the Joker. It's raining, \
and the Batmobile is sitting outside the building ready for Batman to take \
off and return to the Batcave. If the rain's velocity is described by the \
vector field R(x,y,z) = (0,0,-5) meters / second, what is the flux integral \
of this vector field over the Batmobile's windshield? Remember that the \
vehicle is sitting still and the rain is hitting the entire hemisphere, not \
just the front half.\n\n(iv) Batman leaves the building, hops into the \
Batmobile, and takes off. If he's driving at 50 meters / second, what is the \
flux integral of the rain over the windshield? We'll assume that now the rain \
is only hitting the front half of the windshield (this isn't entirely \
accurate, but oh well), so use your parametrization from part (ii).\n\n \
You will have to be careful about how you change the vector field that \
describes the rain; it will depend on the choice you make for the axis the \
Batmobile is travelling along and the parametrization you chose for the front \
half of the windshield. More succinctly: make sure your rain doesn't hit the \
side or back of the windshield."
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]],
Cell[TextData[{
StyleBox["Exercise 2",
FontWeight->"Bold"],
"\n\nHere's another \"driving in the rain\" problem. Batman is chasing the \
Penguin during a rainstorm. Penguin's getaway car doesn't have any wipers, \
so it's vitally important to figure out how much rain is hitting his \
windshield. If it's too much, he won't be able to drive fast enough to \
escape. No super villian (or super hero) would ever be caught dead driving a \
normal automobile, so Penguin's car doesn't have a normal windshield. Like \
the batmobile, his car has a cockpit with a \"bubble\" windshield shaped like \
the top half of the ellipsoid ",
Cell[BoxData[
FormBox[
StyleBox[\(x\^2\/9 + y\^2\/4 + z\^2 = 1\),
FontSize->16], TraditionalForm]]],
".\n\n(i) What is a parametrization of the whole windshield? Assume that \
the center of the ellipsoid is at the origin. Remember, we're only \
interested in the top half.\n\n(ii) While driving, rain will only be hitting \
the front half of the windshield. Describe a parametrization of this portion \
of the windshield. You should be able to just restrict the domain of your \
parameters a bit from your parametrization in (i). (We'll define \"front \
half\" as being the portion of the windshields where ",
Cell[BoxData[
\(TraditionalForm\`x > 0\)]],
". This is the same as in the example above.\n\n(iii) Batman is within \
sight of the Penguin when they both approach intersections with red lights. \
Knowing that Batman is scrupulous to a fault, the Penguin has no qualms about \
stopping; he knows Batman will stop as well, instead of catching up to him. \
If the rain's velocity is described by the vector field R(x,y,z) = (0,0,-5) \
meters / second, how much rain is hitting the Penguin's windshield while he \
sits at the light? Remember he's sitting still, so the rain is hitting his \
entire windshield.\n\n(iv) Once the light turns green, the Penguin takes off. \
His car is rocket propelled, and he's quickly driving 40 meters per second. \
What is the flux integral of the rain over the windshield? We'll assume that \
now the rain is only hitting the front half of the windshield (this isn't \
entirely accurate, but oh well), so use your parametrization from part (ii).\n\
\n You will have to be careful about how you change the vector field that \
describes the rain; it will depend on the choice you make for the axis the \
Batmobile is travelling along and the parametrization you chose for the front \
half of the windshield. More succinctly: make sure your rain doesn't hit the \
side or back of the windshield."
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]],
Cell[TextData[{
StyleBox["Exercise 3\n",
FontWeight->"Bold"],
"\nCommissioner Gordon is in dire need of Batman's help, so he activates \
the Bat symbol and shines it on the clouds above Gotham City. If you aren't \
familiar with what the Bat symbol looks like...well, you should be. There's a \
picture at ",
ButtonBox["http://www.math.umn.edu/~drake/batman.gif",
ButtonData:>{
URL[ "http://www.math.umn.edu/~drake/batman.gif"], None},
ButtonStyle->"Hyperlink"],
" (it's the yellow oval in the middle of the t-shirt).\n\n Let's model the \
surface of the clouds with the function z = sin(x + y). Let's say the \
spotlight for the Bat symbol is shining from the top of City Hall in gritty \
downtown Gotham City upwards to the swirling clouds above the troubled \
metropolis. In more mathematical terms, the spolight shines from the negative \
z-axis upwards onto a disk in the xy plane centered at the origin, of radius \
5. Since it's hard to parametrize the lit-up portion of the Bat symbol (the \
yellow part in the picture), let's simplify the problem and assume the lit-up \
portion of the Bat symbol is actually just a disk with a smaller disk cut out \
of it; this is called an ",
StyleBox["annulus",
FontSlant->"Italic"],
". You can parametrize this 2-dimensional object the exact same way you \
parametrize a disk, but start the radius at something greater than 0. For \
this problem, let's use the annulus centered at the origin whose radius \
ranges from 2 to 5.\n\nBecause of fog (it always seems to be foggy in Gotham \
City), the vector field that describes the intensity of the photons hitting \
the clouds is not constant, but is described by \n\n",
Cell[BoxData[
\(L \((x, y, z)\)\ = \ \((0, 0, \((x + y)\)\^2)\)\)]],
"\n\n(this is a rather poor model, but it makes the integrals reasonable).\n\
\n(i) Describe the parametrization of the surface of the clouds we're \
interested in. The surface of the clouds is described by ",
Cell[BoxData[
\(TraditionalForm\`sin(x + y)\)]],
"; the portion of the xy plane in which we're interested is the annulus \
from 2 to 5.\n\n(ii) Find the surface integral of the above vector field over \
the annulus which we're using instead of the actual Bat symbol. Use an \
upward-pointing normal vector."
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]],
Cell[TextData[{
StyleBox["Exercise 4",
FontWeight->"Bold"],
"\n\nBatman is attempting to sneak into the Penguin's secret underground \
lair. The Penguin, suspecting this, has installed a Batman Detection System \
(BDS) which emits a special form of Batman-detecting radiation that can be \
described by the vector field\n\nB(x,y,z) = (-y,x, -5z/6),\n\nwhere (0,0,0) \
is the center of the Penguin's lair. Our hero is familiar with the Penguin's \
criminal ways, and has brought with him a special Anti-Batman Detection \
System Device (ABDSD) that will deactivate the Batman Detection System for a \
short while. The device is shaped like a torus with r = 1, and R = 2; we're \
using the terminology from problem 30 on page 338 in your text.\n\n(i) Batman \
has placed the ABDSD so that it is centered at (5,5,5) in the Penguin's lair \
and is horizontal (i.e., the vertical line described by x=5, y=5 does not \
intersect any part of the torus; it \"goes through the middle of the doughnut \
hole\"). Parametrize the surface of the ABDSD. You've essentially done this \
problem in your homework.\n\n(Hint: first assume the ABDSD is centered at \
(0,0,0), then translate your answer to center it at (5,5,5).)\n\n(ii) \
Calculate the flux of the Batman-detecting radiation across the ABDSD. Use an \
inward-pointing normal vector.\n\n(iii)The answer from (ii) represents units \
of radiation crossing the ABDSD per second. Let's say the ABDSD somehow \
absorbs all this radiation (thus preventing the Penguin from detecting \
Batman's presence), and that it can store a total of 1000 units of radiation \
before it stops working. How much time does Batman have before the ABDSD \
fails?\n\n(iv) (OPTIONAL) You may want to come back to this after we've \
learned about the Divergence Theorem. Redo part (ii), but this time say the \
ABDSD is centered at the origin. Compare the flux you get in both cases. What \
is unusual or unexpected about these two answers? With the Divergence \
Theorem, we'll be able to explain this quite easily."
}], "Text",
CellFrame->True,
Background->RGBColor[1, 0.498039, 0.498039]],
Cell[TextData[{
"Exercise 5 (This exercise is worth zero points.)\n\nWhich popular culture \
Batman series is better, the campy 60's television series (which featured \
animated \"POW!\" balloons, and the excellent quote with which we began this \
lab), or the 1990 Tim Burton movie, which featured Michael Keaton (at the \
time better known as Mr. Mom or Beetlejuice) and Jack Nicholson (who had come \
a loooong way from ",
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")?\n\nProvide rigorous mathematical reasoning to defend your choice. :)\n\n\
And keep in mind that the Burton movie prominently featured music from the \
most famous resident of the Suburb Currently Known As Chanhassen..."
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The quote at the beginning of this lab is from \
http://www.showcase.ca/action/series/batman_quotes.asp. It's near the bottom \
of the page. Much of the expository material in this lab is by Jonathan \
Rogness; the \"Driving in the Rain\" example is a standard one and no claim \
of originality is made here. (In fact, I spent some time searching on the \
web to determine who first wrote this example down without any luck; at least \
5-10 other math courses throughout the country have posted similar examples, \
though.) The rest of the lab (including the \"Batman-ization\") is by Dan \
Drake. Comments and questions are welcome to either of us.
Update (2004): Exercise 2 added by Jonathan Rogness
This lab is copyright 2002, 2004 by Jonathan Rogness (rogness@math.umn.edu) \
and Dan Drake (drake@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/
Although 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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