4.3 极赤道系统

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  极赤道系统

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  You likely recall that when you graph multiple equations on the same line, you usually end up with locations on the graph where they intersect (unless you are graphing parallel lines!).
  ::您可能会记得, 当您在同一行绘制多个方程式时, 您通常会在图形中找到位置, 位置相互交叉( 除非您正在绘制平行线的图形 !) ) 。

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  The same is true when graphing equations in polar form , and/or on a polar graph. When you graph the intersection of multiple polar equations, you treat them just as you would rectangular equations, graph both and find the areas that are true for both equations.
  ::当用极形和/或极图绘制方程式时,情况也是如此。当用极形图绘制多个极方程式的交叉点时,对待它们的方式与矩形方程式一样,同时绘制两个方程式的图形,并找到两个方程式都真实的区域。

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  Systems of Polar Equations
  ::极赤道系统

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  Polar equations can be graphed using . Graphing two polar equations on the same set of axes may result in having point(s) of intersection.
  ::极方程式可以使用 。在同一组轴上绘制两个极方程式的图形可能会导致交叉点的形成。

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  All points on a polar graph are coordinates that make the equation valid. The coordinates of point(s) of intersection when substituted into each equation will make both of the equations valid.
  ::极图上的所有点都是使方程式有效的坐标。将交叉点的坐标替换为每个方程式时,将使两个方程式都有效。

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  One method to find point(s) of intersection for two polar graphs is by setting the equations equal to each other.
  ::找到两个极图的交叉点的方法之一是将方程对等。

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  Call the first equation r ₁ and the second equation r ₂ .
  ::调用第一个等式 r1 和第二个等式 r2。

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  Points of intersection are when r ₁ = r ₂ , so set the equations equal and then solve the resulting trigonometric equation.
  ::交叉点是 r1 = r2 时的 r1 = r2, 所以将方程设置为等值, 然后解开由此产生的三角方程 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Find the intersection of $\begin{align*}r_1 = 3 \ \mbox{sin} \theta \end{align*}$ and $\begin{align*} r_2=\sqrt{3} \mbox{ cos}\ \theta \end{align*}$ .
  ::查找 r1 = 3 sin 和 r2 = 3 cos 的交叉点。

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  Set the equations equal to each other: $\begin{align*}3\ \mbox{sin}\ \theta =\sqrt{3}\ \mbox{cos}\ \theta \end{align*}$
  ::设定对等方程式: 3 sin *% 3 cos *

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  divide both sides of the equation by θ and 3: $\begin{align*}\frac{3\ \mbox{sin}\ \theta}{3\ \mbox{cos}\ \theta} = \frac{\sqrt{3}\ \mbox{cos}\ \theta}{3\ \mbox{cos}\ \theta}\end{align*}$
  ::将方程两侧除以 + 和 3 : 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪 3 罪

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  Simplify: $\begin{align*}\frac{\mbox{sin}\ \theta}{\mbox{cos}\ \theta} = \frac{\sqrt{3}}{3}\end{align*}$
  ::简化:罪 cos 33

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  Use the identity : $\begin{align*}\frac{\mbox{sin}\ \theta}{\mbox{cos}\ \theta} = \mbox{tan}\ \theta\end{align*}$
  ::使用身份:罪

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  $\begin{align*}\mbox{tan}\ \theta = \frac{\sqrt{3}}{3}\end{align*}$
  ::丹 丹 33

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  $\begin{align*}\theta = \frac{\pi}{6}\end{align*}$ or $\begin{align*}\frac{7 \pi}{6}\end{align*}$
  ::6 或 76

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  substitute $\begin{align*}\frac{\pi}{6}\end{align*}$ in either equation to obtain r = 1.5
  ::获得 r = 1.5 的任一方程式中的 = 6 替代 = 6 = 1.5

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  substitute $\begin{align*}\frac{7\pi}{6}\end{align*}$ in either equation to obtain -1.5
  ::在任一方程式中取用-1.5, 替换 7+6

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  NOTE: the coordinates  $\begin{align*}\left (1.5,\frac{\pi}{6}\right )\end{align*}$ and $\begin{align*}\left (-1.5,\frac{7\pi}{6}\right)\end{align*}$ represent the SAME polar point so there is only one solution to this equation.
  ::注:坐标(1.5, 6)和(-5. 7, 6)代表SAME极点,因此这一方程只有一个解决办法。

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  Are we done? If we look at the graphs of r ₁ and r ₂ , we can see that there is another point of intersection:
  ::说完了吗?如果我们看看r1和r2的图表,我们可以看到还有另一个交叉点:

  

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  when θ = 0, r ₁ = 3 θ = 3 sin(0) = 0
  ::当 = 0 = 0, r1 = 3 = = 3 = 3 sin(0) = 0

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  That means r ₁ = 3 sin θ goes through the pole (0, 0).
  ::意味着R1=3罪 穿过柱子(0,0)

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  For r ₂ : when $\begin{align*} \theta = \frac{\pi}{2}\end{align*}$ , r ₂ = 0 that is $\begin{align*}r_2 = \sqrt{3}\ \mbox{cos}\ \theta\end{align*}$ goes through the point $\begin{align*}\left (0,\frac{\pi}{2}\right )\end{align*}$ .
  ::对于 r2: 当 2, r2 = 0, 即 r2= 3 cos 穿过点 (0,%2) 时, r2 = 0 。

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  Therefore, both graphs go through the pole and the pole is a point of intersection.
  ::因此,两个图表都穿过极,而极是一个交叉点。

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  The pole was NOT revealed as a point of intersection using the first step! (Why? Hint: How many ways are there to represent the pole in polar coordinates?) This shows us that after you use algebraic methods to find intersections at points other than the pole, you should also check for intersections at the pole.
  :为什么? 提示:在极坐标上有多少方法代表极坐标? )这向我们表明,在使用代数法在极以外点找到交叉点之后,你也应该检查在极线上的交叉点。

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  Example 2
  ::例2

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  Find the point(s) of intersection for the two graphs:  r ₁ = 1 and  r ₂ = 2 sin 2 θ.
  ::查找两个图形的 r1 = 1 和 r2 = 2 sin 2 的交叉点 。

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  Set r ₁ = r ₂ and solve:
  ::设置 r1 = r2 并解析 :

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  $\begin{align*}1 = 2\ \mbox{sin}\ 2\theta\end{align*}$
  ::1=2 sin 2

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  $\begin{align*}\frac{1}{2} = \mbox{sin}\ 2\theta\end{align*}$
  ::12=sin 2

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  Use a substitution α = 2 θ to solve
  ::使用替代 α = 2 来解析

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  $\begin{align*}\frac{1}{2} = \mbox{sin}\ \alpha \end{align*}$
  ::12=sin α

  $\begin{align*}\alpha = \frac{\pi}{6}, \frac{5\pi}{6}\end{align*}$

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  Since α = 2 θ , solving for θ gives us
  ::因为α=2,解决 给我们

  $\begin{align*}\theta = \frac{\pi}{12}, \frac{5\pi}{12}\end{align*}$

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  But, recall that θ has the range 0 ≤ θ ≤ 2 π . Since we solved with 0 ≤ α ≤ 2 π we actually need to consider values of θ with 0 ≤ θ ≤ 4 π . Why? Recall that sin(2 θ ) has two cycles between 0 and 2 π , and so we add two more solutions,
  ::但是,请记住 有0 2 。 既然我们用 0 2 解决了, 我们实际上需要考虑 的值 。 为什么? 记住 , 罪有2个周期, 介于 0 和 2 之间, 因此我们增加两个解决方案,

  $\begin{align*}\alpha = \frac{13\pi}{6}, \frac{17\pi}{6}\end{align*}$

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  and since α = 2 θ ,
  ::并且自此以后,α=2,

  $\begin{align*}\theta = \frac{13\pi}{12}, \frac{17\pi}{12}\end{align*}$

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  Finally, we need to consider solutions when r = -1 because r = 1 and r = -1 are the same polar equation. So, solving
  ::最后,我们需要考虑当 r = -1 时的解决方案, 因为 r = 1 和 r = -1 是相同的极方。 所以, 解决

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  $\begin{align*}-\frac{1}{2} = \mbox{sin}\ \alpha\end{align*}$
  ::- 12=sin α

  $\begin{align*}\alpha = \frac{7\pi}{6}, \frac{11\pi}{6}\end{align*}$

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  Again, using α = 2 θ and adding solutions for the repetition gives us four more solutions,
  ::同样,使用α = 2 和添加重复的解决方案给我们提供了另外四个解决方案:

  $\begin{align*}\theta = \frac{7\pi}{12}, \frac{11\pi}{12}, \frac{19\pi}{12}, \frac{21\pi}{12}\end{align*}$

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  So in total, there are eight solutions to this set of equations.
  ::所以总共,这组方程式有八种解决办法。

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  NOTE: Recall that where the angle is θ requires looking at all potential values between 0 and 2π. When the angle is 2θ as it is in this case, be sure to look for all potential values between 0 and 4π. When the angle is 3θ as it is in this case, be sure to look for all potential values between 0 and 6π., and so on.
  ::注:当角度为 ° 时,需要查看 0 和 2 ° 之间的所有潜在值。当角度为 2 ° 时,请务必查找 0 和 4 ° 之间的所有潜在值。当角度为 3 ° 时,请务必查找 0 和 6 ° 之间的所有潜在值。等等。

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  Since r ₁ cannot equal 0, the pole is not on its graph and not a point of intersection.
  ::由于 r1 不能等于 0, 极不在图中, 而不是交叉点 。

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  The graph reveals eight points of intersection which were found earlier.
  ::该图显示了早先发现的八个交叉点。

  

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  Example 3
  ::例3

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  Find point(s) of intersection, if any exist, for the following pair of equations:  r ₁ = 2 and  r ₂ = secθ.
  ::下列方程式的交叉点(如果存在的话)的查找点:r1=2和r2=秒。

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  Here we will use a table of values for each function, solving by quadrant. Recall that the period of sec θ is 2π.
  ::在此, 我们将使用每个函数的数值表, 以象数解 。 回顾 秒 的 = 2 = 2 = 。

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  For the first quadrant:
  ::对于第一个象限 :

  θ (angle)	0	π/6	π/4	π/3	π/2
  r 1 (distance)	2	2	2	2	2
  r 2	1	1.15	1.4	2	und

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  For the second quadrant:
  ::第二个象限:

  θ (angle)	2π/3	3π/4	5π/6	π
  r 1 (distance)	2	2	2	2
  r 2	-2	-1.4	-1.15	-1

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  For the third quadrant:
  ::第三象限:

  θ (angle)	7π/6	5π/4	4π/3	3π/2
  r 1 (distance)	2	2	2	2
  r 2	-1.15	-1.4	-2	und

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  For the fourth quadrant:
  ::第四象限:

  θ (angle)	5π/3	7π/4	11π/6	2π
  r 1 (distance)	2	2	2	2
  r 2	2	1.4	1.15	1

  

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  Note that the 3 ^rd and 4 ^th quadrant repeat 1 ^st and 2 ^nd quadrant values.
  ::请注意,第三和第四象限重复第一和第二象限值。

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  Observe in the table of values that (2, π/3) and (2, 5π/3) are the points of intersection. Look at the curves- the first equation yields a circle while the second yields a line. The maximum number of intersecting points for a line and a circle is 2. The two points have been found.
  ::在数值表中观察作为交叉点的值(2, /3)和(2, /3)。看看曲线 - 第一个方程式产生一个圆,第二个方程式产生一个圆,第二个则产生一个线。一条线和一个圆的最大交叉点数是2。发现两个点。

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  Example 4
  ::例4

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  Find the point(s) of intersection for this pair of Polar equations:  r = 2 + 4 sinθ and θ = 60°.
  ::查找此对极方程的交叉点: r = 2 + 4 sin 和 = 60 °。

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  The equation θ = 60 ^o is a line making a 60° angle with the r axis.
  ::等式 = 60o 是一条以 r 轴为60 度角的线。

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  Make a table of values for r = 2 + 4 sin θ
  ::绘制 r = 2 + 4 sin 的数值表

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  For the first quadrant:
  ::对于第一个象限 :

  θ (angle)	0	30	45	60	90
  R (distance)	2	4	4.83	5.46	6

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  For the second quadrant:
  ::第二个象限:

  θ (angle)	120	135	150	180
  R (distance)	5.46	4.83	4	2

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  For the third quadrant:
  ::第三象限:

  θ (angle)	210	225	240	270
  R (distance)	0	-.83	-1.46	-2

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  For the fourth quadrant:
  ::第四象限:

  θ (angle)	300	315	330	360
  R (distance)	-1.46	-.83	0	2

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  Notice there are two solutions in the table., (60, 5.46) and (240, -1.46) = (60, 1.46). Recall that when r < 0, you plot a point ( r , θ ), by rotating 180 ^o ( or π ).
  ::表格中有两种解决办法。 (60, 5.46) 和 (240,-1.46) = (60, 1.46) = (60, 1.46) 。 请注意,当 r < 0 时,您通过旋转 180o (或 ) 绘制一个点(r, ) 。

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  Finally, we need to check the pole: r = 2 + 4 sin θ passes through the pole for θ = 330 ^o , and θ = 60 ^o also passes through the pole. Thus, the third point of intersection is (0, 0).
  ::最后,我们需要检查杆:r = 2 + 4 sin = = 330o, r = 2 + 4 sin = 通过杆 = 330o, = 60o 也通过杆。 因此,第三个交叉点是 (0 0) 。

  

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  Example 5
  ::例5

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  Find the point(s) of intersection for this pair of Polar equations:  r ₁ = 2 cosθ and  r ₂ = 1.
  ::查找此对极方程的交叉点: r1 = 2 cos 和 r2 = 1。

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  Make a table:
  ::绘制表格:

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  For the first quadrant:
  ::对于第一个象限 :

  θ (angle)	0	π/6	π/4	π/3	π/2
  r 1 (distance)	2	$\begin{align*}\sqrt{3}\end{align*}$	$\begin{align*}\sqrt{2}\end{align*}$	1	0
  r 2	1	1	1	1	1

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  For the second quadrant:
  ::第二个象限:

  θ (angle)	2π/3	3π/4	5π/6	π
  r 1 (distance)	-1	$\begin{align*}-\sqrt{2}\end{align*}$	$\begin{align*}-\sqrt{3}\end{align*}$	-2
  r 2	1	1	1	1

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  For the third quadrant:
  ::第三象限:

  θ (angle)	7π/6	5π/4	4π/3	3π/2
  r 1 (distance)	$\begin{align*}-\sqrt{3}\end{align*}$	$\begin{align*}-\sqrt{2}\end{align*}$	-1	0
  r 2	1	1	1	1

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  For the fourth quadrant:
  ::第四象限:

  θ (angle)	5π/3	7π/4	11π/6	2π
  r 1 (distance)	1	$\begin{align*}\sqrt{2}\end{align*}$	$\begin{align*}\sqrt{3}\end{align*}$	2
  r 2	1	1	1	1

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  So the unique solutions are at $\begin{align*}\theta = \frac{\pi}{3} , \frac{4\pi}{3}\end{align*}$ . There are also two repeated solutions in this set (can you find them?).
  ::唯一的解决方案是 3,4,4,3,3,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2

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  Here is a graph showing the two solutions:
  ::以下是一个图表,显示两种解决办法:

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  Review
  ::回顾

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  The graphs of $\begin{align*}r = 1\end{align*}$ and $\begin{align*}r = 2 cos \theta\end{align*}$ are shown below.
  ::r=1和r=2cos的图示如下。

  

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  1. How many times do they intersect?
     ::它们交叉了多少次?
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  2. In which quadrants do they intersect?
     ::在哪些四分法中,它们相互交错?
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  3. At what points do the intersections occur?
     ::十字路口在哪里发生?

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  Based on the image below and the following information: the intersection of the graphs of $\begin{align*}r = cos \theta\end{align*}$ and $\begin{align*}r = 1 - cos \theta\end{align*}$
  ::基于以下图像和以下信息:r=cos和r=1-cos的图形交叉点。

  

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  4. Identify how many times they intersect?
     ::确定它们交叉了多少次?
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  5. At what points do the intersections occur?
     ::十字路口在哪里发生?

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  Find the points of intersection of the following pairs of curves.
  ::查找以下两条曲线的交叉点。

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  6. $\begin{align*}r = 2\end{align*}$ $\begin{align*}r = 2 cos \theta\end{align*}$
     ::r=2r=2cos
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  7. $\begin{align*}r = sin 2 \theta\end{align*}$ $\begin{align*} r =2 sin \theta\end{align*}$
     ::r=sin2r=2sin
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  8. $\begin{align*}r = 2 + 2 sin \theta\end{align*}$ $\begin{align*} r = 2 - 2 cos \theta\end{align*}$
     ::r=2+2sinr=2-2cos
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  9. $\begin{align*}r = 3 cos \theta\end{align*}$ $\begin{align*} r = 2 - cos \theta\end{align*}$
     ::r=3cosr=2-cos

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  Find the point(s) of intersection for each system of equations. Graph to verify your solution.
  ::查找每个方程式系统中的交叉点。 用于校验您的解决方案的图表 。

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  10. $\begin{align*}r_1 = csc \theta\end{align*}$ $\begin{align*}r_2 = 2 sin \theta\end{align*}$
      ::r1=cscr2=2sin
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  11. $\begin{align*}r_1 = cos \theta\end{align*}$ $\begin{align*}r_2 = 1 + sin \theta\end{align*}$
      ::r1=cosr2=1+sin
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  12. $\begin{align*}r_1 = sin \theta\end{align*}$ $\begin{align*}r_2 = sin 2 \theta\end{align*}$
      ::r1=sinr2=sin2
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  13. $\begin{align*}r_1 = -4 sin \theta\end{align*}$ $\begin{align*}r_2 = -4 cos \theta\end{align*}$
      ::-=YTET -伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=-伊甸园字幕组=- 翻译:

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  Review (Answers)
  ::回顾(答复)

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  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 。