11.11 加速的矢量组件

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  加速的矢量组件

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  There’s a new backwards roller coaster at the local amusement park. Phil wants to ride, but he suffers from back pain and wants to know how the forces will affect his spine. Since force is equal to mass times acceleration , he’d like to find the tangent and normal forces for the ride down the roller coaster’s first big hill. How can he do this?
  ::当地游乐场有一辆新的后向滚车。 菲尔想骑车,但他背痛,想知道部队会如何影响他的脊椎。 由于武力相当于质量时间加速,他想为搭乘过山车的首座大山找到不切实际和正常的力量。 他怎么能这样做呢?

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  Vectors and Acceleration
  ::矢量和加速

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  Acceleration describes the rate of change of an object’s velocity as the object moves along a plane curve.  You can represent this rate of change with a vector, because acceleration has both magnitude and direction. If an object's speed doesn’t change but its direction does, it still has acceleration.
  ::加速度描述一个对象在沿着平面曲线移动时的速度变化速度。您可以用矢量来表示这种变化速度,因为加速度既具有星度,又具有方向。如果一个对象的速度没有变化,但方向却有变化,那么它仍然有加速度。

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  For many applications, you need more information about acceleration than the data given in terms of horizontal and vertical unit vectors. This is because often what’s most important is the direction and magnitude of acceleration relative to the curve, not the direction and speed of magnitude relative to the  $\begin{array}{r}x\end{array}$ and  $\begin{array}{r}y\end{array}$ axes of the coordinate plane.
  ::对于许多应用程序,您需要的加速度信息比水平和垂直单位矢量的数据要多。这是因为,最重要的往往是相对于曲线的加速度方向和幅度,而不是相对于坐标平面的 x 和 y 轴的加速度方向和速度。

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  Since acceleration is the rate of change of velocity, and since velocity is the rate of change of position, acceleration is the second derivative of position. So, if  $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ is a curve that describes the position of an object over time,  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)\end{array}$ is a curve that describes the acceleration of the object over time.
  ::由于加速是速度变化的速率,而且由于速度是位置变化的速率,加速是位置的第二个衍生物。因此,如果F(t)是一个曲线,可以描述一个物体在一段时间内的位置,那么,F(t)是一个曲线,可以描述物体在一段时间内加速的速度。

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  When you divide the acceleration vector into its components relative to a plane curve, you express it in terms of the acceleration tangent to the curve and the acceleration normal (perpendicular) to the curve.  If an object is moving in a circle , the normal acceleration is the acceleration that you use to compute centripetal force.
  ::当将加速度矢量除以与平面曲线相对的成份时,用曲线的加速度正切值和曲线的加速度常数(直角)表示。如果一个对象在一个圆圈中移动,则正常加速度是用来计算百分数力的加速度。

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  When you break the acceleration vector into its tangent and normal components, you find that  $\begin{array}{r}\stackrel{\to }{A}(t)=a_T\stackrel{\to }{T}(t)+a_N\stackrel{\to }{N}(t)\end{array}$ where  $\begin{array}{r}\stackrel{\to }{T}(t)\end{array}$ is the unit tangent vector and  $\begin{array}{r}\stackrel{\to }{N}(t)\end{array}$ is the unit normal vector at time $\begin{array}{r}t\end{array}$ .
  ::当您将加速度向量折断为正切和正常组件时,您会发现 A(t) = ATT(t)+ANN(t) 是单位正切向量,而 N(t) 是单位时的正向量。

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  To find  $\begin{array}{r}{a}_T\end{array}$ and $\begin{array}{r}{a}_N\end{array}$ , you can use the vector-valued functions that represent position and velocity.
  ::要找到 AT 和 AN, 您可以使用代表位置和速度的矢量值函数。

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  Say a car travels along a curve. Its path is modeled by the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(5t,7t^2)\end{array}$ . What is  $\begin{array}{r}{a}_T\end{array}$ when $\begin{array}{r}t=1\end{array}$ ?
  ::说一辆汽车沿曲线行驶。其路径由矢量值函数F(t)=(5t,7t2)。t=1时AT是什么?

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  For an object traveling along a plane curve, $\begin{array}{r}{a}_T=\frac{\stackrel{\to }{A}(t)\cdot \stackrel{\to }{V}(t)}{\Vert \stackrel{\to }{V}(t)\Vert }\end{array}$ .
  ::aT=A(t)V(t)V(t)(t)。

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  Notice that  $\begin{array}{r}{a}_T\end{array}$ is a scalar because the dot product of two vector-valued functions returns a scalar and the magnitude of a vector-valued function is a scalar.
  ::注意 AT 是一个星标, 因为两个矢量值函数的点产物返回一个星标, 而矢量值函数的大小是一个星标。

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  Since velocity is the derivative of  $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ and acceleration is the second derivative of $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ , you can rewrite this equation as $\begin{array}{r}{a}_T=\frac{F{}^{\mathrm{\prime }}(t)\cdot F{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)}{\Vert F{}^{\mathrm{\prime }}(t)\Vert }\end{array}$ . Use the equation to find the tangent component of the acceleration of the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)\end{array}$ .
  ::由于速度是 F(t) 的衍生物,加速度是 F(t) 的第二个衍生物, 您可以重写此方程式为 aT=F(t) F(t) }F(t) 。 使用方程式查找矢量估值函数 F(t) 加速度的正切部分 。

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  $$\begin{array}{r}F(t)=(5t,7t^2)\\ F{}^{\mathrm{\prime }}(t)=(5,14t)\\ \Vert F{}^{\mathrm{\prime }}(t)\Vert =\sqrt{25+196t^2}\\ F{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)=(0,14)\end{array}$$

  
  ::F(t) = (5,7t,7t2) F_(t) = (5,14t) = (5,14t) = F_(t) = 25+1962F}(t) = (0,14)

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  Find the dot product of  $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\end{array}$ and $\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)\end{array}$ .
  ::查找 F(t) 和 F(t) 的点产物 。

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  $$\begin{array}{r}F{}^{\mathrm{\prime }}(t)\cdot F{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)=(0)(5)+(14t)(14)=196t\end{array}$$

  
  ::Fä(t)F(t)=(0)(5)+(14)14=196t

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  Find  $\begin{array}{r}{a}_T\end{array}$ and evaluate when $\begin{array}{r}t=1\end{array}$ .
  ::查找 aT 并评估 t=1 的时间 。

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  $$\begin{array}{r}{a}_T=\frac{F{}^{\mathrm{\prime }}(t)\cdot F{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)}{\Vert F{}^{\mathrm{\prime }}(t)\Vert }\\ a_T=\frac{196t}{\Vert \sqrt{25+196t^2}\Vert }=\frac{196}{\sqrt{221}}\approx 13.18\end{array}$$

  
  ::aT=F_(t)_F_(t)_F_(t)_(t)_(aT=196___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(t)___________(t)__________________________________________________________________________________________________________________________________________________________________________________________________________________________________(t(t)

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  At $\begin{array}{r}t=1\end{array}$ ,  $\begin{array}{r}{a}_T\end{array}$ is about $\begin{array}{r}13.18m/s^2\end{array}$ .
  ::T=1, AT值约为13.18 m/s2。

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  Now, find  $\begin{array}{r}{a}_N\end{array}$ for the same car. You can find  $\begin{array}{r}{a}_N\end{array}$ by using the equation: $\begin{array}{r}{a}_N=\sqrt{\Vert \stackrel{\to }{A}(t){\Vert }^2-(a_T{)}^2}\end{array}$ , which you can rewrite as $\begin{array}{r}{a}_N=\sqrt{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t){\Vert }^2-(a_T{)}^2}\end{array}$ .
  ::现在,为同一辆车寻找 anN。 您可以使用公式来找到 anN: anNA}( t)%2 -( at) 2, 您可以重新写入为 anNF}( t)%2 -( aT) 2。

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  $$\begin{array}{r}{a}_N=\sqrt{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t){\Vert }^2-(a_T{)}^2}\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)=(0,14)\\ \Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)\Vert =\sqrt{0^2+{14}^2}=14\\ a_T^2={\left(\frac{196}{\sqrt{221}}\right)}^2=\frac{38416}{221}\\ a_N=\sqrt{196-\frac{38416}{221}}\approx 4.709\end{array}$$

  
  
  :t) = (0, 14, 412) = (t) 02+142=14aT2= (196221) 2= 38416221aN= 196- 38416221N=4.709

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  Once you have  $\begin{array}{r}{a}_T\end{array}$ and $\begin{array}{r}{a}_N\end{array}$ , you can rewrite  $\begin{array}{r}\stackrel{\to }{A}(t)\end{array}$ in terms of  $\begin{array}{r}\stackrel{\to }{T}(t)\end{array}$ and  $\begin{array}{r}\stackrel{\to }{N}(t)\end{array}$ instead of in terms of  $\begin{array}{r}i\end{array}$ and  $\begin{array}{r}j\end{array}$ at $\begin{array}{r}t=1\end{array}$ . Notice that  $\begin{array}{r}{a}_T\end{array}$ and  $\begin{array}{r}{a}_N\end{array}$ are only the scalar values at $\begin{array}{r}t=1\end{array}$ . If you want to find the equation for acceleration in terms of  $\begin{array}{r}\stackrel{\to }{T}(t)\end{array}$ and  $\begin{array}{r}\stackrel{\to }{N}(t)\end{array}$ at some other point on the curve, you will have to compute new values for  $\begin{array}{r}{a}_T\end{array}$ and $\begin{array}{r}{a}_N\end{array}$ .
  ::一旦您有 AT 和 anN, 您可以重写 A( t) 和 N( t) , 而不是 i 和 j t= 1. 注意 AT 和 anN 仅仅是 t= 1 的 scalar 值。 如果您想要在曲线上其他某个点找到 T( t) 和 N( t) 的加速度方程式, 您就必须为 AT 和 anN 计算新的值 。

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  At $\begin{array}{r}t=1\end{array}$ , the equation for acceleration relative to this curve is $\begin{array}{r}(13.18)\stackrel{\to }{T}(1)+(4.709)\stackrel{\to }{N}(1)\end{array}$ .
  ::t=1时,相对于该曲线的加速度方程式为(13.18)T(1)+(4.709)N(1)。

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  You can use  $\begin{array}{r}{a}_N\end{array}$ to more easily compute the curvature of a plane curve at a particular time. That’s because you can express  $\begin{array}{r}k(t)\end{array}$ in terms of $\begin{array}{r}{a}_N\end{array}$ . This expression is often a bit easier to untangle than the standard equation for curvature, and you may find that it makes your calculations simpler.
  ::您可以使用 anN 来更方便地计算某个特定时间的平面曲线的曲线。 这是因为您可以用 anN 表示 k( t) 。 这个表达式往往比曲线的标准方程式更容易解开, 并且你可能会发现它会简化你的计算。

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  $$\begin{array}{r}k(t)=\frac{a_N}{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t){\Vert }^2}\end{array}$$

  
  :kt)=aNF(t)2

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  Let's find the curvature of the vector-valued function for the car when $\begin{array}{r}t=1\end{array}$ .
  ::让我们找出t=1时汽车矢量值函数的曲线。

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  $$\begin{array}{r}k(t)=\frac{a_N}{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t){\Vert }^2}\\ k(1)=\left(\frac{4.709}{{\left(\sqrt{25+196(1{)}^2}\right)}^2}\right)=\frac{4.709}{221}=.0213\end{array}$$

  
  
  :kt)=aNF(t)2k(1)=(4.709(25+196(1)2)2)=4.709221=.0213)

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  The curvature of this function when  $\begin{array}{r}t=1\end{array}$ is .0213.
  ::3⁄4 ̄ ̧漯B

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find how much force is keeping Phil in his seat on the roller coaster. Phil creates a model of the coaster’s first hill. He uses the vector-valued function  $\begin{array}{r}\stackrel{\to }{F}(t)=(30t,5t^2)\end{array}$ to model the car’s movement along the curve.  To find out how much force is keeping him in his seat and how much is pushing his back into the back of the seat, he’ll need to break acceleration into its tangent and normal components.  He’ll use this information to compute the forces on his body when $\begin{array}{r}t=5\end{array}$ .  He weighs 75 kg.
  ::早些时候,有人要求你找出菲尔在过山车座上有多少武力。 菲尔创造了海岸车首座山的模型。 他用矢量值函数F( t)=( 30t, 5t2)来模拟车轮沿曲线的动向。 要弄清楚他坐在座位上有多少武力,以及把他的背推到座位后座上多少力,他需要将加速度断裂成其正切和正常的部件。 他将利用这些信息来计算他5岁时身上的力。 他体重75公斤。

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  First, Phil computes the tangential acceleration .
  ::首先,菲尔计算了相向加速度。

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  $$\begin{array}{r}{a}_T=\frac{\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)}{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\Vert }\\ \stackrel{\to }{F}(t)=(30t,5t^2)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(30,10t)\\ \Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\Vert =\sqrt{900+100t^2}=\sqrt{100(9+t^2)}=10\sqrt{9+t^2}\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)=(0,10)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)=(30)(0)+(10t)(10)=100t\\ a_T=\frac{\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)}{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\Vert }=\frac{100t}{10\sqrt{9+t^2}}=\frac{10t}{\sqrt{9+t^2}}\end{array}$$

  
  ::aT=F(t)(t)F(t)(t)F(t)(t)=(30,5t2)F(t)=(30,10t)(30,10t)(30,10t)(30)(30)(0)0+900+100t2=100(9+2)(t)=109+t2=109+t2

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  He evaluates it at $\begin{array}{r}t=5\end{array}$ .
  ::他在t=5时评价它。

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  $$\begin{array}{r}\frac{10t}{\sqrt{9+t^2}}=\frac{10(5)}{\sqrt{9+25}}=\frac{50}{\sqrt{34}}m/s^2\end{array}$$

  
  ::10t9+t2=10(5)9+25=5034米/秒2

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  He multiplies it by his weight in kilograms to find the force on his back in joules.
  ::以公斤的重量乘以倍增 以在焦耳斯发现背部的力

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  $$\begin{array}{r}\frac{50}{\sqrt{34}}\times 75\approx 643J\end{array}$$

  
  ::5034-375643 J

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  He uses the tangential acceleration to find the normal acceleration, then multiplies that by his weight to find the normal force. That’s the force that keeps him in his seat as the rollercoaster goes down the hill.
  ::他用相近加速度来寻找正常加速度,然后按重量乘以倍增以找到正常力。 当滚轮车下山时,正是这种力量将他留在座位上。

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  $$\begin{array}{r}{a}_N=\sqrt{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t){\Vert }^2-(a_T{)}^2}\\ \Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)\Vert =10\\ a_T=8.575\\ a_N=\sqrt{100-73.529}=5.145m/s^2\\ 5.145\times 75=385.87J\end{array}$$

  
  :t)2 -(aT)2 -(f)2 -(t)10aT=8,575aN=100-73.529=5.145m/s25.145×75=385.87J

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  Paul weighs 50 kilograms and is riding a roller coaster. The coaster heads down a hill that can be modeled by the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(20t,5t^2)\end{array}$ . For the following examples, use this information.
  ::Paul 体重50公斤,正在骑着一辆过山车。 海岸人向下一个山丘前进, 可以通过矢量值函数F( t) =( 20t, 5t2) 来模拟。 对于以下例子, 请使用此信息 。

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  Example 2
  ::例2

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  Find  $\begin{array}{r}{a}_T\end{array}$ for this curve when $\begin{array}{r}t=2\end{array}$ .
  ::t=2 时为此曲线查找 AT 。

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  First find $\begin{array}{r}{a}_T\end{array}$ :
  ::第一个找到 AT :

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  $$\begin{array}{r}\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)=(20,10t)\\ \Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\Vert =\sqrt{400+100t^2}=\sqrt{100(4+t^2)}=10\sqrt{4+t^2}\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)=(0,10)\\ \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)=(0)(20)+(10)(10t)=100t\\ a_T=\frac{\stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\cdot \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)}{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t)\Vert }=\frac{100t}{10\sqrt{4+t^2}}\end{array}$$

  
  ::F(t) = (20,10t) = F(t) = (20,10t) F(t) F(400+100t2) = 100(4+2) = 104+t2F(t) = (0,10F}(t) F* = (0)(20) +(10) (10t) = 100taT= F(t) F(t) F(t) F* F(t) (t) 100+104+t2

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  Now find  $\begin{array}{r}{a}_T\end{array}$ at $\begin{array}{r}t=2\end{array}$ :
  ::现在在 t= 2: 找到 at t= 2 的 AT :

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  $$\begin{array}{r}{a}_T=\frac{10(2)}{\sqrt{4+4}}=\frac{20}{\sqrt{8}}\approx 7.071\end{array}$$

  
  ::aT=10(2)4+4=2087.71

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  Example 3
  ::例3

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  Find  $\begin{array}{r}{a}_N\end{array}$ for the curve when $\begin{array}{r}t=2\end{array}$ .
  ::t=2 时为曲线查找 anN 。

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  $$\begin{array}{r}{a}_N=\sqrt{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t){\Vert }^2-(a_T{)}^2}\\ \Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}{}^{\mathrm{\prime }}(t)\Vert =10\\ a_N=\sqrt{100-(7.071{)}^2}\approx 7.071\end{array}$$

  
  :t)2 -(aT)2 -(t)___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________(t)________________________________________________________________________(t______________________________________________________________________________________________________________________(7._(7.

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  Example 4
  ::例4

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  What are the normal and tangential forces affecting Paul’s body at $\begin{array}{r}t=2\end{array}$ ?
  ::在t=2影响保罗身体的正常和不相干的力量是什么?

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  $$\begin{array}{r}F=ma\\ F_T=m(a_T)=50(7.071)=353.55J\\ F_N=m(a_N)=50(7.071)=353.55J\end{array}$$

  
  ::F=MAFT=m(aT)=50(7.071=353.55)JFN=m(aN)=50(7.071)=353.55J

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  Example 5
  ::例5

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  What is the curvature of the roller coaster’s path at $\begin{array}{r}t=2\end{array}$ ? What is the radius of the circle that would best fit this curve?
  ::T=2 的过山车路径的曲线是什么? 圆圆的半径最符合这个曲线?

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  $$\begin{array}{r}k(t)=\frac{a_N}{\Vert \stackrel{\to }{F}{}^{\mathrm{\prime }}(t){\Vert }^2}\\ k(2)=\frac{7.071}{{\left(10\sqrt{4+2^2}\right)}^2}=\frac{7.071}{800}=.0088\\ r=\frac{1}{k(2)}=\frac{1}{.0088}=113.6\end{array}$$

  
  ::k( t) =aNF( t) 2k(2) = 7. 71( 104+222) 2 = 7. 71800 =. 0088r = 1k(2) = 1. 0088= 113. 6

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  2 seconds into the hill, the circle of best fit for the hill has a radius of 113.6 meters.
  ::2秒后进入山丘 最适合山丘的圆圈 半径为113.6米

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  Review
  ::回顾

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  For #1-3, use the function $\begin{array}{r}\stackrel{\to }{F}(t)=(t-1,t^2)\end{array}$ .
  ::# 1-3, 请使用 F( t) =( t- 1, t2) 函数 。

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  1. Find  $\begin{array}{r}{a}_T\end{array}$ for this curve when $\begin{array}{r}t=4\end{array}$ .
     ::t=4 时为此曲线查找 AT 。
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  2. Find  $\begin{array}{r}{a}_N\end{array}$ for the curve when $\begin{array}{r}t=4\end{array}$ .
     ::t=4 时为曲线查找 anN 。
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  3. What is the curvature of the path at $\begin{array}{r}t=4\end{array}$ ? What is the radius of the circle that would best fit this curve?
     ::t=4 路径的曲率是多少? 圆的半径是多少? 最适合此曲线的圆的半径是多少?

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  For #4-6, use the function $\begin{array}{r}\stackrel{\to }{G}(t)=(6\sin t,6\cos t)\end{array}$ .
  ::对于# 4-6, 请使用 G( t) =( 6sint, 6cost) 的函数 。

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  4. Find  $\begin{array}{r}{a}_T\end{array}$ for this curve when $\begin{array}{r}t=2\end{array}$ .
     ::t=2 时为此曲线查找 AT 。
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  5. Find  $\begin{array}{r}{a}_N\end{array}$ for the curve when $\begin{array}{r}t=2\end{array}$ .
     ::t=2 时为曲线查找 anN 。
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  6. What is the curvature of the path at $\begin{array}{r}t=2\end{array}$ ? What is the radius of the circle that would best fit this curve?
     ::t=2 路径的曲率是多少? 圆的半径是多少, 最适合此曲线 ?

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  For #7-9, use the function $\begin{array}{r}\stackrel{\to }{K}(t)=(2t^2,t-5)\end{array}$ .
  ::对于 # 7- 9, 请使用 K( t) =( 2t2, t- 5) 函数 。

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  7. Find  $\begin{array}{r}{a}_T\end{array}$ for this curve when $\begin{array}{r}t=1\end{array}$ .
     ::t=1 时为此曲线查找 AT 。
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  8. Find  $\begin{array}{r}{a}_N\end{array}$ for the curve when $\begin{array}{r}t=1\end{array}$ .
     ::t=1 时为曲线查找 anN 。
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  9. What is the curvature of the path at $\begin{array}{r}t=1\end{array}$ ? What is the radius of the circle that would best fit this curve?
     ::t=1 路径的曲率是多少? 圆的半径是多少, 最适合此曲线 ?

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  For #10-12, use the function $\begin{array}{r}\stackrel{\to }{M}(t)=(e^t,t)\end{array}$ .
  ::对于# 10-12, 请使用函数 M( t) = (et, t) 。

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  10. Find  $\begin{array}{r}{a}_T\end{array}$ for this curve when $\begin{array}{r}t=1\end{array}$ .
      ::t=1 时为此曲线查找 AT 。
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  11. Find  $\begin{array}{r}{a}_N\end{array}$ for the curve when $\begin{array}{r}t=1\end{array}$ .
      ::t=1 时为曲线查找 anN 。
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  12. What is the curvature of the path at $\begin{array}{r}t=1\end{array}$ ? What is the radius of the circle that would best fit this curve?
      ::t=1 路径的曲率是多少? 圆的半径是多少, 最适合此曲线 ?

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  For #13-15, use the following information:
  ::对于第13-15号,请使用以下信息:

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  Kristina weighs 54 kilograms and is riding a roller coaster. At one point, the coaster heads down a hill that can be modeled by the vector-valued function $\begin{array}{r}\stackrel{\to }{F}(t)=(5t,15t^2)\end{array}$ .
  ::Kristina重54公斤,正骑着一辆过山车。有一次,海岸人向下一个山丘走去,可以用矢量值函数F(t)=(5t,15t2)。

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  13. Find  $\begin{array}{r}{a}_T\end{array}$ for this curve when $\begin{array}{r}t=3\end{array}$ .
      ::t=3 时为此曲线查找 AT 。
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  14. Find  $\begin{array}{r}{a}_N\end{array}$ for the curve when $\begin{array}{r}t=3\end{array}$ .
      ::t=3 时为曲线查找 anN 。
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  15. What are the normal and tangential forces affecting Kristina's body at $\begin{array}{r}t=3\end{array}$ ?
      ::影响Kristina在t=3的尸体的正常和不相干的力量是什么?

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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 。