课程： 11.9 沿着平板曲线的动议:弧长度

章节大纲

选择节常规

折叠展开
常规

全部折叠展开全部
- 选择活动新闻通告
  
  新闻通告讨论区
选择节沿着平面曲线:弧长度运动

折叠展开
沿着平面曲线:弧长度运动
Ari is driving across the Appalachian Mountains in Pennsylvania on his way to New York City. He wonders if the actual length of the road matches the distances on the road signs. How can he find out how far he’s traveling on a road that goes up and down the mountain?
::阿里正在驶过宾夕法尼亚州的阿巴拉契亚山脉前往纽约市的路上。他想知道这条路的实际长度是否与路标上的距离相匹配。他怎样才能知道他在一条上下山的道路上走多远?

Motion and Arc Length
::运动和弧长度

To compute the length of a line, you can use the Pythagorean Theorem or distance formula. When you find the length of a curve for a vector-valued function , you use a similar formula. This is because what you’re actually doing is finding the sum of infinitely small pieces of the curve, so short that they’re nearly diagonal lines, and then adding them together. The formula for the length of a curve of a vector-valued function $\begin{align*}\vec{F} (t) = (f(t), \ g(t))\end{align*}$ is:
::要计算线条的长度, 您可以使用 Pythagorean Theorem 或距离公式。当找到矢量值函数的曲线长度时, 您会使用类似的公式。这是因为您实际上正在做的是找到曲线无限小部分的总和, 其短到接近对角线, 然后将其相加在一起。矢量值函数 F( t) =( f( t), g( t) ) 的曲线长度公式是 :

$\begin{aligned} L = \int_{a}^{b} \sqrt{f^{'} (t)^{2} + g^{'} (t)^{2}} d t \end{aligned}$
$\begin{align*}L = \int \limits_{a}^{b} \sqrt{f{^\prime}(t)^2 + g{^\prime}(t)^2} dt\end{align*}$
::Labf_(t)2+g_(t)2dt

This equation gives the length in terms of the function's horizontal and vertical changes over time. If you’ve worked with vector-valued functions before, you may notice that the expression $\begin{align*}\sqrt{f {^\prime}(t)^2 + g{^\prime} (t)^2}\end{align*}$ is equivalent to the magnitude of the tangent vector to the curve. This gives you another way to write the formula for the length of the curve of a vector-valued function: $\begin{align*}L = \int \limits_{a}^{b} \Big \| \vec{F}{^\prime}(t) \Big \| \ dt\end{align*}$ .
::此方程式给出函数水平和垂直随时间变化的长度。如果您以前曾使用矢量值函数工作过, 您可能会注意到表达式 f_( t) 2+g_( t) 2 相当于曲线的正切矢量的大小。这为您提供了另一种写矢量值函数曲线长度公式的方法 : LabF( t)\\\\ dt 。

Depending on the curve, these integrals can get very complicated. Sometimes, you may be able to manage a trigonometric identity or $\begin{align*}u\end{align*}$ -substitution. At other times, your only option may be to compute them numerically , using a graphing calculator or an online integral calculation tool. In general, it’s good practice to attempt these by hand, but if you’re using them for physics or engineering applications, you’ll usually use a calculator to approximate a numerical solution.
::根据曲线,这些整体体会变得非常复杂。有时,你可能能够管理三角特征或替代。有时,你唯一的选择可能是用图表计算器或在线整体计算工具进行数字计算。一般来说,手工尝试这些是好的做法,但如果你将它们用于物理或工程应用,你通常会用计算器来估计数字解决方案。

Let's find the length of the curve for the vector-valued function $\begin{align*}\vec{F} (t) = (5t, 2t^2)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t=5\end{align*}$ .
::让我们找到矢量值函数 F(t) = (5t 2,2) 从 t=0 到 t=5 的曲线长度。

First, find $\begin{align*}\vec{F}{^\prime}(t)\end{align*}$ and its magnitude:
::首先, 找到 F( t) 及其大小 :

$\begin{aligned} \vec{F}^{'} (t) = (5, 4 t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{25 + 16 t^{2}} \end{aligned}$
$\begin{align*}\vec{F}{^\prime}(t) = (5, 4t) \\ \Big \| \vec{F}{^\prime}(t) \Big \| = \sqrt{25+16t^2}\end{align*}$
::F(t) = (5,4t) = F(t) = 25+16t2

Set up the integral to find the length:
::设置要查找长度的内装件 :

$\begin{aligned} L = \int_{0}^{5} \sqrt{25 + 16 t^{2}} d t \end{aligned}$
$\begin{align*}L = \int \limits_{0}^{5} \sqrt{25+16t^2} dt\end{align*}$
::L0525+16t2dt

With a little manipulation, you can turn this into an integral of the form $\begin{align*}\int \limits \sqrt{a^2 + u^2} du\end{align*}$ .
::如果稍稍操纵一下,你可以将它变成表A2+u2du的有机组成部分。

$\begin{aligned} L = \int_{0}^{5} \sqrt{(25 + 16 t^{2}) (\frac{16}{16})} d t \\ L = \int_{0}^{5} 4 \sqrt{\frac{25}{16} + t^{2}} d t \\ L = 4 {[\frac{t}{2} \sqrt{\frac{25}{16} + t^{2}} + \frac{25}{32} \ln | t + \sqrt{\frac{25}{16} + t^{2}} |]}_{0}^{5} \end{aligned}$
$\begin{align*}L = \int \limits_{0}^{5} \sqrt{(25+16t^2) \left( \frac{16}{16} \right) } dt \\ L = \int \limits_{0}^{5} 4 \sqrt{\frac{25}{16} + t^2} dt \\ L = 4 \left[ \frac{t}{2} \sqrt{\frac{25}{16}+t^2} + \frac{25}{32} \ln \Bigg| t + \sqrt{\frac{25}{16} + t^2} \Bigg| \right]_{0}^{5}\end{align*}$
::=============================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================================

Now you can substitute and solve.
::现在你可以替代和解决。

$\begin{aligned} L = 4 [\frac{5}{2} \sqrt{\frac{25}{16} + 25} + \frac{25}{32} \ln | 5 + \sqrt{\frac{25}{16} + 25} |] - 4 [0 + \frac{25}{32} \ln | \frac{5}{4} |] = 58.085 \end{aligned}$
$\begin{align*}L = 4 \left[ \frac{5}{2} \sqrt{\frac{25}{16} + 25} + \frac{25}{32} \ln \Bigg| 5 + \sqrt{\frac{25}{16} + 25} \Bigg| \right] - 4 \left[ 0 + \frac{25}{32} \ln \Bigg| \frac{5}{4} \Bigg| \right] = 58.085\end{align*}$
::L=4[522516+25+2532ln @5+2516+25]-4[0+2532ln @54]=580.85

Examples
::实例

Example 1
::例1

Earlier, you were asked how Ari can find the actual length of the route that he is travelling so that he can compare it to the road signs. Ari can model his travel over one mountain with the vector-valued equation $\begin{align*}\vec{F} (t) = \left( t, \sin \left( \frac{t}{10} \right) \right)\end{align*}$ . He uses the formula for length of a curve to compute how far he travels along the curve from 0 to $\begin{align*}20 \pi\end{align*}$ .
::早些时候,有人问过Ari如何找到他旅行路线的实际长度,以便与路标进行比较。Ari可以用矢量值方程式F(t)=(t,sin(t10))来模拟他在一个山上的旅行。他用曲线长度公式计算他从0到20的曲线行走多远。

$\begin{aligned} \vec{F}^{'} (t) = (1, \frac{1}{10} \cos \frac{t}{10}) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{1^{2} + (\frac{1}{100} \cos^{2} \frac{t}{10})} \\ L = \int_{0}^{20 π} \sqrt{1 + (\frac{1}{100} \cos^{2} \frac{t}{10})} d t \end{aligned}$
$\begin{align*}\vec{F} {^\prime} (t) = \left( 1, \frac{1}{10} \cos \frac{t}{10} \right) \\ \Big \| \vec{F} {^\prime} (t) \Big \|= \sqrt{1^2+ \left( \frac{1}{100} \cos^2 \frac{t}{10} \right)} \\ L = \int \limits_{0}^{20 \pi} \sqrt{1 + \left( \frac{1}{100} \cos^2 \frac{t}{10} \right)} dt\end{align*}$
::F(t) =(1,110cost10) F(t) 12+(1,100cos2t10) L0201+(1,100cos2t10) dt

Ari uses his calculator and determines that he’s traveled about 62.9886 miles by going over the top of a mountain. For the same time, his distance on the map is $\begin{align*}20 \pi\end{align*}$ , about 62.8318. That means that his trip is only slightly longer than it would appear from the map.
::Ari使用计算器,确定他穿越山顶大约62.9886英里。同时,地图上的距离是20°,大约62.8318英里。这意味着他的行程只比地图上显示的时间稍长一点。

Example 2
::例2

Find the length of the curve of the vector-valued function $\begin{align*}\vec{F} (t) = (\ln t, t^2)\end{align*}$ from $\begin{align*}t=1\end{align*}$ to $\begin{align*}t=5\end{align*}$ . You may use a calculator or online resource such as WolframAlpha.
::从 t=1 到 t=5 查找矢量值函数 F(t) =( lnt, t2) 的曲线长度。您可以使用计算器或WolframAlpha 等在线资源。

First, find $\begin{align*}\vec{F}{^\prime} (t)\end{align*}$ and its magnitude:
::首先, 找到 F( t) 及其大小 :

$\begin{aligned} \vec{F}^{'} (t) = (\frac{1}{t}, 2 t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{{(\frac{1}{t})}^{2} + (2 t)^{2}} = \sqrt{\frac{1}{t^{2}} + 4 t^{2}} \end{aligned}$
$\begin{align*}\vec{F}{^\prime}(t) = \left( \frac{1}{t}, 2t \right) \\ \Big \| \vec{F}{^\prime}(t) \Big \| = \sqrt{\left( \frac{1}{t} \right)^2 + (2t)^2} = \sqrt{\frac{1}{t^2} + 4t^2}\end{align*}$
::F(t) = (1吨,2吨) @F(t) (t) (1吨) 2+(2吨) 2=1吨2+4吨2

Set up your definite integral to find the length of the curve:
::设置您确定的整体体以查找曲线的长度 :

$\begin{aligned} L = \int_{1}^{5} \sqrt{\frac{1}{t^{2}} + 4 t^{2}} d t . \end{aligned}$
$\begin{align*}L = \int \limits_{1}^{5} \sqrt{\frac{1}{t^2} + 4t^2} dt .\end{align*}$
::L151t2+4t2dt.

Use the calculator or program of your choice to solve. The length of this curve is 24.1176 units.
::使用您选择的计算器或程序解析。此曲线的长度为24.1176 单位。

Example 3
::例3

You can use the length of a curve to find the position vector for an object after it has traversed that section of a curve.
::您可以使用曲线的长度来为一个对象在穿越曲线的该部分后找到位置矢量。

To create a formula for finding position on the curve $\begin{align*}\vec{F} (t)\end{align*}$ based on arc length, you must first find the function $\begin{align*}s(t)\end{align*}$ , where $\begin{align*}s (t) = \int \limits_{0}^{t} \Big \| \vec{F}{^\prime}(u) \Big \| du\end{align*}$ . Then, you evaluate $\begin{align*}s (t)\end{align*}$ and solve the resulting equation for $\begin{align*}t\end{align*}$ . Finally, you plug this equation back into $\begin{align*}\vec{F} (t)\end{align*}$ to reparameterize the original vector-valued function in terms of $\begin{align*}s\end{align*}$ instead of $\begin{align*}t\end{align*}$ . You can see how this process works by examining a relatively simple vector-valued function, one that describes a circular path.
::要创建基于弧长度的曲线 F( t) 定位方程式, 您必须首先找到 s( t) 函数 s, 其中 s( t)\\\\\\\0\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Given the function $\begin{align*}\vec{F} = (5 \sin t, 5 \cos t)\end{align*}$ , find the location of an object when it has traveled $\begin{align*}5 \pi\end{align*}$ units.
::根据 F( 5sint, 5cost) 函数, 当一个对象已行驶 5 单位时, 请找到该对象的位置。

First, find the magnitude of the tangent vector to the function.
::首先,找到函数的正切矢量的大小。

$\begin{aligned} \vec{F}^{'} (t) = (5 \cos t, - 5 \sin t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{(5 \cos t)^{2} + (- 5 \sin t)^{2}} \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{25 \cos^{2} t + 25 \sin^{2} t} \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{25 (\cos^{2} t + \sin^{2} t)} \\ ‖ \vec{F}^{'} (t) ‖ = 5 \end{aligned}$
$\begin{align*}\vec{F}{^\prime} (t) = (5 \cos \ t, -5 \sin \ t) \\ \Big \| \vec{F} {^\prime} (t) \Big \| = \sqrt{ (5 \cos \ t)^2+(-5 \sin \ t)^2} \\ \Big \| \vec{F} {^\prime} (t) \Big \| = \sqrt{25 \cos^2 \ t + 25 \sin^2 \ t} \\ \Big \| \vec{F} {^\prime} (t) \Big \| = \sqrt{25 (\cos^2 \ t + \sin^2 \ t)} \\ \Big \| \vec{F}{^\prime} (t) \Big \| = 5\end{align*}$
::F(t) = (5cos) = (5cos) t,-5sin ) (t) (5cos) (t) 2+(5sin) (t) (t) (25cos) (t) (t) (t) (t) (2cos) t+(sin2) (t) (f) (t) (f) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t)

Now, write out the arc length function, $\begin{align*}s (t)\end{align*}$ .
::现在, 写出弧长度函数, s( t) 。

$\begin{align*}s (t) = \int \limits_{0}^{t} 5 \ du\end{align*}$
:t)%05 日( 秒)

Evaluate $\begin{align*}s(t)\end{align*}$ :
::评价 s(t) :

$\begin{aligned} s (t) = \int_{0}^{t} 5 d u \\ s = {[5 u]}_{0}^{t} = 5 t \end{aligned}$
$\begin{align*}s (t) = \int \limits_{0}^{t} 5 \ du \\ s = \left[ 5 \ u \right]_{0}^{t} = 5 t\end{align*}$
::s( t) 0t5 dus= [ 5 u] 0t=5t

Solve for $\begin{align*}t\end{align*}$ :
::解决 t:

$\begin{aligned} s = 5 t \\ t = \frac{s}{5} \end{aligned}$
$\begin{align*}s = 5t \\ t= \frac{s}{5}\end{align*}$
::s=5t=s5 =5t=s5

Now, reparameterize the original function in terms of $\begin{align*}s\end{align*}$ :
::现在,重新校准原函数的参数,以 s:

$\begin{aligned} \vec{F} (t) = (5 \sin t, 5 \cos t) \\ t = \frac{s}{5} \\ \vec{F} (t) = (5 \sin \frac{s}{5}, 5 \cos \frac{s}{5}) \end{aligned}$
$\begin{align*}\vec{F} (t) = (5 \sin t, 5 \cos t) \\ t = \frac{s}{5} \\ \vec{F} (t) = \left ( 5 \sin \frac{s}{5}, 5 \cos \frac{s}{5} \right )\end{align*}$
::F(t) = (5sint,5cost) t= s5F(t) = (5sins5,5coss5)

If you know how far an object has traveled along the curve $\begin{align*}\vec{F} (t)\end{align*}$ , you can find its position and how much time has passed.
::如果您知道一个对象在曲线F(t)上走多远,可以找到它的方位和时间。

To find out how long it took an object to travel $\begin{align*}5 \pi\end{align*}$ units and where that object is now, simply substitute.
::发现一个物体需要多久才能旅行5°C单位,以及该物体现在的位置,只是取而代之。

$\begin{aligned} t = \frac{s}{5} \\ t = \frac{5 π}{5} \\ t = π \end{aligned}$
$\begin{align*}t = \frac{s}{5} \\ t = \frac{5 \pi} {5} \\ t = \pi\end{align*}$
::t=s5t=5°5t

$\begin{align*}\pi\end{align*}$ units of time have passed, and the object is at:
::时间单位已过,对象为:

$\begin{aligned} \vec{F} (π) = (5 \sin π, 5 \cos π) = (0, - 5) \end{aligned}$
$\begin{align*}\vec{F} (\pi) = (5 \sin \pi, 5 \cos \pi) = (0, -5)\end{align*}$
::F() = (5sin,5cos) =(0,5)

You may use a calculator for the following examples.
::下列示例可使用计算器。

Example 4
::例4

At time 0, an object begins to move along the curve $\begin{align*}\vec{F} (t) = (3 \sin t, 3 \cos t)\end{align*}$ . How much time will have passed once the object has traveled $\begin{align*}\frac{3 \pi}{2}\end{align*}$ units?
::时间 0 时, 对象开始沿着曲线 F( t) =( 3sint, 3cost) 移动。一旦天体经过 32 单位, 时间会过多久 ?

$\begin{aligned} \vec{F}^{'} (t) = (3 \cos t, - 3 \sin t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{(3 \cos t)^{2} + (- 3 \sin t)^{2}} \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{9 \cos^{2} t + 9 \sin^{2} t} \\ ‖ \vec{F}^{'} (t) ‖ = 3 \end{aligned}$
$\begin{align*}\vec{F} {^\prime} (t) = (3 \cos t, -3 \sin t) \\ \Big \| \vec{F}{^\prime} (t) \Big \| = \sqrt{(3 \cos t)^2 + (-3 \sin t)^2} \\ \Big \| \vec{F} {^\prime} (t) \Big \| = \sqrt{9 \cos^2 t+ 9 \sin^2 t} \\ \Big \| \vec{F} {^\prime} (t) \Big \| = 3\end{align*}$
::F(t) = (3cost,-3sint) F(t) (3cost) 2+(-3sint) 2F(t) 9cos2t+9sin2}(t) 3

$\begin{aligned} s (t) = \int_{0}^{t} 3 d u \\ s (t) = 3 t \\ t = \frac{s}{3}, s = \frac{3 π}{2} \\ t = \frac{3 π}{6} = \frac{π}{2} \end{aligned}$
$\begin{align*}s (t) = \int \limits_{0}^{t} 3 \ du \\ s (t) = 3t \\ t = \frac{s}{3}, s = \frac{3 \pi}{2} \\ t= \frac{3 \pi}{6} = \frac{\pi}{2}\end{align*}$
:t)%0t3 dus(t) = 3tt=s3,s= 3}2t= 362

Example 5
::例5

At time 0, an object begins to move along the curve $\begin{align*}\vec{F} (t) = (15 \cos t, 15 \sin t)\end{align*}$ . What is its position when it has traveled $\begin{align*}\frac{15 \pi}{2}\end{align*}$ units?
::时间 0 时, 对象开始沿着曲线 F( t) =( 15cost, 15sint) 移动。当一个物体已移动 152 单位时, 它的位置是什么 ?

$\begin{aligned} \vec{F}^{'} (t) = (- 15 \sin t, 15 \cos t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{225 \sin^{2} t + 225 \cos t^{2}} = 15 \\ s (t) = \int_{0}^{t} 15 d u \\ s = 15 t \\ t = \frac{s}{15} \\ \vec{F} (\frac{s}{15}) = (15 \cos (\frac{s}{15}), 15 \sin (\frac{s}{15})) \\ \vec{F} (\frac{\frac{15 π}{2}}{15}) = (15 \cos \frac{π}{2}, 15 \sin \frac{π}{2}) \\ \vec{F} (\frac{π}{2}) = (0, 15) \end{aligned}$
$\begin{align*}\vec{F} {^\prime} (t) = (-15 \sin t, 15 \cos t) \\ \Big \| \vec{F} {^\prime} (t) \Big \| = \sqrt{225 \sin^2 t + 225 \cos t^2} = 15 \\ s(t) = \int \limits_{0}^{t} 15 \ du \\ s= 15 t \\ t =\frac{s}{15} \\ \vec{F} \left( \frac{s}{15} \right) = \left( 15 \cos \left( \frac{s}{15} \right), 15 \sin \left( \frac{s}{15} \right) \right) \\ \vec{F} \left( \frac{\frac{15 \pi}{2}}{15} \right) = \left( 15 \cos \frac{\pi}{2}, 15 \sin \frac{\pi}{2} \right) \\ \vec{F} \left ( \frac{\pi}{2} \right ) = (0, 15)\end{align*}$
::F(t) = (- 15sint, 15cost) F(t) 225sin2t+225cost2=15s(t) 0t15 dus=15t=S15F}(s15) =(15cos 15,15sin(s15) F}(15_215) =(15cos2,15sin2) F(_2) F}(%2) =(0) 15。

Review
::回顾

You may use a calculator for the following problems.
::您可以使用计算器处理下列问题。

Find the length of the curve for the vector-valued function $\begin{align*}\vec{F} (t) = \left( t, t^{\frac{3}{2}} \right)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t=5\end{align*}$ .
::查找矢量值函数 F(t) =(t,t32) 从 t=0 到 t=5 的曲线长度。

Find the length of the curve for the vector-valued function $\begin{align*}\vec{G} (t) = (6 \sin t, 6 \cos t)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t = \pi\end{align*}$ .
::查找矢量值函数 G( t) =( 6sin, 6cost) 从 t=0 到 t 的曲线长度。

Find the length of the curve for the vector-valued function $\begin{align*}\vec{H} (t) = (\ln \cos t, t)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t= \frac{\pi}{2}\end{align*}$ .
::查找矢量值函数 H(t) =(lncost,t) 从 t=0 到 t2 的曲线长度。

Find the length of the curve for the vector-valued function $\begin{align*}\vec{K} (t) = \left( \frac{1}{2} t^2, \frac{1}{6} (4t+ 4)^{\frac{3}{2}} \right)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t=5\end{align*}$ .
::查找矢量值函数 K(t) =( 12t2, 16( 4t+4) 32) 从 t=0 到 t=5 的曲线长度。

Find the length of the curve for the vector-valued function $\begin{align*}\vec{M} (t) = (e^t \cos t, e^t \sin t)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t=1\end{align*}$ .
::查找矢量值函数 M(t) =(etcost,etsint) 从 t=0 到 t=1 的曲线长度。

Find the length of the curve for the vector-valued function $\begin{align*}\vec{P} (t) = (12 \cos t, 12 \sin t)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t=\frac{\pi}{2}\end{align*}$ .
::查找矢量值函数 P(t) =( 12cos@t, 12sint) 从 t=0 到 t2 的曲线长度。

Find the length of the curve for the vector-valued function $\begin{align*}\vec{D} (t) = (\sin t, \cos t)\end{align*}$ from $\begin{align*}t=0\end{align*}$ to $\begin{align*}t=\pi\end{align*}$ .
::查找矢量值函数 D( t) = (sint, cost) 从 t=0 到 t 的曲线长度。

At time 0, an object begins to move along the curve $\begin{align*}\vec{G} (t) = (6 \sin t, 6 \cos t)\end{align*}$ . How much time will have passed once the object has traveled $\begin{align*}\frac{\pi}{2}\end{align*}$ units?
::时间 0 时, 对象开始沿着曲线 G( t) =( 6sint, 6cost) 移动。一旦对象经过 2 单位, 时间会过去多少 ?

At time 0, an object begins to move along the curve $\begin{align*}\vec{G} (t) = (6 \sin t, 6 \cos t)\end{align*}$ . What is its position when it has traveled $\begin{align*}\frac{\pi}{2}\end{align*}$ units?
::时间 0 时, 对象开始沿着曲线 G( t) =( 6sint, 6cost) 移动。当一个对象已经移动 2 单位时, 它的位置是什么 ?

At time 0, an object begins to move along the curve $\begin{align*}\vec{P} (t) = (12 \cos t, 12 \sin t)\end{align*}$ . How much time will have passed once the object has traveled $\begin{align*}\frac{3 \pi}{2}\end{align*}$ units?
::时间 0 时, 对象开始沿着曲线 P( t) =( 12cost, 12sint) 移动。一旦天体经过 32 单位, 时间会过多久 ?

At time 0, an object begins to move along the curve $\begin{align*}\vec{P} (t) = (12 \cos t, 12 \sin t)\end{align*}$ . What is its position when it has traveled $\begin{align*}\frac{3 \pi}{2}\end{align*}$ units?
::时间 0 时, 对象开始沿着曲线 P( t) =( 12cost, 12sint) 移动。当一个对象已移动 32 单位时, 它的位置是什么 ?

At time 0, an object begins to move along the curve $\begin{align*}\vec{K} (t) = \left( \frac{1}{2} t^2, \frac{1}{6} (4t,+4)^{\frac{3}{2}} \right)\end{align*}$ . How much time will have passed once the object has traveled 10 units?
::时间 0 时, 对象开始沿着曲线 K( t) =( 12t2, 16( 4t, +4) 32) 移动。一旦天体移动了 10 个单位, 时间会过多久 ?

At time 0, an object begins to move along the curve $\begin{align*}\vec{K} (t) = \left( \frac{1}{2} t^2, \frac{1}{6} (4t,+4)^{\frac{3}{2}} \right)\end{align*}$ . What is its position when it has traveled 10 units?
::时间 0 时, 对象开始沿着曲线 K( t) =( 12t2, 16( 4t, +4) 32) 移动。当一个对象已移动 10 个单位时, 它的位置是什么 ?

At time 0, an object begins to move along the curve $\begin{align*}\vec{M} (t) = (e^t \cos t, e^t \sin t)\end{align*}$ . How much time will have passed once the object has traveled 1 unit?
::时间 0 时, 对象开始沿着曲线 M( t) =( etcost,etsint) 移动。一旦天体经过一个单位, 时间会过多久 ?

At time 0, an object begins to move along the curve $\begin{align*}\vec{M} (t) = (e^t \cos t, e^t \sin t)\end{align*}$ . What is its position when it has traveled 1 unit?
::时间 0 时, 对象开始沿着曲线 M( t) =( etcost,etsint) 移动。当一个对象经过一个单位时, 它的位置是什么 ?

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

常规

沿着平面曲线:弧长度运动

Motion and Arc Length ::运动和弧长度

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Example 5 ::例5

Review ::回顾

Review (Answers) ::回顾(答复)

Motion and Arc Length
::运动和弧长度

Examples
::实例

Example 1
::例1

Example 2
::例2

Example 3
::例3

Example 4
::例4

Example 5
::例5

Review
::回顾

Review (Answers)
::回顾(答复)