节: 模拟三角函数 | 7.7 模拟三角函数-interactive

章节大纲

Lesson Objectives
::经验教训目标

Understand and apply the equation for a function.
::理解并应用函数的方程式。

Choose trigonometric functions to model periodic phenomena with specified , frequency, and midline.
::选择三角函数, 以指定、频率和中线来模拟周期现象。

Create equations in two or more variables to represent relationships between quantities.
::以两个或多个变量创建方程式,以表示数量之间的关系。

Introduction: Catapult Confusion
::导言: 弹射熔化

Sarah, Liam, and Viviana are in a physics class that is having a catapult contest. While researching the best angle at which to launch their projectile, they find that a projectile launched with the same initial velocity from ground level at 30° and 60° will land in the same spot . Is this true? Use the interactive below to explore how horizontal distance is impacted by the launch angle.
::Sarah、Liam和Viviana处于一个物理类中,正在进行弹射竞赛。在研究发射弹射的最佳角度时,他们发现从30°和60°的地面发射的最初速度相同的弹射弹将降落在同一地点。这是真的吗?使用下面的交互作用来探索发射角度对水平距离的影响。

INTERACTIVE

Catapult Confusion

Set the launch angle of the watermelon by dragging the red-point. Then, click the button to launch!
::拖曳红点,设定西瓜的发射角度。然后单击按钮启动!

Compare how far the watermelon travels if it is launched at 30° and 60°.
::如果西瓜在30°和60°发射,则与西瓜的距离比较。

Activity 1: Catapult Confusion Continued
::活动1:弹射干扰继续

To determine whether or not the horizontal distance traveled by an object launched at 30° and 60° is the same, model distance as a function of launch angle. In physics class, the students learned the following formula:
::为了确定在30°和60°发射的物体所穿行的水平距离是否相同,将距离作为发射角度的函数来模拟。

\begin{aligned} R (θ) = \frac{{v_{0}}^{2} \sin (2 θ)}{g} \end{aligned}

::R

= v02sin ( 2) g

$\begin{aligned} R (θ) \end{aligned}$ is the horizontal distance traveled, also known as the range, in .
::R 是所穿行的水平距离, 也称为范围。

$\begin{aligned} v_{0} \end{aligned}$ is the initial velocity, 22 per second.
::v0是初始速度,每秒22次。

$\begin{aligned} θ \end{aligned}$ is the angle at which the projectile is launched.
::是发射射弹的角度。

$\begin{aligned} g \end{aligned}$ is the acceleration due to gravity, $\begin{aligned} 9.8 m / s^{2} . \end{aligned}$
::g 指因重力而加速度,9.8m/s2。

Let's first find the horizontal distance of an object launched at 30°.
::让我们首先发现30°发射的物体的水平距离。

\begin{aligned} R (60^{\circ}) & = \frac{{(22)}^{2} \sin (2 (30^{\circ}))}{(9.8)} \\ = \frac{484 \cdot \sin (60^{\circ})}{9.8} \\ \approx 42.77 m \end{aligned}

::R( 60 ) = ( 222) 2sin (2( 30 ) (9. 8) = 484 (60 ) 9.8 42.77m

Now let's compare this to the horizontal distance of an object launched at 60°.
::现在让我们比较一下60度时发射的物体的水平距离。

\begin{aligned} R (60^{\circ}) & = \frac{{(22)}^{2} \sin (2 (60^{\circ}))}{(9.8)} \\ = \frac{484 \cdot \sin (120^{\circ})}{9.8} \\ \approx 42.77 m \end{aligned}

::R( 60) =( 222) 2sin ( 2( 60) (9.8) = 484( 120) 9. 842. 77m

Answer: The projectile will travel 42.77 meters when launched at both 30° and 60°.
::答复:发射时,弹体在30°和60°之间将穿行42.77米。

Use the interactive below to explore why this happens and to determine which angle will launch the projectile the farthest.
::使用下面的交互方式来探究为什么会发生这种情况,并确定哪个角度会发射最远的投射体。

Discussion Questions
::讨论问题讨论问题

Explain why a projectile launched from ground level at 30° and 60° will land in the same spot. ( 120 has a reference angle of 60, so it will have the same as 60.)
::解释为什么从30°和60°的地面发射的射弹将降落在同一地点。 (120的参考角为60,因此与60相同。 )

Would this be the same if the object wasn't launched from ground level? If so, which angle, 30° or 60°, would produce a farther distance from a height of 5 feet above ground level? (The 30- degree launch would go farther because the 30-degree launch has a greater horizontal velocity while the 60-degree launch has a greater vertical velocity.)
::如果天体不是从地面发射的,这是否是一样的?如果是这样,那么哪个角度,即30°或60°,会比地面高度高5英尺远得多? (30度发射会更远,因为30度发射具有更高的水平速度,而60度发射具有更高的垂直速度。 )

Will other launch angles have the same horizontal distance? (Yes, any angle a and another angle 180-2a. In other words, any .)
::其他发射角度是否具有相同的水平距离? (是的,任何角度a和另一个角度180-2a。换句话说,任何.)

Activity 2: Using Transformations to Write Sinusoidal Models
::活动2:利用变型书写小儿科模型

A large Ferris wheel is 100 feet in diameter and has 12 passenger cars. The Ferris wheel rotates counterclockwise and takes 1 minute to complete a full rotation. Riders climb a stairway to board the bottommost car from a platform 10 feet above the ground. In this scenario, the height of a car is the distance between the car and the ground. In other words, if the car was located at an $\begin{aligned} (x, y) \end{aligned}$ coordinate, x is the horizontal distance from the center of the Ferris wheel, and the height is $\begin{aligned} y \end{aligned}$ . To determine the height of Car 1 as the Ferris wheel rotates, consider that after 15 seconds (which is $\begin{aligned} \frac{1}{4} \end{aligned}$ of the way around or $\begin{aligned} \frac{π}{2} \end{aligned}$ radians), the car is at the top of the wheel, which we know is 100 feet above the 10-foot platform. Thus the height of the car is 110 feet.
::大型的Ferris车轮直径100英尺,有12辆客车。 Ferris车轮逆时旋转,需要1分钟完成完全旋转。 Riders爬上楼梯,从地面上10英尺的平台上登上最底层的汽车。在这种情况下,汽车的高度是汽车与地面之间的距离。换句话说,如果汽车位于一个(x,y)坐标处,x是与Ferris车轮中心之间的水平距离,高度是y。为了确定1号车的高度,而Ferris车轮则在旋转,考虑到在15秒后(环绕方向14或2radians),汽车在方向盘的顶部,我们知道在10英尺的平台上方100英尺。因此,汽车的高度是110英尺。

Example
::示例示例示例示例

Using the Ferris wheel scenario from above, create a function, h(x), to model the height of a car as a function of time . Assume you start the ride from car 1.
::使用上面的Ferris轮式假想,创建函数h(x),以时间函数来模拟汽车的高度。假设你从1号车开始骑车。

As time passes, the height of the Ferris wheel will repeatedly rise to the same maximum and fall to the same minimum. Additionally, starting from car 1 means that we are starting from the midline of the minimum and maximum heights. Since the sine function starts from the midline, this will be the parent function that we use. The graph below models the height of the Ferris wheel over the first revolution as a function of the angle based on the starting position.
::随着时间的流逝, Ferris 轮的高度会反复升到相同的最大高度, 并降到相同的最低点。此外, 从 1号车开始意味着我们从最小和最大高度的中线开始。由于正弦函数从中线开始, 这将是我们使用的父函数。下面的图将Ferris 轮的高度乘以第一次革命的高度作为基于起始位置的角度的函数。

There are two transformations taking place in this function:
::此函数中正在发生两种转变:

The graph is shifted up 60 units. We know this because the midline is at 60, which makes sense given that the center of the Ferris wheel is 60 feet off the ground.
::图表向上移动了60个单位。我们知道这一点,因为中间线是60,这很合理,因为Ferris轮的中心离地60英尺。

The is 50. We know this because the distance from the center of the Ferris wheel to its highest point is $\begin{aligned} 110 - 60 = 50. \end{aligned}$
::我们知道这一点,因为从Ferris车轮中心到最高点的距离是110-60=50。

We know from the previous section that a sine function can be transformed using the form $\begin{aligned} f (x) = a \cdot \sin (b (x - h)) + k \end{aligned}$ where a represents the vertical stretch/shift, b represents the horizontal stretch/shift, h represents the horizontal shift, and k represents the vertical shift . To model the vertical shift of 60, we can add 60 to the output of the function. Additionally, since a represents not only the vertical stretch/shrink but also the amplitude, we know that a will be 50. The graph above can be represented using the function $\begin{aligned} h (θ) = 50 \sin θ + 60 \end{aligned}$ .
::我们从上一节中知道,正弦函数可以使用表f(x)=aQsin(b(x-h))+k来转换,该表表示垂直拉伸/轮档,b表示水平拉伸/轮档,h表示水平转变,k表示垂直转变。要模拟60的垂直移动,我们可以在函数输出中增加60个。此外,由于一个函数不仅代表垂直拉伸/递减,而且代表振幅,我们知道将代表50个。以上图可以使用函数h(x)=50sin=60来表示。

While we did write a function to represent height as a function of angle, we were asked to write the height of the Ferris wheel as a function of time. The period of the function represents time. In the previous lesson, we learned that the period of a sinusoidal function is equal to $\begin{aligned} \frac{2 π}{b} . \end{aligned}$ Since we know that the period of the Ferris wheel is 1 minute, we can use the following equation to find the value of b:
::当我们写了一个函数来代表高度作为角度的函数时, 我们被要求写Ferris轮的高度作为时间的函数。函数的期间代表时间。在上一个教训中, 我们学到, 类流函数的期间等于 2°b 。由于我们知道 Ferris轮的期间是 1 分钟, 我们可以使用以下方程式来找到 b 的值 :

\begin{aligned} \frac{2 π}{b} & = 1 \end{aligned}

::2b=1

Multiplying b to both sides will give the value of b and enough information to write the height of the car as a function of time in minutes.
::乘以两边的b 将会给出 b 的值和足够信息来写出车身高度, 以分钟计时函数。

\begin{aligned} b \times \frac{2 π}{b} & = 1 \times b \\ 2 π & = b \end{aligned}

::bb=1xb2bb

Answer: $\begin{aligned} h (x) = 50 \sin (2 π x) + 60 \end{aligned}$
::答复:h(x)=50sin(2xx)+60

Use the quiz lite to practice modeling scenarios using transformations of sinusoidal functions.
::使用测试精度来使用正弦函数的变形来进行模型假设。

Discussion Questions:
::讨论问题:

The Earth revolves around the Sun over a year with a maximum distance of 94,500,000 miles at its aphelion on July 4th and a minimum distance of 91,500,000 miles at its perihelion on January 3rd. Write a function to model the distance of the Earth from the Sun, in miles, as a function of the angle of rotation, in radians. Assume the function begins on January 3rd, at which point the angle of rotation is 0 radians.
::地球在一年中围绕太阳旋转,7月4日的月4日月4日月4日月4日月4日月4日的月球最大距离为94,500,000英里,而1月3日的近地点最低距离为91,500,000英里。写一个函数,作为旋转角度的函数,以弧度来模拟地球与太阳的距离。假设函数始于1月3日的1月3日,旋转角度为0弧度。

For a challenge, write a function to estimate the distance of the Earth from the Sun, in miles, as a function of time, in days. Assume the function begins on January 3rd.
::对于挑战, 请写一个函数来估计地球与太阳的距离, 以英里计, 以时间函数计, 以天数计。假设函数从 1月3日开始。

Activity 3: Using Sinusoidal Functions to Model Data
::活动3:利用科性函数模拟数据

We have seen in previous chapters that data doesn't always perfectly fit a model. In the real world, we often need to choose a model that best fits a collection of data to analyze and make predictions about the data set. It may not be immediately obvious which situations may call for trigonometric functions but look no farther than the weather. A sinusoidal function can model any relationship that repeatedly rises and falls. Data that forms a repeating pattern is known as periodic data. Use the interactive below to practice writing a function to model periodic data .
::我们在前几章中已经看到,数据并不总是完全适合模型。在现实世界中,我们常常需要选择最适合数据收集的模型来分析和预测数据集。可能并不十分明显,哪些情况可能需要三角函数,但看得不远于天气。正弦函数可以模拟反复上升和下降的任何关系。形成重复模式的数据被称为周期数据。使用下面的交互式数据来练习写一个函数来模拟周期数据。

INTERACTIVE

Match the Data

The goal of this interactive is to match the data shown below by creating your own sine graph.
::此互动的目的是通过创建您自己的正弦图来匹配以下显示的数据。

Click the button to start graphing
::单击按钮开始图形化

Drag the sliders to change the corresponding variables of f(x)
::拖动滑动器以更改 f( x) 的相应变量

Wrap-Up: Review Questions
::总结:审查问题

Summary

The general equation for the transformations of a sine function is $\begin{aligned} y = a \sin (b (x - h) + k \end{aligned}$
- $\begin{aligned} a \end{aligned}$ is the vertical stretch, changing the amplitude.
  ::a 是垂直伸展,改变振幅。
- $\begin{aligned} b \end{aligned}$ is the horizontal stretch, changing the period.
  ::b 是水平拉伸,改变周期。
- $\begin{aligned} h \end{aligned}$ is the horizontal shift, or phase shift, moving the graph right or left.
  ::h 是指水平移动,或相向移动,向右或向左移动图形。
- $\begin{aligned} k \end{aligned}$ is the vertical shift, moving the graph up or down.
  ::k 是垂直移动, 向上或向下移动图形。
::正弦函数转换的一般方程式是 y=asin(b(x-h)+k a 是垂直伸缩, 改变振幅。 b 是水平伸缩, 改变周期。 h 是水平移动, 或阶段移动, 向右或向左移动。 k 是垂直移动, 向上或向下移动。

章节大纲

Introduction: Catapult Confusion ::导 言: 弹射熔化

Activity 1: Catapult Confusion Continued ::活动1:弹射干扰继续

Activity 2: Using Transformations to Write Sinusoidal Models ::活动2:利用变型书写小儿科模型

Example ::示例示例示例示例

Activity 3: Using Sinusoidal Functions to Model Data ::活动3:利用科性函数模拟数据