Chapter 8.1, Problem 14E

To determine

To plot: the points and two other polar coordinates.

Answer to Problem 14E

Here, the two polar coordinates of the point are one with $r>0$is $\left(3,1+2\pi \right)=\left(3,7.28\right)$ and the other with $r<0$ is $\left(-3,1+\pi \right)=\left(-3,4.14\right)$ .

Explanation of Solution

Given: $\left(3,1\right)$

Calculation:

plot the point that has the given polar coordinates.

  $\left(3,1\right)$

given the plot in $\left(r,\theta \right)$ form. This means that the point is the distance $r$from the origin(the radius), and the angle formed by the polar axis and the line segment (which connects the origin and the point) is $\theta$. Let’s start by simply graphing the line $\theta =1$ . For reference, 1 radian is very close to $\frac{\pi }{3}=1.047$. The point will lie somewhere on this line.

  

This point is $3$ units from the origin and lies on this line. Draw the circle $r=3$to show the places where the line and circle intersect. The angle $\theta =1$ points into Quadrant I, so our positive radius will also point into Quadrant I.

  

This red dot is the plot of the point $\left(3,1\right)$.

Given two other polar coordinate representations of the point, one with $r<0$ and the other with $r>0$ . To do this, use two formulas given in this section, which state that the point $\left(r,\theta \right)$can also be represented by

  $\left(r,\theta +2\pi n\right)or\left(-r,\theta +\left(2n+1\right)\pi \right)$.

The first formula is true because angles that are separated by multiples of $2\pi$ are congruent. The second formula works because if you add $\pi$ to an angle, you flip it across the across the origin, and when you make the radius negative, you flip it across the origin again. Flipping across the origin twice brings you back to where you started.

Let’s apply these formulas to the point $\left(3,1\right)$ . For the polar coordinate equivalent with $r>0$ , use the first formula, letting $n=1$.

  $\begin{array}{l}\left(3,1\right)\\ \left(3,1+2\pi n\right)\\ \left(3,1+2\pi \left(1\right)\right)\\ \left(3,1+2\pi \right)=\left(3,7.28\right)\end{array}$

For the polar coordinates equivalent with $r<0$ , use the second formula, letting $n=0$.

  $\begin{array}{l}\left(3,1\right)and\left(-r,\theta +\left(2n+1\right)\pi \right)\\ \left(-3,1+\left(2n+1\right)\pi \right)\\ \left(-3,1+\left(2\left(0\right)+1\right)\pi \right)\\ \left(-3,1+\pi \right)\\ \left(-3,1+\pi \right)=\left(-3,4.14\right)\end{array}$

Conclusion:

Therefore, the two polar coordinates of the point are one with $r>0$is $\left(3,1+2\pi \right)=\left(3,7.28\right)$ and the other with $r<0$ is $\left(-3,1+\pi \right)=\left(-3,4.14\right)$ .