11.4 矢量值函数的衍生因素

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  矢量值函数的衍生因素

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  Karl is tracking the motion of a satellite across the surface of the earth. He finds that, during a certain time period, the satellite’s movement can be modeled by the function $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,\sin 2t)\end{array}$ . He’d like to know how the satellite’s movement is changing when  $\begin{array}{r}t=5\end{array}$ seconds. How can he find a vector to describe this rate of change?
  ::Karl正在跟踪卫星在地球表面的移动。 他发现,在一定的一段时间里,卫星的移动可以由函数F(t)=(t2,sin)=(t2,sin)=2t)来模拟。他想知道卫星的移动在t=5秒时是如何变化的。他如何找到一个矢量来描述这种变化速度?

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  Derivatives of Vector-Valued Functions
  ::矢量值函数的衍生因素

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  Taking the derivative of a vector-valued function is a lot like taking the derivative of a scalar function. Vector-valued functions have multiple components, and each component is a function. For example, $\begin{array}{r}\stackrel{\to }{F}(t)=\left(t^3,2t+7,\frac{1}{t}\right)\end{array}$  is a vector-valued function in 3 dimensions. To take the derivative of $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ , you would take the derivative of each component. Your result would be $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=\left(3t^2,2,-\frac{1}{t^2}\right)\end{array}$ .
  ::获取矢量值函数的衍生物与获取星标函数的衍生物非常相似。 矢量值函数包含多个元件, 每个元件都是一个函数。 例如, F( t) = (t3, 2t+7, 1t) 是三维的矢量值函数。 要获取 F( t) 的衍生物, 您将选择每个元件的衍生物。 您的结果将是 F( t) = (3t2, 2- 1t2)。

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  In order for a vector-valued function to be differentiable at a point, all of its component functions must be differentiable at that point.
  ::为使矢量估值的功能在某一点可以区分,其所有组成部分功能在那一点必须是可区分的。

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  Let's find the derivative of the vector-valued function  $\begin{array}{r}\stackrel{\to }{F}(t)=(\sin t,\ln t)\end{array}$ when  $\begin{array}{r}t=\pi \end{array}$ and when $\begin{array}{r}t=0\end{array}$ .
  ::让我们在 t 和 t=0 时找到矢量值函数 F( t) =( sint, lnt) 的衍生物 。

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  First, find the derivative of each component of the vector-valued function.
  ::首先,找到矢量价值函数每个组成部分的衍生物。

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  $$\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=\left(\cos t,\frac{1}{t}\right)\end{array}$$

  
  ::F(t)=(cost,1t)

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  Now, evaluate the derivative for $\begin{array}{r}t=\pi \end{array}$ .
  ::现在,评估衍生物的T。

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  $$\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(\pi )=\left(\cos \pi ,\frac{1}{\pi }\right)=\left(-1,\frac{1}{\pi }\right).\end{array}$$

  
  ::F()=(cos,1)=(1,1)。

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  The vector  $\begin{array}{r}\left(-1,\frac{1}{\pi }\right)\end{array}$ is tangent to the curve when $\begin{array}{r}t=\pi \end{array}$ .
  ::矢量 (-1,1) 在 t 时正切到曲线 。

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  You cannot find a tangent to the curve when $\begin{array}{r}t=0\end{array}$ , because the second component of the derivative function, $\begin{array}{r}\frac{1}{t}\end{array}$ , is undefined at $\begin{array}{r}t=0\end{array}$ . Since you can’t find $\begin{array}{r}\frac{1}{0}\end{array}$ , you can’t find the tangent vector when $\begin{array}{r}t=0\end{array}$ .
  ::t=0 时无法找到曲线的正切值,因为衍生函数的第二个组件 1t 在 t=0 时未定义。由于找不到 10, 在 t=0 时无法找到正切矢量 。

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  You can use the derivative of a vector-valued function at a specific point to find the unit vector tangent to the curve at that point. The unit tangent vector is useful for solving problems where you need to know the direction of change of a vector-valued function, but not the magnitude of that change.
  ::您可以在一个特定点使用矢量估值函数的衍生物来找到该点的曲线的单位矢量正切值。单位正切值矢量对于解决需要了解矢量估值函数变化方向而不是该变化规模的问题非常有用。

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  To find the unit tangent vector for a function, it helps to first rewrite the function in terms of its component vectors. For a function in two dimensions, you can use $\begin{array}{r}i\end{array}$  to represent the horizontal component and  $\begin{array}{r}j\end{array}$  to represent the vertical component .
  ::要找到函数的单位正切矢量,它有助于首先根据其组成矢量重写函数。对于两个维度的函数,您可以使用 i 代表水平元件,j 代表垂直元件。

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  Applying this information, let's find the unit tangent vector for the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,\sin (t))\end{array}$ .
  ::在应用此信息时, 我们为矢量值函数 F( t) =( t2, sin( t)) 找到单位正切矢量。

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  First, rewrite the function in terms of $\begin{array}{r}i\end{array}$  and $\begin{array}{r}j\end{array}$ .
  ::首先,按照i和j重写该函数。

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=t^{2}i+\sin (t)j\end{array}$$

  
  ::F(t) = t2i+sin(t)j

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  Find the derivative.
  ::找到衍生物

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  $$\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=2ti+\cos (t)j\end{array}$$

  
  ::F(t)=2ti+cos(t)j

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  Use the distance formula to find the magnitude of tangent vector.
  ::使用距离公式查找正切矢量的大小。

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  $$\begin{array}{r}\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\Vert =\sqrt{4t^2+{\cos }^2(t)}\end{array}$$

  
  :t)4t2+cos2(t)

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  Divide the tangent vector by its magnitude to find the unit tangent vector.
  ::将正切矢量除以其大小以找到单位正切矢量。

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  $$\begin{array}{r}\stackrel{\to }{T}(t)=\frac{\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)}{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\Vert }\\ \stackrel{\to }{T}(t)=\frac{2ti+\cos (t)j}{\sqrt{4t^2+{\cos }^2(t)}}=\frac{2t\left(\sqrt{4t^2+{\cos }^2(t)}\right)}{4t^2+{\cos }^2(t)}i+\frac{\cos (t)\left(\sqrt{4t^2+{\cos }^2(t)}\right)}{4t^2+{\cos }^2(t)}j\end{array}$$

  
  ::T(t) = F(t) (t) (t) = 2ti+cos(t) j4t2+cos2}(t) = 2t(4t2+cos2(t) ) 4t2+cos2}}(t) 4t2+cos2(t) i+cos(t)(4t) 4t2+cos2(t) 4t2+cos2}(t) j

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  Now, you can evaluate the unit tangent vector at a specific time to find the direction of change at that time. For instance, to evaluate this vector at $\begin{array}{r}t=10\end{array}$ , plug 10 into the formula and solve.
  ::现在, 您可以在特定时间对单位正切矢量进行评价, 以找到该时间的变化方向。 例如, 在 t=10 处对此矢量进行评价, 将 10 插入公式并解析 。

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  $$\begin{array}{rl}\stackrel{\to }{T}(t)& =\frac{2(10)\left(\sqrt{4(10{)}^2+{\cos }^2(10)}\right)}{4(10{)}^2+{\cos }^2(10)}i+\frac{\cos (10)\left(\sqrt{4(10{)}^2+{\cos }^2(10)}\right)}{4(10{)}^2+{\cos }^2(10)}j\\ \stackrel{\to }{T}(t)& =.999i+.049j\end{array}$$

  
  ::T(t) = 2(10) (10)(4(10)) 2+cos2 (10) 4(10) 2+cos2 (10) (10) (10)(4(10)) 2+cos2 (10) (4) (10) 2+cos2 (10) (10)jT(t) =.9999i+.049j

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked how Karl will find a vector that describes the rate of change of the satellite’s position when $\begin{array}{r}t=5\end{array}$ . He’ll start out by taking the derivative of the vector-valued function that describes the satellite’s position with respect to time.
  ::早些时候,有人问Karl如何找到一个矢量来描述卫星在t=5时的位置变化速度。 他将首先从矢量估值函数的衍生物开始,该函数将描述卫星在时间上的位置。

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,\sin 2t)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(2t,2\cos 2t)\end{array}$$

  
  ::F(t) =( t2, sin2t) F(t) =( 2t, 2cos2t)

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  Now, he can evaluate the tangent vector at time $\begin{array}{r}t=5\end{array}$ .
  ::现在,他可以在t=5时对正切矢量进行评估。

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  $\begin{array}{r}{\stackrel{\to }{F}}^{\mathrm{\prime }}(t)=(2(5),2\cos 2(5))=(10,1.97)\end{array}$
  ::F(t)=(2(5),2cos2(5))=(10,1.97)

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  He can use the distance formula to find the magnitude of the vector, and the definition of tangent to find its angle.
  ::他可以使用距离公式来找到矢量的大小, 和正切的定义来找到它的角 。

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  $$\begin{array}{r}\sqrt{{10}^2+{1.97}^2}=\sqrt{103.88}=10.19\\ \tan \theta =\frac{1.97}{10}=.197\\ \theta ={11.14}^{\circ }\end{array}$$

  
  ::102+1.972=103.88=10.19tan=10.19tan=1.9710=.977.11.14=11.14=10.19tan=1.9710=11.

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  The satellite’s position is changing at a rate of $\begin{array}{r}10.19\text{units}/\text{second}\end{array}$  and at an angle of 11.14 degrees.
  ::该卫星的位置正在以每秒10.19个单位和11.14度角的速度变化。

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  Example 2
  ::例2

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  A particular small plane’s flight path after it leaves the airport can be described by the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(\sin t,t^3)\end{array}$ . How far is the plane from the airport when $\begin{array}{r}t=10\end{array}$ ? How is the plane’s position changing at that time? How quickly is the plane moving (the magnitude of the tangent vector) and in what direction?
  ::特定小飞机离开机场后的飞行路线可以用矢量值函数F(t)=(sint,t3)来描述,飞机离机场有多远,t=10?飞机当时的位置如何变化?飞机移动的速度(相向矢量的大小)和方向如何?

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  To find the plane’s distance from the airport, find the magnitude of the vector described by $\begin{array}{r}\stackrel{\to }{F}(10)\end{array}$ .
  ::为了找到飞机距离机场的距离,找到F(10)所述矢量的大小。

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  $\begin{array}{r}\stackrel{\to }{F}(10)=(\sin (10),{10}^3)=(.174,1000)\end{array}$
  ::F(10)=(sin(10),103)=(174,1000)

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  Use the distance formula to find the magnitude.
  ::使用距离公式查找音量。

  $\begin{array}{r}\sqrt{{.174}^2+{1000}^2}\approx 1000\end{array}$

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  At $\begin{array}{r}t=10\end{array}$ , the plane is 1000 units from the airport.
  ::T=10时,飞机来自机场1000个单位。

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  To find the tangent vector, take the derivative of the original vector-valued function:
  ::要找到正切矢量,请使用原矢量估值函数的衍生物:

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(\sin (t),t^3)\end{array}$
  ::F(t) = (sin(t),t3)

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  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(\cos (t),3t^2)\end{array}$
  ::F(t)=(cos(t),3t2)

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  Now, to find how quickly the plane is moving and in what direction, evaluate $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\end{array}$  for $\begin{array}{r}t=10\end{array}$ .
  ::现在,要找到飞机移动的速度和方向, 请对 F( t) t= 10 进行评价 。

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  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(10)=(\cos (10),3(10{)}^2)=(.985,300)\end{array}$
  ::F(10)=(cos(10),3(10)2)=(985 300)

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  Find the magnitude of the tangent vector to find the function’s rate of change. Find the angle of the tangent vector to find the direction of change:
  ::查找正切矢量的大小以查找函数的变速率。 查找正切矢量的角以找到变化的方向 :

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  $$\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(10)=(\cos (10),3(10{)}^2)=(.985,300)\\ \sqrt{{.985}^2+{300}^2}\approx 300.002\\ \tan \theta =\frac{3t^2}{\cos (t)}=\frac{300}{.985}=304.57\\ \theta ={89.81}^{\circ }\end{array}$$

  
  ::F(10)=(cos(10),3(10)2=(985,300). 9852+3002300. 002tan3t2cos(t)=300.985=304.5789.81

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  The plane’s location is changing at a rate of 300.002 units per unit of at an angle of 89.81 degrees with respect to the airport.
  ::飞机位置的变化速度为每单位300.002个单位,相对于机场而言,角度为89.81度。

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  For the following three examples, use these vector-valued functions:
  ::对于以下三个例子,使用这些矢量估值功能:

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  1. $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,-t)\end{array}$
     ::F(t) = (t2,-t)
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  2. $\begin{array}{r}\stackrel{\to }{G}(t)=(\cos t,t+7)\end{array}$
     ::G(t) = (cost,t+7)
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  3. $\begin{array}{r}\stackrel{\to }{H}(t)=(\ln t,2t^3)\end{array}$
     ::H(t) = (lnt,2,2,3)

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  Example 3
  ::例3

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  Find the derivative of each function. Evaluate the derivative of each function at $\begin{array}{r}t=2\end{array}$ .
  ::查找每个函数的衍生物。 评价每个函数的衍生物 att=2。

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  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(2t,-1)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(2)=(2(2),-1)=(4,-1)\end{array}$
  ::F(t) = (2吨,-1) F(2) = (2(2),-1) = (4,-1)

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  $\begin{array}{r}\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(-\sin t,1)\\ \stackrel{\to }{G}{}^{\mathrm{\prime }}(2)=(-\sin (2),1)=(-.035,1)\end{array}$
  ::G(t) = (- 辛, 1) G(2) = (- 辛, 2) = (- 辛, 2, 1) = (- 035, 1)

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  $\begin{array}{r}\stackrel{\to }{H}{}^{\mathrm{\prime }}(t)=\left(\frac{1}{t},6t^2\right)\\ \stackrel{\to }{H}{}^{\mathrm{\prime }}(2)=\left(\frac{1}{2},6(2{)}^2\right)=\left(\frac{1}{2},24\right)\end{array}$
  ::H(t) =(1吨,6吨2,H) = (12,6吨2) = (12,6吨2) = (12,24)

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  Example 4
  ::例4

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  Find the magnitude and direction of change for each function at  $\begin{array}{r}t=2\end{array}$
  ::查找 t=2 的每个函数变化的大小和方向

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  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(2)=(4,-1)\\ \Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(2)\Vert =\sqrt{4^2+(-1{)}^2}=\sqrt{17}\\ \tan \theta =\frac{-1}{4}=-.25\\ \theta =-{14.04}^{\circ }\end{array}$
  ::F(2)=( 4, - 1, )\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  $\begin{array}{r}\stackrel{\to }{G}{}^{\mathrm{\prime }}(2)=(-.035,1)\\ \Vert \stackrel{\to }{G}{}^{\mathrm{\prime }}(2)\Vert =\sqrt{(-.035{)}^2+1^2}=1.001\\ \tan \theta =\frac{-1}{.035}=-{87.995}^{\circ }\end{array}$
  ::G(2)=(-.035,1,1,_G)(2)(-.035)2+12=1.0011.03587.995

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  $\begin{array}{r}\stackrel{\to }{H}{}^{\mathrm{\prime }}(2)=\left(\frac{1}{2},24\right)=(.5,24)\\ \Vert \stackrel{\to }{H}{}^{\mathrm{\prime }}(2)\Vert =\sqrt{{.5}^2+{24}^2}=24.005\\ \tan \theta =\frac{24}{.5}=48\\ \theta ={88.81}^{\circ }\end{array}$
  ::H(2)=(12,24)=(5,24) H(2) .52+242=2,4005tan24.5=4888.81

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  Example 5
  ::例5

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  Find the endpoint for the unit vector tangent to each function at $\begin{array}{r}t=2\end{array}$ . Remember that you’ve already found the magnitude of each tangent vector. This should simplify your calculations.
  ::查找单位矢量切换值的端点到每个函数 att=2. 记住您已经找到了每个正切矢量的大小。 这将简化您的计算 。

  1.  

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  $\begin{array}{r}T(2)=\frac{\stackrel{\to }{F}{}^{\mathrm{\prime }}(2)}{||\stackrel{\to }{F}{}^{\mathrm{\prime }}(2)||}=\frac{4i-j}{\sqrt{17}}=\frac{4\sqrt{17}}{17}i-\frac{\sqrt{17}}{17}j=.97i-.24j\\ (.97,-.24)\end{array}$
  ::T(2)=F(2)F(2)(2)4i-j17=41717i-1717j=.97i-24j(97,-24)

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  $\begin{array}{r}T(2)=\frac{\stackrel{\to }{G}{}^{\mathrm{\prime }}(2)}{||\stackrel{\to }{G}{}^{\mathrm{\prime }}(2)||}=\frac{-.035i+j}{1.001}=-.035i+.999j\\ (-.035,.999)\end{array}$
  ::T(2)=G(2)(2).03hi+j1.001.03hi+.9999j(-.035.9999)

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  Notice that  $\begin{array}{r}\stackrel{\to }{G}{}^{\mathrm{\prime }}(2)\end{array}$ was already very close to being a unit vector because its magnitude is very close to 1.
  ::请注意 G(2) 已经非常接近为单位矢量, 因为它的大小非常接近 1 。

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  $\begin{array}{r}T(2)=\frac{\stackrel{\to }{H}{}^{\mathrm{\prime }}(2)}{\Vert \stackrel{\to }{H}{}^{\mathrm{\prime }}(2)\Vert }=\frac{.5i+24j}{24.005}=.0208i+.9998j\\ (.0208,.9998)\end{array}$
  ::T(2)=H(2)(2)(5)i+24j240005=.0208i+.9998j(.028.9998)

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  Review
  ::回顾

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  For #1-5, find the derivative of the function. Evaluate the derivative of the function at $\begin{array}{r}t=5\end{array}$ .
  ::对于 # 1 5, 找到函数的衍生物 。 在 t= 5 处评价函数的衍生物 。

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  1.   $\begin{array}{r}\stackrel{\to }{F}(t)=\left(t^2,\frac{1}{t}\right)\end{array}$
     ::F(t) = (t2,1吨)
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  2.   $\begin{array}{r}\stackrel{\to }{G}(t)=(6\sin t,3\cos t)\end{array}$
     ::G(t) = (6sint,3cost)
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  3.   $\begin{array}{r}\stackrel{\to }{H}(t)=(t+1,t^2)\end{array}$
     ::H(t) = (t+1,t2)
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  4.   $\begin{array}{r}\stackrel{\to }{K}(t)=(2t^2,t-5)\end{array}$
     ::K(t) =( 2t2, t- 5)
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  5.   $\begin{array}{r}\stackrel{\to }{M}(t)=(\ln t,\sin t)\end{array}$
     ::M(t) = (lnt,sint)

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  For #6-10, find the magnitude and direction of change for the function at $\begin{array}{r}t=5\end{array}$ .
  ::对于# 6- 10, 请在 t= 5 找到函数的更改大小和方向 。

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  6.   $\begin{array}{r}\stackrel{\to }{F}(t)=\left(t^2,\frac{1}{t}\right)\end{array}$
     ::F(t) = (t2,1吨)
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  7.   $\begin{array}{r}\stackrel{\to }{G}(t)=(6\sin t,3\cos t)\end{array}$
     ::G(t) = (6sint,3cost)
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  8.   $\begin{array}{r}\stackrel{\to }{H}(t)=(t+1,t^2)\end{array}$
     ::H(t) = (t+1,t2)
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  9.   $\begin{array}{r}\stackrel{\to }{K}(t)=(2t^2,t-5)\end{array}$
     ::K(t) =( 2t2, t- 5)
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  10.   $\begin{array}{r}\stackrel{\to }{M}(t)=(\ln t,\sin t)\end{array}$
      ::M(t) = (lnt,sint)

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  For #11-15, find the endpoint for the unit vector tangent to the function at $\begin{array}{r}t=5\end{array}$ .
  ::# 11-15, 找到单位矢量的端点, 与函数的正切值为 t=5 。

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  11.   $\begin{array}{r}\stackrel{\to }{F}(t)=\left(t^2,\frac{1}{t}\right)\end{array}$
      ::F(t) = (t2,1吨)
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  12.   $\begin{array}{r}\stackrel{\to }{G}(t)=(6\sin t,3\cos t)\end{array}$
      ::G(t) = (6sint,3cost)
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  13.   $\begin{array}{r}\stackrel{\to }{H}(t)=(t+1,t^2)\end{array}$
      ::H(t) = (t+1,t2)
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  14.   $\begin{array}{r}\stackrel{\to }{K}(t)=(2t^2,t-5)\end{array}$
      ::K(t) =( 2t2, t- 5)
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  15.   $\begin{array}{r}\stackrel{\to }{M}(t)=(\ln t,\sin t)\end{array}$
      ::M(t) = (lnt,sint)

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  Review (Answers)
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