Course: 6.2 两个极地坐标之间的距离

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两个极坐标间的距离
When playing a game of darts with your friend, the darts you throw land in a pattern like this
::和朋友玩飞镖游戏的时候你扔的飞镖像这样的图案

You and your friend decide to find out how far it is between the two darts you threw. If you know the positions of each of the darts in , can you somehow find a formula to let you determine the distance between the two darts?
::你和你的朋友决定找出你扔的两只飞镖之间的距离。如果你知道其中每只飞镖的位置,你能找到一个公式来决定两只飞镖之间的距离吗?

Finding the Distance Between Two Polar Coordinates
::查找两个极坐标间的距离

Just like the Distance Formula for $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ coordinates, there is a way to find the distance between two polar coordinates. One way that we know how to find distance, or length, is the , $\begin{aligned} a^{2} = b^{2} + c^{2} - 2 b c \cos A \end{aligned}$ or $\begin{aligned} a = \sqrt{b^{2} + c^{2} - 2 b c \cos A} \end{aligned}$ . If we have two points $\begin{aligned} (r_{1}, θ_{1}) \end{aligned}$ and $\begin{aligned} (r_{2}, θ_{2}) \end{aligned}$ , we can easily substitute $\begin{aligned} r_{1} \end{aligned}$ for $\begin{aligned} b \end{aligned}$ and $\begin{aligned} r_{2} \end{aligned}$ for $\begin{aligned} c \end{aligned}$ . As for $\begin{aligned} A \end{aligned}$ , it needs to be the angle between the two radii, or $\begin{aligned} (θ_{2} - θ_{1}) \end{aligned}$ . Finally, $\begin{aligned} a \end{aligned}$ is now distance and you have $\begin{aligned} d = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{2} - θ_{1})} \end{aligned}$ .
::就像x 和 y 坐标的距离公式一样, 能找到两个极坐标之间的距离。我们知道如何找到距离或长度的一种方式是, a2=b2+c2-2bccosA 或 a=b2+c2-2bc2-2bccosA。如果我们有两个点( r1, 1) 和 (r2, 2), 我们可以很容易地将 r1 替换为 b 和 r2 。至于 A, 它必须是两个 radi 之间的角, 或 (21) 。最后, 一个是距离, 并且您有 d= 12+r22-2r2r2c2cos * ( _2) 。

Finding the Distance
::寻找距离

1. Find the distance between $\begin{aligned} (3, 60^{\circ}) \end{aligned}$ and $\begin{aligned} (5, 145^{\circ}) \end{aligned}$ .
::1. 找出(3,60)和(5,145)之间的距离。

After graphing these two points, we have a triangle. Using the new Polar Distance Formula, we have $\begin{aligned} d = \sqrt{3^{2} + 5^{2} - 2 (3) (5) \cos 85^{\circ}} \approx 5.6 \end{aligned}$ .
::在绘制这两个点的图表后,我们有一个三角形。使用新的极距公式,我们有 d=32+52-2(3)(5)cos855.6。

2. Find the distance between $\begin{aligned} (9, - 45^{\circ}) \end{aligned}$ and $\begin{aligned} (- 4, 70^{\circ}) \end{aligned}$ .
::2. 找出(9,-45)和(4,70)之间的距离。

This one is a little trickier than the last #1 because we have negatives. The first point would be plotted in the fourth quadrant and is equivalent to $\begin{aligned} (9, 315^{\circ}) \end{aligned}$ . The second point would be $\begin{aligned} (4, 70^{\circ}) \end{aligned}$ reflected across the pole, or $\begin{aligned} (4, 250^{\circ}) \end{aligned}$ . Use these two values of $\begin{aligned} θ \end{aligned}$ for the formula. Also, the radii should always be positive when put into the formula. That being said, the distance is $\begin{aligned} d = \sqrt{9^{2} + 4^{2} - 2 (9) (4) \cos (315 - 250)^{\circ}} \approx 8.16 \end{aligned}$ .
::这比上一个 #1小一点, 因为我们有负值。第一个点将在第四象限中绘制, 相当于( 9, 315) 。第二个点是( 4, 70) , 或( 4, 250) 。对公式使用这两个值。另外, 在公式中插入 radi 时, 弧度应该始终是正值。也就是说, 距离是 d= 92+42- 2( 9) (4)cos ( 315- 250) 8. 16 。

3. Find the distance between $\begin{aligned} (2, 10^{\circ}) \end{aligned}$ and $\begin{aligned} (7, 10^{\circ}) \end{aligned}$ .
::3. 找出(2,10)和(7,10)之间的距离。

This problem is straightforward from looking at the relationship between the points. The two points lie at the same angle, so the straight line distance between them is $\begin{aligned} 7 - 2 = 5 \end{aligned}$ . However, we can confirm this using the distance formula:
::这个问题从观察两点之间的关系可以直截了当。两点位于同一角度, 因此两点之间的直线距离是 7 - 2= 5。然而, 我们可以用距离公式来确认这一点 :

$\begin{aligned} d = \sqrt{2^{2} + 7^{2} - 2 (2) (7) \cos 0^{\circ}} = \sqrt{4 + 49 - 28} = \sqrt{25} = 5 \end{aligned}$ .
::d=22+72-2(2)(7)(7)cos04+49-28=25=5。

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the distance between the two darts.
::早些时候,你被要求找到两只飞镖之间的距离

Using the Distance Formula for points in a polar plot, it is possible to determine the distance between the 2 darts:
::使用距离公式表示极地图中的点数,可以确定两只飞镖之间的距离:

$\begin{aligned} d = \sqrt{3^{2} + 6^{2} - 2 (4) (6) \cos 45^{\circ}} \\ d = \sqrt{9 + 36 - 48 (.707)} \\ d = \sqrt{45 - 33.936} \\ d \approx 11.064 \end{aligned}$

::d=32+62-2(4)(6)cos45d=9+36-48(707d=45-33.936d11.064)

Example 2
::例2

Given $\begin{aligned} P_{1} \end{aligned}$ and $\begin{aligned} P_{2} \end{aligned}$ , calculate the distance between the points.
::给定 P1 和 P2, 计算点之间的距离。

$\begin{aligned} P_{1} (1, 30^{\circ}) \end{aligned}$ and $\begin{aligned} P_{2} (6, 135^{\circ}) \end{aligned}$
::P1(1,30)和P2(6,135)

Use $\begin{aligned} P_{1} P_{2} = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{2} - θ_{1})} \end{aligned}$ .
::使用 P1P2 = r12+r22-2r1r2cos(21)。

$\begin{aligned} P_{1} P_{2} & = \sqrt{1^{2} + 6^{2} - 2 (1) (6) \cos (135^{\circ} - 30^{\circ})} \\ P_{1} P_{2} & \approx 6.33 u n i t s \end{aligned}$

::P1P2=12+62-2(1)(6)cos(13530)P1P26.33单位

Example 3
::例3

Given $\begin{aligned} P_{1} \end{aligned}$ and $\begin{aligned} P_{2} \end{aligned}$ , calculate the distance between the points.
::给定 P1 和 P2, 计算点之间的距离。

$\begin{aligned} P_{1} (2, - 65^{\circ}) \end{aligned}$ and $\begin{aligned} P_{2} (9, 85^{\circ}) \end{aligned}$
::P1(2,-65)和P2(,9,85)

Use $\begin{aligned} P_{1} P_{2} = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{2} - θ_{1})} \end{aligned}$ .
::使用 P1P2 = r12+r22-2r1r2cos(21)。

$\begin{aligned} P_{1} P_{2} & = \sqrt{2^{2} + 9^{2} - 2 (2) (9) \cos 150^{\circ}} \\ = 10.78 \end{aligned}$

::P1P2=22+92-2(2)(9)cos 15010.78

Example 4
::例4

Given $\begin{aligned} P_{1} \end{aligned}$ and $\begin{aligned} P_{2} \end{aligned}$ , calculate the distance between the points.
::给定 P1 和 P2, 计算点之间的距离。

$\begin{aligned} P_{1} (- 3, 142^{\circ}) \end{aligned}$ and $\begin{aligned} P_{2} (10, - 88^{\circ}) \end{aligned}$
::P1(3,142)和P2(2,10,-88)

3. Use $\begin{aligned} P_{1} P_{2} = \sqrt{r_{1}^{2} + r_{2}^{2} - 2 r_{1} r_{2} \cos (θ_{2} - θ_{1})} \end{aligned}$ .
::3. 使用P1P2=r12+r22-2r1r2cos(21)。

$\begin{aligned} P_{1} P_{2} & = \sqrt{3^{2} + 10^{2} - 2 (3) (10) \cos (322 - 272)^{\circ}} \\ = 8.39 \end{aligned}$

::P1P2=32+102-2(3)(10)cos(322-272)8.39

Review
::回顾

Find the distance between each set of points. Round your answer to the nearest decimal if needed.
::查找每组点之间的距离。如果需要, 将您的答复舍入到最接近的小数点。

$\begin{aligned} (1, 150^{\circ}) \end{aligned}$ and $\begin{aligned} (2, 130^{\circ}) \end{aligned}$
:1,150)和(2,130)

$\begin{aligned} (4, 90^{\circ}) \end{aligned}$ and $\begin{aligned} (5, 210^{\circ}) \end{aligned}$
:4,90)和(5,210)

$\begin{aligned} (6, 60^{\circ}) \end{aligned}$ and $\begin{aligned} (2, 90^{\circ}) \end{aligned}$
:6,60)和(2,90)

$\begin{aligned} (2, 120^{\circ}) \end{aligned}$ and $\begin{aligned} (1, 150^{\circ}) \end{aligned}$
:2,120)和(1,150)

$\begin{aligned} (7, 210^{\circ}) \end{aligned}$ and $\begin{aligned} (4, 300^{\circ}) \end{aligned}$
:7,210)和(4,300)

$\begin{aligned} (- 4, 120^{\circ}) \end{aligned}$ and $\begin{aligned} (2, 100^{\circ}) \end{aligned}$
:4,120)和(2,100)

$\begin{aligned} (3, - 90^{\circ}) \end{aligned}$ and $\begin{aligned} (5, 150^{\circ}) \end{aligned}$
:3,-90)和(5,150)

$\begin{aligned} (- 4, - 30^{\circ}) \end{aligned}$ and $\begin{aligned} (3, 250^{\circ}) \end{aligned}$
:4,-30)和(3,250)

$\begin{aligned} (7, - 150^{\circ}) \end{aligned}$ and $\begin{aligned} (4, 130^{\circ}) \end{aligned}$
:7,-150)和(4,130)

$\begin{aligned} (- 2, 300^{\circ}) \end{aligned}$ and $\begin{aligned} (2, 10^{\circ}) \end{aligned}$
:-2,300)和(2,10)

For the following questions keep your answers in terms of $\begin{aligned} π . \end{aligned}$
::对于下列问题,请将答案保留在____________________。

Find the length of the arc between the points $\begin{aligned} (3, 40^{\circ}) \end{aligned}$ and $\begin{aligned} (3, 150^{\circ}) \end{aligned}$ .
::查找点(3,40)和点(3,150)之间的弧长度。

Find the length of the arc between the points $\begin{aligned} (1, 10^{\circ}) \end{aligned}$ and $\begin{aligned} (1, 70^{\circ}) \end{aligned}$ .
::查找点(1,10)和点(1,70)之间的弧长度。

Find the area of the sector created by the origin and the points $\begin{aligned} (4, 20^{\circ}) \end{aligned}$ and $\begin{aligned} (4, 110^{\circ}) \end{aligned}$ .
::查找产地和点(4,20)和点(4,110)创造的部门面积。

Find the area of the sector created by the origin and the points $\begin{aligned} (2, 100^{\circ}) \end{aligned}$ and $\begin{aligned} (2, 180^{\circ}) \end{aligned}$ .
::查找产地和点(2,100)和点(2,180)创造的部门面积。

Find the area of the sector created by the origin and the points $\begin{aligned} (5, 110^{\circ}) \end{aligned}$ and $\begin{aligned} (5, 160^{\circ}) \end{aligned}$ .
::查找产地和点(5,110)和点(5,160)创造的部门面积。

Review (Answers)
::回顾(答复)

Click to see the answer key or go to the Table of Contents and click on the Answer Key under the 'Other Versions' option.
::单击可查看答题键, 或转到目录中, 单击“ 其他版本” 选项下的答题键。

Section outline

General

两个极坐标间的距离

Finding the Distance Between Two Polar Coordinates ::查找两个极坐标间的距离

Finding the Distance ::寻找距离

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)

Finding the Distance Between Two Polar Coordinates
::查找两个极坐标间的距离

Finding the Distance
::寻找距离

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Review
::回顾

Review (Answers)
::回顾(答复)