课程： 11.10 沿着平板曲线测量曲线

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选择节沿平板曲线测量曲线

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沿平板曲线测量曲线
Nina is planning a family hike in a nearby state park. Her grandmother has asked her to find the easiest hike that will get the family to the top of the mountain. In the past, Nina has noticed that curvier trails tend to be more difficult trails. How can Nina compare the curviness of two trails on the map so that she can pick the less curvy trail?
::尼娜计划到附近的州立公园去探亲。她祖母要求她寻找最容易的远足,让家人到山顶。过去,尼娜注意到,弯曲的足迹往往更难走。尼娜如何比较地图上两条足迹的弯曲性,以便她能够选择不太弯曲的足迹?

Measuring Curvature
::测量曲线

Curvature is the mathematical term that describes how quickly and dramatically a curve is changing direction. The radius of curvature lets you compare the change in a particular curve to the change in a circle of a certain radius. So, for instance, if a place on a curve has a radius of curvature equal to 5, it is curving at the same rate that a circle with a radius of 5 curves.
::曲线是一个数学术语,用来描述曲线正在快速和急剧地改变方向。曲线的半径使您可以比较特定曲线的变化与某个半径圆圈的变化。例如,如果曲线上的某个位置的曲线半径等于5,那么曲线的曲线正以与半径为5的圆一样的速度弯曲。

The radius of curvature is defined as $\begin{align*}\frac{1}{k(t)}\end{align*}$ , where $\begin{align*}k(t)\end{align*}$ is the curvature of the plane curve. You can compute the curvature of a vector-valued function $\begin{align*}\vec{F}(t)\end{align*}$ by using the formula $\begin{align*}k(t) = \frac{\Big \| \vec{T}{^\prime}(t) \Big \|}{\Big \| \vec{F}{^\prime}(t) \Big \|} \end{align*}$ , where $\begin{align*}\vec{T}(t)\end{align*}$ is the unit tangent vector to the curve.
::曲线的半径定义为 1k( t) , k( t) 是平面曲线的曲线。您可以使用公式 k( t) \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ k( t)\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ T, T\\\\T\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

Applying the formulas from above, let's find $\begin{align*}k(t)\end{align*}$ for $\begin{align*}\vec{F}(t) = (5 \sin t, 5 \cos t) \end{align*}$
::应用上面的公式时, 我们为 F( t) =( 5sint, 5cost) 找到 k( t) 。

First, find $\begin{align*}\vec{F}{^\prime}(t)\end{align*}$ and use it to find the unit tangent vector to $\begin{align*}\vec{F}(t)\end{align*}$ :
::首先, 找到 F( t) 并用它找到 F( t) 的正切向量 :

$\begin{aligned} \vec{F}^{'} (t) = (5 \cos t, - 5 \sin t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{(5 \cos t)^{2} + (- 5 \sin t)^{2}} = 5 \\ \vec{T} (t) = \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} = \frac{(5 \cos t) i - (5 \sin t) j}{5} = (\cos t) i - (\sin t) j = (\cos t, - \sin t) \end{aligned}$
$\begin{align*}\vec{F}{^\prime}(t) = (5 \cos t, -5 \sin t) \\ \Big\|\vec{F}{^\prime}(t)\Big\|= \sqrt{(5 \cos t)^2 + (-5 \sin t)^2} = 5 \\ \vec{T}(t) = \frac{\vec{F}{^\prime}(t)}{\Big\|\vec{F}{^\prime}(t)\Big\|} = \frac{(5 \cos t)i - (5 \sin t)j}{5} = (\cos t)i - (\sin t)j = (\cos t, - \sin t)\end{align*}$
::F(t) = (5cost,-5sint) = (5cost) 2+(-5cost) 2= (-5sint) 2= 5T(t) = F(t) F(t) (5cost) i - (5cosint) j5=(sosint) i=(sint) j=(cost,-sint)

Next, find $\begin{align*}\vec{T}{^\prime}(t)\end{align*}$ :
::下一步, 找到 T( t) :

$\begin{aligned} \vec{T}^{'} (t) = (- \sin t, - \cos t) \end{aligned}$
$\begin{align*}\vec{T}{^\prime}(t) = (-\sin t, -\cos t)\end{align*}$
::T(t) = (- sinnt,- cost)

Now, find the magnitude of $\begin{align*}\vec{T}{^\prime}(t)\end{align*}$ :
::找到T( t) 的大小 :

$\begin{aligned} ‖ \vec{T}^{'} (t) ‖ = \sqrt{(- \sin t)^{2} + (- \cos t)^{2}} = 1 \end{aligned}$
$\begin{align*}\Big \| \vec{T}{^\prime}(t) \Big \| = \sqrt{(-\sin t)^2 + (-\cos t)^2} = 1\end{align*}$
:t)(-sin(t)2+(-cos(t)2=1)

Divide the magnitude of $\begin{align*}\vec{T}{^\prime}(t) \end{align*}$ by the magnitude of $\begin{align*}\vec{F}{^\prime}(t) \end{align*}$ to find the value of $\begin{align*}k(t)\end{align*}$ :
::将 T( t) 的大小除以 F( t) 的大小以查找 k( t) 的值 :

$\begin{aligned} k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{F}^{'} (t) ‖} = \frac{1}{5} \end{aligned}$
$\begin{align*}k(t) = \frac{\Big\| \vec{T}{^\prime}(t)\Big\|}{\Big\| \vec{F}{^\prime}(t)\Big\|} = \frac{1}{5}\end{align*}$
:kt) T(t) FF(t) 15)

Notice that the original function, $\begin{align*}\vec{F}{^\prime}(t)\end{align*}$ described a circle with a radius of 5, and that its curvature is $\begin{align*}\frac{1}{5}\end{align*}$ .
::注意原始函数 F(t) 描述半径为 5 的圆,其曲线为 15 。

The curvature of a circle with a radius of $\begin{align*}r\end{align*}$ is always $\begin{align*}k(t) = \frac{1}{r}\end{align*}$ . Circles with smaller radii curve more quickly, while circles with larger radii curve more slowly. This means that larger circles have smaller values for $\begin{align*}k(t)\end{align*}$ .
::圆圆半径为 r 圆的曲率总是 k(t) = 1r。圆的弧度较小, 速度更快, 圆的弧度较大, 弧度较大, 速度较慢。这意味着大圆的k(t) 值较小。

Circles have the same curvature at every point on their curve. However, most plane curves have different values of $\begin{align*}k\end{align*}$ at different points along the curve.
::圆圈在曲线的每个点上都有相同的曲率。但是,大多数平面曲线在曲线各点上都有不同的 k 值。

Take the vector-valued function $\begin{align*}\vec{F}(t) = (5t, t^2)\end{align*}$ . Let's find $\begin{align*}k(1)\end{align*}$ and $\begin{align*}k(5)\end{align*}$
::选择矢量值函数 F( t) =( 5t, t2) 。让我们找到 k(1) 和 k(5) 。

First, find the general equation for $\begin{align*}k(t)\end{align*}$ by finding $\begin{align*}\vec{F}{^\prime}(t), \vec{T}(t), \vec{T}{^\prime}(t)\end{align*}$ , and the magnitudes of $\begin{align*}\vec{F}{^\prime}(t)\end{align*}$ and $\begin{align*}\vec{T}{^\prime}(t)\end{align*}$ .
::首先,通过找到 F(t), T(t), T(t), T(t), 以及 F(t) 和 T(t) 的大小, 找到 k(t) 的一般方程。

$\begin{aligned} \vec{F}^{'} (t) = (5, 2 t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{5^{2} + (2 t)^{2}} = \sqrt{25 + 4 t^{2}} \\ \vec{T} (t) = \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} = \frac{5 i + 2 t j}{\sqrt{25 + 4 t^{2}}} \\ \vec{T} (t) = \frac{5 i}{\sqrt{25 + 4 t^{2}}} + \frac{2 t j}{\sqrt{25 + 4 t^{2}}} \\ \vec{T} (t) = ((\frac{5}{\sqrt{25 + 4 t^{2}}}), (\frac{2 t}{\sqrt{25 + 4 t^{2}}})) \end{aligned}$
$\begin{align*}\vec{F}{^\prime}(t) = (5, 2t) \\ \Big \| \vec{F}{^\prime}(t) \Big \| = \sqrt{5^2 + (2t)^2} = \sqrt{25 + 4t^2} \\ \vec{T}(t) = \frac{\vec{F}{^\prime}(t)}{\Big \| \vec{F}{^\prime}(t) \Big \|} = \frac{5i + 2tj}{\sqrt{25 + 4t^2}} \\ \vec{T}(t) = \frac{5i}{\sqrt{25 + 4t^2}} + \frac{2tj}{\sqrt{25 + 4t^2}} \\ \vec{T}(t) = \left(\left(\frac{5}{\sqrt{25 + 4t^2}}\right), \left(\frac{2t}{\sqrt{25 + 4t^2}}\right)\right) \\\end{align*}$
::F( t) = F( 5, 2t) = F( 5, 2t) = 25+4T2T( t) = F( t) ( f) ( t) 5i+2tj25+4T2T}( t) = 5i25+4t2+2tj25+4t2T( t) = ( 525+4t2) = (2t25+4t2) = (2t25+4t2)

After you have $\begin{align*}\vec{T}(t)\end{align*}$ , find $\begin{align*}\vec{T}{^\prime}(t)\end{align*}$ and its magnitude. Since you’ll be plugging in values later on, you don’t need to spend a lot of time simplifying.
::在 T(t) 之后, 找到 T(t) 及其大小。由于您稍后会插入数值, 您不需要花很多时间简化。

$\begin{aligned} \vec{T} (t) = (5 (25 + 4 t^{2})^{- \frac{1}{2}}, 2 t (25 + 4 t^{2})^{- \frac{1}{2}}) \\ \vec{T}^{'} (t) = ((- 20 t) (25 + 4 t^{2})^{- \frac{3}{2}}, - 8 t^{2} (25 + 4 t^{2})^{- \frac{3}{2}} + 2 (25 + 4 t^{2})^{- \frac{1}{2}}) \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{{((- 20 t) (25 + 4 t^{2})^{- \frac{3}{2}})}^{2} + {(- 8 t^{2} (25 + 4 t^{2})^{- \frac{3}{2}} + 2 (25 + 4 t^{2})^{- \frac{1}{2}})}^{2}} \end{aligned}$
$\begin{align*}\vec{T}(t) = \left(5(25+4t^2)^{-\frac{1}{2}}, 2t(25+4t^2)^{-\frac{1}{2}}\right) \\ \vec{T}{^\prime}(t) = \left((-20t)(25+4t^2)^{-\frac{3}{2}}, -8t^2(25+4t^2)^{-\frac{3}{2}} + 2(25+4t^2)^{-\frac{1}{2}}\right) \\ \Big \| \vec{T}{^\prime}(t) \Big \| = \sqrt{\left((-20t)(25+4t^2)^{-\frac{3}{2}}\right)^2 + \left(-8t^2(25+4t^2)^{-\frac{3}{2}} + 2(25+4t^2)^{-\frac{1}{2}}\right)^2}\end{align*}$
::T(t) = (5(25+4t2) - 12,2,2(25+4-2) - 12,2(25+4-2) - 12) T(t) = (20) = (20) = (25+4) - 12,2(25+4-2) - 12) T(t) = (20) = (20(25+4-2) - (25+4-2) - 12) T(t) = ((20) = (25+4) - (5) - 12,2 (25+4-2) - 12) T(25+4-2) - 32 (25+4) - 12 2 2

Set up your formula for $\begin{align*}k(t)\end{align*}$ :
::设置 k( t) 的公式 :

$\begin{aligned} k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{F}^{'} (t) ‖} \\ = \frac{\sqrt{{((- 20 t) (25 + 4 t^{2})^{- \frac{3}{2}})}^{2} + {(- 8 t^{2} (25 + 4 t^{2})^{- \frac{3}{2}} + 2 (25 + 4 t^{2})^{- \frac{1}{2}})}^{2}}}{\sqrt{25 + 4 t^{2}}} \end{aligned}$
$\begin{align*}k(t)=\frac{\Big \| \vec{T}{^\prime}(t)\Big \|}{\Big \| \vec{F}{^\prime}(t)\Big \|} \\ = \frac{\sqrt{\left((-20t)(25+4t^2)^{-\frac{3}{2}}\right)^2 + \left(-8t^2(25+4t^2)^{-\frac{3}{2}} + 2(25+4t^2)^{-\frac{1}{2}}\right)^2}}{\sqrt{25+4t^2}}\end{align*}$
::k( t) T( t) F( t) ( (- 20t)( 25+4t2) - 322+( 8t2( 25+4t2) - 322+2( 25+4t2) - 32+2( 25+4t2) - 12) 225+4t2

Then, evaluate at 1 and 5:
::然后在1点和5点进行评估:

$\begin{aligned} k (t) = \frac{‖ \vec{T}^{'} (1) ‖}{‖ \vec{F}^{'} (1) ‖} \\ = \frac{\sqrt{{((- 20 (1)) (25 + 4 (1)^{2})^{- \frac{3}{2}})}^{2} + {(- 8 (1)^{2} (25 + 4 (1)^{2})^{- \frac{3}{2}} + 2 (25 + 4 (1)^{2})^{- \frac{1}{2}})}^{2}}}{\sqrt{25 + 4 (1)^{2}}} \\ = \frac{10}{29 \sqrt{29}} \approx .0640 \\ k (5) = \frac{‖ \vec{T}^{'} (5) ‖}{‖ \vec{F}^{'} (5) ‖} \\ = \frac{\sqrt{{((- 20 (5)) (25 + 4 (5)^{2})^{- \frac{3}{2}})}^{2} + {(- 8 (5)^{2} (25 + 4 (5)^{2})^{- \frac{3}{2}} + 2 (25 + 4 (5)^{2})^{- \frac{1}{2}})}^{2}}}{\sqrt{25 + 4 (5)^{2}}} \\ = \frac{2}{125 \sqrt{5}} = .00716 \end{aligned}$
$\begin{align*}k(t)=\frac{\Big \| \vec{T}{^\prime}(1)\Big \|}{\Big \| \vec{F}{^\prime}(1)\Big \|} \\ = \frac{\sqrt{\left((-20(1))(25+4(1)^2)^{-\frac{3}{2}}\right)^2 + \left(-8(1)^2(25+4(1)^2)^{-\frac{3}{2}} + 2(25+4(1)^2)^{-\frac{1}{2}}\right)^2}}{\sqrt{25+4(1)^2}} \\ = \frac{10}{29 \sqrt{29}} \approx .0640 \\ k(5)=\frac{\Big \| \vec{T}{^\prime}(5)\Big \|}{\Big \| \vec{F}{^\prime}(5)\Big \|} \\ = \frac{\sqrt{\left((-20(5))(25+4(5)^2)^{-\frac{3}{2}}\right)^2 + \left(-8(5)^2(25+4(5)^2)^{-\frac{3}{2}} + 2(25+4(5)^2)^{-\frac{1}{2}}\right)^2}}{\sqrt{25+4(5)^2}} \\ = \frac{2}{125 \sqrt{5}} = .00716\end{align*}$
::k(t) -T------(((-)-20(1)(25+4(1)2)-2-2-2-2-2-2-2+2+4(1)2-2-2-2-2-2+2+4(1)2-2)-12-225+4(1)2=1029 .0640k(5)----(5) (((-)-20(5)(5)(25+4(5)2)-2-32+-8(5)2(25+4(5)2)-32+2(25+4(5)2)-2-2)-225+4(5)2=21255=.00716)-32+2(25+4(5)2)-2-2)-225+4(5)2=21255=.00716

A higher measurement of curvature means that the curve is changing more rapidly. The curve is changing more rapidly at $\begin{align*}t=1\end{align*}$ than it is at $\begin{align*}t=5\end{align*}$ .
::曲线的测度越高, 曲线的变化速度就越快。曲线在t=1比t=5的变速越快。

Examples
::实例

Example 1
::例1

Earlier, you were asked how Nina compare the curvature of two trails to determine which hike is easier. Nina creates two vector-valued functions to model how her family will hike along the trails.
::早些时候,有人问尼娜如何比较两条小路的曲折,以确定哪条小路更容易徒步。尼娜创造了两个矢量价值的功能,以模拟她的家人如何沿着小路走。

$\begin{align*}\vec{F}(t) = (t, 4t^2) \\ \vec{G}(t) = (2t, \sin t)\end{align*}$
::F(t) = (t,4t2) G(t) = (2t,sin@t)

She decides to compare $\begin{align*}k(t)\end{align*}$ at $\begin{align*}t=20\end{align*}$ to see how difficult each hike will be after the family has been on the trail for about 20 minutes.
::她决定比较t=20的k(t), 以了解家庭在足迹上约20分钟后每次出行有多困难。

First, she finds $\begin{align*}k(t)\end{align*}$ for the trail that she’s modeled with $\begin{align*}\vec{F}(t)\end{align*}$ .
::首先,她找到了K(t),以她用F__(t)做模型的线索。

$\begin{aligned} \vec{F}^{'} (t) = (1, 8 t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{1 + 64 t^{2}} \\ \vec{T} (t) = \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} = ((1 + 64 t^{2})^{- \frac{1}{2}}, (1 + 64 t^{2})^{- \frac{1}{2}} (8 t)) \\ \vec{T}^{'} (t) = (- 64 t (1 + 64 t^{2})^{- \frac{3}{2}}, 8 (1 + 64 t^{2})^{- \frac{3}{2}}) \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{{(- 64 t (1 + 64 t^{2})^{- \frac{3}{2}})}^{2} + {(8 (1 + 64 t^{2})^{- \frac{3}{2}})}^{2}} \\ k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{F}^{'} (t) ‖} = \frac{\sqrt{{(- 64 t (1 + 64 t^{2})^{- \frac{3}{2}})}^{2} + {(8 (1 + 64 t^{2})^{- \frac{3}{2}})}^{2}}}{\sqrt{1 + 64 t^{4}}} \\ k (20) = \frac{\sqrt{{(- 64 (20) (1 + 64 (20)^{2})^{- \frac{3}{2}})}^{2} + {(8 (1 + 64 (20)^{2})^{- \frac{3}{2}})}^{2}}}{\sqrt{1 + 64 (20)^{2}}} \\ k (20) = .000002 \end{aligned}$
$\begin{align*}\vec{F}{^\prime}(t) = (1, 8t) \\ \Big\|\vec{F}{^\prime}(t)\Big\| = \sqrt{1+64t^2} \\ \vec{T}(t)= \frac{\vec{F}{^\prime}(t)}{\Big \| \vec{F}{^\prime}(t)\Big \|} = \left((1+64t^2)^{-\frac{1}{2}}, (1+64t^2)^{-\frac{1}{2}}(8t)\right) \\ \vec{T}{^\prime}(t) = \left(-64t(1+64t^2)^{-\frac{3}{2}}, 8(1+64t^2)^{-\frac{3}{2}}\right) \\ \Big \|\vec{T}{^\prime}(t)\Big \| = \sqrt{\left(-64t(1+64t^2)^{-\frac{3}{2}}\right)^2 + \left(8(1+64t^2)^{-\frac{3}{2}}\right)^2} \\ k(t) = \frac{\Big \|\vec{T}{^\prime}(t)\Big \|}{\Big \| \vec{F}{^\prime}(t)\Big\|}= \frac{\sqrt{\left(-64t(1+64t^2)^{-\frac{3}{2}}\right)^2 + \left(8(1+64t^2)^{-\frac{3}{2}}\right)^2}}{\sqrt{1+64t^4}} \\ k(20) = \frac{\sqrt{\left(-64(20)(1+64(20)^2)^{-\frac{3}{2}}\right)^2 + \left(8(1+64(20)^2)^{-\frac{3}{2}}\right)^2}}{\sqrt{1+64(20)^2}} \\ k(20) = .000002\end{align*}$
::F(t) = (1,8t) F(t) 1+642T(t) = F(t) (1+64t2) -12,(1+642) -12(t)) T(t) =(64t(1+642) - 32,8(1+642) -32) T(t) (1+64t2) 1+642T}(t) 1+642T}(t) 1+642T}(t) =(t) 8(t) (t) * F(t) (t) (64t) _(1+642) - 32,21+64t4k(20) =(64(20) 1+64(20) 2) - 322) 8(1+64(2) - 642) 221+64(202) 2(20) =0.002。

The first path has a curvature of .000002 at $\begin{align*}t=20\end{align*}$ . Nina repeats the process for the path modeled by $\begin{align*}\vec{G}(t)\end{align*}$ .
::第一个路径在 t=20 时的曲线为 .00002。尼娜重复了 G( t) 所建路径的过程。

$\begin{aligned} \vec{G}^{'} (t) = (2, \cos t) \\ ‖ \vec{G}^{'} (t) ‖ = \sqrt{4 + \cos^{2} t} \\ \vec{T} (t) = \frac{\vec{G}^{'} (t)}{‖ \vec{G}^{'} (t) ‖} = (2 (4 + \cos^{2} t)^{- \frac{1}{2}}, \cos t (4 + \cos^{2} t)^{- \frac{1}{2}}) \\ \vec{T}^{'} (t) = (\sin (2 t) (4 + \cos^{2} t)^{- \frac{3}{2}}, - 4 \sin t (4 + \cos^{2} t)^{- \frac{3}{2}}) \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{({(\sin (2 t) (4 + \cos^{2} t)^{- \frac{3}{2}})}^{2} + {(- 4 \sin t (4 + \cos^{2} t)^{- \frac{3}{2}})}^{2})} \\ k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{G}^{'} (t) ‖} \\ k (20) = \frac{‖ \vec{T}^{'} (20) ‖}{‖ \vec{G}^{'} (20) ‖} = .0634 \end{aligned}$
$\begin{align*}\vec{G}{^\prime}(t) = (2, \cos t) \\ \Big\|\vec{G}{^\prime}(t)\Big\| = \sqrt{4 + \cos^2 t} \\ \vec{T}(t)= \frac{\vec{G}{^\prime}(t)}{\Big \| \vec{G}{^\prime}(t)\Big \|} = \left(2(4+\cos^2 t)^{-\frac{1}{2}}, \cos t(4+\cos^2 t)^{-\frac{1}{2}}\right) \\ \vec{T}{^\prime}(t) = \left(\sin(2t)(4 + \cos^2 t)^{-\frac{3}{2}}, -4 \sin t(4 + \cos^2 t)^{-\frac{3}{2}}\right) \\ \Bigg \|\vec{T}{^\prime}(t)\Bigg \| = \sqrt{\left(\left(\sin(2t)(4 + \cos^2 t)^{-\frac{3}{2}}\right)^2 + \left(-4 \sin t(4 + \cos^2 t)^{-\frac{3}{2}}\right)^2\right)} \\ k(t) = \frac{\Big \| \vec{T}{^\prime}(t)\Big \|}{\Big \| \vec{G}{^\prime}(t)\Big \|} \\ k(20) = \frac{\Big \| \vec{T}{^\prime}(20)\Big \|}{\Big \| \vec{G}{^\prime}(20)\Big \|} = .0634\end{align*}$
::G(t) = (2, cos(t) 4+cos2(t) = G(t) (g) (t) (2, 4+cos2(t) - 12, cos(4+cos2(t) 4+cos2(t) -12) T(t) = (sin(2, (t) 4+cos2(t) (t) (t) (4+cos2(t) - 322) k(t) (t) (t) (t) (t) (T) (t) (t) (t) (t) (t) (t) (t) (t) (T) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (T(t) (T(t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t) (t)(t)(t) (t)(t)(T(T(t)(t)(t)(t)(t)(T)(t)(t)(t))(t)(t)(t)(t)(t)(t)(t)(t))))(t)(t)(t)(t)(_)(t)(t)(t)(t)(T(t)(t)(t)(t)(_

$\begin{align*}k(20)\end{align*}$ is larger for $\begin{align*}\vec{G}(t)\end{align*}$ than for $\begin{align*}\vec{F}(t)\end{align*}$ , so $\begin{align*}\vec{G}(t)\end{align*}$ is the curvier path. Nina decides to take the first trail when she goes hiking with her grandmother.
::k( 20) G( t) 大于 F( t) , 所以 G( t) 是弯曲路径。尼娜决定与祖母远足时走第一条路。

Example 2
::例2

Given the curves $\begin{align*}\vec{F}(t) = (t, \ln(t))\end{align*}$ and $\begin{align*}\vec{G}(t) = (t, t^3)\end{align*}$ , which is curvier at $\begin{align*}t=1\end{align*}$ ?
::根据曲线 F(t) = (t,ln(t)) 和 G(t) = (t,t3) , 也就是 curvier at t=1?

Find $\begin{align*}k(1)\end{align*}$ for each vector-valued function and then compare them.
::为每个矢量估值函数查找 k(1), 然后比较它们。

First, find $\begin{align*}\vec{T}(t)\end{align*}$ of $\begin{align*}\vec{F}(t)\end{align*}$ :
::首先,找到F(t)中的 T(t) :

$\begin{aligned} \vec{F}^{'} (t) = (1, \frac{1}{t}) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{1^{2} + {(\frac{1}{t})}^{2}} = \sqrt{1 + \frac{1}{t^{2}}} \\ \vec{T} (t) = \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} = ({(1 + \frac{1}{t^{2}})}^{- \frac{1}{2}}, t^{- 1} {(1 + \frac{1}{t^{2}})}^{- \frac{1}{2}}) \end{aligned}$
$\begin{align*}\vec{F}{^\prime}(t) = \left(1, \frac{1}{t}\right) \\ \Big \| \vec{F}{^\prime}(t)\Big \| = \sqrt{1^2 + \left(\frac{1}{t}\right)^2} = \sqrt{1+ \frac{1}{t^2}} \\ \vec{T}(t) = \frac{\vec{F}{^\prime}(t)}{\Big\| \vec{F}{^\prime}(t)\Big\|} = \left(\left(1+ \frac{1}{t^2}\right)^{-\frac{1}{2}}, t^{-1} \left(1+ \frac{1}{t^2}\right)^{-\frac{1}{2}}\right)\end{align*}$
::F(t) =(1,1,1,1,4) F(t) 12+(1,1) 2=1+1,2t2T(t) =F(t) F(t) ((1,1+1,2) -12,t-1(1,1+1,2) -12)

Then find $\begin{align*}\vec{T}{^\prime}(t)\end{align*}$ and its magnitude for $\begin{align*}\vec{F}(t)\end{align*}$ :
::然后找到 T( t) 及其 F( t) 的大小 :

$\begin{aligned} \vec{T}^{'} (t) = ({(1 + \frac{1}{t^{2}})}^{- \frac{3}{2}} (t^{- 3}), {(1 + \frac{1}{t^{2}})}^{- \frac{3}{2}} t^{- 4} - {(1 + \frac{1}{t^{2}})}^{- \frac{1}{2}} t^{- 2}) \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{{({(1 + \frac{1}{t^{2}})}^{- \frac{3}{2}} (t^{- 3}))}^{2} + {({(1 + \frac{1}{t^{2}})}^{- \frac{3}{2}} t^{- 4} - {(1 + \frac{1}{t^{2}})}^{- \frac{1}{2}} t^{- 2})}^{2}} \end{aligned}$
$\begin{align*}\vec{T}{^\prime}(t) = \left(\left(1+ \frac{1}{t^2}\right)^{-\frac{3}{2}}(t^{-3}), \left(1+\frac{1}{t^2}\right)^{-\frac{3}{2}}t^{-4} -\left(1+ \frac{1}{t^2}\right)^{-\frac{1}{2}}t^{-2}\right) \\ \Big \| \vec{T}{^\prime}(t) \Big \| = \sqrt{\left(\left(1+\frac{1}{t^2}\right)^{-\frac{3}{2}}(t^{-3})\right)^2 + \left(\left(1+\frac{1}{t^2}\right)^{-\frac{3}{2}}t^{-4} - \left(1+\frac{1}{t^2}\right)^{-\frac{1}{2}} t^{-2}\right)^2}\end{align*}$
::T(t) = ((1+1) - 32(t-3),(1+1) - 32(t-3),(1+1) - 32t-2) - 32t-4 - (1+1) - 12t-2) T(t) (1+1) - 32(t-3)) - 2+(1+1) - 32t-2) - 32t-4(1+1) - 12t-2) - 2

Find $\begin{align*}k(t)\end{align*}$ for $\begin{align*}\vec{F}(t)\end{align*}$ and evaluate at $\begin{align*}t=1\end{align*}$ :
::F(t) 查找 k(t) 并按 t= 1 进行评价 :

$\begin{aligned} k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{F}^{'} (t) ‖} = \frac{\sqrt{{({(1 + \frac{1}{t^{2}})}^{- \frac{3}{2}} (t^{- 3}))}^{2} + {({(1 + \frac{1}{t^{2}})}^{- \frac{3}{2}} t^{- 4} - {(1 + \frac{1}{t^{2}})}^{- \frac{1}{2}} t^{- 2})}^{2}}}{\sqrt{1 + \frac{1}{t^{2}}}} \\ k (1) = \frac{‖ \vec{T}^{'} (1) ‖}{‖ \vec{F}^{'} (1) ‖} = \frac{\sqrt{{({(1 + \frac{1}{1^{2}})}^{- \frac{3}{2}} (1^{- 3}))}^{2} + {({(1 + \frac{1}{1^{2}})}^{- \frac{3}{2}} 1^{- 4} - {(1 + \frac{1}{1^{2}})}^{- \frac{1}{2}} 1^{- 2})}^{2}}}{\sqrt{1 + \frac{1}{1^{2}}}} \end{aligned}$
$\begin{align*}k(t)=\frac{\Big\|\vec{T}{^\prime}(t)\Big\|}{\Big\|\vec{F}{^\prime}(t)\Big\|}= \frac{\sqrt{\left(\left(1+\frac{1}{t^2}\right)^{-\frac{3}{2}}(t^{-3})\right)^2 + \left(\left(1+\frac{1}{t^2}\right)^{-\frac{3}{2}}t^{-4} - \left(1+\frac{1}{t^2}\right)^{-\frac{1}{2}} t^{-2}\right)^2}}{\sqrt{1+ \frac{1}{t^2}}} \\ k(1)=\frac{\Big\|\vec{T}{^\prime}(1)\Big\|}{\Big\|\vec{F}{^\prime}(1)\Big\|}= \frac{\sqrt{\left(\left(1+\frac{1}{1^2}\right)^{-\frac{3}{2}}(1^{-3})\right)^2 + \left(\left(1+\frac{1}{1^2}\right)^{-\frac{3}{2}}1^{-4} - \left(1+\frac{1}{1^2}\right)^{-\frac{1}{2}}1^{-2}\right)^2}}{\sqrt{1+ \frac{1}{1^2}}}\end{align*}$
:kt) TT(t) F(t) (1+1t2)-32(t-33))-2+(1+1t2)-32-32t-4-(1+1t2)-(1+1T2)-221+1t2k-(1+12t-2)-(1)-2-(1+112)-11-32(1-3))-2(1+1+112)-32-(1+112)-321+112

The curvature of $\begin{align*}\vec{F}(t)\end{align*}$ when $\begin{align*}t=1\end{align*}$ is $\begin{align*}\frac{\sqrt{2}}{4} \approx .3536\end{align*}$ .
::当 t=1 时 F( t) 的曲率为 24 .3536 。

Now, repeat the process with $\begin{align*}\vec{G}(t)\end{align*}$ .
::现在, 重复 G( t) 的过程。

$\begin{aligned} \vec{G}^{'} (t) = (1, 3 t^{2}) \\ ‖ \vec{G}^{'} (t) ‖ = \sqrt{(1)^{2} + (3 t^{2})^{2}} = \sqrt{1 + 9 t^{4}} \\ \frac{\vec{G}^{'} (t)}{‖ \vec{G}^{'} (t) ‖} = ({(1 + 9 t^{4})}^{- \frac{1}{2}}, 3 t^{2} (1 + 9 t^{4})^{- \frac{1}{2}}) \\ \vec{T}^{'} (t) = ((- 18 t^{3}) (1 + 9 t^{4})^{- \frac{3}{2}}, - 54 t^{5} (1 + 9 t^{4})^{- \frac{3}{2}} + 6 t (1 + 9 t^{4})^{- \frac{1}{2}}) \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{{((- 18 t^{3}) (1 + 9 t^{4})^{- \frac{3}{2}})}^{2} + {(- 54 t^{5} (1 + 9 t^{4})^{- \frac{3}{2}} + 6 t (1 + 9 t^{4})^{- \frac{1}{2}})}^{2}} \\ k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{G}^{'} (t) ‖} \\ = \frac{\sqrt{{((- 18 t^{3}) (1 + 9 t^{4})^{- \frac{3}{2}})}^{2} + {(- 54 t^{5} (1 + 9 t^{4})^{- \frac{3}{2}} + 6 t (1 + 9 t^{4})^{- \frac{1}{2}})}^{2}}}{\sqrt{1 + 9 t^{4}}} \\ k (1) = \frac{\sqrt{{((- 18 (1)^{3}) (1 + 9 (1)^{4})^{- \frac{3}{2}})}^{2} + {(- 54 (1)^{5} (1 + 9 (1)^{4})^{- \frac{3}{2}} + 6 (1) (1 + 9 (1)^{4})^{- \frac{1}{2}})}^{2}}}{\sqrt{1 + 9 (1)^{4}}} \\ k (1) = \frac{3 \sqrt{10}}{50} \approx .1897 \end{aligned}$
$\begin{align*}\vec{G}{^\prime}(t) = (1, 3t^2) \\ \Big\|\vec{G}{^\prime}(t)\Big\| = \sqrt{(1)^2 + (3t^2)^2} = \sqrt{1+9t^4} \\ \frac{\vec{G}{^\prime}(t)}{\Big \| \vec{G}{^\prime}(t) \Big \|} = \left(\left(1+9t^4\right)^{-\frac{1}{2}}, 3t^2(1+9t^4)^{-\frac{1}{2}}\right) \\ \vec{T}{^\prime}(t) = \left((-18t^3)(1+9t^4)^{-\frac{3}{2}}, -54t^5 (1+9t^4)^{-\frac{3}{2}} + 6t(1+9t^4)^{-\frac{1}{2}}\right) \\ \Big\|\vec{T}{^\prime}(t)\Big\| = \sqrt{\left((-18t^3)(1+9t^4)^{-\frac{3}{2}}\right)^2 + \left(-54t^5 (1+9t^4)^{-\frac{3}{2}} + 6t(1+9t^4)^{-\frac{1}{2}}\right)^2} \\ k(t) = \frac{\Big\|\vec{T}{^\prime}(t)\Big\|}{\Big\| \vec{G}{^\prime}(t)\Big\|} \\ = \frac{\sqrt{\left((-18t^3)(1+9t^4)^{-\frac{3}{2}}\right)^2 + \left(-54t^5 (1+9t^4)^{-\frac{3}{2}} + 6t(1+9t^4)^{-\frac{1}{2}}\right)^2}}{\sqrt{1+9t^4}} \\ k(1) = \frac{\sqrt{\left((-18(1)^3)(1+9(1)^4)^{-\frac{3}{2}}\right)^2 + \left(-54(1)^5 (1+9(1)^4)^{-\frac{3}{2}} + 6(1)(1+9(1)^4)^{-\frac{1}{2}}\right)^2}}{\sqrt{1+9(1)^4}} \\ k(1) = \frac{3 \sqrt{10}}{50} \approx .1897\end{align*}$
:t) = (1+9,3,3,2,2) (3,3) = (1,9,3,3,3,2,2,2,1,3,3,2,3,4,2,4,4,2,4,(1,9,3,9,2,2,2,4,(t)) = (1,9,9,3,3,3,3,3,3,(t) =(,1,8,3,3,3,2,) G,(t) =(1,9,3,2,2,2,(t) =(,1,9,3,2,1,1,1,1,9,1,9,9,9,4,4,(1,9,4,4,4,4,)-(t),32,(t),(t),(1,9,9,5,3,3,1,1,9,9,3,1,9,9,9,4,4,4,4,4,4,1,4,4,1,9,9,9,9,9,4,4,4,4,4,4,4,4,4,4,4,4,4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,((()和()和())和()和()和()和())和(()和()))和(()和(())和)(((()和)()()()()()()()()()()()()()()()()()()()()()()))))))))))))和)(()))和(至)()(((()))))))))((至(至(((())))和(()()()((((((()()()()()()()()()()(()))))))(((((())))(())(((())))()()()()()()((

The curvature, $\begin{align*}k(1)\end{align*}$ , of $\begin{align*}\vec{F}(t)\end{align*}$ is larger than $\begin{align*}k(1)\end{align*}$ of $\begin{align*}\vec{G}(t)\end{align*}$ . This means that when $\begin{align*}t=1\end{align*}$ , $\begin{align*}\vec{F}(t)\end{align*}$ is curvier than $\begin{align*}\vec{G}(t)\end{align*}$ .
::F(t) 的曲线, k(1) 大于 G(t) 的 k(1) 。这意味着当 t= 1, F(t) 的曲线大于 G(t) 时。

Example 3
::例3

Give the curvature for the following 3 curves.
::给以下3个曲线的曲率。

$\begin{align*}\vec{F}(t) = (5 \sin t, 5 \cos t)\end{align*}$
::F(t) = (5sint,5cost)

$\begin{align*}\vec{G}(t) = (12 \cos t, 12 \sin t)\end{align*}$
::G(t) =( 12cosT, 12sint)

$\begin{align*}\vec{H}(t) = (100 \cos t, 100 \sin t)\end{align*}$
::H(t) = (100cost 100sint)

All three curves are circles, so their curvatures are equal to $\begin{align*}\frac{1}{r}\end{align*}$ for all values of $\begin{align*}t\end{align*}$ .
::所有三个曲线都是圆圈, 所以它们的曲线是所有值的至1r。

$\begin{align*}\frac{1}{5}\end{align*}$

$\begin{align*}\frac{1}{12}\end{align*}$

$\begin{align*}\frac{1}{100}\end{align*}$

Example 4
::例4

Find the curvature, $\begin{align*}k(t)\end{align*}$ , for the following function at $\begin{align*}t=1\end{align*}$ : $\begin{align*}\vec{F}(t) = (t^2, t^3)\end{align*}$ .
::为 t=1: F(t) = (t2,t3) 的以下函数查找曲线, k(t) 。

$\begin{aligned} \vec{F}^{'} (t) = (2 t, 3 t^{2}) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{(2 t)^{2} + (3 t^{2})^{2}} = \sqrt{4 t^{2} + 9 t^{4}} \\ \vec{T} (t) = \frac{\vec{F}^{'} (t)}{‖ \vec{F}^{'} (t) ‖} = (2 t (4 t^{2} + 9 t^{4})^{- \frac{1}{2}}, 3 t^{2} (4 t^{2} + 9 t^{4})^{- \frac{1}{2}}) \\ \vec{T}^{'} (t) = (2 (4 t^{2} + 9 t^{4})^{- \frac{1}{2}} (8 t + 36 t^{3}) - t (4 t^{2} + 9 t^{4})^{- \frac{3}{2}}, 6 t (4 t^{2} + 9 t^{4})^{- \frac{1}{2}} \\ + - \frac{3 t^{2}}{2} (4 t^{2} + 9 t^{4})^{- \frac{3}{2}} (8 t + 36 t^{3})) \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{{(2 (4 t^{2} + 9 t^{4})^{- \frac{1}{2}} - t (4 t^{2} + 9 t^{4})^{- \frac{3}{2}} (8 t + 36 t^{3}))}^{2}} \\ + \sqrt{{(6 t (4 t^{2} + 9 t^{4})^{- \frac{1}{2}} + - \frac{3 t^{2}}{2} (4 t^{2} + 9 t^{4})^{- \frac{3}{2}} (8 t + 36 t^{3}))}^{2}} \\ k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{F}^{'} (t) ‖} \\ k (1) = \frac{‖ \vec{T}^{'} (1) ‖}{‖ \vec{F}^{'} (1) ‖} = \frac{1.214}{3.606} = .3367 \end{aligned}$
$\begin{align*}\vec{F}{^\prime}(t) = (2t, 3t^2) \\ \Big \| \vec{F}{^\prime}(t) \Big \| = \sqrt{(2t)^2 + (3t^2)^2} = \sqrt{4t^2 + 9t^4} \\ \vec{T}(t) = \frac{\vec{F}{^\prime}(t)}{\Big \|\vec{F}{^\prime}(t)\Big \|} = \left(2t(4t^2+9t^4)^{-\frac{1}{2}}, 3t^2(4t^2+9t^4)^{-\frac{1}{2}}\right) \\ \vec{T}{^\prime}(t) = \left(2(4t^2+9t^4)^{-\frac{1}{2}}(8t+36t^3) - t(4t^2+9t^4)^{-\frac{3}{2}}, 6t(4t^2+9t^4)^{-\frac{1}{2}} \right . \\ \left . + -\frac{3t^2}{2} (4t^2+9t^4)^{-\frac{3}{2}}(8t+36t^3)\right) \\ \Bigg\|\vec{T}{^\prime}(t)\Bigg\| =\sqrt{\left(2(4t^2 + 9t^4)^{-\frac{1}{2}} -t(4t^2 + 9t^4)^{-\frac{3}{2}}(8t+36t^3)\right)^2} \\ + \sqrt{\left(6t(4t^2 + 9t^4)^{-\frac{1}{2}} + -\frac{3t^2}{2}(4t^2 + 9t^4)^{-\frac{3}{2}}(8t+36t^3)\right)^2} \\ k(t) = \frac{\|\vec{T}{^\prime}(t)\|}{\|\vec{F}{^\prime}(t)\|} \\ k(1) = \frac{\|\vec{T}{^\prime}(1)\|}{\|\vec{F}{^\prime}(1)\|} = \frac{1.214}{3.606} = .3367\end{align*}$
::F( t) = (2, 2, 3, 3, 3, 3, 4, 4, 4, 2, 4, 2, 4, 2, + (3, 2) 2, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 3, 3, 4, 3, 4, 4, 3, 3, 3, 3, 3, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 2, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 2, 2, 2, 2, 6, 6, 6, 6, 6, 2, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 4, 4, 4, 6, 6, 6, 6, 4, 6, 6, 4, 4, 4, 6, 4, 4, 6, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 6, 6, 4, 4, 4, 4, 4, 4, 4,

Example 5
::例5

Write the equation for a vector-valued function that is curvier at $\begin{align*}t=2\end{align*}$ than $\begin{align*}\vec{F}(t) = (t, t^2)\end{align*}$ .
::写入矢量估值函数的方程,该函数在 t=2 时为 curvier, 而不是 F(t) = (t) = (t, t2)。

First, find $\begin{align*}k(t)\end{align*}$ for $\begin{align*}\vec{F}(t)\end{align*}$ when $\begin{align*}t=2\end{align*}$ .
::首先, 在 t=2 时找到 k( t) F( t) 的 k( t) 。

$\begin{aligned} \vec{F} (t) = (t, t^{2}) \\ \vec{F}^{'} (t) = (1, 2 t) \\ ‖ \vec{F}^{'} (t) ‖ = \sqrt{1 + 4 t^{2}} \\ \vec{T} (t) = ((1 + 4 t^{2})^{- \frac{1}{2}}, 2 t (1 + 4 t^{2})^{- \frac{1}{2}}) \\ \vec{T}^{'} (t) = (- 4 t (1 + 4 t^{2})^{- \frac{3}{2}}, 2 (1 + 4 t^{2})^{- \frac{3}{2}}) \\ ‖ \vec{T}^{'} (t) ‖ = \sqrt{{(- 4 t (1 + 4 t^{2})^{- \frac{3}{2}})}^{2} + {(2 (1 + 4 t^{2})^{- \frac{3}{2}})}^{2}} \\ k (t) = \frac{‖ \vec{T}^{'} (t) ‖}{‖ \vec{F}^{'} (t) ‖} \\ k (2) = \frac{\sqrt{{(- 4 (2) (1 + 4 (2)^{2})^{- \frac{3}{2}})}^{2} + {(2 (1 + 4 (2)^{2})^{- \frac{3}{2}})}^{2}}}{\sqrt{1 + 4 (2)^{2}}} = .0285 \end{aligned}$
$\begin{align*}\vec{F}(t) = (t, t^2) \\ \vec{F}{^\prime}(t) = (1, 2t) \\ \Big\|\vec{F}{^\prime}(t)\Big\| = \sqrt{1 + 4t^2} \\ \vec{T}(t)= \left((1+4t^2)^{-\frac{1}{2}}, 2t(1+4t^2)^{-\frac{1}{2}}\right) \\ \vec{T}{^\prime}(t) = \left(-4t(1+4t^2)^{-\frac{3}{2}}, 2(1+4t^2)^{-\frac{3}{2}}\right) \\ \Big \|\vec{T}{^\prime}(t)\Big \| = \sqrt{\left(-4t(1+4t^2)^{-\frac{3}{2}}\right)^2 + \left(2(1+4t^2)^{-\frac{3}{2}}\right)^2} \\ k(t) = \frac{\Big \| \vec{T}{^\prime}(t)\Big \|}{\Big \| \vec{F}{^\prime}(t)\Big \|} \\ k(2) = \frac{\sqrt{\left(-4(2)(1+4(2)^2)^{-\frac{3}{2}}\right)^2 + \left(2(1+4(2)^2)^{-\frac{3}{2}}\right)^2}}{\sqrt{1+4(2)^2}} = .0285\end{align*}$
::F(t) = (t) = (t,t2) F(t) (t) 1,2t) (f) (t) 1+4t2T(t) =(1+4t2) - 12,2t(1+4t2) - 12) T(t) =(4t(1+4T2) - 322,2(1+4t2) - 322) (T) +(2+1+4t2) - 322K(t) +(t) (t) (t) (t) (k) =(4(2)(2)1+4(2)2)-32-2(1+4(2)2) - 3221+4(2)2=0285。

Notice that at $\begin{align*}t=2\end{align*}$ , $\begin{align*}k(t)\end{align*}$ is equal to .0285. This means any equation with a curvature greater than .0285 will be curvier than $\begin{align*}\vec{F}(t)\end{align*}$ at $\begin{align*}t=2\end{align*}$ . So, for example, you can write the equation for a circle with a curvature of 1 and that will be curvier than $\begin{align*}\vec{F}(t)\end{align*}$ at $\begin{align*}t=2\end{align*}$ .
::注意 t= 2, k( t) 等于 0.0285 。这意味着任何曲线大于. 0285 的方程式将比 t= 2 的 F ( t) 。因此, 例如, 您可以写一个圆形的方程式, 曲线为 1 , 曲线将比 T = 2 的 F ( t) 曲线为 curvier 。

Since, for a circle, $\begin{align*}k(t)=\frac{1}{r}\end{align*}$ , a circle with a radius of 1 will have a $\begin{align*}k(t)\end{align*}$ of 1. So $\begin{align*}\vec{G}(t)=(\sin t, \cos t)\end{align*}$ is one equation that is curvier at $\begin{align*}t=2\end{align*}$ than $\begin{align*}\vec{F}(t)\end{align*}$ is. There are an infinite number of solutions to this problem.
::因为对于圆, k( t) = 1r, 半径为 1 的圆将有一个 k( t) 。所以 G( t) = ( sint, cost) 是一个方程式, 即 curvier at t= 2 而不是 F( t) 。这个问题有无限数量的解决方案。

Review
::回顾

Use the following vector-valued functions for #1-15.
::在 # 1-15 中使用以下矢量估值函数。

$\begin{align*}\vec{F}(t) = \left(t^2, \frac{1}{t}\right)\end{align*}$
::F(t) = (t2,1吨)

$\begin{align*}\vec{G}(t) = (6 \sin t, 3 \cos t)\end{align*}$
::G(t) = (6sint,3cost)

$\begin{align*}\vec{H}(t) = (t+1, t^2)\end{align*}$
::H(t) = (t+1,t2)

$\begin{align*}\vec{B}(t) = (2t^2, t-5)\end{align*}$
::B(t) = (2吨2,2吨5)

$\begin{align*}\vec{M}(t) = (e^t, \sin t)\end{align*}$
::M(t) =(et,sin*t)

$\begin{align*}\vec{P}(t) = (4 \sin t, 4 \cos t)\end{align*}$
::P(t) = (4sint,4cost)

$\begin{align*}\vec{S}(t) = (3t^2, 5t)\end{align*}$
::S(t) =( 3t2, 5t)

Find $\begin{align*}k(1)\end{align*}$ for $\begin{align*}\vec{F}(t) \end{align*}$ .
::F(t) 的查找 k(1) 。

Find $\begin{align*}k(\pi)\end{align*}$ for $\begin{align*}\vec{G}(t)\end{align*}$ .
::G( t) 的查找 k 。

Find $\begin{align*}k(5)\end{align*}$ for $\begin{align*}\vec{H}(t) \end{align*}$ .
::H(t) 的查找 k(5) 。

Find $\begin{align*}k(3)\end{align*}$ for $\begin{align*}\vec{B}(t)\end{align*}$ .
::B(t) 的查找 k(3) 。

Find $\begin{align*}k(\pi)\end{align*}$ for $\begin{align*}\vec{M}(t)\end{align*}$ .
::M(t) 的查找 k 。

Find $\begin{align*}k(2)\end{align*}$ for $\begin{align*}\vec{P}(t)\end{align*}$ .
::P(t) 的查找 k(2) 。

Find $\begin{align*}k(0.5)\end{align*}$ for $\begin{align*}\vec{S}(t)\end{align*}$ .
::S(t) 的查找 k( 0. 5) 。

Find $\begin{align*}k(3)\end{align*}$ for $\begin{align*}\vec{F}(t)\end{align*}$ .
::F(t) 的查找 k(3) 。

Find $\begin{align*}k \left(\frac{\pi}{2}\right)\end{align*}$ for $\begin{align*}\vec{G}(t)\end{align*}$ .
::G( t) 的查找 k( % 2) 。

Which curve is curvier at $\begin{align*}t=\pi\end{align*}$ , $\begin{align*}\vec{G}(t)\end{align*}$ or $\begin{align*}\vec{P}(t)\end{align*}$ ?
::哪个曲线是 curvier at t, G(t) 或 P(t) ?

Which curve is curvier at $\begin{align*}t=1\end{align*}$ , $\begin{align*}\vec{B}(t)\end{align*}$ or $\begin{align*}\vec{F}(t)\end{align*}$ ?
::t=1, B(t) 或 F(t) 是哪个曲线的曲线 ?

Find the radius of the circle of best fit for $\begin{align*}\vec{S}(t)\end{align*}$ at $\begin{align*}t=0.5\end{align*}$ .
::查找在 t=0.5 处适合 S(t) 的圆圆半径。

Find the radius of the circle of best fit for $\begin{align*}\vec{H}(t)\end{align*}$ at $\begin{align*}t=5\end{align*}$ .
::在 t=5 找到最适合 H(t) 的圆半径。

Find the radius of the circle of best fit for $\begin{align*}\vec{P}(t)\end{align*}$ at $\begin{align*}t=2\end{align*}$ .
::在 t=2 找到最适合 P(t) 的圆圆半径。

Find the radius of the circle of best fit for $\begin{align*}\vec{F}(t)\end{align*}$ at $\begin{align*}t=1\end{align*}$ .
::在 t=1 中查找最适合 F(t) 的圆半径。

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::回顾(答复)

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::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

沿平板曲线测量曲线

Measuring Curvature ::测量曲线

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)

Measuring Curvature
::测量曲线

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)