Course: 10.5 参数等量的应用

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参数等量的应用

A regular function has the ability to graph the height of an object over time. allow you to actually graph the complete position of an object over time. For example, parametric equations allow you to make a graph that represents the position of a point on a Ferris wheel. All the details like height off the ground, direction, and speed of spin can be modeled using the parametric equations.
::普通函数能够绘制一个对象随时间推移的高度。允许您实际绘制一个对象随时间推移的完整位置。例如,参数方程式允许您绘制一个图形,以显示Ferris轮上某个点的位置。所有细节,如在地面上高度、方向和旋转速度,都可以用参数方程式模拟。

What is the position equation and graph of a point on a Ferris wheel that starts at a low point of 6 feet off the ground, spins counterclockwise to a height of 46 feet off the ground, then goes back down to 6 feet in 60 seconds?
::在离地6英尺的低点开始,逆时针旋转到离地46英尺的高点,然后在60秒后返回到6英尺高点。

Applying Parametric Equations
::应用参数等量

There are two types of parametric equations that are typical in real life situations. The first is circular motion as was described in the concept problem. The second is projectile motion.
::在现实生活中,典型的参数等式有两种:第一是概念问题中描述的循环运动;第二是投射运动。

Circular Motion
::通知动议

Parametric equations that describe circular motion will have $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ as periodic functions of sine and cosine. Either $\begin{align*}x\end{align*}$ will be a sine function and $\begin{align*}y\end{align*}$ will be a cosine function or the other way around. The best way to come up with parametric equations is to first draw a picture of the circle you are trying to represent.
::描述圆形运动的参数方程式将具有x和y作为正弦和正弦的周期函数。要么x将是一个正弦函数, y将是一个正弦函数, 要么是相反的函数。提出参数方程式的最佳方式是先绘制您试图代表的圆形图。

Next, it is important to note the starting point, center point and direction. You should already have the graphs of sine and cosine memorized so that when you see a pattern in words or as a graph, you can identify what you see as $\begin{align*}+\sin, -\sin, +\cos, -\cos.\end{align*}$
::接下来,重要的是要注意起点、中点和方向。您应该已经拥有正弦和正弦的图形, 以便当您在文字或图表中看到一个模式时, 可以识别您所看到的 +sin,-sin,+cos,-cos。

Take the example given above with the Ferris Wheel that starts at a low point of 6 feet off the ground, spins counterclockwise to a height of 46 feet off the ground, then goes back down to 6 feet in 60 seconds. The the vertical component starts at a low point of 6, travels to a middle point of 26 and then a height of 46 and back down. This is a $\begin{align*}-\cos\end{align*}$ pattern. The of the $\begin{align*}-\cos\end{align*}$ is 20 and the vertical shift is 26. Lastly, the period is 60. You can use the period to help you find $\begin{align*}b.\end{align*}$
::以上面提到的Ferris轮轮为例,它从地上6英尺的低点开始,逆时针旋转到离地46英尺的高度,然后在60秒后返回到6英尺的高度。垂直部件从6英尺的低点开始,到26英尺的中点,再到46英尺的中点,然后返回到46英尺的高度。这是一个-cos模式。 ─cos模式是20英尺,垂直移动是26英尺。最后, 周期是60英尺, 您可以使用这一时间来帮助您找到b。

\begin{aligned} 60 & = \frac{2 π}{b} \\ b & = \frac{π}{30} \end{aligned}

$\begin{align*}60 &=\frac{2\pi}{b} \\ b &=\frac{\pi}{30}\end{align*}$
::60=230

Thus the vertical parameterization is:
::因此,垂直参数化是:

$\begin{align*}y=-20 \cos \left(\frac{\pi}{30}t\right)+26\end{align*}$
::y 20cos(30t)+26

The horizontal parameterization is found by noticing that the $\begin{align*}x\end{align*}$ values start at 0, go up to 20, go back to 0, then down to -20, and finally back to 0. This is a $\begin{align*}+\sin\end{align*}$ pattern with amplitude 20. The period is the same as with the vertical component.
::水平参数化通过注意到x值开始于 0, 到达20, 返回到 0, 返回到 -20, 最后返回到 0 。这是 +sinpattors , 具有振幅 20 。时间段与垂直组件相同。

Thus parametric equations for the point on the wheel are:
::因此,轮子上点的参数方程是:

\begin{aligned} x & = 20 \sin (\frac{π}{30} t) \\ y & = - 20 \cos (\frac{π}{30} t) + 26 \end{aligned}

$\begin{align*}x &=20 \sin \left(\frac{\pi}{30} t \right) \\ y &=-20 \cos \left(\frac{\pi}{30} t \right) +26\end{align*}$
::x=20sin(% 30t) y20cos(% 30t) +26

Note that horizontal and vertical components of parametric equations are the $\begin{align*}x =\end{align*}$ and $\begin{align*}y =\end{align*}$ functions respectively or the horizontal and vertical parameterization.
::请注意,参数方程的水平和垂直构件分别为x=andy=函数或水平和垂直参数化。

Projectile Motion
::投射动作

Projectile motion has a vertical component that is quadratic and a horizontal component that is linear . This is because there are 3 parameters that influence the position of an object in flight: starting height, initial velocity, and the force of gravity. The horizontal component is independent of the vertical component. This means that the starting horizontal velocity will remain the horizontal velocity for the entire flight of the object.
::投影运动有一个垂直成份是二次形的,一个水平成份是线性的。这是因为有三个参数影响物体在飞行中的位置:起始高度、初始速度和重力。水平成份独立于垂直成份。这意味着起始水平速度将保持物体整个飞行的水平速度。

Note that gravity, $\begin{align*}g,\end{align*}$ has a force of about $\begin{align*}-32 \ ft/s^2 \end{align*}$ or $\begin{align*}-9.81 \ m/s^2.\end{align*}$ The examples and practice questions in this concept will use feet.
::注意重力(g)的威力大约为-32立方英尺/秒2或-9.81米/秒2。

If an object is launched from the origin at a velocity of $\begin{align*}v\end{align*}$ then it has horizontal and vertical components that can be found using basic trigonometry.
::如果一个物体以 v 的速率从源头发射,则该物体具有横向和纵向组件,可用基本三角测量法找到。

\begin{aligned} \sin θ & = \frac{v_{V}}{v} \to v \cdot \sin θ = v_{V} \\ \cos θ & = \frac{v_{H}}{v} \to v \cdot \cos θ = v_{H} \end{aligned}

$\begin{align*}\sin \theta &=\frac{v_V}{v} \rightarrow v \cdot \sin\theta=v_V \\ \cos\theta &=\frac{v_H}{v} \rightarrow v \cdot \cos\theta=v_H\end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}我...

The horizontal component is basically finished. The only adjustments that would have to be made are if the starting location is not at the origin, wind is added or if the projectile travels to the left instead of the right. See Example A.
::水平组件基本已完成。唯一需要做的调整是如果起始位置不是起点、风是增加的, 或者如果投射体向左而不是向右移动。见例A。

$\begin{align*}x=t \cdot v \cdot \cos\theta\end{align*}$
::x=tvc

The vertical component also needs to include gravity and the starting height. The general equation for the vertical component is:
::垂直部件还需要包括重力和起始高度。

$\begin{align*}y=\frac{1}{2} \cdot g \cdot t^2 + t \cdot v \cdot \sin\theta + k\end{align*}$
::y=12 gt2+tvsink

The constant $\begin{align*}g\end{align*}$ represents gravity, $\begin{align*}t\end{align*}$ represents time, $\begin{align*}v\end{align*}$ represents initial velocity and $\begin{align*}k\end{align*}$ represents starting height. You will explore this equation further in calculus and physics. Note that in this concept, most answers will be found and confirmed using technology such as your graphing calculator.
::恒定 g 表示重力, t 表示时间, v 表示初始速度, k 表示起始高度。您可以在微积分和物理中进一步探索这个方程式。请注意, 在这个概念中, 多数答案会使用您的图形计算器等技术找到和确认。

Examples
::实例

Example 1
::例1

A ball is thrown from the point (30, 5) at an angle of $\begin{align*}\frac{4\pi}{9}\end{align*}$ to the left at an initial velocity of $\begin{align*}68 \ ft/s.\end{align*}$ Model the position of the ball over time using parametric equations. Use your graphing calculator to graph your equations for the first four seconds while the ball is in the air.
::从点(30, 5) 向左倾出一个球, 角度为4°9, 最初速度为68英尺/秒。使用参数方程来模拟球的方位。使用您的图形计算器来绘制球在空中前四秒的方程图。

The horizontal component is $\begin{align*}x=-t \cdot 68 \cdot \cos(\frac{4\pi}{9})+30.\end{align*}$ Note the negative sign because the object is traveling to the left and the +30 because the object starts at (30, 5).
::水平组件为 xt68cos(49)+30。请注意负符号,因为对象向左移动,而+30则因为对象起始于(30,5)时。

The vertical component is $\begin{align*}y=\frac{1}{2} \cdot (-32) \cdot t^2+t \cdot 68 \cdot \sin (\frac{4\pi}{9})+5.\end{align*}$ Note that $\begin{align*}g=-32\end{align*}$ because gravity has a force of $\begin{align*}-32 \ ft/s^2\end{align*}$ and the $\begin{align*}+5\end{align*}$ because the object starts at (30, 5).
::垂直元件为y=12(- 32t2+t68sin(49)+5. 请注意,g32,因为重力的强度为-32立方英尺/秒2,而+5,因为天体起始于(30,5)时。

Example 2
::例2

When does the ball from Example 1 reach its maximum and when does the ball hit the ground? How far did the person throw the ball?
::例1的球何时达到最大,球何时击中地面?

To find when the function reaches its maximum, you can find the vertex of the parabola . Analytically this is messy because of the decimal coefficients in the quadratic. Use your calculator to approximate the maximum after you have graphed it. Depending on how small you make your $\begin{align*}T_{step}\end{align*}$ should find the maximum height to be about 75 feet.
::要找到函数达到最大值时, 您可以找到抛物线的顶部。分析来说, 这很乱, 因为四边形中的小数系数。使用您的计算器在绘制了图后可以接近最大值。取决于您使Tstep变小多少, 您应该发现最大高度大约为75英尺。

To find out when the ball hits the ground, you can set the vertical component equal to zero and solve the quadratic equation. You can also use the table feature on your calculator to determine when the graph goes from having a positive vertical value to a negative vertical value. The benefit for using the table is that it simultaneously tells you the $\begin{align*}x\end{align*}$ value of the zero.
::要发现球击中地面时,您可以设置等于零的垂直元件并解析二次方程。您也可以使用计算器上的表格特性来确定图形从正垂直值到负垂直值的时间。使用表格的好处是它同时告诉您零的X值。

After about 4.2588 seconds the ball hits the ground at (-20.29, 0). This means the person threw the ball from (30, 5) to (-20.29, 0), a horizontal distance of just over 50 feet.
::在大约4.2588秒后,球击中地面(20.29,0),这意味着球从(30,5)向(20.29,0)投球的人向(20.29,0)投球,水平距离略高于50英尺。

Example 3
::例3

Kieran is on a Ferris wheel and his position is modeled by the parametric equations:
::基兰在Ferris轮上他的位置以参数方程式为模型

\begin{aligned} x_{K} & = 10 \cdot \cos (\frac{π}{5} t) \\ y_{K} & = 10 \cdot \sin (\frac{π}{5} t) + 65 \end{aligned}

$\begin{align*}x_K &=10 \cdot \cos \left(\frac{\pi}{5}t \right) \\ y_K &=10 \cdot \sin \left(\frac{\pi}{5}t \right)+65\end{align*}$
::xK=10cos( 5t)yK=10sin( 5t)+65

Jason throws the ball modeled by the equation in Example 1 towards Kieran who can catch the ball if it gets within three feet. Does Kieran catch the ball?
::杰森以例1的方程式为模型向基兰投球如果球在三英尺内,他就能抓住球。基兰能抓住球吗?

This question is designed to demonstrate the power of your calculator. If you simply model the two equations simultaneously and ignore time you will see several points of intersection . This graph is shown below on the left. These intersection points are not interesting because they represent where Kieran and the ball are at the same place but at different moments in time.
::这个问题旨在展示您的计算器的力量。如果您简单地同时模拟两个方程式, 忽略时间, 您将会看到多个交叉点。此图显示在左侧。这些交叉点并不有趣, 因为它们代表着基兰和球的位置是在同一地点, 但是在不同的时间点。

When the $\begin{align*}T_{max}\end{align*}$ is adjusted to 2.3 so that each graph represents the time from 0 to 2.3, you get a better sense that at about 2.3 seconds the two points are close. This graph is shown above on the right.
::当将Tmax调整为2.3时,每个图表代表0到2.3的时间,你就会更清楚地感觉到,在大约2.3秒时,两个点接近。这个图表在右边显示。

You can now use your calculator to help you determine if the distance between Kieran and the ball actually does go below 3 feet. Start by plotting the ball’s position in your calculator as $\begin{align*}x_1\end{align*}$ and $\begin{align*}y_1\end{align*}$ and Kieran’s position as $\begin{align*}x_2\end{align*}$ and $\begin{align*}y_2.\end{align*}$ Then, plot a new parametric equation that compares the distance between these two points over time. You can put this under $\begin{align*}x_3\end{align*}$ and $\begin{align*}y_3.\end{align*}$ A calculator can reference internal variables like $\begin{align*}x_1, y_1\end{align*}$ that have already been set in the calculator’s memory to form new variables like $\begin{align*}x_3, y_3\end{align*}$ . Note that you can find the $\begin{align*}x_1, x_2, y_1, y_2\end{align*}$ entries in the vars and parametric menu.
::现在您可以使用您的计算器来帮助您确定基兰与球之间的距离是否真的低于3英尺。首先, 将计算器中的球位置绘制为 x1 和 y1 , 将基兰的位置绘制为 x2 和 y2 。然后, 绘制一个新的参数方程, 比较这两个点之间的时间距离。您可以将此设置在 x3 和 y3. 下。计算器可以参照计算器记忆中已经设置的类似 x1 y1 的内部变量来形成新的变量, 如 x3, y3 。注意您可以在运算和参数菜单中找到 x1, x2, y1, y1, y2 条目。

\begin{aligned} x_{3} & = t \\ y_{3} & = \sqrt{(x_{2} - x_{1})^{2} + (y_{2} - y_{1})^{2}} \end{aligned}

$\begin{align*}x_3 &=t \\ y_3 &=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\end{align*}$
::x3=ty3=(x2- x1)2+(y2-y1)2

Now when you graph, you should change your window settings and let $\begin{align*}t\end{align*}$ vary between 0 and 4, the $\begin{align*}x\end{align*}$ window show between 0 and 4 and the $\begin{align*}y\end{align*}$ window show between 0 and 5. This way it should be clear if the distance truly does get below 3 feet.
::当您图形时, 您应该更改窗口设置, 并在 0 和 4 之间更改, x 窗口显示在 0 和 4 之间, y 窗口显示在 0 和 5 之间。这样一来, 如果距离真的在 3 英尺以下, 距离应该很明显。

Depending on how accurate your $\begin{align*}T_{step}\end{align*}$ is, you should find that the distance is below 3 feet. Kieran does indeed catch the ball.
::基兰确实抓住了球

Example 4
::例4

At what velocity does a football need to be thrown at a $\begin{align*}45^\circ\end{align*}$ angle in order to make it all the way across a football field?
::为了让足球横跨足球场需要以45°C的角度投出多少速度?

A football field is 100 yards or 300 feet. The parametric equations for a football thrown from (300, 0) back to the origin at speed $\begin{align*}v\end{align*}$ are:
::足球场为100码或300英尺。

\begin{aligned} x & = - t \cdot v \cdot \cos (\frac{π}{4}) + 300 \\ y & = \frac{1}{2} \cdot (- 32) \cdot t^{2} + t \cdot v \cdot s i n (\frac{π}{4}) \end{aligned}

$\begin{align*}x &=-t \cdot v \cdot \cos \left(\frac{\pi}{4} \right)+300 \\ y &=\frac{1}{2} \cdot (-32) \cdot t^2 + t \cdot v \cdot sin \left(\frac{\pi}{4} \right)\end{align*}$
::\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\・\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Substituting the point (0, 0) in for $\begin{align*}(x, y)\end{align*}$ produces a system of two equations with two variables $\begin{align*}v, t\end{align*}$ .
::将点(0,0)替换为(x,y)产生一个由两个变数和两个变数组成的系统。

\begin{aligned} 0 & = - t \cdot v \cdot \cos (\frac{π}{4}) + 300 \\ 0 & = \frac{1}{2} \cdot (- 32) \cdot t^{2} + t \cdot v \cdot \sin (\frac{π}{4}) \end{aligned}

$\begin{align*}0 &=-t \cdot v \cdot \cos \left(\frac{\pi}{4} \right)+300 \\ 0 &=\frac{1}{2} \cdot (-32) \cdot t^2 + t \cdot v \cdot \sin \left(\frac{\pi}{4} \right)\end{align*}$
::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

You can solve this system many different ways.
::你可以用很多不同的方法解决这个系统。

$\begin{align*}t =\frac{5 \sqrt{3}}{2} \approx 4.3 \ seconds, v=40\sqrt{6} \approx 97.98 \ ft/s\end{align*}$
::t=5324.3秒,v=40697.98英尺/秒

In order for someone to throw a football at a $\begin{align*}45^\circ\end{align*}$ angle all the way across a football field, they would need to throw at about $\begin{align*}98 \ ft/s\end{align*}$ which is about 66.8 mph.
::球员必须投出大约98英尺/秒, 大约66.8英里。

$\begin{align*}\frac{98 \ feet}{1 \ sec} \cdot \frac{3600 \ sec}{1 \ hour} \cdot \frac{1 \ mile}{5280 \ feet} \approx \frac{66.8 \ miles}{1 \ hour}\end{align*}$
::98英尺1 秒3600 秒1小时1 英寸1 英里5280 英寸66.8 英里1 小时1

Example 5
::例5

Nikki got on a Ferris wheel ten seconds ago. She started 2 feet off the ground at the lowest point of the wheel and will make a complete cycle in four minutes. The ride reaches a maximum height of 98 feet and spins clockwise. Write parametric equations that model Nikki’s position over time. Where will Nikki be three minutes from now?
::Nikki在10秒前上了Ferris车轮。她从地面最低的方向盘开始离地面2英尺,四分钟后将完成整个车轮。驾驶车身最高高度为98英尺,顺时针旋转。写出模拟Nikki位置的参数方程式。 Nikki三分钟后会在哪里?

Don’t let the 10 second difference confuse you. In order to deal with the time difference, use $\begin{align*}\left(t + \frac{1}{6}\right)\end{align*}$ instead of $\begin{align*}t\end{align*}$ in each equation. When $\begin{align*}t=0\end{align*}$ , ten seconds $\begin{align*}\left(\frac{1}{6}\right.\end{align*}$ of a minute $\begin{align*}\left.\right)\end{align*}$ have already elapsed.
::不要让10秒的差别混淆你。为了解决时间差问题,请使用(t+16)而不是在每一个方程式中使用 t。当 t=0, 10秒(每分钟16秒)已经过去。如果t=0, 10秒(每分钟16秒)已经过去了。

\begin{aligned} x & = - 48 \cdot \sin (\frac{π}{2} (t + \frac{1}{6})) \\ y & = - 48 \cdot \cos (\frac{π}{2} (t + \frac{1}{6})) + 50 \end{aligned}

$\begin{align*}x &= -48 \cdot \sin\left(\frac{\pi}{2}\left(t+\frac{1}{6}\right)\right) \\ y &= -48 \cdot \cos\left(\frac{\pi}{2}\left(t+\frac{1}{6}\right)\right)+50 \end{align*}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}...

At $\begin{align*}t=3, x \approx 46.36\end{align*}$ and $\begin{align*}y \approx 37.58\end{align*}$
::Tat=3,x46.36 y37.58

Summary

Two types of parametric equations are typical in real-life situations: circular motion and projectile motion.
::在现实环境中,典型的参数等式有两种:循环运动和投射运动。

Circular motion parametric equations have $\begin{align*}x\end{align*}$ and $\begin{align*}y\end{align*}$ as periodic functions of sine and cosine, with either $\begin{align*}x\end{align*}$ being a sine function and $\begin{align*}y\end{align*}$ being a cosine function or vice versa.
::圆形运动参数方程式具有正弦和正弦的x和y的周期函数,以x为正弦和正弦函数,y为正弦函数,y为正弦函数,y为正弦函数,y为正弦函数,y为正弦函数,y为正弦函数,y为正弦和正弦的周期函数。

Projectile motion has a vertical component that is quadratic and a horizontal component that is linear, with three parameters influencing the position: starting height, initial velocity, and the force of gravity.
::投影运动有一个垂直成份,即二次成份,一个水平成份是线性成份,有三个参数影响位置:起始高度、初始速度和重力。

Review
::回顾

Candice gets on a Ferris wheel at its lowest point, 3 feet off the ground. The Ferris wheel spins clockwise to a maximum height of 103 feet, making a complete cycle in 5 minutes.
::坎迪斯在离地面3英尺的最低点乘着摩天轮。摩天轮顺时速旋转到最高高度103英尺,在5分钟内完成整个循环。

1. Write a set of parametric equations to model Candice’s position.
::1. 编写一套模拟Candice位置的参数方程。

2. Where will Candice be in two minutes?
::2. 两分钟后Candice会在哪里?

3. Where will Candice be in four minutes?
::3. 四分钟后Candice会在哪?

One minute ago Guillermo got on a Ferris wheel at its lowest point, 3 feet off the ground. The Ferris wheel spins clockwise to a maximum height of 83 feet, making a complete cycle in 6 minutes.
::一分钟前,吉列尔莫在离地面3英尺的最低处乘着摩天轮。摩天轮顺时速旋转到最高高度83英尺,6分钟后完成整个循环。

4. Write a set of parametric equations to model Guillermo’s position.
::4. 编写一套参数方程,以模拟吉列尔莫的位置。

5. Where will Guillermo be in two minutes?
::5. 两分钟后吉列尔莫会在哪?

6. Where will Guillermo be in four minutes?
::6. 四分钟后吉列尔莫会在哪?

Kim throws a ball from (0, 5) to the right at 50 mph at a $\begin{align*}45^\circ \end{align*}$ angle.
::金从(0, 5)向右投球, 球在50mph, 角度是45。

7. Write a set of parametric equations to model the position of the ball.
::7. 编写一套参数方程,以模拟球的位置。

8. Where will the ball be in 2 seconds?
::8. 2秒后球会在哪里?

9. How far does the ball get before it lands?
::9. 球落地前有多远?

David throws a ball from (0, 7) to the right at 70 mph at a $\begin{align*}60^\circ \end{align*}$ angle. There is a 6 mph wind in David’s favor.
::大卫将球从(0, 7 ) 投向右方70mph , 投向60°Q角。大卫有6mph风。

10. Write a set of parametric equations to model the position of the ball.
::10. 编写一套参数方程,以模拟球的位置。

11. Where will the ball be in 2 seconds?
::11. 2秒后球会在哪里?

12. How far does the ball get before it lands?
::12. 球落地前有多远?

Suppose Riley stands at the point (250, 0) and launches a football at 72 mph at an angle of $\begin{align*}60^\circ \end{align*}$ towards Kristy who is at the origin. Suppose Kristy also throws a football towards Riley at 65 mph at an angle of $\begin{align*}45^\circ \end{align*}$ at the exact same moment. There is a 6 mph breeze in Kristy’s favor.
::假设Riley站在(250,0)点,在72英里处向原产地的Kristy发射一支橄榄球,角度为60英里。假设Kristy也在65英里处向Riley发射一支橄榄球,角度为45英里。克里斯蒂的帮助是6英里微风。

13. Write a set of parametric equations to model the position of Riley’s ball.
::13. 编写一套参数方程,以模拟莱利球的位置。

14. Write a set of parametric equations to model the position of Kristy’s ball.
::14. 编写一套参数方程,以模拟Kristy球的位置。

15. Graph both functions and explain how you know that the footballs don’t collide even though the two graphs intersect.
::15. 绘制两个功能的图,并解释你如何知道足球不会相撞,即使两个图表相互交错。

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::回顾(答复)

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::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

参数等量的应用

Applying Parametric Equations ::应用参数等量

Circular Motion ::通知动议

Projectile Motion ::投射动作

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)