节: 沿着平面曲线运动: 切向和普通矢量 | 11.8 沿着平面曲线的动议:切和正常矢量

章节大纲

Moira designs roller coasters for theme parks. She’s in the early stages of designing a new coaster, and she needs to understand how various forces will be acting on the riders over the course of the ride. Right now, she doesn’t need to know how much force is involved; she just needs to know direction. How can she find out how the direction of the coaster is changing when $\begin{aligned} t = 1 \end{aligned}$ ?
::莫伊拉为主题公园设计过山车。她正处于设计新海岸车的早期阶段,她需要了解各种部队如何在骑车过程中对骑车者采取行动。现在,她不需要知道有多少部队参与其中;她只需要知道方向。当 t = 1 时,她如何知道海岸车的方向正在改变?

Tangent and Normal Vectors of Curves
::曲线的切向和正常矢量

Sometimes you need to understand the shape of an object’s path, but you don’t really need to know how quickly the object is traveling along the path. In these cases, a unit tangent vector can let you see change in direction without showing you change in speed. A unit tangent vector is a vector with a magnitude of one that is tangent to a vector-valued function at a particular point. The unit tangent has the same direction as a typical tangent vector, but has a magnitude of 1.
::有时您需要理解一个对象路径的形状, 但您并不真的需要知道该对象沿路径行驶的速度有多快。在这种情况下, 单位正切矢量可以在不显示速度变化的情况下让您看到方向变化。单位正切矢量是一种矢量, 其大小与特定点的矢量值函数相切。单位正切方向与典型正切矢量相同, 但大小为 1 。单位正切方向与典型的正切矢量相同, 大小为 1 。

To find the unit tangent vector, you find the tangent vector and then divide by the magnitude of the vector. This leaves you with a vector that has direction, but a magnitude of 1. For a vector-valued function $\begin{aligned} \vec{F} (t) \end{aligned}$ , the unit tangent vector $\begin{aligned} \vec{T} (t) \end{aligned}$ can be described as:
::要找到单位正切矢量,您可以找到正切矢量,然后除以矢量的大小。这为您留下一个带有方向的矢量,但值为 1 的矢量。对于矢量估值函数F(t),单位正切矢量T(t)可描述为:

$\begin{aligned} \vec{T} (t) = \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} . \end{aligned}$

::T(t) = F(t) F(t) (t) 。

Let's find the equation for the unit tangent vector of the function $\begin{aligned} \vec{F} (t) = (t^{2}, 5 t) \end{aligned}$ and evaluate for $\begin{aligned} t = 5 \end{aligned}$ .
::让我们为函数 F( t) = ( t2, 5t) 的单位正切矢量寻找方程式, 并对 t= 5 进行评价。

First, find the derivative of the function.
::首先,找到函数的衍生物。

$\begin{aligned} \vec{F} (t) = (t^{2}, 5 t) \\ \vec{F}^{'} (t) = (2 t, 5) \end{aligned}$
::F(t) = (t2,5t) F(t) = (2t,5)

Now, find the magnitude of $\begin{aligned} \vec{F}^{'} (t) \end{aligned}$ :
::找到F( t) 的大小 :

$\begin{aligned} ‖ \vec{F}^{'} (t) ‖ = \sqrt{(2 t)^{2} + 5^{2}} = \sqrt{4 t^{2} + 25} \end{aligned}$

::@F(t)(2t)2+52=4t2+25

Separate $\begin{aligned} \vec{F}^{'} (t) \end{aligned}$ into its horizontal and vertical components and divide by the magnitude to find the unit tangent, $\begin{aligned} \vec{T} (t) \end{aligned}$ .
::将 F( t) 分隔成其水平和垂直构件, 并按大小分隔, 以找到单位正切值, T( t) 。

$\begin{aligned} \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} = \frac{2 t i + 5 j}{\sqrt{4 t^{2} + 25}} \\ = \frac{2 t \sqrt{4 t^{2} + 25}}{4 t^{2} + 25} i + \frac{5 \sqrt{4 t^{2} + 25}}{4 t^{2} + 25} j \end{aligned}$

::F(t)F(t)2ti+5j4t2+25=2t4t2+254t2+25i+54t2+254t2+25j

For every $\begin{aligned} t \end{aligned}$ , $\begin{aligned} \vec{T} (t) \end{aligned}$ will give the unit tangent vector for that point on the curve. When you evaluate the unit tangent vector at a specific time, $\begin{aligned} t \end{aligned}$ , you will find the direction that the object is moving at that particular time, but you will not find its speed.
::对于每 t, T( t) 将给出曲线上该点的单位正切矢量。当您在特定时间评价单位正切矢量时, t, 将会找到该对象在特定时间移动的方向, 但找不到它的速度。

To evaluate at $\begin{aligned} t = 5 \end{aligned}$ , substitute 5 for $\begin{aligned} t \end{aligned}$ .
::以 t=5 代替 t = 5 进行评价。

$\begin{aligned} \vec{T} (5) = \frac{2 (5) \sqrt{4 (5)^{2} + 25}}{4 (5)^{2} + 25} i + \frac{5 \sqrt{4 (5)^{2} + 25}}{4 (5)^{2} + 25} j \\ \vec{T} (5) = \frac{2 \sqrt{5}}{5} i + \frac{\sqrt{5}}{5} j \\ \vec{T} (5) = (\frac{2 \sqrt{5}}{5}, \frac{\sqrt{5}}{5}) \end{aligned}$

::T(5)=2(5)4(5)2+254(5)2+251(5)2+54(5)2+254(5)2+252(5)2+25jT(5)=255i+55jT(5)=(255,55)

Once you’ve found the unit tangent vector to a vector-valued function, you can also find the unit normal vector . The unit normal vector is a unit vector that is perpendicular to the curve at a given point. The unit normal vector is especially useful when you’re trying to compute the normal force on a moving object.
::一旦您发现向量值函数的单位正切矢量, 您也可以找到单位正向矢量。单位正向矢量是单位正向矢量, 与给定点的曲线垂直。当您试图计算移动对象的正常力时, 单位正向矢量特别有用。

Because the unit normal vector is perpendicular to the curve, it is also perpendicular to the unit tangent vector. This means that the dot product of the unit normal vector and the unit tangent vector at a certain point should equal 0.
::由于单位正常矢量与曲线垂直,也与单位正切矢量垂直。这意味着单位正常矢量和单位正切矢量在某一点的圆点产物应等于0。

You can find the unit normal vector $\begin{aligned} \vec{N} (t) \end{aligned}$ by dividing the derivative of the unit tangent vector by the magnitude of the derivative of the unit tangent vector. That is,
::您可以通过将单位正切矢量的衍生物除以单位正切矢量的衍生物大小来找到单位正向矢量N(t)。也就是说,

$\begin{aligned} \vec{N} (t) = \frac{\vec{T}^{'} (t)}{‖ \vec{T}^{'} (t) ‖} . \end{aligned}$

::N(t) = T(t) (t) (t) (t) 。

Let's find the unit normal vector for the vector-valued function $\begin{aligned} \vec{F} (t) = (t, t^{2}) \end{aligned}$ .
::让我们为矢量值函数 F(t) =(t,t2) 找到单位正常向量。

First, find the unit tangent vector for the function.
::首先,找到函数的单位正切矢量。

$\begin{aligned} \vec{T} (t) = \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} \\ \vec{T} (t) = \frac{i + 2 t j}{\sqrt{1^{2} + (2 t)^{2}}} = \frac{i + 2 t j}{\sqrt{4 t^{2} + 1}} \\ \vec{T} (t) = \frac{1}{\sqrt{4 t^{2} + 1}} i + \frac{2 t}{\sqrt{4 t^{2} + 1}} j \end{aligned}$

::T(t) = F(t) (t) (t) (t) (t) (t) =i+2tj12+(2t) 2=i+2tj4t2+1T}(t) =14t2+1i+2t4t2+1j

To find the normal vector, find the for the horizontal and vertical components of $\begin{aligned} \vec{T} (t) \end{aligned}$ .
::要找到普通矢量, 请为 T( t) 的水平和垂直组件查找。

$\begin{aligned} \vec{T} (t) = \frac{1}{\sqrt{4 t^{2} + 1}} i + \frac{2 t}{\sqrt{4 t^{2} + 1}} j \\ \vec{T}^{'} (t) = [- \frac{1}{2} (4 t^{2} + 1)^{- \frac{3}{2}} (8 t)] i + [2 (4 t^{2} + 1)^{- \frac{1}{2}} + 2 t (- \frac{1}{2} (4 t^{2} + 1)^{- \frac{3}{2}}) (8 t)] j \\ \vec{T}^{'} (t) = [- 4 t (4 t^{2} + 1)^{- \frac{3}{2}}] i + [2 (4 t^{2} + 1)^{- \frac{1}{2}} - 8 t^{2} ((4 t^{2} + 1)^{- \frac{3}{2}})] j \end{aligned}$

::T( t) = 14t2+1i+2t4T2+1jT( t) = [- 12( 4t2+1) - 32( 8t)] i+[ 2( 4t2+1) - 12+2( 12( 4t2+1) - 32( 8t) jT( t) = [- 4t( 4t2+1+1) - 32] i+[ 2( 4t2+1) - 12T2( 2( 4t2+1) - 32)]jT( t) = [2( 4t2+1) - 32] i+[2( 4t2+1) - 12t2( 2( 4t2+1) - 32)] jj

Find the magnitude of $\begin{aligned} \vec{T}^{'} (t) \end{aligned}$ , then divide. Your final equation will be pretty messy, but it’s fine to leave it in its un-simplified form.
::找到 T( t) 的大小, 然后除以。您的最后方程式将相当混乱, 但将其保留在不简化的形式是好的。

$\begin{aligned} ‖ \vec{T}^{'} (t) ‖ = \sqrt{{[- 4 t (4 t^{2} + 1)^{- \frac{3}{2}}]}^{2} + {[2 (4 t^{2} + 1)^{- \frac{1}{2}} - 8 t^{2} ((4 t^{2} + 1)^{- \frac{3}{2}})]}^{2}} \\ \vec{N} (t) = \frac{[- 4 t (4 t^{2} + 1)^{- \frac{3}{2}}] i + [2 (4 t^{2} + 1)^{- \frac{1}{2}} - 8 t^{2} ((4 t^{2} + 1)^{- \frac{3}{2}})] j}{\sqrt{{[- 4 t (4 t^{2} + 1)^{- \frac{3}{2}}]}^{2} + {[2 (4 t^{2} + 1)^{- \frac{1}{2}} - 8 t^{2} ((4 t^{2} + 1)^{- \frac{3}{2}})]}^{2}}} \end{aligned}$

:t)[-4t(4t2+1)-32]2+[2(4t2+1)-12-8t2((4t2+1)-32)]2N(t)=[[-4t(4t2+1)-32]i[2(4t2+1)-12-8t2((4t2+1)-32)]j[-4t(4t2+1)-32]2+[2(2(4t2+1)-1)-12(2)(4t2+1)-32]2]2+[2(2)(4t2+1)-(3-2)2]2]]

Now, let's evaluate the unit normal vector for $\begin{aligned} t = 5 \end{aligned}$ .
::现在,让我们来评估单位正常向量 t=5。

This will be a pretty messy equation, and it’s fine to use your calculator to solve it. In general, the unit normal vector is used in physics and engineering applications, and in a professional setting, you would use a calculator or similar program to compute its value.
::这将是一个相当混乱的方程式,使用您的计算器解决它也很好。一般来说,单位正常矢量被用于物理和工程应用,在专业环境中,您会使用计算器或类似程序来计算其值。

$\begin{aligned} \vec{N} (5) = \frac{[- 4 (5) (4 (5)^{2} + 1)^{- \frac{3}{2}}] i + [2 (4 (5)^{2} + 1)^{- \frac{1}{2}} - 8 (5)^{2} ((4 (5)^{2} + 1)^{- \frac{3}{2}})] j}{\sqrt{{[- 4 (5) (4 (5)^{2} + 1)^{- \frac{3}{2}}]}^{2} + {[2 (4 (5)^{2} + 1)^{- \frac{1}{2}} - 8 (5)^{2} ((4 (5)^{2} + 1)^{- \frac{3}{2}})]}^{2}}} \\ \vec{N} (5) \approx - .9950 i + .0995 j . \end{aligned}$

::N(5)=[-4(5)(5)2+1-32]i+[(2(5)2+1)-1-12-8(5)2(4((5)2+1)-32)]j[-4(5)(4(5)2+1)-32]2+[2(4((5)2+1)+1)-12(5)(4(5)2+1)-8(5)(2(5)+1)][2N(5)(5)(5).9950i+.0995j。

Your answer may vary slightly depending on how and where you rounded. However, you can check to see if you’re close by using the distance formula to make sure that the length of your unit normal vector is 1, or within a rounding error of 1.
::您的答案可能略有不同, 取决于您四舍五入的方式和位置。但是, 您可以使用距离公式来确保单位正常矢量的长度为 1, 或者在圆形误差为 1 的范围内, 检查您是否接近。

Examples
::实例

Example 1
::例1

Earlier, you were asked how Moira can make calculations about her roller coaster to help her in designing it. One of the hills on Moira’s new coaster can be modeled by the vector-valued function $\begin{aligned} \vec{F} (t) = (\frac{t}{4 π}, 23 \sin t) \end{aligned}$ where angles are in radians. She’d like to know how the direction of the coaster is changing when $\begin{aligned} t = 1 \end{aligned}$ . First, she finds the unit tangent vector for the curve.
::早些时候,有人问莫伊拉如何计算她的过山车来帮助她设计它。莫伊拉新海岸车的一座山可以用矢量值函数F(t)=(t4,23sint)为模型,角度在弧度中。她想知道当t=1时,海岸车的方向会如何变化。首先,她找到了曲线的单位切向矢量。

$\begin{aligned} \vec{F} (t) = (\frac{t}{4 π}, 23 \sin t) \\ \vec{F}^{'} (t) = \frac{1}{4 π} i + (23 \cos t) j \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{\frac{1}{16 π^{2}} + 529 \cos^{2} t} \\ \vec{T} (t) = \frac{1}{4 π \sqrt{\frac{1}{16 π^{2}} + 529 \cos^{2} t}} i + \frac{(23 \cos t)}{\sqrt{\frac{1}{16 π^{2}} + 529 \cos^{2} t}} j \end{aligned}$

::F(t) = (t4, 23sint) F(t) = 14i+(23cost) j(t) 1162+529cos2tT(t) = 141162+529cos2+529cos2ti+(23cost) 1162+529cos2}tj

She evaluates it at $\begin{aligned} t = 1 \end{aligned}$ :
::她在t=1:1时评价它:

$\begin{aligned} \vec{T} (t) = \frac{1}{4 π \sqrt{\frac{1}{16 π^{2}} + 529 \cos^{2} 1}} i + \frac{(23 \cos 1)}{\sqrt{\frac{1}{16 π^{2}} + 529 \cos^{2} 1}} j \end{aligned}$

::T( t) = 141162+529cos21i+( 23cos1) 1162+529cos21j

and finds that, at that time, the unit tangent vector is approximately $\begin{aligned} T (t) = (.00640, .99998) \end{aligned}$ . This vector gives the instantaneous direction of movement at $\begin{aligned} t = 1 \end{aligned}$ . Moira can use this information to figure out the momentum on passengers and to ensure that the roller coaster can be ridden safely.
::并发现,当时单位切向矢量大约为T(t) = (. 00640. 99998) 。该矢量在 t=1. Moira 可以使用此信息了解乘客的动向,并确保过山车安全登上。

For the following examples , use the function $\begin{aligned} \vec{F} (t) = (2 t, t^{3}) \end{aligned}$ .
::以下示例使用函数F(t) =( 2t, t3) 。

Example 2
::例2

Find the equation for the unit tangent vector of $\begin{aligned} \vec{F} (t) \end{aligned}$ .
::F(t) 的正切向量的单位方程式查找方程式。

$\begin{aligned} \vec{F}^{'} (t) = 2 i + 3 t^{2} j \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{2^{2} + (3 t^{2})^{2}} = \sqrt{4 + 9 t^{4}} \\ \vec{T} (t) = \frac{2}{\sqrt{4 + 9 t^{4}}} i + \frac{3 t^{2}}{4 + 9 t^{4}} j \end{aligned}$
Example 3
::F( t) = 2i+3T2jF( t) 22+( 3t2) 2= 4+9t4T( t) = 24+9t4i+3t24+9t4j例 3

Find the equation for the unit normal vector of $\begin{aligned} \vec{F} (t) \end{aligned}$ .
::F(t) 的单位正态矢量查找方程式。

$\begin{aligned} \vec{T}^{'} (t) = [- 36 t^{3} (4 + 9 t^{4})^{- \frac{3}{2}}] i \\ + [6 t (4 + 9 t^{4})^{- \frac{1}{2}} (3 t^{2}) (- \frac{1}{2}) (4 + 9 t^{4})^{- \frac{3}{2}} (36 t^{3})] j \\ \vec{T}^{'} (t) = [- 36 t^{3} (4 + 9 t^{4})^{- \frac{3}{2}}] i + [24 t (4 + 9 t^{4})^{- \frac{3}{2}}] j \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{{[- 36 t^{3} (4 + 9 t^{4})^{- \frac{3}{2}}]}^{2} + {[24 t (4 + 9 t^{4})^{- \frac{3}{2}}]}^{2}} \\ \vec{N} (t) = \frac{\vec{T}^{'} (t)}{‖ \vec{T}^{'} (t) ‖} \\ \vec{N} (t) = \frac{[- 36 t^{3} (4 + 9 t^{4})^{- \frac{3}{2}}] i}{\sqrt{{[- 36 t^{3} (4 + 9 t^{4})^{- \frac{3}{2}}]}^{2} + {[24 t (4 + 9 t^{4})^{- \frac{3}{2}}]}^{2}}} \\ + \frac{[24 t (4 + 9 t^{4})^{- \frac{3}{2}}] j}{\sqrt{{[- 36 t^{3} (4 + 9 t^{4})^{- \frac{3}{2}}]}^{2} + {[24 t (4 + 9 t^{4})^{- \frac{3}{2}}]}^{2}}} \end{aligned}$

:t) = [-36t3 (4+94+94)(4+94)-32) i+ +[6(4+9)(4+94)(4+94)(4+94)(3)(12)(4+94)(12)(4+94)-12)(4+94)(4+94)(t)-12)(4+94)-32) i+(4+9(9)(4+93)(4+93)(4+3)(4+363)(4+94)(4+94)(4)-(4)(4)(4+9+94)-(32) 2+[24(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(94)(4)(9)(4)(4)(4)(9)(4)(4)(4)(9)(9)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)-32)-32)(2)_(3); j(4(4(4)(4)(3);2)(2))2(4(4(4(4(4)(4)(4)(4)(4)(4)(4)(4)(4)(4)(3))和9)(3)和9)(4)(j)(j)(j)(j)(j)(j)(j)(j)(j))))))))和)(j)(j)(j)(j)(j)(j)(j)(j)(j)(j)(j)(j))(2))(2))(2))(2))(2))(2)——————————————————————————————(4)(2))(2))(2))

Example 4
::例4

Find the approximate value for the unit tangent vector when $\begin{aligned} t = 2 \end{aligned}$ .
::查找t=2时单位正切矢量的大致值。

$\begin{aligned} \vec{T} (2) = \frac{2}{\sqrt{4 + 9 (2)^{4}}} i + \frac{3 (2)^{2}}{\sqrt{4 + 9 (2)^{4}}} j \approx . 1644 i + .9864 j \\ \vec{T} (2) \approx (.1644, .9864) \end{aligned}$

::T(2)=24+9(2)4i+3(2)24+9(2)4j.1644i+.9864j(2)(1644,9864)

Example 5
::例5

Find the approximate value for the unit normal vector when $\begin{aligned} t = 2 \end{aligned}$ .
::查找t=2时单位正常矢量的大致值。

$\begin{aligned} \vec{N} (2) = \frac{[- 36 (2)^{3} (4 + 9 (2)^{4})^{- \frac{3}{2}}] i}{\sqrt{{[- 36 (2)^{3} (4 + 9 (2)^{4})^{- \frac{3}{2}}]}^{2} + {[24 (2) (4 + 9 (2)^{4})^{- \frac{3}{2}}]}^{2}}} \\ + \frac{[24 (2) (4 + 9 (2)^{4})^{- \frac{3}{2}}] j}{\sqrt{{[- 36 (2)^{3} (4 + 9 (2)^{4})^{- \frac{3}{2}}]}^{2} + {[24 (2) (4 + 9 (2)^{4})^{- \frac{3}{2}}]}^{2}}} \\ \vec{N} (2) \approx . - 9864 i + .1644 j \\ \vec{N} (t) \approx (- .9864, .1644) \end{aligned}$

::N(2)=[-36(2)3(4+9(2)(4)-32)i[-36(2)3(4+9(2)(4)-32)2+[24(2)(2)(4)+9(2)(4)-32]2+[24(2)(2)(4)+9(2)(4)-4)-32]j[-36(2)(3)(4+9(2)(4)-32)2+[24(2)(2)(2)(4)+9(2)(4)+9(2)(4)-4(4)-32]2+[24(2)(2)(2)(4)+9(2)(4)+9(2)(4)-32]2N(2)。

Review
::回顾

For #1-4, use the function $\begin{aligned} \vec{F} (t) = (t - 1, t^{2}) \end{aligned}$ .
::对于 # 1 4, 请使用 F( t) =( t- 1, t2) 函数。

Find the equation for the unit tangent vector of $\begin{aligned} \vec{F} (t) \end{aligned}$ .
::F(t) 的正切向量的单位方程式查找方程式。

Find the equation for the unit normal vector of $\begin{aligned} \vec{F} (t) \end{aligned}$ .
::F(t) 的单位正态矢量查找方程式。

Find the approximate value for the unit tangent vector when $\begin{aligned} t = 4 \end{aligned}$ .
::查找t=4时单位正切矢量的大致值。

Find the approximate value for the unit normal vector when $\begin{aligned} t = 4 \end{aligned}$ .
::查找t=4时单位正向矢量的约值。

For #5-8, use the function $\begin{aligned} \vec{G} (t) = (6 \sin t, 6 \cos t) \end{aligned}$ .
::对于# 5-8, 请使用 G( t) =( 6sin t, 6cos t) 的函数。

Find the equation for the unit tangent vector of $\begin{aligned} \vec{G} (t) \end{aligned}$ .
::查找 G( t) 的单位正切矢量的方程。

Find the equation for the unit normal vector of $\begin{aligned} \vec{G} (t) \end{aligned}$ .
::查找 G( t) 单位正态矢量的方程式。

Find the value for the unit tangent vector when $\begin{aligned} t = π \end{aligned}$ .
::当 t 时查找单位正切矢量的值。

Find the value for the unit normal vector when $\begin{aligned} t = π \end{aligned}$ .
::当 t 时查找单位正向矢量的值。

For #9-12, use the function $\begin{aligned} \vec{K} (t) = (2 t^{2}, t - 5) \end{aligned}$ .
::对于# 9-12, 请使用 K( t) =( 2t2, t- 5) 函数。

Find the equation for the unit tangent vector of $\begin{aligned} \vec{K} (t) \end{aligned}$ .
::查找 K( t) 的单位正切矢量的方程。

Find the equation for the unit normal vector of $\begin{aligned} \vec{K} (t) \end{aligned}$ .
::为 K( t) 的单位正态矢量查找方程式。

Find the approximate value for the unit tangent vector when $\begin{aligned} t = 2 \end{aligned}$ .
::查找t=2时单位正切矢量的大致值。

Find the approximate value for the unit normal vector when $\begin{aligned} t = 2 \end{aligned}$ .
::查找t=2时单位正常矢量的大致值。

For #13-16, use the function $\begin{aligned} \vec{M} (t) = (e^{t}, t) \end{aligned}$ .
::对于# 13-16, 请使用函数 M( t) = (et, t) 。

Find the equation for the unit tangent vector of $\begin{aligned} \vec{M} (t) \end{aligned}$ .
::查找 M( t) 的单位正切矢量的方程。

Find the equation for the unit normal vector of $\begin{aligned} \vec{M} (t) \end{aligned}$ .
::为 M( t) 的单位正态矢量查找方程式。

Find the value for the unit tangent vector when $\begin{aligned} t = 1 \end{aligned}$ .
::查找t=1时单位正切矢量的值。

Find the value for the unit normal vector when $\begin{aligned} t = 1 \end{aligned}$ .
::查找t=1时单位正向矢量的值。

Review (Answers)
::回顾(答复)

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。

章节大纲

Tangent and Normal Vectors of Curves ::曲线的切向和正常矢量

Examples ::实例

Example 1 ::例1

Example 2 ::例2

F → ′ ( t ) = 2 i + 3 t 2 j ‖ F → ′ ( t ) ‖ = 2 2 + ( 3 t 2 ) 2 = 4 + 9 t 4 T → ( t ) = 2 4 + 9 t 4 i + 3 t 2 4 + 9 t 4 j Example 3 ::F( t) = 2i+3T2jF( t) 22+( 3t2) 2= 4+9t4T( t) = 24+9t4i+3t24+9t4j例 3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Tangent and Normal Vectors of Curves
::曲线的切向和正常矢量

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)