Chapter 9.5, Problem 83E

a.

To determine

To prove: The distance between the points $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$ is $\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}$

Explanation of Solution

Given information:

The distance between the points $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$ is $\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}$

The point $\left(r_1,{\theta }_1\right)$ represents vector $\overrightarrow{r_1}$ with magnitude $r_1$ .

Point $\left(r_2,{\theta }_2\right)$ represents vector $\overrightarrow{r_2}$ with magnitude $r_2$ .

Therefore,

The angle between two vectors is ${\theta }_2-{\theta }_1$

The distance between two points is the magnitude of the displacement vector $\overrightarrow{r_1}-\overrightarrow{r_2}$

  $\Rightarrow$ Distance $=\left|\overrightarrow{r_1}-\overrightarrow{r_2}\right|$

  $\Rightarrow$ Distance $=\sqrt{{\left|\overrightarrow{r_1}\right|}^2+{\left|\overrightarrow{r_2}\right|}^2-2\left|\overrightarrow{r_1}\right|\left|\overrightarrow{r_2}\right|\cos \left({\theta }_2-{\theta }_1\right)}$

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

Hence proved.

b.

To determine

To find: Distance formula for ${\theta }_1={\theta }_2$ .

Answer to Problem 83E

Distance $=\left|r_1-r_2\right|$

Here $\left|\right|$ represents absolute value of distance.

Explanation of Solution

Given information:

The distance between the points $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$ is $\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}$

  ${\theta }_1={\theta }_2$

  ${\theta }_1={\theta }_2$

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

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

  $\Rightarrow$ Distance $=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left(0\right)}$

  $\Rightarrow$ Distance $=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2}$

  $\Rightarrow$ Distance $=\sqrt{{\left(r_1-r_2\right)}^2}$

  $\Rightarrow$ Distance $=\left|r_1-r_2\right|$ (as distance cannot be negative)

Yes,

Since ${\theta }_1={\theta }_2$ Both points lie on Same line.

Therefore, the distance between them is difference of magnitudes of both vectors.

i.e Distance $=\left|\overrightarrow{r_1}\right|-\left|\overrightarrow{r_2}\right|$

  $\Rightarrow$ Distance $=\left|r_1-r_2\right|$

c.

To determine

To find: Distance formula for ${\theta }_1={\theta }_2$ .

Answer to Problem 83E

Distance $=\sqrt{r_1{}^2+r_2{}^2}$

Explanation of Solution

Given information:

The distance between the points $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$ is $\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}$

  ${\theta }_1-{\theta }_2={90}^0$

  ${\theta }_1-{\theta }_2={90}^0$

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

  $\Rightarrow$ Distance $=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\cos \left({90}^0\right)}$

  $\Rightarrow$ Distance $=\sqrt{r_1{}^2+r_2{}^2-2r_1{r}_2\left(0\right)}$

  $\Rightarrow$ Distance $=\sqrt{r_1{}^2+r_2{}^2-0}$

  $\Rightarrow$ Distance $=\sqrt{r_1{}^2+r_2{}^2}$

The points $\left(0,0\right)$ , $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$ indicates the vertices of a right angled triangle.

Therefore, the distance is the length of Hypotenuse of the triangle formed.

Therefore,

  $\Rightarrow$ Distance $=$ Hypotenuse $=\sqrt{r_1{}^2+r_2{}^2}$