11.6 矢量值函数的整合

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• 选择节矢量值函数整合

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  矢量值函数整合

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  Raymond launches a weather balloon with a camera attached to it. He plans to snap photos of the earth from the stratosphere. He’ll need to retrieve his camera when it returns to earth, so he’ll need to know its position throughout the journey.  He’s created a vector-valued function to model his camera’s velocity with respect to time. Can he use that function to find another function that describes his camera’s position?
  ::雷蒙德推出了一个配有相机的气象气球。他计划从平流层拍摄地球的照片。 当相机返回地球时,他需要收回相机,因此他需要在整个旅程中了解其位置。 他创造了一个矢量价值的功能来模拟相机的时间速度。 他能否利用这一功能找到另一个描述相机位置的功能?

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  Integration of Vector-Valued Functions
  ::矢量值函数整合

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  You can integrate vector-valued functions using the same techniques that you use to integrate scalar functions and parametric functions. When you integrate a vector-valued function, you integrate the horizontal and vertical components separately. The result of integration will be a new vector-valued function, or, if you compute a definite integral, a new vector.
  ::您可以使用与您用来整合天平函数和参数函数相同的技术整合矢量估值函数。当您整合了矢量估值函数和参数函数时,您可以将水平和垂直组成部分分别整合。整合的结果将是一个新的矢量估值函数,或者,如果您计算了一个确定的整体函数,则将一个新的矢量。

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  So, if  $\begin{array}{r}\stackrel{\to }{F}(t)=(f(t),g(t))\end{array}$ and  $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ is a vector-valued function, then $\begin{array}{r}\int \stackrel{\to }{F}(t)=\left(\int f(t)dt,\int g(t)dt\right)\end{array}$ .
  ::所以,如果 F(t) =(f(t),g(t)) 和 F(t) 是一个矢量估值函数, 那么 ( F) (t) = ((f) (t) dt, (g) (t) dt) 。

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  As with scalar and parametric functions, the vector-valued function  $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ is the integral of the vector-valued function  $\begin{array}{r}\stackrel{\to }{G}(t)\end{array}$ if and only if the derivative of  $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ is equal to $\begin{array}{r}\stackrel{\to }{G}(t)\end{array}$ .
  ::与标度和参数函数一样,矢量估值函数F(t)是矢量估值函数G(t)的有机组成部分,条件是并且只有在F(t)的衍生物等于G(t)的情况下。

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  Let's find the integral of the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,-\cos t)\end{array}$ .
  ::让我们找到矢量值函数 F( t) =( t2, - cos t) 的组成部分 。

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    $\begin{array}{r}\int \stackrel{\to }{F}(t)=\left(\int t^2,\int -\cos t\right)\end{array}$
  
  ::F(t) = (2,COST)

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  Find the integrals of the horizontal and vertical components:
  ::查找水平和垂直构件的内装件:

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  $$\begin{array}{r}\int t^{2}dt=\frac{1}{3}{t}^3+C\\ \int -\cos tdt=-\sin t+D\end{array}$$

  
  ::@ t2dt=13t3+Ccost dtsint+D

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  So, you have
  ::所以,你有

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  $$\begin{array}{r}\int \stackrel{\to }{F}(t)=\left(\frac{1}{3}{t}^3+C,-\sin t+D\right)\end{array}$$

  
  ::F(t) =(13t3+C,-sint+D)

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  Remember, you can use all of the techniques you learned in order to integrate scalar functions to integrate each component of a vector-valued function.
  ::记住, 您可以使用你学到的所有技术, 整合 scal 函数, 整合矢量值函数的每个组件 。

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  To find the definite integral of a vector-valued function, evaluate each component of the function separately. Your answer will be a single vector.
  ::要找到矢量估值函数的确定组成部分, 请单独评估函数的每个组成部分 。 您的答案将是单个矢量 。

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  Now, let's find the value of the definite integral  $\begin{array}{r}\underset{0}{\overset{2}{\int }}\stackrel{\to }{F}(t)dt\end{array}$ when $\begin{array}{r}\stackrel{\to }{F}(t)=\left(2t,\frac{1}{2}{t}^2\right)\end{array}$ .
  ::现在,让我们在 F( t) =( 2t, 12t2) 时, 找到确定的整体值 02F( t) dt 。

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  First, find the definite integrals for the horizontal and vertical components.
  ::首先,找到横向和纵向构件的确定构件。

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  $$\begin{array}{r}\underset{0}{\overset{2}{\int }}2tdt={\left[t^2\right]}_0^2=2^2-0^2=4\\ \underset{0}{\overset{2}{\int }}\frac{1}{2}{t}^{2}dt=\frac{1}{2}\underset{0}{\overset{2}{\int }}{t}^{2}dt=\frac{1}{2}{\left[\frac{t^3}{3}\right]}_0^2=\frac{1}{2}\left[\frac{2^3-0^3}{3}\right]=\frac{8}{6}=\frac{4}{3}\end{array}$$

  
  ::022t dt=[t2,022=22-02=40212t2 dt=1202t2 dt=12[t33]02=12[23-033]=86=43

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  The definite integral of this function is the vector $\begin{array}{r}\left(4,\frac{4}{3}\right)\end{array}$ .
  ::这一功能的决定性组成部分是矢量(4,43)。

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  The integrals of vector-valued functions are very useful for engineers, physicists, and other people who deal with concepts like force, work, momentum, velocity, and movement.  For instance, the velocity of an object can be described as the integral of the vector-valued function that describes the object's acceleration .  This is because acceleration is defined as the rate of change of an object's velocity. Acceleration is the derivative of velocity, and velocity is the integral of acceleration.
  ::矢量价值函数的元件对于工程师、物理学家和其他处理强力、工作、动力、速度和移动等概念的人非常有用。例如,一个物体的速度可以被描述为说明物体加速度的矢量价值函数的有机体。这是因为加速度的定义是物体速度的变速率。加速是速度的衍生物,速度是加速的有机体。

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  Say an object’s acceleration can be described by the vector-valued function (-1, 5). What vector-valued function that describes its velocity?
  ::说一个物体的加速度可以用矢量估值函数(-1, 5)来描述。 说明其速度的矢量估值函数是什么?

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  $$\begin{array}{r}\stackrel{\to }{A}(t)=(-1,5)\\ \stackrel{\to }{V}(t)=\int \stackrel{\to }{a}(t)dt\end{array}$$

  
  ::A(t) =(-1,5,5)V(t) a(t) dt

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  So, to find the velocity function, find the integrals of the horizontal and vertical components of the acceleration function.
  ::因此,为了找到速度函数, 找到加速函数的水平和垂直组成部分的内装件。

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  $$\begin{array}{r}\int \stackrel{\to }{A}(t)dt=\left(\int -1dt,\int 5dt\right)=\left(-t+C,5t+D\right)\\ \stackrel{\to }{V}(t)=(-t+C,5t+D)\end{array}$$

  
  :t) dt= (1 dt, 5 dt) = (~~~~+C,5t+D) V(t) = (~~~~+C,5t+D)

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  You now have a vector-valued function that describes the velocity of the object at a given time, based on its acceleration.
  ::您现在有一个矢量估值函数, 根据其加速度来描述特定时间天体的速度 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked about how Raymond can find another function to model the position  of a weather balloon from his velocity function. Velocity describes how an object’s position changes with respect to time. This means that velocity shows the rate and direction of change for a position vector, so velocity is the derivative of position. From the definition of an integral, this means that a vector-valued function describing the position of an object is the integral of the vector-valued function that describes velocity of the same object.
  ::早些时候,有人询问雷蒙德如何从速度函数中找到另一个功能来模拟气象气球的位置。速度描述一个天体相对于时间的位置变化。这意味着速度显示位置矢量的变化速度和方向,因此速度是位置的衍生物。从一个集成的定义中,这意味着描述一个天体位置的矢量估值函数是描述同一天体速度的矢量估值函数的组成部分。

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  Raymond has modeled his camera’s velocity after liftoff as $\begin{array}{r}\stackrel{\to }{V}(t)=(t+12,-6t^2+18t)\end{array}$ . He can integrate to find a vector-valued function that will describe the path that the camera will take. He’s measuring time in hours.
  ::Raymond在起飞后将相机的速度模拟为 V(t) =(t+12,- 6t2+18t ) 。 他可以集成到一个矢量值函数,描述相机将走的道路。 他用小时来测量时间 。

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  So, $\begin{array}{r}\stackrel{\to }{D}(t)=\int \stackrel{\to }{V}(t)dt\end{array}$ . He integrates the horizontal and vertical components separately and finds that:
  ::所以,D(t)V(t) dt. 他将水平和垂直组件分开整合,发现:

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  $$\begin{array}{r}\int t+12dt=\frac{1}{2}{t}^2+12t+C\\ \int -6t^2+18tdt=-2t^3+9t^2+K\end{array}$$

  
  ::++12dt=12t2+12t+C6t2+18t dt2t3+9t2+K

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  Therefore, $\begin{array}{r}\stackrel{\to }{D}(t)=\left(\frac{1}{2}{t}^2+12t+C,-2t^3+9t^2+K\right)\end{array}$ .
  ::因此,D(t)=(12t2+12t+C,-2t3+9t2+K)。

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  Example 2
  ::例2

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  Given the following vector-valued function, find the integral.
  ::考虑到以下矢量价值的功能,请找到有机体。

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=(e^{2t},4\sin 2t)\end{array}$$

  
  ::F(t) = (e2t,4sin2t)

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=(e^{2t},4\sin 2t)\\ \int \stackrel{\to }{F}(t)dt=\left(\int e^{2t}dt,\int 4\sin 2tdt\right)\\ \int e^{2t}dt=\frac{1}{2}{e}^{2t}+C\\ \int 4\sin 2tdt=-2\cos 2t+K\\ \int \stackrel{\to }{F}(t)dt=\left(\frac{1}{2}{e}^{2t}+C,-2\cos 2t+K\right)\end{array}$$

  
  ::F(t) = (e2t,4sin2t) F(t) dt= (e2tdt, 4sin2t dt) @e2tdt=12e2t+C4sin2t dt2cos2t+K) (t) dt=(12e2t+C, -2cos}2t+K)

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  Example 3
  ::例3

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  Find the definite integral of the following vector-valued function.
  ::查找以下矢量估值函数的确定组成部分。

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=\left(3t^2,\frac{1}{t}\right)\\ \underset{.1}{\overset{e}{\int }}\stackrel{\to }{F}(t)dt\end{array}$$

  
  ::F(t) = (3t2,1t) .1eF(t) dt

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=\left(3t^2,\frac{1}{t}\right)\\ \underset{.1}{\overset{e}{\int }}\stackrel{\to }{F}(t)dt=\left(\underset{.1}{\overset{e}{\int }}3t^{2}dt,\underset{.1}{\overset{e}{\int }}\frac{1}{t}dt\right)\\ \underset{.1}{\overset{e}{\int }}3t^{2}dt=[t^3{]}_{.1}^e=e^3-{.1}^3\approx 20.08\\ \underset{.1}{\overset{e}{\int }}\frac{1}{t}dt=[\ln t{]}_{.1}^e=1-(-2.3)=3.3\\ \underset{.1}{\overset{e}{\int }}\stackrel{\to }{F}(t)dt=(20.08,3.3)\end{array}$$

  
  ::F(t) = (3t2,1t) .1eF(t) dt= (.1e3t2 dt,.1e1tdt) .1e3t2 dt=[t3,1e=_1320.081e1tdt=[lnt] 1e=1-(2-2.3) =3.31eF}(t) dt=(20.08,3.3)

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  Example 4
  ::例4

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  The vector-valued function  $\begin{array}{r}\stackrel{\to }{A}(t)=(t^2,\cos t)\end{array}$ describes the acceleration of an object. What is the vector-valued function to describe the object’s velocity?
  ::矢量估值函数 A(t) =( t2, cos & t) 描述对象的加速度。 矢量估值函数描述对象的速度是什么 ?

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  $$\begin{array}{r}\stackrel{\to }{A}(t)=(t^2,\cos t)\\ \stackrel{\to }{V}(t)=\int \stackrel{\to }{A}(t)dt\\ \int \stackrel{\to }{A}(t)dt=\left(\int t^{2}dt,\int \cos tdt\right)\\ \int t^{2}dt=\frac{1}{3}{t}^3+C\\ \int \cos tdt=\sin t+K\\ \stackrel{\to }{V}(t)=\left(\frac{1}{3}{t}^3+C,\sin t+K\right)\end{array}$$

  
  ::A(t) = (t2,cost) V(t) A(t) dA(t) d(t) dt= (t2 dt,cost dt) t2 dt=13t3+Ccost dt=sin@t) =(13t3+C,sint+K)

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  Review
  ::回顾

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  or #1-10, u se the following vector-valued functions:
  ::或#1-10,使用以下矢量估值函数:

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

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  For #1-5, find the indefinite integral of the function.
  ::对于1 -5,找到该功能的无限组成部分。

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  1.   $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$
     ::F__(t)
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  2.   $\begin{array}{r}\stackrel{\to }{G}(t)\end{array}$
     :t) G__________________________________________________________________________________________________________________________________________________________________(吨)
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  3.   $\begin{array}{r}\stackrel{\to }{H}(t)\end{array}$
     ::H(t)
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  4.   $\begin{array}{r}\stackrel{\to }{K}(t)\end{array}$
     ::K(t)
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  5.   $\begin{array}{r}\stackrel{\to }{M}(t)\end{array}$
     ::M(t)

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  For #6-10, calculate each definite integral.
  ::对于#6-10,计算每个确定的组成部分。

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  6.   $\begin{array}{r}\underset{2}{\overset{5}{\int }}\stackrel{\to }{F}(t)dt\end{array}$
     ::25F(t) dt
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  7.   $\begin{array}{r}\underset{0}{\overset{\frac{\pi }{2}}{\int }}\stackrel{\to }{G}(t)dt\end{array}$
     ::02G(t) dt
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  8.   $\begin{array}{r}\underset{0}{\overset{5}{\int }}\stackrel{\to }{H}(t)dt\end{array}$
     ::05H(t) dt
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  9.   $\begin{array}{r}\underset{0}{\overset{10}{\int }}\stackrel{\to }{K}(t)dt\end{array}$
     ::010K(t) dt
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  10.   $\begin{array}{r}\underset{0}{\overset{\pi }{\int }}\stackrel{\to }{M}(t)dt\end{array}$
      ::0M(t) dt

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  For #11-12, the vector-valued function  $\begin{array}{r}\stackrel{\to }{A}(t)=(5t,t^2)\end{array}$ describes the acceleration of an object.
  ::对于# 11-12, 矢量值函数 A(t) =( 5t, t2) 描述一个对象的加速度 。

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  11.  What is the vector-valued function to describe the object’s velocity?
      ::矢量估值函数如何描述天体的速度?
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  12.  What is the velocity of the object at time  $\begin{array}{r}t=2\end{array}$ if the initial velocity of the object is $\begin{array}{r}\stackrel{\to }{V}(0)=(0,0)\end{array}$ ?
      ::如果天体的初始速度是V(0)=(0)=(0),那么时间t=2时天体的速度是多少?

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  For #13-14, the vector-valued function  $\begin{array}{r}\stackrel{\to }{V}(t)=(t^3,\sin t)\end{array}$ describes the velocity of an object.
  ::对于 # 13-14, 矢量估值函数 V(t) = (t3, sin@t) 描述一个对象的速度 。

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  13.  What is the vector-valued function to describe the object’s position?
      ::矢量估值函数如何描述物体的位置?
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  14.  Where is the object at time  $\begin{array}{r}t=\pi \end{array}$ if $\begin{array}{r}\stackrel{\to }{D}(0)=(2,5)\end{array}$ ?
      ::如果 D(0) =(2,5) , 时间 t 对象在哪里 ?
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  15.  If the vector-valued function  $\begin{array}{r}\stackrel{\to }{A}(t)=(\sin t,5)\end{array}$ describes the acceleration of an object with  $\begin{array}{r}\stackrel{\to }{V}(0)=(0,0)\end{array}$ and $\begin{array}{r}\stackrel{\to }{D}(0)=(0,0)\end{array}$ , where is the object at time $\begin{array}{r}t=\frac{\pi }{2}\end{array}$ ?
      ::如果矢量估值函数 A(t) =(sint,5) 表示一个对象的加速度, 其值为 V( 0) = (0,0) 和 D( 0) = (0,0) = (0) , 那么该对象在时间 t2 中的位置 ?

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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 。