Chapter 8.1, Problem 69E

(a)

To determine

To prove: the distance between the polar points .

Answer to Problem 69E

Here, It is proved that the distance between the polar points $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$is

  $d=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$

Explanation of Solution

Given: $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$is $d=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$

Calculation:

Use the Law of Cosines to prove that the distance between the polar points $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$ is

  $d=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$.

Let’s first make an example plot. Point $1$is in red, and Point $2$is in blue.

Each forms an angle with the positive $x-axis$ , $\theta$ , and is a distance $r$from the origin.

  

If connect the two points, form a triangle made of this segment and the two radii. The angle between the two radii is the difference of the two angles , ${\theta }_1-{\theta }_2$ . Now, use the Law of Cosines to find the third side, $d$ . The Law of Cosines is

  $c^2=a^2+b^2-2ab\cos C$

Since two sides, $r_1$ and $r_2$ , which represents $a$ and $b$ , and $C$ , which is the angle between $a$ and $b$ , solve for solve for $c$ , the side opposite angle $C$ . Let’s plug in the values

  $\begin{array}{l}a=r_1,\\ b=r_2\\ \cos C=\cos \left({\theta }_1-{\theta }_2\right),\\ and\\ c=d\\ c^2=a^2+b^2-2ab\cos C\\ d^2=r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)\\ d=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}\end{array}$

Note that because $\cos \left(-\theta \right)=\cos \theta ,\cos \left({\theta }_1-{\theta }_2\right)$is equivalent to $\cos \left({\theta }_2-{\theta }_1\right)$ . Rewrite our expression as

  $d=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$

This is the equation asked to prove at the beginning.

Conclusion:

Hence, It is proved that the distance between the polar points $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$is

  $d=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$

(b)

To determine

To find: the distance between the points of the given polar coordinates

Answer to Problem 69E

Here, the distance between the points whose polar coordinates are $\left(3,3\pi /4\right)$ and $\left(1,7\pi /6\right)$ is $2.906$ .

Explanation of Solution

Calculation:

Find the distance between the points whose polar coordinates are $\left(3,3\pi /4\right)$ and $\left(1,7\pi /6\right)$, using the formula from part (a).

Let’s make Point $1$

  $\left(3,3\pi /4\right)$ and Point $2$

  $\left(1,7\pi /6\right)$ . Then, our values are

  $r_1=3,{\theta }_1=\frac{3\pi }{4},r_2=1,{\theta }_2=\frac{7\pi }{6}$

The formula from part (a) states that the distance between two polar points is

  $d=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}$

Plug in the assigned values

  $\begin{array}{l}d=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2-2\left(3\right)\left(1\right)\cos \left(\frac{7\pi }{6}-\frac{3\pi }{4}\right)}\\ d=\sqrt{9+1-6\cos \left(\frac{5\pi }{12}\right)}\\ d=\sqrt{10-6\cos \left(\frac{5\pi }{12}\right)}\end{array}$

It will actually be easier to find the value of the cosine if the cosine difference formula is used, which states that

  $\cos \left(s-t\right)=\cos s\cos t+\sin s\sin t$

So,

  $\begin{array}{l}\cos \left(\frac{7\pi }{6}-\frac{3\pi }{4}\right)=\cos \frac{7\pi }{6}\cos \frac{3\pi }{4}+\sin \frac{7\pi }{6}\sin \frac{3\pi }{4}\\ \cos \left(\frac{7\pi }{6}-\frac{3\pi }{4}\right)=\left(-\frac{\sqrt{3}}{2}\right)\left(-\frac{\sqrt{2}}{2}\right)+\left(-\frac{1}{2}\right)\left(-\frac{\sqrt{2}}{2}\right)\\ \cos \left(\frac{7\pi }{6}-\frac{3\pi }{4}\right)=\left(-\frac{\sqrt{6}}{4}\right)+\left(-\frac{\sqrt{2}}{4}\right)\\ \cos \left(\frac{5\pi }{12}\right)=\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{2}-\left(\sqrt{3}-1\right)}{4}\end{array}$

Place this value back into the distance formula.

  $\begin{array}{l}d=\sqrt{10-6\cos \left(\frac{5\pi }{12}\right)}\\ d=\sqrt{10-6\left(\frac{\sqrt{6}-\sqrt{2}}{4}\right)}\\ d=\sqrt{10-\frac{3\left(\sqrt{6}-\sqrt{2}\right)}{2}}\end{array}$

This is the exact value of the distance between Points $1$ and $2$ . The decimal value is about $2.906$ .

Conclusion:

Hence, the distance between the points whose polar coordinates are $\left(3,3\pi /4\right)$ and $\left(1,7\pi /6\right)$ is $2.906$ .

(c)

To determine

To convert: the points to rectangular coordinates and distance between them.

Answer to Problem 69E

Here, the decimal value is about $2.906$

Explanation of Solution

Calculation:

Convert the two points from part (b) to regular coordinates, and then find the distance between them (using the regular rectangular formula).

Let’s start with Point $1$

find the x- and y- coordinates by using two formulas relating polar and rectangular coordinates.

  $x=r\cos \theta andy=r\sin \theta$

Point $1$

  $\left(3,3\pi /4\right)$

  $\begin{array}{l}{x}_1=3\cos \frac{3\pi }{4}\\ x_1=3\left(-\frac{\sqrt{2}}{2}\right)\\ x_1=-\frac{3\sqrt{2}}{2}\\ y_1=3\sin \frac{3\pi }{4}\\ y_1=3\left(\frac{\sqrt{2}}{2}\right)\\ y_1=\frac{3\sqrt{2}}{2}\end{array}$

Point $1$ is

  $\left(-\frac{3\sqrt{2}}{2},\frac{3\sqrt{2}}{2}\right)$

In rectangular coordinates.

Point $2$

  $\left(1,7\pi /6\right)$

  $\begin{array}{l}{x}_2=1\cos \frac{7\pi }{6}\\ x_2=1\left(-\frac{\sqrt{3}}{2}\right)\\ x_2=-\frac{\sqrt{3}}{2}\\ y_2=1\sin \frac{7\pi }{6}\\ y_2=1\left(-\frac{1}{2}\right)\\ y_2=-\frac{1}{2}\end{array}$

Point $2$ is

  $\left(-\frac{\sqrt{3}}{2},-\frac{1}{2}\right)$

In rectangular coordinates.

  $d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}$

Let’s put in the coordinates found for the two points.

  $\begin{array}{l}d=\sqrt{{\left(\left(-\frac{\sqrt{3}}{2}\right)-\left(-\frac{3\sqrt{2}}{2}\right)\right)}^2+{\left(\left(-\frac{1}{2}\right)-\left(\frac{3\sqrt{2}}{2}\right)\right)}^2}\\ d=\sqrt{{\left(\frac{-\sqrt{3}+3\sqrt{2}}{2}\right)}^2+{\left(-\frac{1+3\sqrt{2}}{2}\right)}^2}\end{array}$

This expression can be further simplified, or just enter the entire expression into a calculator. If this is done, find that the decimal value is about $2.906$ . Note that this is the same value found in Part (b).

Conclusion:

Hence, the decimal value is about $2.906$