(************** Content-type: application/mathematica **************
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Notebook[{
Cell[BoxData[
\(<< Calculus`VectorAnalysis`\)], "Input",
InitializationCell->True],
Cell[BoxData[
\(<< Graphics`\)], "Input",
InitializationCell->True],
Cell[BoxData[
\(<< RealTime3D`\)], "Input",
InitializationCell->True],
Cell[CellGroupData[{
Cell["\<\
Calculus IV
Lab 2\
\>", "Title"],
Cell["Non-Cartesian coordinate systems", "Subtitle"],
Cell[CellGroupData[{
Cell["Introduction", "Section"],
Cell[TextData[{
StyleBox["Mathematica",
FontSlant->"Italic"],
" supports a very wide range of 3-dimensional coordinate systems. \
(Double-click on VectorAnalysis and then hit function key F1 to go to the \
place in the Help Browser where these are discussed.) This can be very \
useful, but it can also be very confusing. How do you know what these \
different coordinate systems \"look like\" and when to use each one?\n\nWell, \
this lab won't teach you that. However, it will show you some techniques to \
investigate this question and how to deal with these alternate coordinate \
systems in ",
StyleBox["Mathematica",
FontSlant->"Italic"],
".\n\nFirst, we will work through an investigation of some properties of \
different coordinate systems using the two main non-Cartesian systems that we \
are most familiar with. Then, you will apply these techniques to some less \
familiar ones. (For all the work that follows, you must load the library \
module Calculus`VectorAnalysis`. In this notebook, it is done automatically \
when you execute the first command.)\n\nSince we are in 3 dimensions here, we \
need 3 coordinates to specify a point in space. To find out what names ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" gives to these coordinates by defaulty, use:"
}], "Text"],
Cell[BoxData[
\(Coordinates[Cylindrical]\)], "Input"],
Cell["\<\
So, what are the transformation equations to convert from Cylindrical to \
Cartesian coordinates?\
\>", "Text"],
Cell[BoxData[
\(CoordinatesToCartesian[{r, \[Theta], z}, Cylindrical]\)], "Input"],
Cell["\<\
(Notice that I used my own variable names here. This makes the formulae look \
nicer. It can get you in trouble on occasion, however. More on that later.)
So, what about Spherical coordinates?\
\>", "Text"],
Cell[BoxData[
\(Coordinates[Spherical]\)], "Input"],
Cell[BoxData[
\(CoordinatesToCartesian[{\[Rho], \[Phi], \[Theta]},
Spherical]\)], "Input"],
Cell[TextData[{
"Just a warning here: For some reason, ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" reverses the standard roles of \[Theta] and \[Phi] in spherical \
coordinates. In ",
StyleBox["Mathematica",
FontSlant->"Italic"],
", \[Theta] stands for the angle from the Z axis, while \[Phi] is the angle \
of the projection into the XY plane from the X axis. This isn't all that \
hard to deal with (just always remember to enter points in {\[Rho],\[Phi],\
\[Theta]} format, rather than the more usual {\[Rho],\[Theta],\[Phi]} form, \
using our standard \[Theta] and \[Phi]), but you do have to be careful to \
remember it."
}], "Text"]
}, Open ]],
Cell[CellGroupData[{
Cell["Visualization", "Section"],
Cell[TextData[{
"So, how can you visualize what these alternate coordinate systems \"look \
like\"? This can be a bit tricky, so I will show you one way to do this. \n\
\nConsider a coordinate system {u,v,w}. If you fix w and let u and v vary, \
you will get a surface (why?). If you do this for different values of w, you \
will get a bunch of surfaces that are \"parallel\" in some sense (literally, \
in Cartesian coordinates). These can be thought of as \"level surfaces\" for \
the coordinate system: everything on one of these surfaces has the same w \
coordinate. You can do the same thing for u and v by letting each of them \
stay constant. Putting all these together give a pretty good picture of how \
the coordinate system behaves. Let's take a look at this for Cartesian and \
Spherical coordinates:\n\nFirst, you use the Table function to generate your \
\"level surfaces\" for each combination of variables. Notice that the Table \
function works like:\n\nTable[ Stuff to do for different values of k, {k, \
kStart, kEnd, kStep}]\n\nwhere:\nkStart = The beginning value to plug in for \
k\nkEnd = The final value to plug in for k\nkStep = The amount to increment k \
by each time\n\nSo, when I use the command:\nTable[ParametricPlot3D[ \
stuff...], {k,-4,4,2}]\n\nI am telling it to make a series of \
ParametricPlot3D's, letting k = -4, -2, 0, 2, 4.\n\n(You can ignore the \
\"spelling error\" messages. ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is trying to be helpful, but not succeding very well here.)"
}], "Text"],
Cell[BoxData[
\(zSlices =
Table[ParametricPlot3D[{u, v, k}, {u, \(-5\), 5}, {v, \(-5\),
5}], {k, \(-4\), 4, 2}]\)], "Input"],
Cell[BoxData[
\(ySlices =
Table[ParametricPlot3D[{u, k, v}, {u, \(-5\), 5}, {v, \(-5\),
5}], {k, \(-4\), 4, 2}]\)], "Input"],
Cell[BoxData[
\(xSlices =
Table[ParametricPlot3D[{k, u, v}, {u, \(-5\), 5}, {v, \(-5\),
5}], {k, \(-4\), 4, 2}]\)], "Input"],
Cell["\<\
Notice that in each of these, u and v are used as variables for graphing the \
surface, while k always stands for the variable that will get held constant \
for each surface. Now, let's put all this together in a useful picture:\
\>", "Text"],
Cell[BoxData[
\(Show[xSlices, ySlices, zSlices]\)], "Input"],
Cell[TextData[{
"Not surprisingly, we see that Cartesian coordinates can be thought of as a \
bunch of intersecting planes. You locate points at the intersection of the \
three perpindicular planes. This is boring, but makes for a good start.\n\nA \
couple of things you should notice about how I set this up: \n\nI didn't let \
k run over as large a domain as I did u and v. This was so that when I put \
everything together at the end (using the Show command), we could actually \
see the intersecting planes. (If you let k run from -5 to 5, like u and v, \
all you see is a cube. Perfectly correct, but not very informative.) \n\n\
You will probably have to experiment with the domain for u and v, as well as \
the range for k (and the number of slices) in different coordinate systems to \
come up with an informative final graph.\n\nLet's do the same thing for \
Spherical coordinates. Notices that it is very similar, though we have to \
convert to Cartesian coordinates before we can plot anything \
(ParametricPlot3D plots only in Cartesian). Also, notice the use of Evaluate \
before the change of coordinates; this is just to stop ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" complaining (it compiles the function before plotting)."
}], "Text"],
Cell[BoxData[
\(\[Rho]Slices =
Table[ParametricPlot3D[
Evaluate[
CoordinatesToCartesian[{k, \[Phi], \[Theta]},
Spherical]], {\[Phi],
0, \[Pi]}, {\[Theta], \[Pi]\/6, \(11 \[Pi]\)\/6}], {k, 1, 5,
2}]\)], "Input"],
Cell["\<\
Notice that I didn't complete the spheres (I restricted \[Theta]). I did \
this so that when we put the graphs together, you can get a \"cut-away\" view \
inside (see below).\
\>", "Text"],
Cell[BoxData[
\(\[Theta]Slices =
Table[ParametricPlot3D[
Evaluate[
CoordinatesToCartesian[{\[Rho], \[Phi], k}, Spherical]], {\[Phi],
0, \[Pi]}, {\[Rho], 0, 6}], {k, \[Pi]\/3, \(5 \[Pi]\)\/3, \[Pi]/
3}]\)], "Input"],
Cell[BoxData[
\(\[Phi]Slices =
Table[ParametricPlot3D[
Evaluate[
CoordinatesToCartesian[{\[Rho], k, \[Theta]},
Spherical]], {\[Theta], \[Pi]\/6, \(11 \[Pi]\)\/6}, {\[Rho],
0, 6}], {k, 0, \[Pi], \[Pi]/6}]\)], "Input"],
Cell[BoxData[
\(Show[\[Rho]Slices, \[Theta]Slices, \[Phi]Slices]\)], "Input"],
Cell["\<\
Again, you locate points at the intersection of the three types of surfaces \
(planes, spheres, and cones).\
\>", "Text"],
Cell[CellGroupData[{
Cell["Problem:", "Subsubsection"],
Cell[TextData[{
"Pick two other non-standard coordinate systems that ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" supports (listed in the Help Browser page I mention above; Cartesian, \
Cylindrical, and Spherical are all out) and generate visualizations of those \
systems like I did above. Also, have ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" give you the formula to translate from these other coordinate systmes \
into Cartesian. Choose your domains, etc., so that your final picture is \
easy to interpret.\n\nWarning: Some of the coordinate systems in the list \
seem to have more than 3 coordinates. This isn't really true; the first 3 \
variables listed are always the coordinates. The \"extra\" variables are \
actually constant parameters; you get a slightly different coordinate system \
for each choice of those parameters. You can use the default values for each \
one for this exercise."
}], "Text",
CellMargins->{{87.3125, Inherited}, {Inherited, Inherited}}]
}, Open ]]
}, Open ]],
Cell[CellGroupData[{
Cell["Computations", "Section"],
Cell[TextData[{
"While alternate coordinate systems can be really useful at times \
(especially in integration), you do have to be careful when working with \
them. Standard vector operations work differently in non-Cartesian \
coordinate systems. However, you can always transform everything into \
Cartesian coordinates and do the work there if necessary. Fortunately, ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" automagically does most of the grunt work for you on this. ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" also has special versions of these operations that work in these \
different coordinate systems."
}], "Text"],
Cell[CellGroupData[{
Cell["Vector operations", "Subsection"],
Cell["\<\
So, for example, to take the dot product of two vectors in spherical \
coordinates, we would do the following (warning: remember that these are in {\
\[Rho],\[Phi],\[Theta]} form):\
\>", "Text"],
Cell[BoxData[
\(DotProduct[{1, \[Pi]\/2, \[Pi]\/6}, {2, \[Pi]\/2, \(2 \[Pi]\)\/3},
Spherical]\)], "Input"],
Cell["If we use Cartesian coordinates, we get:", "Text"],
Cell[BoxData[
\(DotProduct[{1, \[Pi]\/2, \[Pi]\/6}, {2, \[Pi]\/2, \(2 \[Pi]\)\/3},
Cartesian]\)], "Input"],
Cell[BoxData[
\(CrossProduct[{1, \[Pi]\/2, \[Pi]\/6}, {2, \[Pi]\/2, \(2 \[Pi]\)\/3},
Spherical]\)], "Input"],
Cell[BoxData[
\(CrossProduct[{1, \[Pi]\/2, \[Pi]\/6}, {2, 0, \(2 \[Pi]\)\/3},
Spherical]\)], "Input"],
Cell[CellGroupData[{
Cell["Problem", "Subsubsection"],
Cell[TextData[{
"Explain why the dot product in spherical coordinates is 0 above. Why is \
the second dot product (in Cartesian coordinates, by default) ",
StyleBox["not",
FontSlant->"Italic"],
" 0? \n\nWhat coordinate system is the answer to the cross product given \
in? (This is ",
StyleBox["important",
FontSlant->"Italic"],
". Don't continue until you have figured this out.) How do you know?"
}], "Text",
CellMargins->{{78.625, Inherited}, {Inherited, Inherited}},
TextAlignment->Left]
}, Open ]]
}, Open ]],
Cell[CellGroupData[{
Cell["Differentiation operators", "Subsection"],
Cell["\<\
There are also built-in functions to find the gradient, divergence, and curl \
(among other things):\
\>", "Text"],
Cell[BoxData[
\(Grad[x\^2 + y\ Cos[z]]\)], "Input"],
Cell[TextData[{
"Notice, we have a problem here. The gradient of the given functions is \
certainly ",
StyleBox["not",
FontSlant->"Italic"],
" the 0 vector. So, what is going on? This goes back to the issue I \
mentioned in the introduction. This gradient was taken in the default \
coordinate system (Cartesian), but it was taken with the ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" default variables for that coordinate system. Let's find out what those \
are:"
}], "Text"],
Cell[BoxData[
\(CoordinateSystem\)], "Input"],
Cell[BoxData[
\(Coordinates[]\)], "Input"],
Cell[TextData[{
"So, as far as ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is concerned, the variables in this coordinate system are Xx, Yy, and Zz \
(strange but true). That means that when we took the gradient above, it \
treated x, y, and z as ",
StyleBox["constants",
FontSlant->"Italic"],
". There are two ways to fix this:\n\nWe could conform:"
}], "Text"],
Cell[BoxData[
\(Grad[Xx\^2 + Yy\ Cos[Zz]]\)], "Input"],
Cell[TextData[{
"Personally, I find this a bit bizarre to try to read. The other way is to \
tell ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" to use ",
StyleBox["our",
FontSlant->"Italic"],
" preferred variables:"
}], "Text"],
Cell[BoxData[
\(Grad[x\^2 + y\ Cos[z], \ Cartesian[x, y, z]]\)], "Input"],
Cell["\<\
Or, if you need to do a lot of work in Cartesian coordinates with your \
preferred variables, you could set the default coordinate system this way:\
\>", "Text"],
Cell[BoxData[
\(SetCoordinates[Cartesian[x, y, z]]\)], "Input"],
Cell[BoxData[
\(Grad[x\^2 + y\ Cos[z]]\)], "Input"],
Cell[BoxData[
\(Grad[x\^2 + y\ Cos[z], \ Cartesian[x, y, z]]\)], "Input"],
Cell[TextData[{
"Thus, it is extremely important that, before you start doing computations \
in any coordinate system using ",
StyleBox["Mathematica",
FontSlant->"Italic"],
"'s built-in functions, you are aware of what variables ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is expecting (or you set them like you want them).\n\nOf course, a more \
interesting question is, how do these things work in other coordinate \
systems? Let's look at the gradient in spherical coordinates:"
}], "Text"],
Cell[BoxData[
\(Coordinates[Spherical]\)], "Input"],
Cell[BoxData[
\(SetCoordinates[Spherical[\[Rho], \[Phi], \[Theta]]]\)], "Input"],
Cell["\<\
Warning: The default coordinate system is now set to spherical until you \
change it again. The safe thing to do is to always specify the coordinate \
system when you perform a computation. Also, notice that I am taking this \
opportunity to rename the variables so they fit the ones we are used to \
(though they still need to be entered in {\[Rho],\[Phi],\[Theta]} order).\
\>", "Text"],
Cell[BoxData[
\(Grad[\(\[Rho]\^2\) Sin[\[Phi]] Cos[\[Theta]]]\)], "Input"],
Cell["\<\
Or, we can get a general formula for the gradient of a function f[\[Rho],\
\[Phi],\[Theta]] whose coordinates are in spherical coordinates:\
\>", "Text"],
Cell[BoxData[
\(Grad[f[\[Rho], \[Phi], \[Theta]]]\)], "Input"],
Cell[TextData[{
"In this expression, ",
Cell[BoxData[
RowBox[{
SuperscriptBox["f",
TagBox[\((1, 0, 0)\),
Derivative],
MultilineFunction->None], "[", \(\[Rho], \[Phi], \[Theta]\), "]"}]],
"Output"],
" stands for the partial derivative of f with respect to the first \
coordinate (\[Rho]), etc. Notice that, in addition to the expected partial \
derivatives, you get some extra coefficients, in this case ",
Cell[BoxData[
\(TraditionalForm\`1\/\[Rho]\)]],
" and ",
Cell[BoxData[
\(TraditionalForm\`Csc[\[Phi]]\/\[Rho]\)]],
" (usually written as ",
Cell[BoxData[
\(TraditionalForm\`1\/\(\[Rho]\ Sin[\[Phi]]\)\)]],
"). A complete treatment of where these factors come from is beyond the \
scope of this lab, but basically they are the reciprocals of the coefficients \
needed to convert a change in that variable into an actual change in \
arc-length (called the \"Scale Factors\"). So, in other words, if you \
increase \[Rho] by ",
Cell[BoxData[
\(TraditionalForm\`d\[Rho]\)]],
", then it just gets longer by ",
Cell[BoxData[
\(TraditionalForm\`d\[Rho]\)]],
", so the scale factor is 1. However, if you increase \[Phi] by ",
Cell[BoxData[
\(TraditionalForm\`d\[Phi]\)]],
", the actual change in distance (i.e., arc-length) covered is ",
Cell[BoxData[
\(TraditionalForm\`\[Rho]\ d\[Phi]\)]],
", so the scale factor is \[Rho]. (This is easier to see in polar \
coordinates. If you increase \[Theta] by ",
Cell[BoxData[
\(TraditionalForm\`d\[Theta]\)]],
", the actual change in arc-length is ",
Cell[BoxData[
\(TraditionalForm\`r\ d\[Theta]\)]],
", so the scale factor would be ",
Cell[BoxData[
\(TraditionalForm\`r\)]],
".) To see the scale factors for a particular coordinate system, you can \
use:"
}], "Text"],
Cell[BoxData[
\(ScaleFactors[Spherical[\[Rho], \[Phi], \[Theta]]]\)], "Input"],
Cell[BoxData[
\(ScaleFactors[Cylindrical[r, \[Theta], z]]\)], "Input"],
Cell[TextData[{
"Now, you only have to deal with this when you need your output vectors \
given in your non-Cartesian coordinate system. In the cases we looked at \
above, the gradients are given as vectors in ",
StyleBox["spherical coordinates ",
FontSlant->"Italic"],
"(i.e., these are the components of unit vectors in the \[Rho], \[Phi], and \
\[Theta] directions, respecitvely). This is mainly useful if you are doing \
all your work in that coordinate system. (The good news is that all the \
built-in differential operators, like grad, div, and curl, automatically take \
care of this for you, if you specify your coordinate system.)\n\nOn the other \
hand, if you really needed to work with a function from an alternate \
coordinate system in Cartesian coordinates (for example if you wanted to \
actually graph the function or its gradient vector field or tangent plane in \
",
StyleBox["Mathematica",
FontSlant->"Italic"],
"), one way to handle this is to convert the function to Cartesian \
coordintes to begin with.\n\nIf you had the function (in spherical \
coordinates):"
}], "Text"],
Cell[BoxData[
\(f[{\[Rho]_, \[Phi]_, \[Theta]_}] := \(\[Rho]\^2\) Sin[\[Phi]]
Cos[\[Theta]]\)], "Input"],
Cell[TextData[{
"You could first transform it into a function of ",
Cell[BoxData[
\(TraditionalForm\`{x, y, z}\)]],
":"
}], "Text"],
Cell[BoxData[
\(fCart[{x_, y_, z_}] =
f[CoordinatesFromCartesian[{x, y, z}, Spherical]] //
FullSimplify\)], "Input"],
Cell["\<\
From then on, you would treat this as a standard Cartesian function \
(admittedly ugly). Notice the liberal use of FullSimplify in this work; it \
is used to automatically apply things like trig identities, etc.\
\>", "Text"],
Cell[BoxData[
\(Grad[fCart[{x, y, z}], Cartesian[x, y, z]] // FullSimplify\)], "Input"],
Cell[TextData[{
"This gives you the gradient in terms of your standard ",
StyleBox["i, j, ",
FontSlant->"Italic"],
"and ",
StyleBox["k",
FontSlant->"Italic"],
" unit vectors. You can get a much nicer looking gradient by taking it \
directly in spherical:"
}], "Text"],
Cell[BoxData[
\(sphGrad =
Grad[f[{\[Rho], \[Phi], \[Theta]}],
Spherical[\[Rho], \[Phi], \[Theta]]] // FullSimplify\)], "Input"],
Cell[TextData[{
"But this gives you the gradient vector in terms of the unit vectors in the \
\[Rho], \[Phi], and \[Theta] directions. Which method you use depends on \
what you are trying to do. (Of course, with spherical and cylindrical \
coordinates, ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" is perfectly capable of graphing them \"as-is\", but that isn't true \
about other more esoteric coordinate systems. Also, if you were working with \
different functions in different coordinate systems, converting everything to \
Cartesian makes it possible to work with them all together.)"
}], "Text"],
Cell[BoxData[
\(SetCoordinates[Cartesian[x, y, z]]\)], "Input"]
}, Open ]],
Cell[CellGroupData[{
Cell["Integration operations", "Subsection"],
Cell[TextData[{
"If you are computing a change of variables for a new coordinate system \
(i.e., in a triple integral), ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" has some built in functions that are useful for that as well:"
}], "Text"],
Cell[BoxData[
\(JacobianDeterminant[Spherical[\[Rho], \[Phi], \[Theta]]]\)], "Input"],
Cell[BoxData[
\(JacobianDeterminant[Cylindrical]\)], "Input"],
Cell["\<\
As usual, be sure you know what the default variables of your coordinate \
system are before using this:\
\>", "Text"],
Cell[BoxData[
\(JacobianDeterminant[Cylindrical[r, \[Theta], z]]\)], "Input"],
Cell[BoxData[
\(SetCoordinates[Cartesian[x, y, z]]\)], "Input"],
Cell[TextData[{
"Also, if you are doing a path integral, you will need to be able to find \
the \"arc length factor\", which is the factor you have to introduce to \
convert an integral ",
StyleBox["ds",
FontSlant->"Italic"],
" into and integral ",
StyleBox["dt",
FontSlant->"Italic"],
" (or whatever your independent variable is). Of course, since we know \
that ",
Cell[BoxData[
\(TraditionalForm\`ds = \(\(\[VerticalSeparator]\)\(\(c\& \[Rule] \)' \
\((t)\)\)\(\[VerticalSeparator]\)\(dt\)\)\)]],
", this \"arc length factor\" is just ",
Cell[BoxData[
\(TraditionalForm\`\(\(\[VerticalSeparator]\)\(\(c\& \[Rule] \)' \
\((t)\)\)\(\[VerticalSeparator]\)\)\)]],
". Now, you could certainly compute that yourself, but ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" has a built-in function for this:"
}], "Text"],
Cell[BoxData[
\(ArcLengthFactor[{2 Sin[t], 2 Cos[t], t}, t] //
FullSimplify\)], "Input"],
Cell[TextData[{
"Now, even though we haven't really discussed this possibility in class, \
there isn't any reason you can't have curves defined in other coordinate \
systems. So, for example, you could define a parametric curve in cylindrical \
coordinates as: ",
Cell[BoxData[
\(TraditionalForm\`r = 1, \ \[Theta] = t, \ z = t\)]],
" or, in vector form:"
}], "Text"],
Cell[BoxData[
\(curveCylindrical = {2, t, t}\)], "Input"],
Cell["\<\
Now, if you just apply the ArcLengthFactor directly to this curve, you get:\
\>", "Text"],
Cell[BoxData[
\(ArcLengthFactor[{2, t, t}, t]\)], "Input"],
Cell[TextData[{
"Notice, that this is actually the ",
StyleBox["same",
FontSlant->"Italic"],
" curve I defined above in Cartesian coordinates, a helix with a radius of \
2 about the z axis (moving at the same \"speed\"). Therefore, they ought to \
have the same factors. The way to fix this, of course, is to tell ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" that your curve is actually specified in cylindrical coordinates (the \
above command assumed Cartesian):"
}], "Text"],
Cell[BoxData[
\(ArcLengthFactor[{2, t, t}, t, Cylindrical]\)], "Input"],
Cell["If you are curious about the actual formula:", "Text"],
Cell[BoxData[
\(ArcLengthFactor[{x[t], y[t], z[t]}, t, Cartesian]\)], "Input"],
Cell[BoxData[
\(ArcLengthFactor[{r[t], \[Theta][t], z[t]}, t, Cylindrical]\)], "Input"]
}, Open ]],
Cell[CellGroupData[{
Cell["Graphing", "Subsection"],
Cell[TextData[{
"You can also graph functions and parametric equations in these coordinate \
systems. In Lab 1, you wrote your own translation functions to do this for \
spherical and cylindrical coordinates. Let's look at another way this could \
be done, using the built-in ",
StyleBox["Mathematica",
FontSlant->"Italic"],
" coordinate systems.\n\nLet's graph ",
Cell[BoxData[
\(TraditionalForm\`\[Rho] = Sin[\[Theta]] Cos[\[Phi]]\)]],
" (in spherical coordinates):"
}], "Text"],
Cell[BoxData[
\(\[Rho] = Sin[\[Theta]] Cos[\[Phi]]\)], "Input"],
Cell[BoxData[
\(plotSurface =
ParametricPlot3D[
Evaluate[
CoordinatesToCartesian[{\[Rho], \[Phi], \[Theta]},
Spherical]], {\[Phi], 0, \[Pi]}, {\[Theta], 0, 2 \[Pi]},
PlotPoints \[Rule] 40]\)], "Input"],
Cell[TextData[{
"To find the equation of a tangent plane to a surface at a point, remember \
that if you can express the equation as ",
Cell[BoxData[
\(TraditionalForm\`f[x, y, z] = 0\)]],
", you can find the equation of the tangent plane by setting ",
Cell[BoxData[
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Let's graph this. There are a number of possibilities here. The easiest way \
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Find equations in Cartesian coordinates for planes through the following sets \
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Actually, let's leave off one of the sides (which face is that?):\
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