Transformations of $\RR^2$
Coordinate transformations let us view or discuss the plane
or space in a different way. A 2d transformation $\vec
T(u,v)=(x,y)$ tells us how to transform a $(u,v)$ axes into an
$(x,y)$ axes—the two outputs of $\vec T$ are considered the $x$
and $y$ coordinates corresponding to the inputs $u$ and $v$.
In the previous problem, you saw how we can think of 2D
transformations as mapping the points of one plane onto another.
Another way of thinking about transformations is to view them as
giving additional coordinates to points on the plane, i.e., if
$\vec T(u,v)=(x,y)$, then the point $(x,y)$ also can be called
the point $(u,v)$ in the coordinate system associated with $\vec
T$. The next problem investigates this way of thinking about
transformations.
See Larson 10.4.
The transformation $\vec T(r,\theta) =
(r\cos\theta,r\sin\theta)$ is called the
polar
coordinate transformation. We will use this transformation to
answer the following questions.
 $\vec T(2,\pi/6)=(a,b)$. What is $(a,b)$? Draw the vector
$(a,b)$ on the $(x,y)$ axes, starting at the origin.
 Show that the length of the vector $(a,b)$ is 2. This is
the “radius” of the point $(a,b)$.
 Show that the angle between the positive $x$axis and the
vector $(a,b)$ is $\pi/6$. This angle is called the azimuth
angle.
 Show that if $\vec T(r,\theta)=(x,y)$, then the
“radius” of the point $(x,y)$ is $r$.
 Show that if $\vec T(r,\theta)=(x,y)$, then the azimuth
angle of the vector $(x,y)$ is $\theta$.
Sometimes a transformation may associate multiple coordinates
with the same $(x,y)$ point on the plane. In this problem,
again use the polar coordinate transform $\vec T(r,\theta) =
(r\cos\theta,r\sin\theta)$. Find 5 different $(r,\theta)$ so
that $\vec T(r,\theta)=(\sqrt{3},1)$ (each of these is a
different set of polar coordinates for the same point
$(x,y)=(\sqrt{3},1)$). Make at least one of your $(r,\theta)$
coordinates have a negative $r$, and at least one have a
negative $\theta$.
Consider the coordinate transformation $$\vec
T(a,\omega)=(a\cos\omega,a^2\sin \omega).$$

Let $a=3$; graph the curve $\vec
T(3,\omega)=(3\cos\omega,9\sin\omega)$ for
$\omega\in[0,2\pi]$.

Use Sage to check your answer.
Let $\omega =\frac{\pi}{4}$ and then, on the same
axes as above, add the graph of $\vec
T\left(a,\frac{\pi}{4}\right)=\left(a\frac{\sqrt 2}{2},a^2
\frac{\sqrt 2}{2}\right)$ for $a\in[0,4]$.

Use Sage to check your answer.
To the same axes as above, add the graphs of $\vec
T(1,\omega), \vec T(2,\omega), \vec T(4,\omega)$ for
$\omega\in[0,2\pi]$ and $\vec T(a,0), \vec T(a,\pi/2), \vec
T(a,\pi/6)$ for $a\in[0,4]$.
[Hint: when you're done, you should have a bunch of
parabolas and ellipses.]
Transformations of $\RR^3$
In 3 dimensions, the most common coordinate systems are
cylindrical and spherical. The equations for these coordinate
systems are:
 Cylindrical Coordinates: $x=r\cos\theta,\, y=r\sin\theta,\, z=z$
 Spherical Coordinates: $x=\rho\sin\phi\cos\theta,\, y=\rho\sin\phi\sin\theta,\,
z=\rho\cos\phi$
\marginparbmw{See page 893.}
See Larson 11.7.
Let $P=(x,y,z)$ be a point in space. This point lies on a
cylinder of radius $r$, where the cylinder has the $z$ axis as
its axis of symmetry. The height of the point is $z$ units up
from the $xy$ plane. The point casts a shadow in the $xy$ plane
at $Q=(x,y,0)$. The angle between the ray $\vec{OQ}$ and the
$x$axis is $\theta$.
 Explain why $x=r\cos\theta$, $y=r\sin\theta$, and
$z=z$.
 What are bounds on $r$, $\theta$, and $z$ that will give
all points on the surface of a cylinder of radius 1 wrapped
around the $z$ axis between the $xy$ plane and $z=1$? [Hint:
the bounds on $r$ are $r=1$.]
 What are bounds on $r$, $\theta$, and $z$ that will give
all points inside a solid cylinder of radius 2 wrapped around
the $z$axis extending from 1 unit below the $xy$ plane to 1
unit above the $xy$ plane?
\marginparbmw{See page 897.}
See Larson 11.7.
Let $P=(x,y,z)$ be
a point in space. This point lies on a sphere of radius $\rho$
(“rho”), where the sphere's center is at the
origin $O=(0,0,0)$. The point casts a shadow in the $xy$ plane
at $Q=(x,y,0)$. The angle between the ray $\vec{OQ}$ and the
$x$axis is $\theta$, and is called the azimuth angle. The
angle between the ray $\vec{OP}$ and the $z$ axis is $\phi$
("phi"), and is called the inclination angle,
polar angle, or zenith angle.
 Explain why $x=\rho\sin\phi\cos\theta$,
$y=\rho\sin\phi\sin\theta$, and $z=\rho\cos\phi$.
 What are bounds for $\rho$, $\theta$, and $\phi$ that
will give all the points on the surface a sphere of radius 1?
[Hint: the bounds for $\rho$ are $\rho=1$.]
 What are bounds on $\rho$, $\theta$, and $\phi$ that will
give all the points on or above the $xy$ plane inside a solid
sphere of radius 1?
 What are bounds on $\rho$, $\theta$, and $\phi$ that will
give all the points on the surface of a sphere of radius 2
above the plane $z=1$ and where the $y$ coordinates are
positive?
There is some disagreement between different fields about
the notation for spherical coordinates. In some fields (like
physics), $\phi$ represents the azimuth angle and $\theta$
represents the inclination angle. In some fields, like geography,
instead of the inclination angle, the elevation angle is
given—the angle from the $xy$plane (for example, lines of lattitude are
from the elevation angle). Additionally, sometimes the
coordinates are written in a different order. You should always
check the notation for spherical coordinates before communicating
using them.
See Larson 11.7:89–94,
111–114.
Consider the spherical coordinates transformation
$$\vec
T(\rho,\phi,\theta)=(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi),$$
$$\vec
T(\rho,\theta,\phi)=(\rho\sin\phi\cos\theta,\rho\sin\phi\sin\theta,\rho\cos\phi),$$
Graphing this transformation requires $3+3=6$ dimensions. In
this problem we'll construct parts of this graph by graphing
various surfaces. We did something similar for the polar
coordinate transformation already in
this problem.

Graph the surface $T(2,\theta,\phi)$ (in other
words, the surface $\rho=2$) where $\theta\in [0,2\pi]$,
$\phi\in [0,\pi]$.

Graph $T(\rho,\theta,\pi/4)$ for $\rho\geq 0$,
$\theta\in [0,2\pi]$ (in other words, all points where
$\phi=\pi/4$). What happens if $\rho$ can be negative
(i.e., $\rho\in\mathbb{R}$)?
 Graph $T(\rho, \theta, \pi/2)$ for $\rho\in\mathbb{R}$,
$\theta\in [0,2\pi]$ (in other words, all points where
$\phi=\pi/2$).
 Graph $T(\rho,\pi/4, \phi)$ for $\rho\geq 0$, $\phi\in
[0,\pi]$ (in other words, all points where
$\theta=\pi/4$).
Transformation Functions and Parametric Surfaces
We saw in this problem that if we make one of the input variables a constant in a transformation function, we get a surface. We can use this idea to find parametric equations for a given surface.
For example, in a problem above, $T$ was the spherical coordinate transformation, and we set $\rho=2$: $T(2, \theta, \phi)=(2\sin\phi\cos\theta,\,2\sin\phi\sin\theta,\,2\cos\phi)$ to get a sphere of radius 2. Since we were eliminating one of the input variables by setting it to be a number, we could have written this same function as $\vec q(\theta,\phi)=(2\sin\phi\cos\theta,\,2\sin\phi\sin\theta,\,2\cos\phi)$. This function $\vec q\colon \RR^2\to\RR^3$ is then a parametric equation for a sphere of radius 2.
Use
Sage or
Wolfram
Alpha
to plot your parametrization.
See 16.5:
116 for more practice.
See Larson 15.5:21–30 and 15.5,
Example 3.
Consider the surface $z=9x^2y^2$ plotted in
this problem. We'll use the rectangular coordinate transformation $\vec T(x,y,z)=(x,y,z)$ (this transformation does nothing to 3d space).
 We're trying to find a parametric equation $\vec r(x,y)=(?,?,?)$ for this surface. If we set the input variable $z$ in $\vec T$, we can give a parametrization $\vec r\colon
\mathbb{R}^2\to\mathbb{R}^3$ of the surface. In other words, $\vec r(x,y)=(x,y,?)$ so that the parametric surface $\vec r$ is the surface $z=9x^2y^2$? [Hint: Use the surface equation
to eliminate the input variable $z$ in $T$.]
 What bounds must you place on $x$ and $y$ to obtain the
portion of the surface above the plane $z=0$?
 If $z=f(x,y)$ is any surface, give a parametrization of
the surface (i.e., $x=?, y=?, z=?$ or $\vec r
(?,?)=(?,?,?)$.)
See 16.5: 116 for more practice.
See Larson 15.5:1–10
Recall the spherical coordinate transformation
$$\vec T(\rho,\phi,\theta) =
(\rho\sin\phi\cos \theta, \rho\sin\phi\sin \theta,\rho \cos
\phi).$$
$$\vec
T(\rho,\theta, \phi) = (\rho\sin\phi\cos \theta,
\rho\sin\phi\sin \theta,\rho \cos \phi).$$
This is a
function of the form $\vec T\colon
\mathbb{R}^3\to\mathbb{R}^3$. If we eliminate one of the inputs (for example, if we hold it constant), then we have a function of the form $\vec
q\colon \mathbb{R}^2\to\mathbb{R}^3$, which is a parametric
surface.

Give a parametrization of the sphere of radius 4, $\vec q(\theta,\phi)=(?,?,?)$. Make sure your bounds on $\phi$ and $\theta$ hit
every point on the sphere.
 Give a parameterization of the part of the sphere of radius 4 that is above the plane $z=2$. [Hint: the only thing that changes from the previous part is the bounds on $\phi$ and $\theta$.]
Use
Sage or
Wolfram Alpha to plot
your parametrization with your bounds.
See 16.5: 116 for more practice.
See Larson 15.5:1–10
Consider the surface $z=9x^2y^2$.
 Using cylindrical coordinates, $\vec T(r,\theta,z) =
(r\cos \theta, r\sin\theta, z)$, obtain a parametrization
$\vec q(r,\theta)=(?,?,?)$ of the surface using the input
variables $r$ and $\theta$. [Hint: we already know $x=r\cos\theta$ and $y=r\sin\theta$; just write $z=9x^2y^2$ in
terms of $r$ and $\theta$.]
 What bounds must you place on $r$ and $\theta$ to obtain
the portion of the surface above the plane $z=0$?
Sometimes you'll have to invent your own coordinate system
when constructing parametric equations for a surface. If you
notice that there are lots of circles parallel to one of the
coordinate planes, try using a modified version of cylindrical
coordinates. For example, instead of circles in the $xy$ plane
($x=r\cos\theta,y=r\sin\theta,z=z$), maybe you need circles in
the $yz$plane ($x=x,y=r\cos\theta,z=r\sin\theta$).
See Larson 15.5:21–30.
Find a parametric equation $\vec q(?,?)=(?,?,?)$ for the surface $x^2+z^2=9$.
[Hint: read the paragraph above.]

What bounds on the inputs should you use to obtain the portion of
the surface between $y=2$ and $y=3$?
 What bounds on the inputs should you use to obtain the portion of the
surface above $z=0$?
 What bounds on the inputs should you use to obtain the portion of the
surface with $x\geq 0$ and $y\in[2,5]$?
Construct a graph of the surface $z =
x^2y^2$. Do so in 2 ways. (1) Construct a 3D surface plot. (2)
Construct a contour plot, which is a graph with several level
curves. Which level curve passes through the point $(3,4)$? Use
Wolfram Alpha to know if you're right. Just type “plot
z=x\^2y\^2.”
Construct a plot of the vector field $$\vec
F(x,y) = (x+y, x+1)$$ by graphing the field at many integer
points around the origin (I generally like to get the 8 integer
points around the origin, and then a few more). Then explain
how to modify your graph to obtain a plot of the vector field
$$\hat F(x,y) = \frac{(x+y,
x+1)}{\sqrt{(x+y)^2+(1x)^2}}.$$
Polar Coordinates
See Larson section 10.4 and 10.5 for more
reference material on the things we cover in this
section.
In this section, we'll explore polar coordinates a bit more. One of the things to take away from this section is that using a natural coordinate system for your problem can make the problem much simpler.
To construct a graph of a
polar curve, just create an $r,\theta$ table. Choose values for
$\theta$ that will make it easy to compute any trig functions
involved. Then connect the points in a smooth manner, making sure
that your radius grows or shrinks appropriately as your angle
increases.
See
Sage. Use Sage to check your answers in the other graphing problems too.
See 11.4: 120.
Graph the polar curve $r=2+2\cos\theta$. This curve is called a
cardiod (see also
here). The Cartesian equation for this curve is $(x^2+y^22x)^2=4(x^2+y^2)$. Which is simpler, the polar equation or the Cartesian equation?
Graph the polar curve $r=3\sin 2\theta$ and identify which points on the curve came from negative $r$ values and which came from positive $r$ values. This curve is called a
rose (see also
here). The Cartesian equation for this curve is $(x^2+y^2)^3=36x^2y^2$. Which equation is simpler?
Graph the polar curve $r=2\cos 3\theta$ and identify which points on the curve came from negative $r$ values and which came from positive $r$ values. This curve is also called a rose. The Cartesian equation for this curve is $(x^2+y^2)^2=2(x^33xy^2)$. Which equation is simpler?
See Larson 10.5:17–26
Find the points of intersection of $r=33\cos\theta$ and
$r=3\cos\theta$. (If you don't graph the curves, you'll
probably miss a few points of intersection.)
Here's another intersection problem.
Find the points of intersection of $r=2\cos 2\theta $ and
$r=\sqrt 3$. (If you don't graph the curves, you'll probably
miss a few points of intersection.)
See 11.3:510.
The following points are given using their polar
coordinates. Plot the points in the Cartesian plane, and give
the Cartesian (rectangular) coordinates of each point. The
points are $$ (1,\pi), \ds \left( 3,\frac{5\pi}{4}\right), \ds
\left( 3,\frac{\pi}{4}\right),\text{ and } \ds \left(
2,\frac{\pi}{6}\right).$$
The next problem provides general formulas for converting
between the Cartesian (rectangular) and polar coordinate systems.
See page 647.
Suppose that $Q$ is a point in the plane with Cartesian
coordinates $(x,y)$ and polar coordinates $(r,\theta)$.
 Write formulas for $x$ and $y$ in terms of $r$ and
$\theta$: $x=?$, $y=?$
 Write a formula to find the distance $r$ from $Q$ to the
origin (in terms of $x$ and $y$): $r=?$
 Write a formula to find the angle $\theta$ between the
$x$axis and a line connecting $Q$ to the origin, in terms of
$x$ and $y$: $\theta = ?$. [Hint: A picture of a triangle
will help here.]
See 11.3: 5366.
Each of the following equations is written in the
Cartesian (rectangular) coordinate system. Convert each to an
equation in polar coordinates, and then solve for $r$ so that
the equation is in the form $r=f(\theta)$. Judge which equation is simpler, Cartesian or polar.
 $x^2+y^2=7$
 $2x+3y=5$
 $x^2=y$
See 11.3: 2752. I strongly suggest that
you do many of these as practice.
Each of the following equations is written in the polar
coordinate system. Convert each to an equation in the Cartesian
coordinates. Judge which equation is simpler, Cartesian or polar.
 $r=9\cos\theta$ [Hint: multiply both sides by $r$.]
 $\ds r=\frac{4}{2\cos\theta+3\sin\theta}$
 $\theta = 3\pi/4$
Recall that for
parametric curves $\vec q(t) = (x(t),y(t))$, to find the slope of
the curve we just compute $$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}.$$
A polar curve of the form $r=f(\theta)$ can be thought of as just
the parametric curve $\vec q(\theta) =
(f(\theta)\cos\theta,f(\theta)\sin\theta)$. So you can find the
slope by computing
$$\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}.$$
See 11.2: 114.
Consider the polar curve $r=1+2\cos \theta$.
 Sketch the curve.
 Compute both $dx/d\theta$ and $dy/d\theta$.
 Find the slope $dy/dx$ of the curve at
$\theta=\pi/2$.
 Give both a vector equation of the tangent line, and a
Cartesian equation of the tangent line at
$\theta=\pi/2$.