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:
      1-16 for more practice.
      See Larson 15.5:21–30 and 15.5,
      Example 3.
    
Consider the surface $z=9-x^2-y^2$ 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 $z=9-x^2-y^2$? [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: 1-16 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: 1-16 for more practice.
        See Larson 15.5:1–10
    
Consider the surface $z=9-x^2-y^2$.
    
      - 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 $z=9-x^2-y^2$ 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 $z =
    x^2-y^2$. 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\^2-y\^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
    $$\hat F(x,y) = \frac{(x+y,
    -x+1)}{\sqrt{(x+y)^2+(1-x)^2}}.$$
  
  
  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: 1-20.
       
    
      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 $(x^2+y^2-2x)^2=4(x^2+y^2)$.  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 $(x^2+y^2)^2=2(x^3-3xy^2)$.  Which equation is simpler?
  
  
      See Larson 10.5:17–26
    Find the points of intersection of $r=3-3\cos\theta$ 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 11.3:5-10.
    
    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: 53-66.
    
    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: 27-52. 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: 1-14.
    
      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
      $\theta=\pi/2$.