11.3 平板中的矢量值函数

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  平面中的矢量估价函数

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  Arianna’s robotics team is building a robot to navigate an obstacle course. The competition guidelines say that the robot will have to ascend a ramp with a grade between 3 degrees and 25 degrees. How much force will the robot’s engine need to exert so that the machine can climb the grade? What sort of relationship is there between the force and the angle of the grade?
  ::亚里安娜的机器人团队正在建立一个机器人来绕过障碍线。 竞争准则规定机器人必须攀登一个高度在3度至25度之间的斜坡。 机器人的引擎需要用多少力才能让机器攀登等级? 力量与等级角度之间有什么关系?

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  Vector-Valued Functions
  ::矢量估价函数

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  A vector-valued function receives a scalar as an input and returns a vector as an output. These functions usually take the form $\begin{array}{r}\stackrel{\to }{F}(t)=\left(f(t),g(t)\right)\end{array}$ . You can work with vector-valued functions on a two dimensional plane or in three dimensional space. They’re especially helpful for dealing with complicated physics problems involving concepts like momentum, force, velocity, and revolution. These functions are useful because each point in the function is the endpoint of a vector with a unique direction and magnitude . A vector-valued function describes the set of vectors that are solutions to a specific set of equations.
  ::矢量估值函数作为输入接收一个星标, 并返回矢量输出。 这些函数通常使用 F( t) =( f( t), g( t) ) 格式。 您可以在二维平面或三维空间使用矢量估值函数。 它们特别有助于处理复杂的物理问题, 包括动力、 力、 速度和革命等概念。 这些函数非常有用, 因为函数中的每个点都是矢量的终点, 具有独特的方向和数量。 矢量估值函数描述一组矢量, 它们是一套特定方程式的解决方案 。

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  Consider the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(t+2,t^2)\end{array}$ . This vector-valued function describes all the vectors that begin at the origin and have an endpoint at  $\begin{array}{r}(t+2,t^2)\end{array}$ for every real number $\begin{array}{r}t\end{array}$ .
  ::考虑矢量估值函数 F(t) = (t+2, t2) 。 此矢量估值函数描述所有从源开始且每个实际数字 t 的端点( t+2, t2) 在端点( t+2, t2) 的矢量。

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  If you sketch a few of these vectors, you can see that as t moves from  $\begin{array}{r}-\mathrm{\infty }\end{array}$ to $\begin{array}{r}\mathrm{\infty }\end{array}$ , the endpoints of the vectors trace out a parabola from left to right.
  ::如果您勾画出其中的一些矢量, 可以看到当 t 从 移动到 时, 矢量的端点会从左向右追踪到 parbola 。

  

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  If you explore a bit more, you’ll discover that the traced parabola is the same as $\begin{array}{r}y=(x-2{)}^2\end{array}$ . What makes the two different is that in the vector-valued function, the parabola is expressed in terms of two functions, both dependent on  $\begin{array}{r}t\end{array}$ .
  ::如果再多探索一点,你会发现追踪到的抛物线与y=(x-2)2相同。 使两者不同的是,在矢量估值函数中,抛物线以两个函数表示,两者都取决于 t。

  $\begin{array}{r}t\end{array}$	$\begin{array}{r}t+2\end{array}$	$\begin{array}{r}{t}^2\end{array}$	$\begin{array}{r}x\end{array}$	$\begin{array}{r}(x-2{)}^2\end{array}$
  -2	0	4	0	4
  -1	1	1	1	1
  0	2	0	2	0
  1	3	1	3	1
  2	4	4	4	4

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  In the function $\begin{array}{r}y=(x-2{)}^2\end{array}$ , each point simply represents a point in space. In the function $\begin{array}{r}\stackrel{\to }{F}(t)=(t+2,t^2)\end{array}$ , each point represents a vector, so you can use the information on the graph to find magnitudes and angles of direction, or to perform operations on vectors.
  ::在 y = (x-2) 2 函数中,每个点仅代表一个空间点。在 F(t) = (t+2, t2) 函数中,每个点代表矢量,这样您就可以使用图上的信息查找方向的大小和角度,或者在矢量上执行操作。

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  Now consider the curve traced by the endpoints of the vectors in the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(3\sin t,5\cos t)\end{array}$ . Identify and sketch the curve.
  ::现在考虑在矢量值函数 F( t) =( 3sint, 5cost) 中的矢量端点所追踪的曲线。 识别并勾画曲线 。

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  You should realize that this vector-valued function looks like the parametric equation for an ellipse centered at the origin. The ellipse has a major axis along the $\begin{array}{r}y\end{array}$ -axis with a length of 10, and a minor axis along the $\begin{array}{r}x\end{array}$ -axis with a length of 6.
  ::您应该意识到, 这个矢量值函数看起来像一个以原为主的椭圆的参数方程。 椭圆沿 y 轴有一个大轴, 长度为 10 , 沿 X 轴有一个小轴, 长度为 6 。

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  If you don’t recognize the form of the equation, you’ll want to sketch the curve using a few values of $\begin{array}{r}t\end{array}$ . To make it easier, use $\begin{array}{r}t=(0,\frac{\pi }{6},\frac{\pi }{4},\frac{\pi }{3},\frac{\pi }{2})\end{array}$ .  Together, these sets of values for  $\begin{array}{r}t\end{array}$ will trace a quarter period from 0 to $\begin{array}{r}2\pi \end{array}$ .
  ::如果您不承认方程的形式, 您将会想要使用 t 的几处值来绘制曲线。 要让曲线更加简单, 请使用 t = (0, 6, 6, 4, 4, 3, 3) 。 , 这些 t 的值将会在 0 到 2 之间追踪四分之一的时间段 。

  $\begin{array}{r}t\end{array}$	$\begin{array}{r}3\sin t\end{array}$	$\begin{array}{r}5\cos t\end{array}$
  0	0	5
  $\begin{array}{r}\frac{\pi }{6}\end{array}$	$\begin{array}{r}\frac{3}{2}\end{array}$	$\begin{array}{r}\frac{5\sqrt{3}}{2}\end{array}$
  $\begin{array}{r}\frac{\pi }{4}\end{array}$	$\begin{array}{r}\frac{3\sqrt{2}}{2}\end{array}$	$\begin{array}{r}\frac{5\sqrt{2}}{2}\end{array}$
  $\begin{array}{r}\frac{\pi }{3}\end{array}$	$\begin{array}{r}\frac{3\sqrt{3}}{2}\end{array}$	$\begin{array}{r}\frac{5}{2}\end{array}$
  $\begin{array}{r}\frac{\pi }{2}\end{array}$	3	0

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  Sketching these vectors will give you a quarter of an ellipse. You can then use the property of symmetry to sketch the other 3 sections of the ellipse.
  ::切除这些矢量会给你四分之一的椭圆。 然后您可以使用对称属性来绘制椭圆的其他3个部分。

  

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  To find the domain for a vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=\left(f(t),g(t)\right)\end{array}$ , you must find the values of  $\begin{array}{r}t\end{array}$ for which  $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ is continuous at both  $\begin{array}{r}f(t)\end{array}$ and $\begin{array}{r}g(t)\end{array}$ .
  ::要找到矢量值函数 F(t) =(f(t),g(t)) 的域, 您必须在 f(t) 和 g(t) 中找到 F(t) 连续的 t 值 。

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  For practice, let's find the domain of the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=\left(\frac{1}{t},\ln (t)\right)\end{array}$ .
  ::用于实践,让我们找到矢量值函数 F(t) =( 1t, ln(t)) 的域 。

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    $\begin{array}{r}\frac{1}{t}\end{array}$ is undefined when  $\begin{array}{r}t\end{array}$ is 0.  The function $\begin{array}{r}\ln (t)\end{array}$ is undefined when $\begin{array}{r}t\le 0\end{array}$ . So, the vector-valued function  $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ is undefined when $\begin{array}{r}t\le 0\end{array}$ . The domain of the function is all real numbers  $\begin{array}{r}t\end{array}$ such that $\begin{array}{r}t>0\end{array}$ .
  ::1t 在 t为 0 时未定义。 函数 IN( t) 时未定义。 因此, 矢量估值函数 F( t) 时未定义。 函数的域是所有真实数字 t , 即 t> 0 。

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  All vector-valued functions can be rewritten as . This transformation allows you to work with them as you would parametric equations and can make some operations easier. To rewrite a vector-valued function as a parametric equation, put it in terms of  $\begin{array}{r}x(t)\end{array}$ and $\begin{array}{r}y(t)\end{array}$ .
  ::所有的矢量估值函数都可以被重写为 。 此转换允许您与它们一起工作, 正如您可以对等方程式一样, 并且可以使某些操作更容易。 重写矢量估值函数作为参数方程式, 请用 x( t) 和 y( t) 来表示 。

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  Let's rewrite  $\begin{array}{r}\stackrel{\to }{F}(t)=\left(t^4,\frac{1}{t}\right)\end{array}$ as a parametric equation. You get:

  $$\begin{array}{r}F(t)=\left((x(t),y(t)\right)\\ x(t)=t^4\\ y(t)=\frac{1}{t}\end{array}$$

  
  ::让我们重写 F}( t) = (t 4, 1t) 作为参数方程 。 您得到 : F (t) = ((x( t) ,y (t) ) x (t) = t4y( t)=1t

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were told about  Arianna building a robot to move up a ramp. Arianna must design a robot that is powerful enough to ascend a ramp that has a grade between 3 and 25 degrees. Arianna decides to define her problem in terms of a vector-valued function. Her robot weighs 3 kilograms. She estimates the acceleration due to gravity as  $\begin{array}{r}g=-9.8m/s^2\end{array}$ . In general, force equals mass times acceleration $\begin{array}{r}(F=ma)\end{array}$ . Her robot is climbing an inclined plane, so Arianna needs to take into account the grade of the surface when considering acceleration. For her robot, the acceleration is equal to $\begin{array}{r}a=g\cdot \sin (t)\end{array}$ .
  ::早些时候,您被告知了阿丽安娜建造一个机器人来移动斜坡。 阿丽安娜必须设计一个强大到足以攀升一个等级在3到25度之间的斜坡的机器人。 阿丽安娜决定用矢量值函数来界定她的问题。她的机器人体重为3公斤。她估计重力加速度为g9.8 m/s2。 一般来说, 强制等于质量乘积加速度( F=ma ) 。 她的机器人正在攀升一个倾斜的平面, 所以在考虑加速时, 阿丽安娜需要考虑到表面的等级。 对于她的机器人来说, 加速度等于=gsin ( t) 。

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  Therefore, the vector that describes the force of gravity pulling her robot down the slope is $\begin{array}{r}F=mg\sin (t)\end{array}$ , where $\begin{array}{r}t\end{array}$ is the angle of the slope that the robot climbs. Her robot will have to exceed this force in order to move up the slope.
  ::因此,描述将其机器人拖下斜坡引力的矢量是F=mgsin(t), t是机器人攀爬的斜坡角。她的机器人必须超过这一力才能向斜坡移动。

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  Once she substitutes 3 kg in for the mass and  $\begin{array}{r}-9.8m/s^2\end{array}$ in for acceleration due to gravity, she’ll be able to graph the vectors for the force of gravity on her robot for all the slopes that may be used during the competition.
  ::一旦她将质量的3千克和因重力加速的-9.8 m/s2代之以3千克,她将能够用图解显示在竞赛期间可能使用的所有斜坡的机器人的引力矢量。

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  Her function is $\begin{array}{r}\stackrel{\to }{F}(t)=(3)(-9.8)\sin (t)\end{array}$ , which she simplifies to $\begin{array}{r}\stackrel{\to }{F}(t)=-29.4\sin (t)\end{array}$ .
  ::她的功能是 F(t) =(3)-9.8sin(t), 她将其简化为 F(t) 29.4sin(t) 。

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  Arianna realizes that her domain for  $\begin{array}{r}t\end{array}$ is $\begin{array}{r}{3}^{\circ }\le t\le {25}^{\circ }\end{array}$ , because the contest rules told her that the slope of the ramp would fall somewhere between these numbers.
  ::Arianna意识到她对 t 的属地是 3t25, 因为竞争规则告诉她, 坡坡坡的斜坡会落下 介于这些数字之间。

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  However, Arianna has a problem. The equation she’s written only gives her the magnitude of the force vector.  To graph it accurately, she’ll need to use trigonometry to break it down into its  $\begin{array}{r}x\end{array}$ and  $\begin{array}{r}y\end{array}$ components.
  ::然而,阿丽安娜有一个问题。 她所写的方程式只给了她力量矢量的大小。 要精确地绘制图表,她需要使用三角测量法将其分解成 x 和 y 组件。

  

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  Each vector in this vector-valued function will be the hypotenuse of a right triangle.  Arianna must use the sine and cosine functions to find the  $\begin{array}{r}x\end{array}$ and  $\begin{array}{r}y\end{array}$ components of the vectors.
  ::在这一矢量估值函数中,每个矢量都将是右三角形的下限。 Arianna 必须使用正弦和余弦函数来找到矢量的 x 和 y 组件 。

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  The  $\begin{array}{r}x\end{array}$ component of her vector-valued function is:
  ::她矢量估价功能的x组成部分是:

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  $$\begin{array}{r}\cos (t)=\frac{\stackrel{\to }{x}(t)}{\stackrel{\to }{F}(t)}\\ \cos (t)=\frac{\stackrel{\to }{x}(t)}{-29.4\sin (t)}\\ \stackrel{\to }{x}(t)=-29.4\sin (t)\cos (t)\end{array}$$

  
  :t) =x(t) F(t) (t) (t) =x(t) - 29.4sin(t) x(t) (t) 29.4sin(t) (t) (t) (t) (t) (t) (t) (t)

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  The  $\begin{array}{r}y\end{array}$ component of her vector-valued function is:
  ::她矢量估价功能的Y部分是:

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  $$\begin{array}{r}\sin (t)=\frac{\stackrel{\to }{y}(t)}{\stackrel{\to }{F}(t)}\\ \sin (t)=\frac{\stackrel{\to }{y}(t)}{-29.4\sin (t)}\\ \stackrel{\to }{y}(t)=-29.4{\sin }^2(t)\end{array}$$

  
  :t) =y(t) F(t) sin(t) =y(t)- 29.4sin(t)y(t) 29.4sin2(t)

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  Therefore, her vector-valued function is:
  ::因此,她的病媒估价功能是:

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  $$\begin{array}{r}\stackrel{\to }{F}(t)=(-29.4\sin (t)\cos (t),-29.4{\sin }^2(t))\end{array}$$

  
  ::F(t) = (- 29.4sin(t) cos(t),- 29.4sin2(t))

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  Arianna finds the value for the lowest bound of her domain, for the highest bound of her domain, and for a few values in between, just so that she can get an idea of the way that force varies depending on the angle of the ramp. The values in the second column on her chart are the magnitudes of the force vector- how much force the robot must overcome at various angles. The values in the third and fourth columns give the points to plot in order to sketch the vector-valued function.
  ::Arianna 找到了她域内最低界限的值, 域内最高界限的值, 以及介于两者之间的几处值, 只是为了让她能够根据斜坡角度来了解力的变化方式。 她的图表第二列中的值是力矢量的大小, 即机器人在不同角度必须克服多大的力。 第三和第四列中的值给出了绘图点, 以便绘制矢量值函数的草图 。

  $\begin{array}{r}t\end{array}$	$\begin{array}{r}(-29.4)\sin (t)\end{array}$	$\begin{array}{r}(-29.4)\sin (t)\cos (t)\end{array}$	$\begin{array}{r}(-29.4){\sin }^2(t)\end{array}$
  3	-1.538	-1.537	-0.081
  5	-2.562	-2.553	-0.223
  10	-5.105	-5.028	-0.887
  15	-7.609	-7.350	-1.969
  20	-10.055	-9.449	-3.439
  25	-12.425	-11.261	-5.251

  

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  Arianna’s robot must have a motor capable of producing more than 12.425 Newtons of force to ascend all possible ramps in the contest. When she sketches the curve, she notices that the relationship between the angle of the ramp and the force necessary to ascend is not linear . In fact, it resembles part of a circle .
  ::亚里安娜的机器人必须拥有能够生产超过12.425牛顿的发动机才能在比赛中攀升所有可能的斜坡。 当她绘制曲线时,她注意到斜坡角度与攀升必要力之间的关系不是线性。 事实上,它与圆圈的一部分相似。

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  For the following examples, consider the following three vector-valued functions:
  ::对于以下例子,考虑以下三种矢量估值功能:

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  1. $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,\ln (t))\end{array}$
     ::F(t) = (t2,ln(t))
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  2. $\begin{array}{r}\stackrel{\to }{F}(t)=(4\sin t,4\cos t)\end{array}$
     ::F(t) = (4sint,4cost)
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  3. $\begin{array}{r}\stackrel{\to }{F}(t)=(t,5t+3)\end{array}$
     ::F(t) = (t,5t+3)

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  Example 2
  ::例2

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  Convert the functions into parametric form .
  ::将函数转换成参数形状。

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  1. $\begin{array}{r}F(t)=(x(t),y(t))\\ x(t)=t^2\\ y(t)=\ln (t)\end{array}$  
     ::F(t) = (x(t) y(t) ) x(t) = t2y(t) = ln(t)

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  2. $\begin{array}{r}F(t)=(x(t),y(t))\\ x(t)=4\sin t\\ y(t)=4\cos t\end{array}$  
     ::F( t) = (x( t) y( t) ) x( t) = 4sin_ ty( t) = 4cos* t

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  3. $\begin{array}{r}F(t)=(x(t),y(t))\\ x(t)=t\\ y(t)=5t+3\end{array}$  
     ::F(t) = (x(t) y(t) ) x(t) = ty(t) = 5t+3

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  Example 3
  ::例3

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  Find the domain of each function.
  ::查找每个函数的域 。

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  1. Since  $\begin{array}{r}\ln (t)\end{array}$ is only defined for positive numbers, the domain of the function is all real numbers  $\begin{array}{r}t\end{array}$ such that $\begin{array}{r}t>0\end{array}$ .
     ::由于 In( t) 仅定义正数, 函数的域是所有真实数字, 如此等於 0 。
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  2.   $\begin{array}{r}\sin (t)\end{array}$ and  $\begin{array}{r}\cos (t)\end{array}$ are defined for all real numbers, so the domain of this function is all real numbers.
     ::sin(t) 和 cos(t) 定义了所有真实数字, 所以此函数的域是全部真实数字 。
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  3. The domain of this function is all real numbers.
     ::此函数的域是所有实际数字 。

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  Example 4
  ::例4

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  Identify the basic shape of each function and then graph. You may use a graphing calculator, an app, or a website such as Fooplot.
  ::指定每个函数和图形的基本形状。 您可以使用图形计算器、 应用程序或 Fooplot 等网站 。

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  1.  The curve resembles the natural log function.
     ::曲线类似于自然日志函数 。

  

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  2. The curve is a circle, centered at the origin, with a radius of 4.
     ::曲线是一个圆形,以原点为中心,半径为4。

  

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  3. The curve is a line with a slope of 5 and a $\begin{array}{r}y\end{array}$ -intercept of 3.
     ::曲线是5的斜坡线和3的 Y 探测线。

  

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  Example 5
  ::例5

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  Find the magnitude and angle of  $\begin{array}{r}\stackrel{\to }{F}(1)\end{array}$ for functions 1 and 3.
  ::查找函数 1 和 3 F(1) 的大小和角度。

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  1. $\begin{array}{r}\stackrel{\to }{F}(1)=(1,0)\\ \Vert \stackrel{\to }{F}(1)\Vert =\sqrt{1^2+0^2}=1\\ \tan \theta =\frac{0}{1}=0\\ \theta =0^{\circ }\end{array}$  
     ::F(1)=(1,0)F(1)12+02=1tan01=00

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  3. $\begin{array}{r}\stackrel{\to }{F}(1)=(1,5(1)+3)=(1,8)\\ \Vert \stackrel{\to }{F}(1)\Vert =\sqrt{1^2+8^2}=\sqrt{65}\\ \tan \theta =\frac{8}{1}=8\\ \theta ={82.87}^{\circ }\end{array}$  
     ::F(1)=(1,5(1)+3)=(1,8)F(1)12+82=65tan81=882.87

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  Review
  ::回顾

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  For #1-4, convert each function into parametric form.
  ::在 # 1 4, 将每个函数转换为参数形式 。

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  1.   $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,\frac{1}{t})\end{array}$
     ::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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  For #5-8, find the domain of each function.
  ::对于# 5-8, 找到每个函数的域 。

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

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  For #9-12, identify the basic shape of each function and then graph.
  ::对于# 9-12, 指定每个函数的基本形状, 然后绘制图形 。

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  9.   $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,\frac{1}{t})\end{array}$
     ::F(t) = (t2,1吨)
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  10.   $\begin{array}{r}\stackrel{\to }{G}(t)=(6\sin t,3\cos t)\end{array}$
      ::G(t) = (6sint,3cost)
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  11.   $\begin{array}{r}\stackrel{\to }{H}(t)=(t+1,t^2)\end{array}$
      ::H(t) = (t+1,t2)
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  12.   $\begin{array}{r}\stackrel{\to }{K}(t)=(2t^2,t-5)\end{array}$
      ::K(t) =( 2t2, t- 5)
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  13.  Find the magnitude and angle of  $\begin{array}{r}\stackrel{\to }{F}(t)=(t^2,\frac{1}{t})\end{array}$ for $\begin{array}{r}t=5\end{array}$ .
      ::为 t=5 查找 F(t) =( t2, 1t) 的大小和角度。
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  14.  Find the magnitude and angle of  $\begin{array}{r}\stackrel{\to }{G}(t)=(6\sin t,3\cos t)\end{array}$ for $\begin{array}{r}t=0\end{array}$ .
      ::查找 t=0 的 G( t) =( 6sin, 3cost) 的大小和角度。
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  15.  Find the magnitude and angle of  $\begin{array}{r}\stackrel{\to }{H}(t)=(t+1,t^2)\end{array}$ for $\begin{array}{r}t=4\end{array}$ .
      ::为 t=4 查找 H(t) = (t+1,t2) 的大小和角度。
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  16.  Find the magnitude and angle of  $\begin{array}{r}\stackrel{\to }{K}(t)=(2t^2,t-5)\end{array}$ for $\begin{array}{r}t=1\end{array}$ .
      ::查找 t= 1 的 K( t) =( 2t2, t- 5) 的大小和角度 。

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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 。