Chapter 10, Problem 8RWDT

To determine

To find: If $r=f\left(\theta \right)$ is a continuous function of $\theta$ for $0\le \theta \le \pi$ ,is it necessary to follow that the corresponding polar curve is the graph of $y$ as a continuous function of $x$

Answer to Problem 8RWDT

It is not necessary to follow that the corresponding polar curve is the graph of $y$ as a continuous function of $x$ .

Explanation of Solution

Given information:

The polar curve $r=f\left(\theta \right)$ is continuous for $0\le \theta \le \pi$ .

In the Cartesian coordinate system each point $\left(x,y\right)$ can be represented as an ordered pair $\left(r,\theta \right)$ in the polar coordinate system. The first coordinate is called radical coordinate and second is called the angular coordinate.

In the cartecian coordinate sysytem the equation of the function is represented as $y=f\left(x\right)$ and in polar coordinate it is represented by $r=f\left(\theta \right)$ .

If in polar coordinate system curve $r=f\left(\theta \right)$ is continuous for $0\le \theta \le \pi$ .

  $0\le \theta \le \pi$ represents that in Cartesian system that points are between positive $x$ axis and negative $x$ axis

  $\theta$ can be represented in terms of Cartesian coordinate as $\theta ={\tan }^{-1}\left(\frac{y}{x}\right)$ .

Let take an example

By the definition of continuity for functions of a single variable a function $f$ is said to be continuous if:

For every $c$ in the domain of $\left(f,f\left(c\right)\right)$is defined(i.e. has a finite value) and $x\to cf\left(x\right)=f\left(c\right)$ .

If the function is single variable the statement is true but if the function is multivariable then the statement is not always true.

In order to shoe that let take an example:

  $(x,y)\to (0,0)\frac{x^2-y^2}{\left(x^2+y^2\right)}$

If this limit is converted into polar coordinates, it becomes $\left(r,\theta \right)\to \left(0,0\right){\cos }^2\theta -{\sin }^2\theta =\cos 2\theta$ and it’s evident that the function oscillates between-1 and 1 in a certain way when circle around the origin, independent of it can be noticed by approaching the origin on along some lines (if coordinates are not changed) to see that the limit at the origin doesn’t exist.

Like substitute $y=mx$ the limit becomes:

  $\begin{array}{l}x\to 0\frac{x^2-{\left(mx\right)}^2}{\left(x^2+{\left(mx\right)}^2\right)}\\ x\to 0\frac{x^2\left(1-m^2\right)}{1+m^2}=0\end{array}$

The limit exists but if the path is change it is not necessary that the limit exists. For a function being continuous the limit should exists following every path.

With polar coordinates it is just "parametrizing" all the point in the neighborhood of $\left(0,0\right)$ through their distance from $\left(0,0\right)$ and their angle w.r.t. the positive x -axis. The limit $r\to 0$ is along any path connecting the starting point $\left(0,0\right)$ and not (only) the straight lines. If such limit depends on the path itself, like in the case of the limit $(x,y)\to (0,0)\frac{x^2-y^2}{\left(x^2+y^2\right)}$ then it can be said that the function is not continuous at the limit point.

Therefore if $r=f\left(\theta \right)$ is a continuous function of $\theta$ for $0\le \theta \le \pi$ , it is not necessary to follow that the corresponding polar curve is the graph of $y$ as a continuous function of $x$ .