Chapter 6.4, Problem 61E

Solution

a)

To write: The formula for the distance between $P_1$ and $P_2$ .

The formula for the distance between $P_1$ and $P_2$ is $d=\left|r_1+r_2\right|$ or $d=\left|r_1-r_2\right|$ .

Given information:

The polar coordinates are $\left(r_1,{\theta }_1\right)$ and $\left(r_2,{\theta }_2\right)$ .

Formula used:

Use the law of cosines.

Calculation:

There are two cases to be considered.

1. ${\theta }_1-{\theta }_2$ is an odd integer multiple of $\pi$
2. ${\theta }_1-{\theta }_2$ is an even integer multiple of $\pi$

Also, use the use the Law of Cosines. It is also known that the odd integer multiple of $\pi$ with $\cos \theta$ is equal to $-1$ .

When ${\theta }_1-{\theta }_2$ is an odd integer multiple of $\pi$ ,

  $\begin{array}{c}d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}\\ =\sqrt{r_1^2+r_2^2-2r_1{r}_2\left(-1\right)}\\ =\sqrt{r_1^2+r_2^2+2r_1{r}_2}\\ =\sqrt{{\left(r_1+r_2\right)}^2}\end{array}$

Simplify further.

  $d=\left|r_1+r_2\right|$

It is also known that the even integer multiple of $\pi$ with $\cos \theta$ is equal to $1$ .

When ${\theta }_1-{\theta }_2$ is an even integer multiple of $\pi$ ,

  $\begin{array}{c}d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}\\ =\sqrt{r_1^2+r_2^2-2r_1{r}_2\left(1\right)}\\ =\sqrt{r_1^2+r_2^2-2r_1{r}_2}\\ =\sqrt{{\left(r_1-r_2\right)}^2}\end{array}$

Simplify further.

  $d=\left|r_1-r_2\right|$

Therefore, the formula for the distance between $P_1$ and $P_2$ is $d=\left|r_1+r_2\right|$ or $d=\left|r_1-r_2\right|$ .

b)

To prove: That the distance between $P_1$ and $P_2$ is $d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}$ .

Calculation:

Consider the following figure.

  

Find the distance $d=P_1{P}_2$ by using the Law of Cosines in the $\Delta O{P}_1{P}_2$ .

  $\begin{array}{c}{d}^2=O{P}_1^2+O{P}_2^2-2O{P}_{1}O{P}_2\cos P_{1}O{P}_2\\ d^2=r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)\\ d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_2-{\theta }_1\right)}\\ d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}\end{array}$

Therefore, it is proved that the distance between $P_1$ and $P_2$ is $d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left({\theta }_1-{\theta }_2\right)}$ .

c)

To find: Whether the formula in part (b) agrees with the formula(s) found in part (a)

Yes, the formula in part (b) agrees with the formulas found in part (a).

Calculation:

Consider the following formula.

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

Let ${\theta }_1-{\theta }_2=k\pi$ . Where, $k$ is an integer.

Apply on the above formula.

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

When $k=2n$ ,

  $\begin{array}{c}d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left(2n\pi \right)}\\ =\sqrt{r_1^2+r_2^2-2r_1{r}_2}\\ =\sqrt{{\left(r_1-r_2\right)}^2}\\ =\left|r_1-r_2\right|\end{array}$

When $k=2n+1$ ,

  $\begin{array}{c}d=\sqrt{r_1^2+r_2^2-2r_1{r}_2\cos \left(\left(2n+1\right)\pi \right)}\\ =\sqrt{r_1^2+r_2^2+2r_1{r}_2}\\ =\sqrt{{\left(r_1+r_2\right)}^2}\\ =\left|r_1+r_2\right|\end{array}$

So the formula in part (b) agrees with the formulas found in part (a).

Therefore, yes, the formula in part (b) agrees with the formulas found in part (a).