Consider the coordinate transformation $$\overrightarrow{T}(r,\theta )=(r\cos \theta ,r\sin \theta ).$$ In other words, according to $\overrightarrow{T}$, $x=r\cos \theta$ and $y=r\sin \theta$. This is called the polar coordinate transformation.

1. Draw two 2D axes, one having horizontal axis $r$ and vertical axis $\theta$, and the other axes having horizontal axis $x$ and vertical axis $y$. In the next few parts, we'll see how $\overrightarrow{T}$ will tell us how points on the $(r,\theta )$ axes transform to points on the $(x,y)$ axes.
2. Plot the point $(r,\theta )=(1,\pi /2)$ on the $(r,\theta )$ axes. According to $\overrightarrow{T}$, this point is transformed to what $(x,y)$ point? Plot this corresponding point on the $(x,y)$ axes.
3. See Sage to check your answer.
   Plot the line segment $r=3$ (for $\theta \in [0,2\pi ]$) on the $(r,\theta )$ axes. Plot the corresponding points on the $(x,y)$ axes by letting $r=3$ in $\overrightarrow{T}$ and graphing $\overrightarrow{T}(3,\theta )=(3\cos \theta ,3\sin \theta )$ for $\theta \in [0,2\pi ]$ (hint: this is just a parametric curve, like we've been plotting).
4. See Sage. Notice that you can add two plots together to draw them both.
   Plot the line segment $\theta =\pi /4$ (for $r\in [0,5]$) on the $(r,\theta )$ axes. Plot the corresponding points on the $(x,y)$ axes by letting $\theta =\frac{\pi }{4}$ and then, on the same axes as above, add the graph of $\overrightarrow{T}(r,\frac{\pi }{4})=(r\frac{\sqrt{2}}{2},r\frac{\sqrt{2}}{2})$ for $r\in [0,5]$.
5. Use Sage to check your answer. Add to the $(r,\theta )$ axes the plots of $r=1$, $r=2$, $r=3$, and $r=4$ for $\theta \in [0,2\pi ]$ and the plots of $\theta =0$, $\theta =\pi /2$, $\theta =3\pi /4$, and $\theta =\pi$ for $r\in [0,5]$. Add the corresponding plots to the $(x,y)$ axes.

The transformation $\overrightarrow{T}(r,\theta )=(r\cos \theta ,r\sin \theta )$ is called the polar coordinate transformation. We will use this transformation to answer the following questions.

1. $\overrightarrow{T}(2,\pi /6)=(a,b)$. What is $(a,b)$? Draw the vector $(a,b)$ on the $(x,y)$ axes, starting at the origin.
2. Show that the length of the vector $(a,b)$ is 2. This is the “radius” of the point $(a,b)$.
3. Show that the angle between the positive $x$-axis and the vector $(a,b)$ is $\pi /6$. This angle is called the azimuth angle.
4. Show that if $\overrightarrow{T}(r,\theta )=(x,y)$, then the “radius” of the point $(x,y)$ is $r$.
5. Show that if $\overrightarrow{T}(r,\theta )=(x,y)$, 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 $\overrightarrow{T}(r,\theta )=(r\cos \theta ,r\sin \theta )$. Find 5 different $(r,\theta )$ so that $\overrightarrow{T}(r,\theta )=(\sqrt{3},1)$ (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 $$\overrightarrow{T}(a,\omega )=(a\cos \omega ,a^2\sin \omega ).$$

1. See Sage.
   Let $a=3$; graph the curve $\overrightarrow{T}(3,\omega )=(3\cos \omega ,9\sin \omega )$ for $\omega \in [0,2\pi ]$.
2. Use Sage to check your answer.
   Let $\omega =\frac{\pi }{4}$ and then, on the same axes as above, add the graph of $\overrightarrow{T}(a,\frac{\pi }{4})=(a\frac{\sqrt{2}}{2},a^2\frac{\sqrt{2}}{2})$ for $a\in [0,4]$.
3. Use Sage to check your answer.
   To the same axes as above, add the graphs of $\overrightarrow{T}(1,\omega ),\overrightarrow{T}(2,\omega ),\overrightarrow{T}(4,\omega )$ for $\omega \in [0,2\pi ]$ and $\overrightarrow{T}(a,0),\overrightarrow{T}(a,\pi /2),\overrightarrow{T}(a,-\pi /6)$ for $a\in [0,4]$.

[Hint: when you're done, you should have a bunch of parabolas and ellipses.]

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 $\overrightarrow{OQ}$ and the $x$-axis is $\theta$.

1. Explain why $x=r\cos \theta$, $y=r\sin \theta$, and $z=z$.
2. 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 $r=1$.]
3. 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?

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 $\overrightarrow{OQ}$ and the $x$-axis is $\theta$, and is called the azimuth angle. The angle between the ray $\overrightarrow{OP}$ and the $z$ axis is $\varphi$ ("phi"), and is called the inclination angle, polar angle, or zenith angle.

1. Explain why $x=\rho \sin \varphi \cos \theta$, $y=\rho \sin \varphi \sin \theta$, and $z=\rho \cos \varphi$.
2. What are bounds for $\rho$, $\theta$, and $\varphi$ that will give all the points on the surface a sphere of radius 1? [Hint: the bounds for $\rho$ are $\rho =1$.]
3. What are bounds on $\rho$, $\theta$, and $\varphi$ that will give all the points on or above the $xy$ plane inside a solid sphere of radius 1?
4. What are bounds on $\rho$, $\theta$, and $\varphi$ 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?

Consider the spherical coordinates transformation $$\overrightarrow{T}(\rho ,\varphi ,\theta )=(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi ),$$ $$\overrightarrow{T}(\rho ,\theta ,\varphi )=(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi ),$$ 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.

1. See Sage or Wolfram Alpha.
   Graph the surface $T(2,\theta ,\varphi )$ (in other words, the surface $\rho =2$) where $\theta \in [0,2\pi ]$, $\varphi \in [0,\pi ]$.
2. See Sage or Wolfram Alpha.
   Graph $T(\rho ,\theta ,\pi /4)$ for $\rho \ge 0$, $\theta \in [0,2\pi ]$ (in other words, all points where $\varphi =\pi /4$). What happens if $\rho$ can be negative (i.e., $\rho \in \mathbb{R}$)?
3. Graph $T(\rho ,\theta ,\pi /2)$ for $\rho \in \mathbb{R}$, $\theta \in [0,2\pi ]$ (in other words, all points where $\varphi =\pi /2$).
4. Graph $T(\rho ,\pi /4,\varphi )$ for $\rho \ge 0$, $\varphi \in [0,\pi ]$ (in other words, all points where $\theta =\pi /4$).

Recall the spherical coordinate transformation $$\overrightarrow{T}(\rho ,\varphi ,\theta )=(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi ).$$ $$\overrightarrow{T}(\rho ,\theta ,\varphi )=(\rho \sin \varphi \cos \theta ,\rho \sin \varphi \sin \theta ,\rho \cos \varphi ).$$ This is a function of the form $\overrightarrow{T}:{\mathbb{R}}^3\to {\mathbb{R}}^3$. If we eliminate one of the inputs (for example, if we hold it constant), then we have a function of the form $\overrightarrow{q}:{\mathbb{R}}^2\to {\mathbb{R}}^3$, which is a parametric surface.

1. Use Sage or Wolfram Alpha to plot each parametrization.
   Give a parametrization of the sphere of radius 4, $\overrightarrow{q}(\theta ,\varphi )=(?,?,?)$. Make sure your bounds on $\varphi$ and $\theta$ hit every point on the sphere.
2. 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 $\varphi$ and $\theta$.]

Consider the surface $z=9-x^2-y^2$.

1. Using cylindrical coordinates, $\overrightarrow{T}(r,\theta ,z)=(r\cos \theta ,r\sin \theta ,z)$, obtain a parametrization $\overrightarrow{q}(r,\theta )=(?,?,?)$ 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 $\theta$.]
2. What bounds must you place on $r$ and $\theta$ to obtain the portion of the surface above the plane $z=0$?

Find a parametric equation $\overrightarrow{q}(?,?)=(?,?,?)$ for the surface $x^2+z^2=9$. [Hint: read the paragraph above.]

1. Use Sage or Wolfram Alpha to plot your parametrization.
   What bounds on the inputs should you use to obtain the portion of the surface between $y=-2$ and $y=3$?
2. What bounds on the inputs should you use to obtain the portion of the surface above $z=0$?
3. What bounds on the inputs should you use to obtain the portion of the surface with $x\ge 0$ and $y\in [2,5]$?

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 $(x^2+y^2{)}^3=36x^2{y}^2$. 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 $(x^2+y^2{)}^2=2(x^3-3x{y}^2)$. Which equation is simpler?

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.)

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.

1. $x^2+y^2=7$
2. $2x+3y=5$
3. $x^2=y$

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.

1. $r=9\cos \theta$ [Hint: multiply both sides by $r$.]
2. $\text{ds}r=\frac{4}{2\cos \theta +3\sin \theta }$
3. $\theta =3\pi /4$

Consider the polar curve $r=1+2\cos \theta$.

1. Sketch the curve.
2. Compute both $dx/d\theta$ and $dy/d\theta$.
3. Find the slope $dy/dx$ of the curve at $\theta =\pi /2$.
4. Give both a vector equation of the tangent line, and a Cartesian equation of the tangent line at $\theta =\pi /2$.