Course: 12.11 参数反数 | The Way To Learn

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参数反函数
Introduction
::导言

A graph and its inverse are reflections of each other across the line $\begin{aligned} y = x . \end{aligned}$
::图表及其反射是横跨y=x线的反射。

To find an inverse algebraically, you can substitute $\begin{aligned} x \end{aligned}$ for $\begin{aligned} y \end{aligned}$ and then solve for $\begin{aligned} y . \end{aligned}$ are based on a 3rd variable, so we need to explore how to find the inverse of parametric equations.
::要找到反代数, 您可以用 x 代替 y, 然后用 y. 解决问题, 以第三个变量为基础, 所以我们需要探索如何找到对应方程的反义。

Is the inverse of a function always a function?
::函数的反向总是函数吗 ?

Parametric Inverses
::参数反函数

To find the inverse of a parametric equation , you must switch the function for $\begin{aligned} x \end{aligned}$ with the function for $\begin{aligned} y . \end{aligned}$ This will switch all the points from $\begin{aligned} (x, y) \end{aligned}$ to $\begin{aligned} (y, x), \end{aligned}$ and also have the effect of visually reflecting the graph over the line $\begin{aligned} y = x . \end{aligned}$
::要找到参数方程的反向, 您必须用 y 函数切换 x 的函数。这将将所有点从 (x, y) 切换到 (y, x) , 并具有在 y =x 线上直观反映图形的效果。

Similar to the inverses of regular functions, the inverses of parametric equations are often restricted so they are also functions. Take the following parametric equations:
::与正常函数的反差相似,参数方程的反差往往受到限制,因此它们也是函数。

$\begin{aligned} x & = 2 t \\ y & = t^{2} - 4 \end{aligned}$

::x=2ty=t2-4

To find and graph the inverse of the parametric function on the domain $\begin{aligned} - 2 < t < 2, \end{aligned}$ first switch the $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ functions and graph.
::要查找和图形显示域- 2 < t < 2, 先切换 x 和 y 函数和图形上的参数函数的反向。

$\begin{aligned} x & = t^{2} - 4 \\ y & = 2 t \end{aligned}$

::x=t2 - 4y=2t

The original function is shown in blue and the inverse is shown in red.
::原函数以蓝色显示,反面以红色显示。

Play, Learn, and Explore Parametric Inverses:
::播放、学习和探索参数反差:

The following video shows how to graph parametric equations using the TI-84:
::以下视频展示了如何用TI-84来绘制参数方程图:

In Desmos, graphing a parametric equation is as easy as plotting an ordered pair. Instead of numerical coordinates, you'll need to use expressions in terms of t, such as $\begin{aligned} (\cos t, \sin t) . \end{aligned}$
::在 Desmos 中, 绘制参数方程式的图解和绘制定购对方图一样容易。您需要使用 t 的表达式, 如( cost, sint) , 而不是数字坐标。

The following video demonstrates how to graph parametric equations using Desmos:
::以下影片展示了如何用德摩斯绘制参数方程图:

Examples
::实例

Example 1
::例1

Does the coordinate point (4, 8) satisfy the following function or its inverse?
::坐标点(4,8)是否满足下列函数或其反函数?

$\begin{aligned} x & = 2 t^{2} - 2 \\ y & = t^{2} - 1 \end{aligned}$

::x=2t2--2y=t2-2-1

Solution:
::解决方案 :

Substitute the point into the original function and solve for $\begin{aligned} t \end{aligned}$ .
$\begin{aligned} x & = 2 t^{2} - 2 and & y = t^{2} - 1 \\ 4 & = 2 t^{2} - 2 & 8 = t^{2} - 1 \\ 6 & = 2 t^{2} & 9 = t^{2} \\ 3 & = t^{2} & \pm 3 = t \\ \pm \sqrt{3} & = t \end{aligned}$

::将点替换为 t.x=2t2-2-2andy=t2-14=2t2-2-28=t2-16=2t29=t23=t2&3=t3=t

Since the value solved for $\begin{aligned} t \end{aligned}$ differs, the point does not satisfy the original function.
::由于 t 的解析值不同,该点不符合原始函数。

Substitute the point into the inverse and solve for $\begin{aligned} t \end{aligned}$ .
$\begin{aligned} x & = t^{2} - 1 and & y = 2 t^{2} - 2 \\ 4 & = t^{2} - 1 & 8 = 2 t^{2} - 2 \\ 5 & = t^{2} & 10 = 2 t^{2} \\ \pm \sqrt{5} & = t & 5 = t^{2} \\ \pm \sqrt{5} = t \end{aligned}$
Since the value solved for $\begin{aligned} t \end{aligned}$ is the same, the point satisfies the inverse function.
::将 t. x=t2 - 1andy=2t2 - 24=t2 - 18=2t2 - 25=t210=2t2&2=5=t5=t2+5=t2=t2=t2+5=tSint t的倒数和解算点替换为 t. x=t2 - 1andy=2t2 - 24=t2 - 18=2 - 2t2=2- 25=t2=210=2t2t2&5=t5=t2+5=tSince t的解算值是相同的, 点满足了反函数。

Example 2
::例2

Parameterize the following function and then graph the function and its inverse:
::参数化以下函数,然后绘制函数及其反向图示:

$\begin{aligned} y = x^{2} + x - 4 \end{aligned}$

::y=x2+x- 4 y=x2+x- 4

Solution:
::解决方案 :

For the original function, the parameterization is
::对于原始函数,参数化是

$\begin{aligned} x & = t \\ y & = t^{2} + t - 4. \end{aligned}$

::x=ty=t2+t-4。

The inverse is
::倒数是

$\begin{aligned} x & = t^{2} + t - 4 \\ y & = t . \end{aligned}$

::x=t2+t-4y=t.

Example 3
::例3

Find the points of intersection of the function and its inverse from Example 2.
::从例2中查找函数的交叉点及其反向。

Solution:
::解决方案 :

The parameterized function is
$\begin{aligned} x_{1} & = t \\ y_{1} & = t^{2} + t - 4. \end{aligned}$
The inverse is
::参数化函数为 x1=ty1=t2+t-4。逆值为

$\begin{aligned} x_{2} & = t^{2} + t - 4 \\ y_{2} & = t . \end{aligned}$

::x2=t2+t-4y2=t.

An intersection for two sets of parametric equations happens when the points exist at the same $\begin{aligned} x, y \end{aligned}$ , and $\begin{aligned} t . \end{aligned}$ To find where these intersect, set $\begin{aligned} x_{1} = x_{2} \end{aligned}$ and $\begin{aligned} y_{1} = y_{2} \end{aligned}$ and solve.
::两组参数方程式的交叉点发生于点位于相同的 x、y 和 t。要找到这些交叉点的位置, 请设置 x1=x2 和 y1=y2 并解析。

$\begin{aligned} t & = t^{2} + t - 4 \\ t^{2} & = 4 \\ t & = \pm 2 \end{aligned}$

::t = t2+t- 4t2=4t2

You can tell from the graph in Example 2 that there seem to be four points of intersection. Since $\begin{aligned} t \end{aligned}$ can mean time, the question of intersection is more complicated than simply overlapping. It means that the points are at the same $\begin{aligned} x - \end{aligned}$ and $\begin{aligned} y - \end{aligned}$ coordinate at the same time. Note what the graphs look like when $\begin{aligned} - 1.8 < t < 1.8. \end{aligned}$
::您可以从例2中的图表中看到,似乎有四个十字路口点。由于时间不代表时间,十字路口问题比简单的重叠复杂得多。这意味着各点同时处于相同的 x - 和 y - 坐标。请注意当- 1. 8 < 1. 8 < 1. 8 时这些图表的外观。

Note what the graphs look like when $\begin{aligned} t > 2.2 \end{aligned}$ or $\begin{aligned} t < - 2.2 \end{aligned}$ .
::注意图在 t>2.2 或 t2.2 时看起来像什么。

Notice how when these partial graphs are examined, there is no intersection at anything besides $\begin{aligned} t = \pm 2 \end{aligned}$ and the points (2, 2) and (-2, -2). While the paths of the graphs intersect in four places, they intersect at the same time only twice.
::当检查这些部分图表时, 注意除了 t2 和点(2, 2) 和点(2, 2) 和点(2, 2) 外, 没有其他任何处没有交叉点, 虽然图形的路径在四个地方交叉, 但同时它们只交叉两次。

Example 4
::例4

Recall the question from the Introduction: Is the inverse of a function always a function?
::回顾导言中的问题:函数的反向总是函数吗?

Solution:
::解决方案 :

The inverse of a function is not always a function. To see whether the inverse of a function will be a function, you must perform the horizontal line test on the original function. If the function passes the horizontal line test, then the inverse will be a function. If the function does not pass the horizontal line test, then the inverse produces a relation rather than a function. To find the inverse of a parametric equation, switch the $\begin{aligned} x \end{aligned}$ function with the $\begin{aligned} y \end{aligned}$ function.
::函数的反向不一定总是函数。要查看函数的反向是否是一个函数,您必须在原始函数上进行水平线测试。如果函数通过水平线测试,则反向为函数。如果函数没有通过水平线测试,则反向产生关系而不是函数。要找到参数方程的反向,请用 Y 函数切换 x 函数。

Example 5
::例5

Does the coordinate point (-2, 6) satisfy the following function or its inverse?

$\begin{aligned} x & = t^{2} - 10 \\ y & = \frac{t}{2} - 4 \end{aligned}$

::坐标点 (-2, 6) 是否满足以下函数或其反函数? x=t2 - - 10y=t2 - 4

Solution:
::解决方案 :

Substitute the point into the original function and solve for $\begin{aligned} t \end{aligned}$ .
$\begin{aligned} - 2 & = t^{2} - 10 and & 6 = \frac{t}{2} - 4 \\ 8 & = t^{2} & 10 = \frac{t}{2} \\ \pm 2 \sqrt{2} & = t & 20 = t \end{aligned}$

::将点替换为 t.-2=t2=t2-10和6=t2-48=t210=t222=t20=t

Since the value solved for $\begin{aligned} t \end{aligned}$ differs, the point does not satisfy the original function.
::由于 t 的解析值不同,该点不符合原始函数。

Substitute the point into the inverse and solve for $\begin{aligned} t \end{aligned}$ .
$\begin{aligned} 6 & = t^{2} - 10 and & - 2 = \frac{t}{2} - 6 \\ 16 & = t^{2} & 4 = \frac{t}{2} \\ \pm 4 & = t & 8 = t \end{aligned}$
Since the value solved for $\begin{aligned} t \end{aligned}$ differs, the point does not satisfy the inverse function either.
::将点替换为 t. 6= t2 - 10 和 - 2= t2 - 616= t24= t2+4= t8= t 由于 t. 的解析值不同, 该点也不符合反函数。

Example 6
::例6

Identify where the following parametric function intersects with its inverse.
::确定下列参数函数与其反向相交之处。

$\begin{aligned} x = 4 t \end{aligned}$
::x=4tx=4吨

$\begin{aligned} y = t^{2} - 16 \end{aligned}$
::y=t2-16

Solution:
::解决方案 :

The inverse is
$\begin{aligned} x_{2} & = t^{2} - 16 \\ y_{2} & = 4 t . \end{aligned}$
Solve for $\begin{aligned} t \end{aligned}$ when $\begin{aligned} x_{1} = x_{2} \end{aligned}$ and $\begin{aligned} y_{1} = y_{2} . \end{aligned}$
::x2 = t2 - 16y2 = 4t. solve for t 当 x1=x2 和 y1=y2 时, 倒数为 x2 = 16y2 = 4t. solve for t。

$\begin{aligned} 4 t & = t^{2} - 16 \\ 0 & = t^{2} - 4 t - 16 \\ t & = \frac{4 \pm \sqrt{16 - 4 \cdot 1 \cdot (- 16)}}{2} = \frac{4 \pm \sqrt{80}}{2} = \frac{4 \pm 4 \sqrt{5}}{2} = 2 \pm 2 \sqrt{5} \end{aligned}$

::4t=t2-160=t2-4t-16t=416-41(-16)2=4802=4452=225

The points that correspond to these two times are
::与这两次对应的点数是

$\begin{aligned} x & = 4 (2 + 2 \sqrt{5}), y = (2 + 2 \sqrt{5})^{2} - 16 \\ x & = 4 (2 - 2 \sqrt{5}), y = (2 - 2 \sqrt{5})^{2} - 16. \end{aligned}$

::x=4(2+25),y=(2+25),y=(2+25),y=16x=4(2 -25),y=(2 - 25),y=16。

Summary
::摘要

Two functions are inverses if for every point $\begin{aligned} (a, b) \end{aligned}$ on the 1st function, there exists a point $\begin{aligned} (b, a) \end{aligned}$ on the 2nd function.
::如果对于第一个函数的每一个点(a,b),在第二个函数上存在一个点(b,a),则两个函数是反向的。

An intersection for two sets of parametric equations happens when the points exist at the same $\begin{aligned} x, y, \end{aligned}$ and $\begin{aligned} t . \end{aligned}$
::两组参数方程式的交叉点发生于点点位于相同的x、y和t。

Review
::回顾

Use the function $\begin{aligned} x = t - 4; y = t^{2} + 2 \end{aligned}$ for numbers 1-3 below.
::使用函数 x=t- 4; y=t2+2, 用于以下数字 1-3 。

1. Find the inverse of the function.
::1. 查找函数的反向。

2. Does the coordinate point (-2, 6) satisfy the function or its inverse?
::2. 协调点(-2,6)是否满足该功能或其反向?

3. Does the coordinate point (0, 1) satisfy the function or its inverse?
::3. 坐标点(0,1)是否满足该函数或其反向?

Use the relation $\begin{aligned} x = t^{2}; y = 4 - t \end{aligned}$ for 4-6 below.
::使用关系 x=t2;y=4-t 用于以下 4-6 。

4. Find the inverse of the relation.
::4. 找出这种关系的反面。

5. Does the coordinate point (4, 0) satisfy the relation or its inverse?
::5. 坐标点(4,0)是否满足关系或其反差?

6. Does the coordinate point (0, 4) satisfy the relation or its inverse?
::6. 坐标点(0, 4)是否满足关系或其反差?

Use the function $\begin{aligned} x = 2 t + 1; y = t^{2} - 3 \end{aligned}$ for 7-9.
::7- 9 使用函数 x=2t+1; y=t2- 3。

7. Find the inverse of the function.
::7. 查找函数的反向。

8. Does the coordinate point (1, 5) satisfy the function or its inverse?
::8. 协调点(1,5)是否满足该功能或其反向?

9. Does the coordinate point (9, 13) satisfy the function or its inverse?
::9. 坐标点(9,13)是否满足该函数或其反差?

Use the function $\begin{aligned} x = 3 t + 14; y = t^{2} - 2 t \end{aligned}$ for 10-11.
::10- 11 时使用函数 x=3t+14;y=t2-2t。

10. Find the inverse of the function.
::10. 查找函数的反向。

11. Identify where the parametric function intersects with its inverse.
::11. 查明参数函数与其反相交错之处。

Use the relation $\begin{aligned} x = t^{2}; y = 4 t - 4 \end{aligned}$ for 12-13.
::在 12- 13 中使用关系 x= t2;y= 4t- 4。

12. Find the inverse of the relation.
::12. 找出相互关系的反面。

13. Identify where the relation intersects with its inverse.
::13. 查明关系与其反向交错之处。

14. Parameterize $\begin{aligned} f (x) = x^{2} + x - 6 \end{aligned}$ and then graph the function and its inverse.
::14. 参数f(x)=x2+x-6,然后绘制函数及其反向图。

15. Parameterize $\begin{aligned} f (x) = x^{2} + 3 x + 2 \end{aligned}$ and then graph the function and its inverse.
::15. f(x)=x2+3x+2的参数,然后绘制函数及其反向图。

Review (Answers)
::回顾(答复)

Please see the Appendix.
::请参看附录。

Section outline

General

参数反函数

Introduction ::导言

Parametric Inverses ::参数反函数

Examples ::实例

Summary ::摘要

Review ::回顾

Review (Answers) ::回顾(答复)

Introduction
::导言

Parametric Inverses
::参数反函数

Examples
::实例

Summary
::摘要

Review
::回顾

Review (Answers)
::回顾(答复)