Course: 10.2 圆形和椭圆的参数等量

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圆形和椭圆的参数等量

The cross country team from Forest Park High is practicing in a nearby forest preserve. The female members of the team run a circular loop while the males run along an elliptical path. The paths are two different lengths and the teams are running at two different speeds. Will the teams meet? You can use parametric equations to help you find out.
::森林公园高中的跨国家小组正在附近的森林保护区练习。团队的女性成员环绕环绕环绕,而男性则沿椭圆路径运行。路径有两种不同的长度,团队以两种不同的速度运行。团队会相会吗?您可以使用参数方程来帮助您查找。

Parametric Equations for Circles and Ellipses
::圆形和椭圆的参数等量

In the past, you’ve learned that an ellipse is a rounded shape with two foci . Every coordinate on the ellipse can be described by its distance from the two foci. While a given point’s distance from each focus is unique, the sum of the two distances is the same for every point on the ellipse.
::过去,你已经知道椭圆是一个圆形的形状,有两根面。椭圆上的每一个坐标都可以用它与两根面的距离来描述。虽然一个点与每个焦点的距离是独一无二的,但两条距离的总和对椭圆上的每一个点是一样的。

The standard equation for an ellipse is $\begin{aligned} \frac{(x - h)^{2}}{a^{2}} + \frac{(y - k)^{2}}{b^{2}} = 1 \end{aligned}$ , where $\begin{aligned} (h, k) \end{aligned}$ is the center of the ellipse, and $\begin{aligned} 2 a \end{aligned}$ and $\begin{aligned} 2 b \end{aligned}$ are the lengths of the axes of the ellipse. The longer axis is called the major axis , while the shorter axis is called the minor axis.
::椭圆的标准方程式是 (x-h) 2a2+(y-k) 2b2=1, 其中(h,k) 是椭圆的中心, 2a 和 2b 是椭圆轴的长度。较长的轴被称为主轴, 较短的轴则被称为小轴。

A circle is a special type of ellipse where $\begin{aligned} a \end{aligned}$ is equal to $\begin{aligned} b \end{aligned}$ . You write the standard equation for a circle as $\begin{aligned} (x - h)^{2} + (y - k)^{2} = r^{2} \end{aligned}$ , where $\begin{aligned} r \end{aligned}$ is the radius of the circle and $\begin{aligned} (h, k) \end{aligned}$ is the center of the circle.
::a 圆是一个特殊的椭圆类型,其中,a 等于 b。您将圆的标准方程式写成为 (x-h)2+(y-k)2=r2, r 是圆的半径, (h,k) 是圆的中心。

The parametric form for an ellipse is $\begin{aligned} F (t) = (x (t), y (t)) \end{aligned}$ where $\begin{aligned} x (t) = a \cos (t) + h \end{aligned}$ and $\begin{aligned} y (t) = b \sin (t) + k \end{aligned}$ .
::椭圆的参数表为 F(t) = (x(t),y(t) ),其中 x(t) = acos(t)+h 和 y(t) = bsin(t)+k。

Since a circle is an ellipse where both foci are in the center and both axes are the same length, the parametric form of a circle is $\begin{aligned} F (t) = (x (t), y (t)) \end{aligned}$ where $\begin{aligned} x (t) = r \cos (t) + h \end{aligned}$ and $\begin{aligned} y (t) = r \sin (t) + k \end{aligned}$ .
::由于圆是一个椭圆,其中两个方子都位于中间,两个轴长度相同,圆的参数形式是F(t)=(x(t))=(x(t))=(rcos*(t)+h)和y(t)=rsin(t)+k。

When $\begin{aligned} x (t) = a \cos (t) \end{aligned}$ and $\begin{aligned} y (t) = b \sin (t) \end{aligned}$ , it takes an object $\begin{aligned} 2 π \end{aligned}$ units of time to trace the entire ellipse once. By altering the equation, you can model objects traveling at different speeds. For instance, if $\begin{aligned} x (t) = a \cos (2 t) \end{aligned}$ and $\begin{aligned} y (t) = a \sin (2 t) \end{aligned}$ , the object will complete one lap around the ellipse twice as fast, in $\begin{aligned} π \end{aligned}$ units of time.
::当 x( t) = acos( t) 和 y( t) = bsin( t) 时, 需要时间的天体 2 单位来追踪整个椭圆一次。通过改变方程式, 您可以以不同的速度模拟物体的运行。例如, 如果 x( t) = acos( 2t) 和 y( t) = asin( 2t) , 该天体将会在时间单位中以 + 倍的速度在椭圆周围完成一圈。

Take the ellipse defined by the equation $\begin{aligned} \frac{x^{2}}{25} + \frac{y^{2}}{81} = 1 \end{aligned}$ . Using the information from above, let's write a parametric equation for the ellipse where an object makes one revolution every $\begin{aligned} 8 π \end{aligned}$ units of time.
::使用方程 x225+y281=1 定义的椭圆。使用上面的信息, 让我们为椭圆写一个参数方程, 对象在时间的8- 秒内进行一次革命。

The equation $\begin{aligned} \frac{x^{2}}{25} + \frac{y^{2}}{81} = 1 \end{aligned}$ is of the form $\begin{aligned} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{aligned}$ . Take the square roots of the denominators to find that $\begin{aligned} a \end{aligned}$ is 5 and $\begin{aligned} b \end{aligned}$ is 9.
::x225+y281=1的方程式为 x2a2+y2b2=1。以分母的平方根为单位, 发现一个是5, b是9。

To put this equation in parametric form, you’ll need to recall the parametric formula for an ellipse:
::若要将这个等式以参数形式表示, 您需要记住对椭圆的参数公式 :

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = a \cos (t) \\ y (t) & = b \sin (t) \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = acos(t) y(t) = bsin(t)

So, substituting in the values of $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ for this ellipse, you’ll get:
::所以,用a和b的值代替这个椭圆, 你会得到:

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 5 \cos (t) \\ y (t) & = 9 \sin (t) \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = 5cos(t) y(t) = 9sin(t)

Now, you’ll need to adjust the equation to account for the speed of the object. Normally, an object will trace the complete ellipse in the period of the sine and cosine functions – that is $\begin{aligned} 2 π \end{aligned}$ . However, this object is traversing the route in $\begin{aligned} 8 π \end{aligned}$ units of time. This means that it takes four times as long to traverse the ellipse as a standard object.
::现在,您需要调整方程以计算对象的速度。通常,一个对象将追踪正弦和余弦函数(即2)期间的全椭圆。然而,该对象正在以8时段穿行路线。这意味着将椭圆作为标准对象需要四倍的时间穿行。

To account for this, you need to adjust the equations to reflect the slower speed. So:
::为此,您需要调整方程式以反映慢速。所以:

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 5 \cos (\frac{t}{4}) \\ y (t) & = 9 \sin (\frac{t}{4}) \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = 5cos(t4)y(t) = 9sin(t4)

Now, given the parametric , let's practice converting the equation to standard form.
::现在,考虑到这个参数,让我们实践一下, 将方程转换成标准形式。

Take the following parametric equation of an ellipse
::采取以下椭圆的参数方程

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 4 \cos (t) \\ y (t) & = 10 \sin (t) \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = 4cos(t) y(t) = 10sin(t)

The standard equation for an ellipse is $\begin{aligned} \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1 \end{aligned}$ .
::椭圆的标准方程式为 x2a2+y2b2=1。

Substitute $\begin{aligned} a \end{aligned}$ and $\begin{aligned} b \end{aligned}$ from the parametric equation to get:
::以参数方程的 a 和 b 替换 :

\begin{aligned} \frac{x^{2}}{16} + \frac{y^{2}}{100} = 1 \end{aligned}

::x216+y2100=1

You can now sketch the major and minor axes and label the $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ intercepts of the ellipse. The foci always lie on the major axis. Find them using the following equation: $\begin{aligned} a^{2} - c^{2} = b^{2} \end{aligned}$ where $\begin{aligned} a \end{aligned}$ is the distance of the edge of the ellipse from the center along the major (longer) axis, $\begin{aligned} b \end{aligned}$ is the distance along the minor axis, and $\begin{aligned} c \end{aligned}$ is the distance of the foci from the center of the ellipse.
::您现在可以绘制主轴和小轴,并标出椭圆的 x 和截取。方块总是在主轴上。使用以下方程式查找它们: a2- c2=b2, 其中方程式是椭圆边缘与主轴( 长轴) 中心之间的距离, b 是小轴的距离, c 是椭圆中间的方块距离。

In this ellipse, the vertical axis is the major axis because it is longer. To find the foci along the $\begin{aligned} y \end{aligned}$ -axis:
::在此椭圆中, 垂直轴是主轴, 因为时间较长。要在 y 轴上找到角 :

$\begin{aligned} a^{2} - c^{2} & = b^{2} \\ 100 - c^{2} & = 16 \\ - c^{2} & = - 84 \\ c & = \pm 2 \sqrt{21} \end{aligned}$

::a2-c2=b2100-c2=16-c2=284c=221

Now you can sketch the ellipse.
::现在你可以画出椭圆了

Examples
::实例

Example 1
::例1

Earlier, you were asked about a the cross country team from Forest Park High who are practicing in a nearby forest preserve. The female members of the team run a circular loop while the males run along an elliptical path. The paths are two different lengths and the teams are running at two different speeds. Suppose the girls’ path can be described by the parametric equation
::早些时候,有人问过你一个来自森林公园高中的跨国家小组的情况,他们正在附近的森林保护区里练习。团队的女性成员环绕环绕环绕,而男性则沿着椭圆路径运行。路径有两种不同的长度,团队以两种不同的速度运行。假设女孩的路径可以用参数方程式来描述。

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 100 \cos (t) \\ y (t) & = 100 \sin (t) \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = 100cos @(t) y(t) = 100sin(t)

And the boys’ path can be described by
::男孩的道路可以描述为:

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 200 \cos (\frac{t}{2}) \\ y (t) & = 50 \sin (\frac{t}{2}) \end{aligned}

::F( t) = (x( t) ,y( t) ) x( t) = 200cos( t2) y( t) = 50sin( t2)

Will the teams meet during the first three laps of their runs?
::球队将在前三圈赛跑中相遇吗?

To find out where the paths cross, put each equation into standard form.
::要找出路径的交汇点, 将每个方程式都按标准格式排列。

The equation for the girls’ path becomes:
::女孩之路的方程式变成:

\begin{aligned} x^{2} + y^{2} = (100)^{2} \end{aligned}

::x2+y2=(100)2

The equation for the boys’ path becomes:
::男孩们之路的方程式变成:

\begin{aligned} \frac{x^{2}}{(200)^{2}} + \frac{y^{2}}{(50)^{2}} = 1 \end{aligned}

::x2(200)2+y2(502=1

Put the first equation in terms of $\begin{aligned} y^{2} \end{aligned}$ :
::将第一个方程以 y2 表示 :

\begin{aligned} y^{2} = (100)^{2} - x^{2} \end{aligned}

::y2=( 1002) - x2

Substitute for $\begin{aligned} y^{2} \end{aligned}$ in the second equation:
::第二个方程式中的 y2 替代 :

\begin{aligned} \frac{x^{2}}{(200)^{2}} + \frac{(100)^{2} - x^{2}}{(50)^{2}} = 1 \end{aligned}

::x2(200)2+(1002)-2-x2(502=1)

Solve for $\begin{aligned} x \end{aligned}$ :
::解决 x:

\begin{aligned} \frac{x^{2}}{(200)^{2}} + \frac{(100)^{2} - x^{2}}{(50)^{2}} & = 1 \\ \frac{x^{2}}{40000} + \frac{10000 - x^{2}}{2500} & = 1 \\ x^{2} (2500) + 40000 (10000 - x^{2}) & = (40000) (2500) \\ 2500 x^{2} + 400000000 - 40000 x^{2} & = 100000000 \\ - 37500 x^{2} & = - 300000000 \\ x^{2} & = 8000 \\ x & = \pm 89.44 \end{aligned}

::x2(200)2+(1002)2+(1002-x2(502)2=1x240000+100000-x22500=1x2(2500)+40000(10000-x2)=(4000)(2500-25002)2+4002+4000000-40000-400000x2=100000000-37500x2=300000x2300000x2=80000x_89)。

Now, solve one of the equations for $\begin{aligned} y \end{aligned}$ .
::现在,为y解开一个方程式。

\begin{aligned} x^{2} + y^{2} & = (100)^{2} \\ 8000 + y^{2} & = 100^{2} \\ 8000 + y^{2} & = 10000 \\ y^{2} & = 2000 \\ y & = \pm 44.72 \end{aligned}

::x2+y2 = (100228000+y2=10028000+y2=10028000+y2=100000y2=2000y44.72

So, the circular and the elliptical paths cross at 4 different points:
::圆形和椭圆路径在4个不同点交叉:

$\begin{aligned} (89.44, 44.72), (- 89.44, 44.72), (- 89.44, - 44.72) \end{aligned}$ and $\begin{aligned} (89.44, - 44.72) \end{aligned}$
:89.44.72),(-89.44.72),(-89.44.72),(-89.44,-44.72),(-89.44,-44.72)和(89.44,-44.72)

Now that you know where the objects cross, you’ll need to find out when they cross. For the purposes of this problem, it’s fine to use 3.14 for $\begin{aligned} π \end{aligned}$ .
::既然你知道物体经过哪里,你就需要知道它们经过什么时候。为了这一问题的目的,用3.14来表示是好的。

To find the times that each team reaches the intersections, You’ll need to substitute each set of points back into the original parametric equations. Remember, the teams are each running 3 laps, so you’ll need to find three times for each intersection . You can do this by adding the period of the function to the initial value for each point.
::要找到每个小队到达十字路口的时间, 您需要将每组点替换为原始的参数方程。记住, 每个小队在跑三圈, 所以每个十字路口需要找三次。您可以将函数的时段加到每个点的初始值中。

Start with the girls’ paths.
::从女孩的道路开始。

Remember that both teams are traveling counterclockwise on their paths. To make sense of the data, it helps to find out where the girls are at $\begin{aligned} t = 0 \end{aligned}$ . So:
::记住,两支队伍都在逆时针穿梭在路上。为了了解数据,它有助于找出女孩们在t=0的位置。所以:

\begin{aligned} x (t) & = 100 \cos (t) \\ x (0) & = 100 \cos (0) \\ x (0) & = 100 \end{aligned}

t) = 100cos(t) x(0) = 100cos(0) x(0) = 100

And
::还有

\begin{aligned} y (t) & = 50 \sin (t) \\ y (0) & = 50 \sin (0) \\ y (0) & = 0 \end{aligned}

::y(0)=50sin(0)=50sin(0)y(0)y(0)y(0)y(0)=0

The girls begin their journey at (100, 0) and travel counterclockwise from that point.
::女孩从(100,0)起步,从那时起逆时针旅行。

Now, you can find the times at which the girls reach each intersection. The cosine function has a period of $\begin{aligned} 2 π \end{aligned}$ . By taking into account the function’s period, you only need to solve for two different intersections. You can use simple arithmetic to find the other times.
::现在,您可以找到女孩到达每个十字路口的时间。余弦函数的周期为 2。考虑到函数的周期, 您只需要解决两个不同的十字路口。您可以使用简单的算术来寻找其它的时间。

Start with (89.44, 44.72).
::从(89.44、44.72)开始。

\begin{aligned} x (t) & = 100 \cos (t) \\ 89.44 & = 100 \cos (t) \\ .8944 & = \cos (t) \\ \cos^{- 1} (.8944) & = \cos^{- 1} (\cos (t)) \\ .464 & = t \end{aligned}

t) = 100cos(t)89.44 = 100cos(t).8944 = cos(t)cos- 1((.8944) = cos- 1(cos(t).464 = t)

The period of the cosine function is $\begin{aligned} 2 π \end{aligned}$ , so the girls will also reach that intersection at $\begin{aligned} t = .464 + 2 π \end{aligned}$ and $\begin{aligned} t = .464 + 4 π \end{aligned}$ . Substituting 6.28 for $\begin{aligned} 2 π \end{aligned}$ and 12.48 for $\begin{aligned} 4 π \end{aligned}$ lets you see that they’ll reach the first intersection at $\begin{aligned} t = .464, 6.744 \end{aligned}$ , and $\begin{aligned} 13.204 \end{aligned}$ minutes into the run.
::接弦功能的时间是 2, 所以女孩也会到达十字路口 t= 464+2和 t= 464+4。替换 6.28 和 12.48 和 12. 48 , 让你看到她们会到达第一个十字路口 t= 464, 6. 744 和 13. 204 分钟。

Since they’re traveling counterclockwise, they’ll reach the second intersection next, (-89.44, 44.72).
::由于他们逆时针旅行, 他们将到达第二个十字路口, (-89.44、44.72) 。

\begin{aligned} x (t) & = 100 \cos (t) \\ - 89.44 & = 100 \cos (t) \\ - .8944 & = \cos (t) \\ \cos^{- 1} (- .8944) & = \cos^{- 1} (\cos (t)) \\ 2.68 & = t \\ 2.68 + 2 π & = 8.96 \\ 2.68 + 4 π & = 15.24 \end{aligned}

t) = 100cos(t)-89.44 = 100cos(t)-8944 = cos(t) cos- 1((-) 8944 ) = cos- 1(cos(t) ) 2.68 = t2.68+28.962.68+415.24

They’ll reach the second intersection at $\begin{aligned} t = 2.68, 8.96 \end{aligned}$ and $\begin{aligned} 15.24 \end{aligned}$ .
::他们将到达第二个十字路口 t=2.68,8.96和15.24。

To find out when the runners will reach the third intersection, (-89.44, -44.72), it helps to remember that the cosine function is symmetrical and periodic. So the third intersection, where $\begin{aligned} \cos (t) = - .8944 \end{aligned}$ , occurs the same distance before $\begin{aligned} 2 π \end{aligned}$ as the second intersection occurred after 0. Thus, intersection number three occurs at $\begin{aligned} t = 2 π - 2.68 \end{aligned}$ , or $\begin{aligned} t = 3.6 \end{aligned}$ . To find the times for the next two laps, add 6.28 and 12.56 to get 9.88 and 16.16.
::要知道赛跑者何时会到达第三个十字路口, (-89.44, - 44.72) 它可以帮助记住余弦函数是对称和周期的。因此第三个十字路口, 即cos{( t)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\T=3. \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Finally, use the symmetry of the cosine function to find the times at which the girls’ team will reach the fourth intersection. They reached the first intersection .464 minutes after they started. That means they’ll reach the fourth intersection at .464 minutes before $\begin{aligned} 2 π \end{aligned}$ , or at 5.82 minutes into their run. Since they’re running 3 laps, they’ll also reach the intersection at 12.1 minutes and 18.38 minutes into the run.
::最后,使用余弦函数的对称性来寻找女孩队到达第四个十字路口的时间。她们到达第一个十字路口的时间是开始的464分钟。这意味着她们到达第四个十字路口的时间是2° 之前的464分钟,或者跑的5-82分钟。由于她们跑三圈,她们也到达十字路口的时间是12.1分钟,跑18.38分钟。

Now, complete the same substitution process with the boys’ laps on the elliptical path. Keep in mind that the boys are completing one loop every $\begin{aligned} 4 π \end{aligned}$ minutes.
::现在,完成同样的替代过程,让男孩们在椭圆路径上大腿。记住男孩们每4 - 3 分钟完成一个循环。

\begin{aligned} x (t) & = 200 \cos (\frac{t}{2}) \\ 89.44 & = 200 \cos (\frac{t}{2}) \\ .4472 & = \cos (\frac{t}{2}) \\ \cos^{- 1} (.4472) & = \cos^{- 1} (\cos (\frac{t}{2})) \\ 1.11 & = \frac{t}{2} \\ 2.22 & = t \end{aligned}

t)=200cos(t2)89.44=200cos(t2)4472=cos(t2)cos-1((4472)=cos-1(cos(t2))1.11=t22.22=t

Now, the boys take twice as long to make a loop as the girls do. So, the boys will pass the first intersection at 2.22, 14.78, and 27.34. Using the symmetry of the cosine function, you’ll find that they pass the fourth intersection at 2.22 minutes before $\begin{aligned} 4 π \end{aligned}$ , or 10.34. They’ll pass it on their second and third laps at 22.9 and 35.46.
::现在,男孩们比女孩们要花两倍的时间来绕圈。因此,男孩们将在2.22、14.78和27.34处通过第一个十字路口。使用余弦函数的对称,你会发现他们在4或10.34之前的2.22分钟通过第四个十字路口。他们将在22.9和35.46处的第二圈和第三圈通过。

Now, solve for the second intersection.
::现在,解决第二个十字路口。

\begin{aligned} x (t) & = 200 \cos (\frac{t}{2}) \\ - 89.44 & = 200 \cos (\frac{t}{2}) \\ - .4472 & = 200 \cos (\frac{t}{2}) \\ \cos^{- 1} (- .4472) & = \cos^{- 1} (\cos (\frac{t}{2})) \\ 2.03 & = \frac{t}{2} \\ t & = 4.06 \end{aligned}

t) =200cos(t2)-89.44=200cos(t2)- 4472=200cos(t2)- (c2)- 1(- 4472) =cos- 1(cos(t2))- 2.03=t2t=4.06

Add $\begin{aligned} 4 π \end{aligned}$ and $\begin{aligned} 8 π \end{aligned}$ to get the times for laps 2 and 3. The boys cross the second intersection at 4.06, 16.62, and 29.18. Use the symmetry of the cosine function to find the times for the third intersection. Since the boys reached the second intersection at 4.06 and it takes them $\begin{aligned} 4 π \end{aligned}$ minutes to complete a loop, they’ll reach the third intersection at $\begin{aligned} 4 π \end{aligned}$ -4.06, or 8.5. Add $\begin{aligned} 4 π \end{aligned}$ and $\begin{aligned} 8 π \end{aligned}$ to get the times for laps 2 and 3. The boys reach the third intersection at 8.5, 21.06, and 33.62 minutes.
::添加 4 和 8 和 8 和 4 和 8 来获得 2 圈 2 和 3 圈的时间。男孩通过第二个十字路口的时间为 4.06 、 16.62 和 29.18 。使用余弦函数的对称来找到第三个十字路口的时间。由于男孩到达第二个十字路口的时间为 4.06 和 4 和 4 分钟来完成一个环圈, 他们到达第三个十字路口的时间为 4 - 4.06 或 8.5 。 Add 4 和 8 和 8 , 4 和 8 来获得 2 和 3. 3 圈的时间。男孩到达第三个十字路口的时间为 8.5 、 21.06 和 33.62 分钟。

You may want to create a table to organize your data, so that you can easily see when each team arrives at each intersection.
::您可能想要创建一个表格来组织您的数据, 这样您就可以很容易地看到每个团队到达每个十字路口。

	Girls lap 1 ::女生第一圈	Girls lap 2 ::女孩2圈	Girls lap 3 ::女孩三圈	Boys lap 1 ::男孩1圈	Boys lap 2 ::男孩2圈	Boys lap 3 ::男孩大腿3
(89.44, 44.72)	.464	6.744	13.205	2.22	14.78	27.34
(-89.44, 44.72)	2.68	8.96	15.24	4.06	16.62	29.18
(-89.44, -44.72)	3.6	9.88	16.16	8.5	21.06	33.62
(89.44, -44.72)	5.82	12.1	18.38	10.34	22.9	35.46

As you can see, the teams don’t meet during the first three laps of the practice.
::训练队在训练前三圈不见面。

Example 2
::例2

Find a parametric equation to describe the movement of an object along the following circular path. Assume the object completes a lap every $\begin{aligned} π \end{aligned}$ units of time.
::查找一个参数方程来描述物体沿着以下圆形路径的动向。假设该物体每时单位完成一圈。

\begin{aligned} x^{2} + y^{2} = 16 \end{aligned}

::x2+y2=16

Notice that the circle has a radius of 4. Also notice that the object completes a lap in $\begin{aligned} π \end{aligned}$ units of time which is half the normal time. So the equation is:
::注意圆圆半径为 4 。也注意, 对象在 / 时间单位内完成一圈, 时间单位是正常时间的一半。方程式是 :

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 4 \cos (2 t) \\ y (t) & = 4 \sin (2 t) \end{aligned}

::F( t) = (x( t) ,y( t) ) x( t) = 4cos( 2t) y( t) = 4sin( 2t)

Example 3
::例3

Given the following parametric equation of a circle, find the standard equation.
::根据圆形的以下参数方程,找到标准方程。

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 12 \cos (4 π t) \\ y (t) & = 12 \sin (4 π t) \end{aligned}

::F( t) = (x( t) ,y( t) ) x( t) = 12cos *( 4t) y( t) = 12sin ( 4t)

Notice that the radius is 12. When you put the equation in standard form, you lose the information about how fast the object travels along the path. Instead, you just have an equation for the path. Therefore, you don't need to consider the $\begin{aligned} 4 π \end{aligned}$ when writing in standard form. The equation in standard form is:
::注意半径为 12 。当您将方程式设置为标准格式时, 您会丢失关于天体沿路径行驶速度的信息。相反, 您只需要有一个路径的方程式。因此, 在以标准格式写入时, 您不需要考虑 4 。标准格式的方程式是 :

\begin{aligned} x^{2} + y^{2} = 144 \end{aligned}

::x2+y2=144 键

Example 4
::例4

Write the equation for a circle centered at (4, 2) with a radius of 5 in both standard and parametric form.
::写出圆的方程,以(4,2)为中心,半径为5,以标准形式和参数形式写出。

The standard equation for a circle is with a center at (0, 0) is $\begin{aligned} x^{2} + y^{2} = r^{2} \end{aligned}$ , where r is the radius of the circle. For a circle centered at (4, 2) with a radius of 5, the standard equation would be $\begin{aligned} (x - 4)^{2} + (y - 2)^{2} = 25 \end{aligned}$ . The parametric form for the equation of a circle is:
::圆的标准方程式是圆的中心值( 0, 0) 是 x2+y2=r2, 圆的半径为 r。圆的中心值为 5, 圆的圆( 4, 2), 标准方程式为 (x- 4) 2+(y- 2) 2=25。圆的方程式的参数表为 :

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = r \cos (t) + h \\ y (t) & = r \sin (t) + k \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = rcos(t) +hy(t) = rsin(t)+k

So the parametric equation for this circle would be:
::这个圆的参数方程是:

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 5 \cos (t) + 4 \\ y (t) & = 5 \sin (t) + 2 \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = 5cos(t)+4y(t) = 5sin(t)+2

Example 5
::例5

A planet is traveling in an elliptical orbit described by the equation $\begin{aligned} \frac{x^{2}}{36} + \frac{y^{2}}{25} = 1 \end{aligned}$ . The planet makes one orbit every $\begin{aligned} 2 π \end{aligned}$ units of time. An asteroid belt extends along the line $\begin{aligned} y = \frac{1}{2} x - 3 \end{aligned}$ . At what times will the planet pass through the asteroid belt during its first orbit?
::一个行星正在以方程式 x236+y225=1. 描述的椭圆轨道上飞行,该行星每2°C时就设定一个轨道,小行星带沿y=12x-3 线延伸,小行星带在第一个轨道上何时穿过小行星带?

First, solve the equations to find out the locations where the asteroid belt intersects the orbit. Use substitution and the quadratic formula in order to find the coordinates:
::首先,解开方程式以找出小行星带相交轨道的位置。使用替代方程式和二次方程式寻找坐标:

\begin{aligned} \frac{x^{2}}{36} + \frac{y^{2}}{25} & = 1 \\ y & = \frac{1}{2} x - 3 \\ \frac{x^{2}}{36} + \frac{{(\frac{1}{2} x - 3)}^{2}}{25} = 1 \\ 25 x^{2} + 9 x^{2} - 108 x + 324 = 900 \\ 34 x^{2} - 108 x - 576 = 0 \end{aligned}

::x236+y225=1y=12x-3x236+(12x-33)225=125x2+9x2-108x+324=90034x2-108x-576=0

\begin{aligned} x & = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} \\ x & = \frac{108 \pm \sqrt{900^{2} - 4 (34) (576)}}{2 (34)} \\ x & = 6, x = \frac{- 48}{17} \end{aligned}

::xbb2 - 4ac2ax=1089002-4(34)(576)(34)(34)(34)x=6,x4817

Now substitute $\begin{aligned} x = 6 \end{aligned}$ into one of the equations to find the corresponding $\begin{aligned} y \end{aligned}$ -coordinate:
::现在将 x=6 替换为一个方程式, 以找到相应的 Y 坐标 :

\begin{aligned} y & = \frac{1}{2} x - 3 \\ y & = \frac{1}{2} (3) - 3 \\ y & = 0 \end{aligned}

::y=12x-3y=12(3)--3y=0

The first point of intersection is (6, 0).

::第一个交叉点是(6,0)。

Now, substitute $\begin{aligned} x = - \frac{48}{17} \end{aligned}$ into one of the equations to find the corresponding $\begin{aligned} y \end{aligned}$ -coordinate:
::现在, 将 x4817 替换为一个方程式, 以找到相应的 Y 坐标 :

\begin{aligned} y & = \frac{1}{2} x - 3 \\ y & = \frac{1}{2} (\frac{- 48}{17}) - 3 \\ y & = \frac{- 75}{17} \end{aligned}

::y=12x-3y=12(- 4817)-3y=7517

The second point of intersection is $\begin{aligned} (- \frac{48}{17}, - \frac{75}{17}) \end{aligned}$ .
::第二个交叉点是(-4817,-7517)。

Next, write a parametric equation for the ellipse.
::下一个,写一个对数方程用于椭圆。

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 6 \cos (t) \\ y (t) & = 5 \sin (t) \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = 6cos(t) y(t) = 5sin(t)

Finally, substitute in the two sets of coordinates in order to find the times when the planet will pass through the asteroid belt.
::最后,用两组坐标取代这两组坐标,以便找到行星穿过小行星带的时间。

\begin{aligned} (6, 0) \\ 6 & = 6 \cos (t) \\ 1 & = \cos (t) \\ t & = 0 \end{aligned}

6,0)6=6cos(t)1=cos(t)t=0)

\begin{aligned} (\frac{- 48}{17}, \frac{- 75}{17}) \\ - \frac{48}{17} = 6 \cos (t) \\ \frac{- 6}{17} = \cos t \\ t = 2.06 \end{aligned}

-4817,-7517-4817=6cos(t)-617=cost=2.06)

The planet will pass through the asteroid belt at $\begin{aligned} t = 0 \end{aligned}$ and $\begin{aligned} t = 2.06 \end{aligned}$ during its first orbit.
::该行星将在其第一个轨道上以t=0和t=2.06通过小行星带。

Review
::回顾

Write a parametric equation for the ellipse defined by the equation $\begin{aligned} \frac{x^{2}}{36} + \frac{y^{2}}{4} = 1 \end{aligned}$ , where an object makes one revolution every $\begin{aligned} 4 π \end{aligned}$ units of time.
::为方程式 x236+y24=1 定义的椭圆写参数方程,其中对象每4- 单位时间进行一次革命。

Write a parametric equation for the circle defined by the equation $\begin{aligned} x^{2} + y^{2} = 16 \end{aligned}$ , where an object makes one revolution every $\begin{aligned} 8 π \end{aligned}$ units of time.
::为方程式 x2+y2=16 定义的圆形写一个参数方程,其中对象每8- 单位时间进行一次革命。

Write a parametric equation for the ellipse defined by the equation $\begin{aligned} \frac{x^{2}}{100} + \frac{y^{2}}{256} = 1 \end{aligned}$ , where an object makes one revolution every $\begin{aligned} 2 π \end{aligned}$ units of time.
::为方程式 x2100+y2256=1 定义的椭圆写参数方程,其中对象每2°C的时间单位发生一次革命。

Write a parametric equation for the ellipse defined by the equation $\begin{aligned} \frac{x^{2}}{400} + \frac{y^{2}}{196} = 1 \end{aligned}$ , where an object makes one revolution every $\begin{aligned} 10 π \end{aligned}$ units of time.
::为方程式 x2400+y2196=1 定义的椭圆写参数方程,其中对象每10°C的时间单位进行一次革命。

Given the following parametric equation of an ellipse, write the equation in standard form. Then find the foci of the ellipse and sketch the ellipse.
::根据以下的椭圆的参数方程, 请以标准格式写出方程。然后找到椭圆的方程, 并绘制椭圆的草图。

\begin{aligned} F (t) & = (x (t), y (t)) \\ x (t) & = 14 \cos (\frac{t}{4}) \\ y (t) & = 8 \sin (\frac{t}{4}) \end{aligned}

::F(t) = (x(t) y(t) ) x(t) = 14cos(t4)y(t) = 8sin(t4)

Given the following parametric equation of an ellipse, write the equation in standard form. Then find the foci of the ellipse and sketch the ellipse.
::根据以下的椭圆的参数方程, 请以标准格式写出方程。然后找到椭圆的方程, 并绘制椭圆的草图。

\begin{aligned} G (t) & = (x (t), y (t)) \\ x (t) & = \cos (\frac{t}{3}) \\ y (t) & = 2 \sin (\frac{t}{3}) \end{aligned}

t) = (x(t),y(t))x(t) = cos(t3)y(t) = 2sin(t3)

Given the following parametric equation of a circle, write the equation in standard form. Sketch the circle.
::根据圆的以下参数方程,以标准格式写出方程。绘制圆形。

\begin{aligned} H (t) & = (x (t), y (t)) \\ x (t) & = 5 \cos (3 t) \\ y (t) & = 5 \sin (3 t) \end{aligned}

t)=(x(t),y(t))x(t)=5cos(3t)y(t)=5sin(3t)

Write the equation for a circle centered at (1, 7) with a radius of 8 in both standard and parametric form.
::写一个圆形的方程,以(1、7)为中心,半径为8,以标准形式和参数形式同时写出。

Write the equation for a circle centered at (6, -9) with a radius of 12 in both standard and parametric form.
::写出圆心( 6, 9) 的方程, 半径为12, 以标准形式和参数形式。

Write the following parametric equation in standard form. Then, make a sketch of the curve.
::以标准形式写下以下的参数方程。然后绘制曲线的草图。

\begin{aligned} M (t) & = (x (t), y (t)) \\ x (t) & = \cos (t) \\ y (t) & = - 4 + 2 \sin (t) \end{aligned}

t) = (x(t),y(t))x(t) = cos(t)y(t) 4+2sin(t)

Find parametric equations for the circle $\begin{aligned} x^{2} + y^{2} - 4 x = 0 \end{aligned}$ .
::查找圆 x2+y2-4x=0 的参数方程。

Find parametric equations for the ellipse $\begin{aligned} x^{2} + 4 y^{2} - 2 x = 15 \end{aligned}$
::查找椭圆 x2+4y2-2-2x=15的参数方程

For #13-#15: A track team is traveling in an elliptical path around a track. The path can be described by the equation $\begin{aligned} \frac{x^{2}}{16} + \frac{y^{2}}{49} = 1 \end{aligned}$ . The track team makes one lap every $\begin{aligned} π \end{aligned}$ minutes. A very long hose extends across the entire track along the line $\begin{aligned} y = 2 x + 1 \end{aligned}$ .
::# 13- # 15 : 轨道团队正在一条轨道周围的椭圆路径中行走。路径可以用方程 x216+y249=1 来描述。轨道团队每 + 分钟做一圈。一条非常长的软管沿着 y= 2x+1 线贯穿整个轨道。

At what locations will the track team have to jump over the hose? Give approximate answers rounded to the nearest thousandth.
::轨道小组在哪个地点必须跳过水管?

Write a parametric equation to model the path that the track team makes.
::写一个参数方程以模拟轨道队的路径

At what times will the track team have to jump over the hose during their first three laps? Round answers to the nearest hundredth.
::轨道队什么时候才能在前三圈跳过水管?

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

圆形和椭圆的参数等量

Parametric Equations for Circles and Ellipses ::圆形和椭圆的参数等量

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)