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Polar Coordinates and Parametric Curves


The use of parametric equations and polar coordinates allows for the analysis of families of curves difficult to handle through rectangular coordinates. If a curve is a rectangular coordinate graph of a function, it cannot have any loops since, for a given $x$ value there can be at most one corresponding $y$ value. However, using polar coordinates, curves with loops can appear as graphs of functions.

Getting Started

To assist you, there is a worksheet associated with this lab that contains examples similar to some of the exercises.On your Maple screen go to File - Open then type the following in the white rectangle:
You can copy the worksheet now, but you should read through the lab before you load it into Maple. Once you have read through the exercises, start up Maple, load the worksheet, and go through it carefully. Then you can start working on the exercises.

Vector functions/Parametric curves

You can evaluate this function at any value of t in the usual way.
This is how to access a single component.

Derivatives and the slope

The graph of a parametric curve may not have a slope at every point on the curve. When the slope exists, it is given by the formula:

\begin{displaymath}\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \end{displaymath}

In Maple, the slope of a parametric curve can be calculated using the $\texttt{diff}$ command as in the example below.
> r := -> [3*sin(t),3*cos(t)];
> slope := diff(r(t)[2],t)/diff(r(t)[1],t);

The parametric form of a line

The vector equation for the line passing through the point $\mathbf{r_0}=(x_0,y_0)$ parallel to the vector $\mathbf{v}=[v_1,v_2]$ is given by:

\begin{displaymath}\mathbf{r}(t) = \mathbf{r}_0 + t\mathbf{v} \end{displaymath}

Below is an example in Maple using this parametric form of a line that is tangent to the curve $\mathbf{r}(t)$ defined above at $\displaystyle t=\frac{\pi}{4}$. The plot of the curve and the line on the same graph verifies that the line is tangent at the given point.
Since the slope at $\displaystyle \frac{\pi}{4}$ is -1, we want the line through the point $\displaystyle \mathbf{r}(\frac{\pi}{4})$, parallel to the vector $[1,-1]$.

Plotting Polar Curves

When you graph curves in polar coordinates, you are really working with parametric curves. The basic idea is that you want to plot a set of points by giving their coordinates in $(x,y)$ pairs. When you use polar coordinates, you are defining the points $(x,y)$ in terms of polar coordinates $(r,\theta )$. When you plot polar curves, you are usually assuming that $r$ is a function of the angle $\theta $ and $\theta $ is the parameter that describes the curve. In Maple you have to put square brackets around the curve and add the specification coords=polar. Maple assumes that the first coordinate in the parametric plot is the radius $(r)$ and the second coordinates is the angle $(\theta )$.

Cardioids, Limaçons, and Roses

These are three types of well-known graphs in polar coordinates. The table below will allow you to identify the graphs in the exercises.

Name Equation
cardioid $r = a(1 \pm \cos(\theta))$ or $r = a(1 \pm \sin(\theta))$
limaçon $r = b \pm a \cos(\theta)$ or $r = b \pm a \sin(\theta)$
rose $r=a \cos(n \theta)$ or $r=a \sin(n \theta)$

Below is an example of a cardiod.


Area in Polar Coordinates

The relationship between area and integrals in polar coordinates is a little strange; the area inside a circle given (in polar coordinates) by $r = a$ is NOT just $ \displaystyle \int_{0}^{2\pi} r d\theta$. Here is the rule: Area inside $r = f(\theta )$ is given by $ \displaystyle \frac{1}{2} \int_{\theta_1}^{\theta_2} r^2 d\theta = \frac{1}{2} \int_{\theta_1}^{\theta_2} f(\theta )^2 d\theta $. This comes from the fact that the area in a thin wedge with radius $r$ and angle $\Delta \Theta $ is $\frac{1}{2} r^2 \Delta \Theta $. Note that this gives you the right answer for a circle: $A = \pi r^2$. So to find the area of the cardiod use the following command.
>Area1:=1/2*int((1-cos(theta))^2, theta=0..2*Pi);


  1. Consider the vector function $\mathbf{r}(t) =
(2\cos(t) + 4\cos^2(t))\mathbf{i} + (2\sin(t) + 4\sin(t)\cos(t))\mathbf{j}$ for $-\pi \leq t \leq \pi$.
    1. Calculate the $(x,y)$ coordinate point on the curve at $\displaystyle t =-\frac{\pi}{2}$ and the slope of the curve at $\displaystyle t =-\frac{\pi}{2}$.
    2. Define the vector equation of the line through the point above tangent to the curve at that point.
    3. Plot the graph of $\mathbf{r}(t)$ and this tangent line on the same graphover the interval $-\pi \leq t \leq \pi$. Identify the graph as a cardioid, limaçon, or rose.

  2. Find all points of intersection for each pair of curves in polar coordinates.
    1. $r = 1+\cos(\theta)$ and $r=3/2$ for $0 \leq \theta \leq 2\pi$.
    2. $r = 1-\sin(\theta)$ and $r=3/2+\cos(\theta)$ for $0 \leq \theta \leq 2\pi$.

  3. Find the angles that create only one petal of the five petal rose given by the equation $r=2\cos(5\theta)$. Plot only one petal and find the area of that petal.

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Next: About this document ... Up: lab_template Previous: lab_template
Dina J. Solitro-Rassias