Up to now, we have most often given the location of a point (or coordiantes of a vector) by stating the \((x,y)\) coordinates. These are called the Cartesian (or rectangular) coordinates. Some problems are much easier to work with if we know how far a point is from the origin, together with the angle between the \(x\)-axis and a ray from the origin to the point.
Consider the point \(P\) with Cartesian (rectangular) coordinates \((2,1)\text{.}\)
Find the distance \(r\) from \(P\) to the origin.
Consider the ray \(\stackrel{\rightarrow}{OP}\) from the origin through \(P\text{.}\) Find an angle between \(\vec{OP}\) and the \(x\)-axis.
Use a triangle and trigonometry
Suppose that a point \(Q=(a,b)\) is 6 units from the origin, and the angle the ray \(\vec{OQ}\) makes with the \(x\)-axis is \(\pi/4\) radians. Find the Cartesian (rectangular) coordinates \((a,b)\) of \(Q\text{.}\)
Let \(Q\) be be a point in the plane with Cartesian coordinates \((x,y)\text{.}\) Let \(O=(0,0)\) be the origin. We define the polar coordinates of \(Q\) to be the ordered pair \((r,\theta)\) where \(r\) is the displacement from the origin to \(Q\text{,}\) and \(\theta\) is an angle of rotation (counter-clockwise) from the \(x\)-axis to the ray \(\vec {OP}\text{.}\)
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
\begin{equation*} (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). \end{equation*}The next problem provides general formulas for converting between the Cartesian (rectangular) and polar coordinate systems.
Suppose that \(Q\) is a point in the plane with Cartesian coordinates \((x,y)\) and polar coordinates \((r,\theta)\text{.}\)
Write formulas for \(x\) and \(y\) in terms of \(r\) and \(\theta\text{.}\)
Write a formula to find the distance \(r\) from \(Q\) to the origin (in terms of \(x\) and \(y\))
Write a formula to find the angle \(\theta\) between the \(x\)-axis and a line connecting \(Q\) to the origin.
A picture of a triangle will help here.
In Exercise 4, you should have obtained the equations
\begin{equation*} x=r\cos\theta, y=r\sin\theta. \end{equation*}We can write this in vector notation as \(\begin{pmatrix}x\\y \end{pmatrix} =\begin{pmatrix}r\cos\theta\\ r\sin\theta \end{pmatrix}\text{.}\) This is a vector equation in which you input polar coordinates \((r,\theta)\) and get out Cartesian coordinates \((x,y)\text{.}\) So you input one thing to get out one thing, which means that we have a function. We could write \(\vec T(r,\theta) = (r\cos\theta,r\sin\theta)\text{,}\) where we've used the letter \(T\) as the name of the function because it is a transformation between coordinate systems. To emphasize that the domain and range are both two dimensional systems, we could also write \(T:\mathbb{R}^2\to\mathbb{R}^2\text{.}\) In the next chapter, we'll spend more time with this notation.
The following problem will show you how to graph a coordinate transformation. When you're done, you should essentially have polar graph paper.
Consider the coordinate transformation
\begin{equation*} \vec T(r,\theta) = (r\cos\theta,r\sin\theta). \end{equation*}For this problem, you are just drawing many parametric curves. This is what we did in the previous chapter.
Let \(r=3\) and then graph \(\vec T(3,\theta)=(3\cos\theta,3\sin\theta)\) for \(\theta\in[0,2\pi]\text{.}\)
Let \(\theta=\frac{\pi}{4}\) and then, on the same axes as above, add the graph of \(\vec T\left(r,\frac{\pi}{4}\right)=\left(r\frac{\sqrt 2}{2},r \frac{\sqrt 2}{2}\right)\) for \(r\in[0,5]\text{.}\)
To the same axes as above, add the graphs of \(\vec T(1,\theta), \vec T(2,\theta), \vec T(4,\theta)\) for \(\theta\in[0,2\pi]\)
To the same axes as above, add the graphs of \(\vec T(r,0), \vec T(r,\pi/2), \vec T(r,3\pi/4), \vec T(r,\pi)\) for \(r\in[0,5]\text{.}\)
In the plane,
Graph the curve \(y=\sin x\) for \(x\in[0,2\pi]\) (make an \(x,y\) table)
Graph the curve \(r=\sin\theta\) for \(\theta\in[0,2\pi]\) (an \(r,\theta\) table).
The graphs should look very different. If one looks like a circle, you're on the right track.
Each of the following equations is written in the Cartesian (rectangular) coordinate system. Convert each to an equation in polar coordinates by substituting in the formulas from Exercise 4, and then solve for \(r\) so that the equation is in the form \(r=f(\theta)\text{.}\)
\(x^2+y^2=7\)
\(2x+3y=5\)
\(x^2=y\)
Each of the following equations is written in the polar coordinate system. Convert each to an equation in the Cartesian coordinates. Again, you will probably want to use the formulas from Exercise 4.
\(r=9\cos\theta\)
Don't substitute directly for
\begin{equation*} \theta\text{.} \end{equation*}Instead try substituting for
\begin{equation*} \cos\theta \end{equation*}and
\begin{equation*} \cos\theta \end{equation*}\(\ds r=\frac{4}{2\cos\theta+3\sin\theta}\)
\(\theta = 3\pi/4\)
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.
Graph the polar curve \(r=2\sin 3\theta\text{.}\)
Graph the polar curve \(r=3\cos 2\theta\text{.}\)
Be sure you actually plot out the next two problems, otherwise you'll probably miss a few points of intersection.
Find the points of intersection of \(r=3-3\cos\theta\) and \(r=3\cos\theta\text{.}\)
Find the points of intersection of \(r=2\cos 2\theta\) and \(r=\sqrt 3\text{.}\)
Recall that for parametric curves \(\vec r(t) = (x(t),y(t))\text{,}\) to find the slope of the curve we compute
\begin{equation*} \frac{dy}{dx}=\frac{dy/dt}{dx/dt}. \end{equation*}A polar curve of the form \(r=f(\theta)\) can be thought of as the parametric curve \((x,y) = (f(\theta)\cos\theta,f(\theta)\sin\theta)\text{.}\) So you can find the slope by computing
\begin{equation*} \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}. \end{equation*}Consider the polar curve \(r=1+2\cos \theta\text{.}\) (It wouldn't hurt to provide a quick sketch of the curve.)
Compute both \(dx/d\theta\) and \(dy/d\theta\text{.}\)
Find the slope \(dy/dx\) of the curve at \(\theta=\pi/2\text{.}\)
Give both a vector equation of the tangent line, and a Cartesian equation of the tangent line at \(\theta=\pi/2\text{.}\)
You can find arc length for parametric curves using the formula:
\begin{equation*} s=\int_a^b\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}. \end{equation*}If we replace \(t\) with \(\theta\text{,}\) this becomes a formula for arc length in polar coordinates. However, the formula can be simplified.
Recall that \(x=r\cos\theta\) and \(y=r\sin\theta\text{.}\) Suppose that \(r=f(\theta)\) for \(\theta\in[\alpha,\beta]\) is a continuous function, and that \(f'\) is continuous. Show that the arc length formula can be simplified to
\begin{equation*} s=\int_{\alpha}^{\beta}\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2} = \int_{\alpha}^{\beta}\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} . \end{equation*}the product rule and Pythagorean identity will help.
Set up (do not evaluate) an integral formula to compute the length of
the rose \(r=2\cos 3\theta\text{,}\) and
the rose \(r=3\sin 2\theta\text{.}\)
There are multiple copies of several exercises for your practice. You don't need to do them all. Do as many as it takes to get one right quickly.
These practice problems test your ability to find/identify bounds using polar coordinates
These practice problems test your ability to find arc-lengths