11.5 矢量值函数的差别属性

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  矢量值函数的差别属性

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  Tessa, an engineer who dabbles in steampunk creations, has built a clockwork horse. She’d like it to be able to pull a float in her town’s annual parade, so she needs to calculate how much work the clockwork engine needs to do. Since both the force and the distance may vary at different points during the parade, she also needs to understand how the work being done will change at different points during the parade route. Can she use the differentiation properties of vector-valued functions to solve her problem?
  ::Tessa是一位在蒸汽池创造中挥霍的工程师,他造了一匹时钟马。 她希望能够在其城镇的年度游行中拉出浮标,因此她需要计算时钟引擎需要做多少工作。 由于在游行期间,力量和距离可能在不同地点有所不同,她也需要了解在游行路线的不同地点正在开展的工作将如何改变。 她能否使用矢量值功能的不同属性来解决她的问题?

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  Differentiation Properties of Vector-Valued Functions
  ::矢量值函数的差别属性

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  The of vector-valued functions follow rules similar to the derivatives of scalar functions and . You can use these properties of derivatives to differentiate complex functions quickly. The following problems  will walk you through a few of the most useful properties for working with vector-valued functions.
  ::矢量估值函数遵循类似于 运算符函数衍生物的规则 。 您可以使用这些衍生物的属性快速区分复杂的函数 。 以下问题将显示您在使用矢量估值函数时使用的一些最有用的属性 。

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  Property: The derivative of the sum of two vector-valued functions is equal to the sum of each function's derivative. That is: $\begin{array}{r}\left(\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\end{array}$ .
  ::属性:两个矢量估值函数总和的衍生物等于每个函数的衍生物的总和。 也就是说 F(t)+G(t)) (t)+F(t)+G(t)) 。

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  Consider these vector-valued functions:
  ::考虑这些矢量估值功能:

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,5t)\\ \stackrel{\to }{G}(t)=(\sin t,\cos t)\end{array}$
  ::F(t) = (t2,5t) G(t) = (sint,cost)

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  Let's show that $\begin{array}{r}\left(\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\end{array}$ .
  ::让我们来显示 (F(t)+G(t)) @(t)+G(t)+G(t)) 。

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  If you add the functions together, you get:
  ::如果将函数加在一起,将获得:

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  $$\begin{array}{r}\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)=(t^2,5t)+(\sin t,\cos t)=(t^2+\sin t,5t+\cos t).\end{array}$$

  
  ::F(t)+G(t)=(t2,5t)+(sint,cost)=(t2+sint,5t+cost)。

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  If you take the derivative of the sum of the functions, you have:
  ::如果取自函数总和的衍生物,则有:

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  $$\begin{array}{r}\left(\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=(2t+\cos t,5-\sin t).\end{array}$$

  
  :F(t)+G(t)) (2t+cos@t,5-sint)。

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  On the other hand, if you take the derivatives first and then add the derivatives together, you get:
  ::另一方面,如果你先取衍生物,然后将衍生物加在一起,你就会得到:

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  $$\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(2t,5)\\ \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(\cos t,-\sin t)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(2t+\cos t,5-\sin t)\end{array}$$

  
  ::F(t) = (2t,5,G) = (cost,-sint) F(t) +G(t) = (2t+cost,5-sint)

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  So, $\begin{array}{r}\left(\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)+{\stackrel{\to }{G}}^{\mathrm{\prime }}(t)\end{array}$ .
  ::所以,(F(t)+G(t)) F(t)+G(t) 。

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  You can use this identity in cases where you know the derivative of one of the component functions and of the combined function, but not of the other component function.
  ::如果您知道一个组件函数和合并函数的衍生物,但不知道另一个组件函数的衍生物,您可以使用此身份。

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  A scalar function takes an input and returns a single output instead of a vector.
  ::scalar 函数需要输入, 返回单个输出, 而不是矢量 。

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  Property: If you multiply a scalar function by a vector-valued function , the derivative of their product will follow the product rule for derivatives. $\begin{array}{r}\left(f(t)\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=f{}^{\mathrm{\prime }}(t)\stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)f(t)\end{array}$ .
  ::属性:如果您将一个星标函数乘以矢量估值函数,则其产品的衍生物将遵循衍生物的产品规则。 (f(t) G(t) {f}(t) g(t)+G(t)f(t)。

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  Consider the functions:
  ::考虑以下功能:

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  $\begin{array}{r}f(t)=t^2\\ \stackrel{\to }{G}(t)=(t^4,\ln t)\end{array}$
  :t) = t2G(t) = (t4, ln t)

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  Show that $\begin{array}{r}\left(f(t)\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=f{}^{\mathrm{\prime }}(t)\stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)f(t)\end{array}$ .
  ::显示 (f( t) G(t) ) {f} (t) G(t) + G(t) f(t) 。

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  If you multiply the functions together and take the derivative, you’ll see that:
  ::如果将函数乘在一起并取出衍生物,你会看到:

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  $$\begin{array}{r}f(t)\stackrel{\to }{G}(t)=(t^6,t^2\ln t)\\ \left(f(t)\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\left(6t^2,2t\ln t+\frac{t^2}{t}\right)=(6t^5,2t\ln t+t)\end{array}$$

  
  :t) G(t) = (t) 6,t2ln(t) G(t) (t) (x) (6,2,2tln) (t) +T2t) = (6,5,2tln(t) +t) = (6,5,2tln(t)

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  Notice that since $\begin{array}{r}{t}^2\ln t\end{array}$  is a product of two functions, you need to use the product rule to differentiate it.
  ::请注意,由于 t2lnt 是两个函数的产物, 您需要使用产品规则来区分它 。

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  If you used the product rule to find the derivative of the product of  $\begin{array}{r}f(t)\end{array}$ and $\begin{array}{r}\stackrel{\to }{G}(t)\end{array}$ , you have:
  ::如果您使用产品规则查找 f(t) 和 G(t) 产品的衍生物,您有:

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  $$\begin{array}{r}f(t)=t^2\\ \stackrel{\to }{G}(t)=(t^4,\ln t)\\ \left(f(t)\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=f{}^{\mathrm{\prime }}(t)\stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)f(t)\\ \left(f(t)\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=2t(t^4,\ln t)+\left(4t^3,\frac{1}{t}\right)(t^2)\end{array}$$

  
  :t) = t2G(t) = (t4,ln) t(f) (t) G(t) (t) f}(t) g(t) + G(t) f(t) f(t) g(t) G(t) ) (t) (t) t(t) t(t4,ln) t(t) +(4t3,1t)(t2)

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  Which simplifies to:
  ::简化为:

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  $$\begin{array}{r}(2t^5,2t\ln t)+\left(4t^5,\frac{t^2}{t}\right)\\ =(6t^5,2t\ln t+t)\end{array}$$

  
  :2T5,2tlnt)+(4T5,T2t)=(6T5,2tlnt+t)

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  Finding the derivative of a vector-valued function can also help you to determine if the function is a circle centered at the origin. 
  ::查找矢量估值函数的衍生物也可以帮助您确定该函数是否是一个以源代码为中心圆的圆。

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  Property: If the derivative function of a vector-valued function is perpendicular to the original function – that is, if the angle between the two vectors is always 90 degrees, then the magnitude of the vectors that make up the original function is a constant, and the vector-valued function is a circle.
  ::属性:如果矢量估值函数的衍生函数与原始函数密切相关,也就是说,如果两个矢量之间的角始终为90度,则构成原始函数的矢量的大小是一个常数,而矢量估值函数是一个圆。

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  You can use the dot product of the vector-valued function and its derivative to see if the two are perpendicular. Remember that if the dot product of two functions is zero, then those functions are perpendicular to each other. So, if you compute the dot product of a vector-valued function and its derivative, you can see if the two are perpendicular. If they are, then the vector-valued function is a circle.
  ::您可以使用矢量值函数及其衍生物的圆点产物来查看这两个函数是否垂直。 记住, 如果两个函数的圆点产物为零, 那么这两个函数是垂直的。 所以, 如果您计算了矢量值函数及其衍生物的圆点产物, 您可以看到这两个函数是否垂直。 如果这两个函数是垂直的, 那么矢量值函数是一个圆形 。

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  Consider the function $\begin{array}{r}\stackrel{\to }{F}(t)=(\sin t,\cos t)\end{array}$  and show that this function is a circle.
  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(\cos t,-\sin t)\end{array}$ .
  ::考虑 F(t) = (sint,cost) 并显示此函数是一个圆。 F(t) = (cost,-sint) 。

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  If you find the dot product of the two vector-valued functions, you’ll see that:
  ::如果找到两个矢量价值函数的点产物,你会看到:

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  $$\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=\sin t\cos t+(-\sin t\cos t)=0.\end{array}$$

  
  ::F(t) (t) =sin(t) (t) =0。

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  The dot product is zero, so $\begin{array}{r}\stackrel{\to }{F}(t)\mathrm{\perp }\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\end{array}$ . This means that $\begin{array}{r}\Vert \stackrel{\to }{F}(t)\Vert \end{array}$  is a constant. Since all the points on $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$  are a constant distance from the origin, it must describe a circle centered at the origin. Notice that if a circle wasn’t centered at the origin, the magnitude of the vectors wouldn’t be constant since magnitude is a distance from the origin to the vector’s end point.
  ::点产值为零, 所以 F( t) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\F\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

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  For instance, take the circle $\begin{array}{r}\stackrel{\to }{F}(t)=(5+\cos t,2+\sin t)\end{array}$ .
  ::例如,取 F(t) = (5+cost,2+sint) 圆。

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  The derivative of this vector-valued function is $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(-\sin t,\cos t)\end{array}$ .
  ::此矢量估值函数的衍生物为 F(t) = (- sint,cost)。

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  The dot product, $\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\end{array}$  is:
  ::F(t)F(t)是:

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  $$\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(5+\cos t)(-\sin t)+(2+\sin t)(\cos t)=\\ -5\sin t+-\sin t\cos t+2\cos t+\sin t\cos t=2\cos t-5\sin t\end{array}$$

  
  ::F(t) (t) = (5+cos) (-sin) +(2+sin) +(2+sin) +(cos) +5sin(t) (t) (t) +sin(t) +2cos(t) =2cos(t) -5sin}

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  The derivative of the function is not perpendicular to the function, therefore the magnitude of $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$  is not constant. However, this vector-valued function does describe a circle – it just describes a circle centered at (5, 2) instead of at the origin.
  ::函数的衍生物与函数无关, 因此F( t) 的大小不是恒定的。 但是, 这个矢量估值函数的确描述了一个圆, 它只是描述了一个以5, 2为中心的圆, 而不是以原为中心 。

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  Property: The derivative of the function that is the dot product of two vector-valued functions can be described as $\begin{array}{r}\left(\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)\end{array}$ .
  ::属性:两个矢量值函数的点产函数的衍生物可描述为(F(t)G(t)(t)F(t)G(t)+G(t)F(t)。

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  Consider the vector-valued functions:
  ::考虑矢量估值功能:

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,3t)\\ \stackrel{\to }{G}(t)=\left(\frac{1}{t},\sin t\right)\end{array}$
  ::F(t) = (t2,3t) G(t) = (1t,sint)

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  Let's show that $\begin{array}{r}\left(\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)\end{array}$ .
  ::让我们来显示 (F(t) (t) (t) (F) (t) (t) (t) (t) (t) (t) (F(t) (t) )。

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  Note that $\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)=\frac{t^2}{t}+3t\sin t=t+3t\sin t\end{array}$ . So,
  ::注意 F(t) G(t) = t2t+3tsin@t=t+3tsint。 所以,

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  $$\begin{array}{r}\left(\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=1+3\sin t+3t\cos t.\end{array}$$

  
  :F(t)G(t))1+3sint+3tcost。

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  Now compute $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)=\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)\end{array}$ :
  ::现在计算 F(t) G(t) = G(t) F(t) :

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,3t)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(2t,3)\\ \stackrel{\to }{G}(t)=\left(\frac{1}{t},\sin t\right)\\ \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=\left(-\frac{1}{t^2},\cos t\right)\end{array}$
  ::F(t) = (t2,3t) F(t) = (2t,3) G(t) = (1t,sin(t) G(t) = (-1t2,cos(t) )

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  So you have:
  ::因此,你有:

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  $$\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)=(2t,3)\cdot \left(\frac{1}{t},\sin t\right)+(t^2,3t)\cdot \left(-\frac{1}{t^2},\cos t\right)\\ =\left[\left(\frac{2t}{t}\right)+3\sin t\right]+\left[-\frac{t^2}{t^2}+3t\cos t\right]=[2+\sin t]+[-1+3t\cos t]\\ =1+3\sin t+3t\cos t\end{array}$$

  
  ::F(t)+G(t)+G(t){F}}(t)=(2t,3)(1t,sin(t)+(t2,3t)}+(t2,2t)+3sin}}+[(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t){F}(t)+(t)=(2)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)+(t)

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  So $\begin{array}{r}\left(\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)\end{array}$
  ::所以 (F(t) G(t) (t) G(t) +G(t) F(t) (t)

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked how Tessa can use the derivatives of vector-valued functions to determine how much work her engine needs to be able to do. Tessa plans to have the clockwork horse start out slow and speed up over the course of the route, which winds through a small town. Work is the product of force and distance. Force is the product of an object’s mass and its acceleration . The clockwork horse weighs about 50 kg. Tessa models the force used to move it along the route with the equation:
  ::早些时候,有人问Tessa如何使用矢量值函数的衍生物来确定她的引擎需要做多少工作。Tessa计划让时钟马在经过小城镇的路线上缓慢地加速前进。工作是力量和距离的产物。力是物体质量和加速力的产物。时钟马的重量约为50公斤。Tessa计划用方程式来模拟它沿路线移动的力量:

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=(50t,10t^2)\end{array}$$

  
  ::F(t) = (50t,10t2)

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  And the distance that it has traveled at a certain time  $\begin{array}{r}t\end{array}$ as:
  ::以及它在某一时间的距离:

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  $$\begin{array}{r}\stackrel{\to }{D}(t)=(3t,5t^2)\end{array}$$

  
  ::D(t) =( 3吨, 5吨2)

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  She wants to see how quickly the amount of work needed to move the horse is changing when time is 1, 5, and 10 minutes into the parade. That way, she can make sure that the engine can handle the workload required to move the clockwork horse at various key points in the route.
  ::她想看看在游行时间1、5和10分钟后,马匹移动所需的工作量变化有多快。 这样,她可以确保引擎能够处理在路线的各个关键点移动时钟马所需的工作量。

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  Work can be described as the dot product of the force vector and the work vector. However, since she wants to know how the amount of work is changing, she’ll need to find the derivative of this dot product.
  ::工作可以被描述为动力矢量和工作矢量的点产物。 但是,由于她想知道工作量是如何变化的,她需要找到这一点产品的衍生物。

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(50t,10t^2)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(50,20t)\\ \stackrel{\to }{D}(t)=(3t,5t^2)\\ \stackrel{\to }{D}{}^{\mathrm{\prime }}(t)=(3,10t)\end{array}$
  ::F(t) = (50t,10t2) F(t) = (50,20t) D(t) = (3,10t) = (3,10t)

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  $$\begin{array}{r}\mathrm{\Delta }W=\left(\stackrel{\to }{F}(t)\cdot \stackrel{\to }{D}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{D}(t)+\stackrel{\to }{D}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)\\ \\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{D}(t)=(50)(3t)+(20t)(5t^2)=150t+100t^3\\ \stackrel{\to }{D}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)=(3)(50t)+(10t)(10t^2)=150t+100t^3\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{D}(t)+\stackrel{\to }{D}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)=300t+200t^3\\ \mathrm{\Delta }W=300t+200t^3\end{array}$$

  
  ::{{(t)}D}}{(t){(t){(t)}}{(d)}}{(t){(t){(t){(t)}}}{(t)}}{(t)}}{(t)}=(50){(3){(t){(5)}}=150t+100t3}(t){(t)}{(t){(3){(3){(d)}}}+(10){(t)}=150t+100t_(t){(d)}(t){(t){(t)}}}(d}(t){(t)}}(t){(t){(t)}(t)}(3}(t)+(20t)(5){(5)}}}=150t+100t}(d}(t){(t){(t)}(t)=(3)(50t){(50t){(50t){(50t)}(50t)}+(50t+(30(t)}(t)}(10(10)(10)+(10)+(10)(10)(10)})}=1(t)=(t)}(t)=(t)}(t)}(t(t(t)_(t)}(t)}(t)}(t(t(t(t)}(t)}(t)}(t)}(t)}(t)}(t(t)}(t(t)}(t)}(t)}(t)}(t)}(t)}(t)}(t)}(t)}(t){(t){(t){(t){(t)}(t)}(t)}(t)+(t)}(t)}(t)}(t)}(t)}(t)}(t(t)}(t)}(t)}(t)}(t)}(t)}(t)}(t)}(t)}(t)}(t

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  After she finds the derivative of the dot product, she can plug in values of  $\begin{array}{r}t\end{array}$ to find how quickly work is changing at different points in time.
  ::在她发现点产品衍生物后,她可以插入 t 的值,以找出在不同时间点工作变化的速度。

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  $$\begin{array}{r}t=1\\ \mathrm{\Delta }W=300(1)+200(1{)}^3=500J\\ t=5\\ \mathrm{\Delta }W=300(5)+200(5{)}^3=26,500J\\ t=10\\ \mathrm{\Delta }W=300(10)+200(10{)}^3=203,000J\end{array}$$

  
  ::t=1W=300(1)+200(1)3=500Jt=5W=300(5)+200(5)3=26,500Jt=10W=300(10)+200(10)3=203000J

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  Tessa can use these figures to figure out how much energy her engine needs to produce, and how quickly it will need to produce that energy.
  ::Tessa可以利用这些数字来决定她的引擎需要多少能源来生产, 以及它需要多快来生产这种能源。

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  Example 2
  ::例2

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  Find the derivative of the vector-valued function produced when you add the functions
  ::查找添加函数时生成的矢量价值函数的衍生物

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(2t+7,t^2)\\ \stackrel{\to }{G}(t)=(\sin 4t,t^3)\end{array}$
  ::F(t) = (2t+7,t2) G(t) = (sin4t,t3)

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  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(2,2t)\\ \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(4\cos 4t,3t^2)\end{array}$
  ::F(t) = (2,2t) G(t) = (4cos) =(4cos) 4t,3t2)

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  $\begin{array}{r}\left(\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\right){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(2+4\cos 4t,2t+3t^2)\end{array}$
  :F(t)+G(t))+(t)+(t)=(2+4cos4t,2t+3t2)

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  Example 3
  ::例3

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  Use to determine whether the following vector-valued functions are circles with centers at the origin.
  ::用于确定下列矢量价值的功能是否是由原中心组成的圆圈。

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  • $\begin{array}{r}\stackrel{\to }{F}(t)=(5\sin t,5\cos t)\end{array}$
    ::F(t) = (5sint,5cost)
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  • $\begin{array}{r}\stackrel{\to }{G}(t)=(3\sin t,5\cos t)\end{array}$
    ::G(t) = (3sint,5cost)
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  • $\begin{array}{r}\stackrel{\to }{H}(t)=(4+\sin t,\cos t)\end{array}$
    ::H(t) = (4+sinT,cosT)
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  • $\begin{array}{r}\stackrel{\to }{I}(t)=(12\cos t,12\sin t)\end{array}$
    ::I(t) = (12cost,12sint)

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(5\sin t,5\cos t)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(5\cos t,-5\sin t)\end{array}$
  ::F(t) = (5sint,5cost) F(t) = (5cost,-5sint)

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  $\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(5\sin t)(5\cos t)+(5\cos t)(-5\sin t)=0\end{array}$
  ::F(t) (t) = (5sin}) (5cos) + (5cos) + (5cos) (-5sin(t) =0

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  The dot product is zero, so the function is a circle centered at (0, 0).
  ::点产值为零, 函数是一个圆圆, 以 0, 0为中心 。

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  $\begin{array}{r}\stackrel{\to }{G}(t)=(3\sin t,5\cos t)\\ \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(3\cos t,-5\sin t)\end{array}$
  ::G(t) = (3sint,5cost) G(t) = (3cost,-5sint)

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  $\begin{array}{r}\stackrel{\to }{G}(t)\cdot \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(3\sin t)(3\cos t)+(5\cos t)(-5\sin t)\end{array}$
  ::G(t) (t) = (3sin(t) 3cos(t) +(5cos(t) ~5sin(t) )

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  The dot product is not zero, so the magnitude of $\begin{array}{r}\stackrel{\to }{G}(t)\end{array}$  is not a constant and the function is not a circle centered at the origin.
  ::点值产品不是零, 所以 G( t) 的大小不是一个常数, 函数不是以原点为中心的圆 。

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  $\begin{array}{r}\stackrel{\to }{H}(t)=(4+\sin t,\cos t)\\ \stackrel{\to }{H}{}^{\mathrm{\prime }}(t)=(\cos t,-\sin t)\end{array}$
  ::H(t) = (4+sint,cost) H(t) =(cost,-sint)

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  $\begin{array}{r}\stackrel{\to }{H}(t)\cdot \stackrel{\to }{H}{}^{\mathrm{\prime }}(t)=(4+\sin t)(\cos t)-(\cos t)(\sin t)\\ =4\cos t+\cos t\sin t-\cos t\sin t=4\cos t\end{array}$
  ::H(t) = (4+sin(cos)-(cos(sin(t)) = 4cos(t) +cos(t) (t) (t) (t) (t) (t) =(4+sin(t) =4cos@t)

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  The dot product is not zero, so the magnitude of $\begin{array}{r}\stackrel{\to }{H}(t)\end{array}$  is not a constant and the function is not a circle centered at the origin.
  ::点值产品不是零, 所以H( t) 的大小不是一个常数, 函数不是以原点为中心的圆圈 。

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  $\begin{array}{r}\stackrel{\to }{I}(t)=(12\cos t,12\sin t)\\ \stackrel{\to }{I}{}^{\mathrm{\prime }}(t)=(-12\sin t,12\cos t)\end{array}$
  ::I(t) =( 12cost, 12sint) I(t) =( - 12sint, 12cost)

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  $\begin{array}{r}\stackrel{\to }{I}(t)\cdot \stackrel{\to }{I}{}^{\mathrm{\prime }}(t)=(12\cos t)(-12\sin t)+(12\sin t)(12\cos t)=0\end{array}$
  ::I(t) = (12cos) (- 12sin) +( 12sin) +( 12sin) (12cos) =0

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  The dot product is 0. This means the magnitude of the vector-valued function is 0 and it represents a circle with its center at the origin.
  ::点值产品为 0。 这意味着矢量值函数的大小为 0, 它代表一个圆, 其中心在原点 。

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  Example 4
  ::例4

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  Find the derivative of the scalar function  $\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)\end{array}$ when
  ::查找 scalar 函数 F( t)\\\ GG( t) 的导出函数 F( t) 当

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(3t^2,4t^3)\\ \stackrel{\to }{G}(t)=(e^t,t^2)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(6t,12t^2)\\ \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(e^t,2t)\end{array}$
  ::F(t) = (3t2,4t3) G(t) = (et,t2) F(t) = (6t,12t2) G(t) = (et,2t)

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  $\begin{array}{r}(\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)){}^{\mathrm{\prime }}=\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)=6t{e}^t+12t^4\\ \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)=3t^2{e}^t+8t^4\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{G}(t)+\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}(t)=20t^4+3t^2{e}^t+6t{e}^t\end{array}$
  :F(t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) = 6tet+12t4G}(t) (t) = 3t2et+8t4F}(t) (t) (t) (t) (f) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) =20t4+3t2et++6tet

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  Review
  ::回顾

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  For all problems, consider the following vector-valued functions:
  ::针对所有问题,考虑以下矢量估值功能:

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  $\begin{array}{r}\stackrel{\to }{F}(t)=(2\sin t,2\cos t)\\ \stackrel{\to }{G}(t)=\left(t^2+1,\frac{1}{t}\right)\\ \stackrel{\to }{H}(t)=(1-\sin t,1-\cos t)\\ \stackrel{\to }{I}(t)=\left(e^t,\frac{1}{t^2}\right)\end{array}$
  ::F(t) = (2sint,2cost) G(t) = (t2+1,1t) H(t) =(1-sint,1-cost) I(t) =(et,1t2)

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  1. Find the derivative of $\begin{array}{r}\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\end{array}$ .
     ::查找 F(t) +G(t) 的衍生物。
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  2. Find the derivative of $\begin{array}{r}\stackrel{\to }{G}(t)-\stackrel{\to }{H}(t)\end{array}$ .
     ::查找 G(t)-H(t) 的衍生物。
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  3. Find the derivative of $\begin{array}{r}\stackrel{\to }{I}(t)+\stackrel{\to }{H}(t)\end{array}$ .
     ::查找 I(t) +H(t) 的衍生物。
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  4. Find the derivative of $\begin{array}{r}\stackrel{\to }{H}(t)-\stackrel{\to }{F}(t)\end{array}$ .
     ::查找 H(t)-F(t) 的衍生物。
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  5. Find the derivative of $\begin{array}{r}\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)-\stackrel{\to }{H}(t)\end{array}$ .
     ::查找 F(t)+G(t)-H(t)的衍生物。
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  6. Calculate $\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\end{array}$ . What does this tell you about $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ ?
     :t)\\\\\\\\\\\\\\\\\\\\\\\\F\\\\\\\\\\\\\\\\F\\\\\\\\\\\\\\\\\\\F\\\\F\\\\\\\\\\\\\\\\\\\\F\\\\\F\\\\\\\\\T。这告诉您什么 F\\\\(t)?
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  7. Calculate $\begin{array}{r}\stackrel{\to }{G}(t)\cdot \stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\end{array}$ . What does this tell you about $\begin{array}{r}\stackrel{\to }{G}(t)\end{array}$ ?
     ::计算 G(t) G(t) 。 这告诉您 G(t) 的情况 ?
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  8. Calculate $\begin{array}{r}\stackrel{\to }{H}(t)\cdot \stackrel{\to }{H}{}^{\mathrm{\prime }}(t)\end{array}$ . What does this tell you about $\begin{array}{r}\stackrel{\to }{H}(t)\end{array}$ ?
     ::计算 H(t) H(t) 。 这告诉你H(t) 的什么?
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  9. Calculate $\begin{array}{r}\stackrel{\to }{I}(t)\cdot \stackrel{\to }{I}{}^{\mathrm{\prime }}(t)\end{array}$ . What does this tell you about $\begin{array}{r}\stackrel{\to }{I}(t)\end{array}$ ?
     ::计算 I(t) I(t) 。 这告诉您关于 I(t) 是什么 ?
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  10. Find the derivative of the scalar function $\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{G}(t)\end{array}$ .
      ::查找 scalar 函数 F(t) G(t) 的衍生物。
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  11. Find the derivative of the scalar function $\begin{array}{r}\stackrel{\to }{F}(t)\cdot \stackrel{\to }{H}(t)\end{array}$ .
      ::查找 scalar 函数 F(t) H(t) 的衍生物。
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  12. Find the derivative of the scalar function $\begin{array}{r}\stackrel{\to }{H}(t)\cdot \stackrel{\to }{I}(t)\end{array}$ .
      ::查找 scalar 函数 H(t) I(t) 的衍生物。
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  13. If $\begin{array}{r}\left(\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=(1+4t,18)\end{array}$   and $\begin{array}{r}\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(4t,15)\end{array}$ , what is $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\end{array}$ ?
      ::如果(F(t)+G(t))(1+4t,18)和G(t)=(4t,15),什么是F(t)?
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  14. If $\begin{array}{r}\left(\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}=\left(3t^2+6t,\frac{3}{t}\right)\end{array}$   and $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=\left(6t,\frac{1}{t}\right)\end{array}$ , what is $\begin{array}{r}\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)\end{array}$ ?
      ::如果(F(t)+G(t))(3t2+6t,3t)和F(t)=(6t,1t),G(t)是什么?
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  15. If $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(2t,\sin t)\end{array}$ and $\begin{array}{r}\stackrel{\to }{G}{}^{\mathrm{\prime }}(t)=(e^t,\cos t)\end{array}$ , what is $\begin{array}{r}\left(\stackrel{\to }{F}(t)+\stackrel{\to }{G}(t)\right){}^{\mathrm{\prime }}\end{array}$ ?
      ::如果F(t) =(2t,sin(t) 和G(t) =(et,cos(t) ),那(F(t)+G(t)) 是什么?

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  Review (Answers)
  ::回顾(答复)

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  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 。