11.5 利用激进主义的申请

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• 选择节使用激进软件的应用

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  使用激进软件的应用

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  Applications Using Radicals 
  ::使用激进软件的应用

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  Mathematicians and physicists have studied the motion of pendulums in great detail because this motion explains many other behaviors that occur in nature. This type of motion is called simple harmonic motion and it is important because it describes anything that repeats periodically. Galileo was the first person to study the motion of a pendulum, around the year 1600. He found that the time it takes a pendulum to complete a swing doesn’t depend on its mass or on its angle of swing (as long as the angle of the swing is small). Rather, it depends only on the length of the pendulum.
  ::数学家和物理学家们非常详细地研究了钟摆运动,因为这一运动解释了自然中发生的许多其他行为。 这种运动被称为简单的调和运动,非常重要,因为它描述了任何定期重复的内容。 伽利略是最早研究钟摆运动的人,大约在1600年左右。 他发现,要完成秋摆需要多少钟摆的时间并不取决于其质量或秋千角度(只要秋千的角度很小 ) 。 相反,它只取决于钟摆的长度。

  

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  The time it takes a pendulum to complete one whole back-and-forth swing is called the period of the pendulum. Galileo found that the period of a pendulum is proportional to the square root of its length: $\begin{align*}T = a \sqrt{L}\end{align*}$ . The proportionality constant , $\begin{align*}a\end{align*}$ , depends on the acceleration of gravity: $\begin{align*}a = \frac{2 \pi}{\sqrt{g}}\end{align*}$ . At sea level on Earth, acceleration of gravity is $\begin{align*}g = 9.81 \ m/s^2\end{align*}$ (meters per second squared). Using this value of gravity, we find $\begin{align*}a = 2.0\end{align*}$ with units of $\begin{align*}\frac{s}{\sqrt{m}}\end{align*}$ (seconds divided by the square root of meters).
  ::伽利略发现,一个钟摆的周期与其长度的平方根(T=aL)成正比。比例常数,a,取决于重力加速度:a=2g。在地球上的海平面上,重力加速度为g=9.81m/s2(每秒二平方公尺计数)。使用这一重力值,我们发现一个等于2.0的单位为sm(以平方根为单位)。

  

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  Up until the mid $\begin{align*}20^{th}\end{align*}$ century, all clocks used pendulums as their central time keeping component.
  ::直到20世纪中叶,所有钟头都用钟表作为保持时间的核心组成部分。

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  Real-World Application: Pendulums 
  ::真实世界应用程序:

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  Graph the period of a pendulum of a clock swinging in a house on Earth at sea level as we change the length of the pendulum. What does the length of the pendulum need to be for its period to be one second?
  ::当我们改变钟的长度时,在海平面绘制一个钟钟摆在地球上房子的钟摆的钟摆的时段。 钟摆的时长需要多少时间才能再过一秒钟呢?

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  The function for the period of a pendulum at sea level is $\begin{align*}T = 2 \sqrt{L}\end{align*}$ .
  ::海平面钟摆期间的函数为T=2L。

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  We start by making a table of values for this function:
  ::我们首先为此函数绘制一个值表 :

  $\begin{align*}L\end{align*}$	$\begin{align*}T = 2 \sqrt{L}\end{align*}$
  0	$\begin{align*}T = 2 \sqrt{0} = 0\end{align*}$
  1	$\begin{align*}T = 2 \sqrt{1} = 2\end{align*}$
  2	$\begin{align*}y = 2 \sqrt{2} = 2.8\end{align*}$
  3	$\begin{align*}y = 2 \sqrt{3} = 3.5\end{align*}$
  4	$\begin{align*}y = 2 \sqrt{4} = 4\end{align*}$
  5	$\begin{align*}y = 2 \sqrt{5} = 4.5\end{align*}$

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  Now let's graph the function. It makes sense to let the horizontal axis represent the length of the pendulum and the vertical axis represent the period of the pendulum.
  ::现在让我们来绘制函数。让水平轴代表钟摆的长度,而垂直轴代表钟摆的期间,是有道理的。

  

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  We can see from the graph that a length of approximately $\begin{align*}\frac{1}{4}\end{align*}$ meters gives a period of 1 second. We can confirm this answer by using our function for the period and plugging in $\begin{align*}T = 1 \ second\end{align*}$ :
  ::从图中可以看出,大约14米的长度给出了1秒的时间。我们可以使用我们在这段时期的功能来确认这个答案,并插入T=1秒:

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  $$\begin{align*}T & = 2 \sqrt{L} \Rightarrow 1 = 2 \sqrt{L}\end{align*}$$
  ::T=2L1=2L

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  $$\begin{align*}&\text{Square both sides of the equation:} && 1 = 4L\\ &\text{Solve for} \ L: && L = \frac{1}{4} \ meters\end{align*}$$
  ::方形两侧方形平方形: 1=4LSolve, L:L=14米

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  Real-World Application: TV Screens 
  ::真实世界应用程序:电视屏幕

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  “Square” TV screens have an aspect ratio of 4:3; in other words, the width of the screen is $\begin{align*}\frac{4}{3}\end{align*}$ the height. TV “sizes” are traditionally represented as the length of the diagonal of the television screen. Graph the length of the diagonal of a screen as a function of the area of the screen. What is the diagonal of a screen with an area of $\begin{align*}180 \ in^2\end{align*}$ ?
  ::“平方”电视屏幕的宽度为高度43。 电视“ 尺寸” 传统上以电视屏幕的对角长度表示。 “ 平方” 屏幕的对角长度是屏幕面积的函数。 屏幕的对角值是多少? 屏幕的对角值是多少, 面积是 180 英寸 2 ?

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  Let $\begin{align*}d =\end{align*}$ length of the diagonal, $\begin{align*}x =\end{align*}$ width
  ::Let d= 对角长度, x=宽度

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  Then 4 $\begin{align*}\times\end{align*}$ height = 3 $\begin{align*}\times\end{align*}$ width
  ::然后4x高度 = 3x宽度

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  Or, height = $\begin{align*}\frac{3}{4}x\end{align*}$ .
  ::或者,身高=34x。

  

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  The area of the screen is: $\begin{align*}A =\end{align*}$ length $\begin{align*}\times\end{align*}$ width or $\begin{align*} A = \frac{3}{4} x^2\end{align*}$
  ::屏幕区域为: A= 长度 × 宽度 或 A= 34x2

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  Find how the diagonal length relates to the width by using the Pythagorean theorem :
  ::使用 Pytagorean 定理查找对角长度与宽度的关系 :

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  $$\begin{align*}x^2 + \left(\frac{3}{4} x \right)^2 & = d^2\\ x^2 + \frac{9}{16}x^2 & = d^2\\ \frac{25}{16}x^2 & = d^2 \Rightarrow x^2 = \frac{16}{25}d^2 \Rightarrow x = \frac{4}{5}d\end{align*}$$
  ::x2+( 34x) 2=d2x2+916x2=d22516x2=d22516x2=d2_x2=d2_x2=#x2=1625d2=%x=45d

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  Therefore , the diagonal length relates to the area as follows: $\begin{align*} A = \frac{3}{4} \left( \frac{4}{5}d \right)^2 = \frac{3}{4} \cdot \frac{16}{25}d^2 = \frac{12}{25}d^2\end{align*}$ .
  ::因此,对角长度与以下区域有关:A=34(45d)2=341625d2=1225d2。

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  We can also flip that around to find the diagonal length as a function of the area: $\begin{align*}d^2 = \frac{25}{12} A\end{align*}$ or $\begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*}$ .
  ::我们还可以翻转这个区域,以找到区域函数的对角长度:d2=2512A或d=523A。

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  Now we can make a graph where the horizontal axis represents the area of the television screen and the vertical axis is the length of the diagonal. First let’s make a table of values:
  ::现在我们可以绘制一个图, 水平轴代表电视屏幕的面积, 垂直轴代表对角线的长度。 首先让我们绘制一个数值表 :

  $\begin{align*}A\end{align*}$	$\begin{align*}d = \frac{5}{2 \sqrt{3}} \sqrt{A}\end{align*}$
  0	0
  25	7.2
  50	10.2
  75	12.5
  100	14.4
  125	16.1
  150	17.6
  175	19
  200	20.4

  

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  From the graph we can estimate that when the area of a TV screen is 180 $\begin{align*}in^2\end{align*}$ the length of the diagonal is approximately 19.5 inches. We can confirm this by plugging in $\begin{align*}A = 180\end{align*}$ into the formula that relates the diagonal to the area: $\begin{align*}d = \frac{5}{2\sqrt{3}} \sqrt{A} = \frac{5}{2 \sqrt{3}} \sqrt{180} = 19.4 \ inches\end{align*}$ .
  ::从图中我们可以估计,当电视屏幕面积为180英寸2时,对角线的长度约为19.5英寸。我们可以通过将A=180插入将对角线与面积相联系的公式:d=523A=523180=19.4英寸来确认这一点。

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  Radicals often arise in problems involving areas and volumes of geometrical figures.
  ::在涉及几何数字的面积和数量的问题中,往往会出现激进分子。

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  Real-World Application: Pool Dimensions 
  ::现实世界应用:组合层面

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  A pool is twice as long as it is wide and is surrounded by a walkway of uniform width of 1 foot. The combined area of the pool and the walkway is 400 square feet. Find the dimensions of the pool and the area of the pool.
  ::游泳池的宽度是其宽度的两倍,周围环绕着1英尺的一条统一宽度的行走道。游泳池和行走道的总合面积是400平方英尺。 找到游泳池的尺寸和游泳池的区域。

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  Make a sketch:
  ::绘制一张草图:

  

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  Let $\begin{align*}x =\end{align*}$ the width of the pool. Then:
  ::x = 池的宽度。 然后 :

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  Area $\begin{align*}=\end{align*}$ length $\begin{align*}\times\end{align*}$ width
  ::面积=长度x宽度

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  Combined length of pool and walkway $\begin{align*}= 2x + 2\end{align*}$
  ::池和人行道的综合长度=2x+2

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  Combined width of pool and walkway $\begin{align*}= x + 2\end{align*}$
  ::集合宽度和行道=x+2

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  $\begin{align*}\text{Area} = (2x + 2)(x + 2)\end{align*}$
  ::区域=( 2x+2)( x+2)

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  Since the combined area of pool and walkway is $\begin{align*}400 \ ft^2\end{align*}$ we can write the equation
  ::由于池和人行道的综合面积是400平方英尺,我们可以写方程式

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  $\begin{align*}(2x + 2)(x + 2) = 400\end{align*}$
  :2x+2(x+2)=400)

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  $$\begin{align*}\text{Multiply in order to eliminate the parentheses:} && 2x^2 + 4x + 2x + 4 & = 400\\ \text{Collect like terms:} && 2x^2 + 6x + 4 &= 400\\ \text{Move all terms to one side of the equation:} && 2x^2 + 6x - 396 & = 0\\ \text{Divide all terms by 2:} && x^2 + 3x - 198 & = 0\\ \text{Use the quadratic formula:} && x & =\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\\ && x & = \frac{-3 \pm \sqrt{3^2 - 4(1)(-198)}}{2(1)}\\ && x & = \frac{-3 \pm \sqrt{801}}{2} = \frac{-3 \pm 28.3}{2}\\ && x & = 12.65 \ feet\end{align*}$$
  ::为了删除括号 : 2x2+4x+2x+4x+4x+4=400, 括号如下: 2x2+6x+4+4=400, 将所有条件移到方程的一边: 2x2+6x-3x-396=0Divide所有条件, 乘以 2:x2+3x- 198=0 使用二次方程: xbb _4ac2- 4ac2x_3QQQQ_33-4(1)(- 1982)(1)x}33_8012}328.32x=12.65英尺

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  (The other answer is negative, so we can throw it out because only a positive number makes sense for the width of a swimming pool.)
  :另一个答案是否定的, 所以我们可以把它扔掉, 因为游泳池宽度只有正数才有意义。 )

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  So the dimensions of the pool are: $\begin{align*}length = 12.65\end{align*}$ and $\begin{align*}width = 25.3\end{align*}$ (since the width is 2 times the length)
  ::因此,池的尺寸是:长度=12.65和宽=25.3(因为宽度是长度的2倍)

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  That means that the area of just the pool is $\begin{align*}A = 12.65 \cdot 25.3 \to 320 ft^2\end{align*}$
  ::这意味着,仅池的面积就为A=12.65.65.25.3-320ft2。

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  Check by plugging the result in the area formula:
  ::通过在区域公式中插入结果来检查结果 :

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  Area $\begin{align*}= (2(12.65) + 2)(12.65 + 2) = 27.3 \cdot 14.65 = 400 \ ft^2.\end{align*}$
  ::区域 = (2(12.65)+2 (12.65+2)= 27.314.65=400平方英尺。

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  The answer checks out.
  ::答案检查出来。

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  Example
  ::示例示例示例示例

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  Example 1
  ::例1

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  The volume of a soda can is $\begin{align*}355 \ cm^3\end{align*}$ . The height of the can is four times the radius of the base. Find the radius of the base of the cylinder.
  ::苏打罐的体积是355厘米3,罐头的高度是基底半径的四倍。找到圆筒底部的半径。

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  Make a sketch:
  ::绘制一张草图:

  

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  Let $\begin{align*}x =\end{align*}$ the radius of the cylinder base. Then the height of the cylinder is $\begin{align*}4x\end{align*}$ .
  ::x = 圆筒基数的半径。 然后圆筒的高度为 4x 。

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  The volume of a cylinder is given by $\begin{align*}V = \pi R^2 \cdot h\end{align*}$ ; in this case, $\begin{align*}R\end{align*}$ is $\begin{align*}x\end{align*}$ and $\begin{align*}h\end{align*}$ is $\begin{align*}4x\end{align*}$ , and we know the volume is 355.
  ::VR2h给出了气瓶的体积;在这种情况下,R为x和h为4x,我们知道气瓶的体积为355。

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  Solve the equation:
  ::解决方程式:

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  $$\begin{align*}355 & = \pi x^2 \cdot(4x)\\ 355 & = 4 \pi x^3\\ x^3 & = \frac{355}{4 \pi}\\ x & = \sqrt[3]{\frac{355}{4 \pi}} = 3.046 \ cm\end{align*}$$
  ::

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  Check by substituting the result back into the formula:
  ::将结果重置到公式中 :

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  $$\begin{align*}V = \pi R^2 \cdot h = \pi (3.046)^2 \cdot (4 \cdot 3.046) = 355 \ cm^3\end{align*}$$
  ::VR2h(3.046)2(43.046)=355厘米3

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  So the volume is $\begin{align*}355 \ cm^3\end{align*}$ . The answer checks out.
  ::体积是355厘米 答案已经查出来了

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  Review 
  ::回顾

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  1. If a certain model of a laptop has a diagonal of 15.4 inches and a length of 14.35 inches, find the width.
     ::如果某一型号的笔记本电脑有15.4英寸的对角和14.35英寸的长度,请找到宽度。
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  2. If a certain model of a laptop has a width of 12.78 inches and an area of 114.25 inches squared, find the diagonal.
     ::如果某台笔记本电脑的宽度为12.78英寸,面积为114.25英寸方形,请找到对角。
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  3. The acceleration of gravity can also given in feet per second squared. It is $\begin{align*}g = 32 \ ft/s^2\end{align*}$ at sea level.
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     1. Graph the period of a pendulum with respect to its length in feet.
        ::按钟摆的长度绘制钟摆的时长图。
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     2. For what length in feet will the period of a pendulum be 2 seconds?
        ::钟摆的长度是2秒?
     
     ::重力加速度也可以以每秒平方英尺的速度给出。 这是 g=32 ft/ s2at海平面。 按其长度绘制一个钟摆的时长。 钟摆的时长是2秒?
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  4. The acceleration of gravity on the Moon is $\begin{align*}1.6 \ m/s^2\end{align*}$ .
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     1. Graph the period of a pendulum on the Moon with respect to its length in meters.
        ::根据月球的长度,绘制月球的钟摆时长图。
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     2. For what length, in meters, will the period of a pendulum be 10 seconds?
        ::钟摆的长度,以米计,是10秒吗?
     
     ::月球重力加速度为1.6m/s2. 绘制月球上钟摆的长度图,以米计,以米计,钟摆的长度是10秒?
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  5. The acceleration of gravity on Mars is $\begin{align*}3.69 \ m/s^2\end{align*}$ .
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     1. Graph the period of a pendulum on the Mars with respect to its length in meters.
        ::绘制火星上一个钟摆的长度,以米计。
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     2. For what length, in meters, will the period of a pendulum be 3 seconds?
        ::钟摆的长度,以米计,是3秒吗?
     
     ::火星重力加速度为3.69米/秒。 绘制火星上一个钟摆的长度( 以米计) 。 钟摆的长度( 以米计) 将是 3 秒 ?
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  6. The acceleration of gravity on the Earth depends on the latitude and altitude of a place. The value of $\begin{align*}g\end{align*}$ is slightly smaller for places closer to the Equator than places closer to the poles and the value of $\begin{align*}g\end{align*}$ is slightly smaller for places at higher altitudes that it is for places at lower altitudes. In Helsinki the value of $\begin{align*}g = 9.819 \ m/s^2\end{align*}$ , in Los Angeles the value of $\begin{align*}g = 9.796 \ m/s^2\end{align*}$ and in Mexico City the value of $\begin{align*}g = 9.779 \ m/s^2\end{align*}$ .
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     1. Graph the period of a pendulum with respect to its length for all three cities on the same graph.
        ::在同一图中绘制三个城市的钟摆长度图。
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     2. Use the formula to find for what length, in meters, will the period of a pendulum be 8 seconds in each of these cities?
        ::使用公式来找出每个城市的钟摆时间是8秒吗? 以米计的长度是多少?
     
     ::地球重力加速度取决于一个地方的纬度和高度。对于靠近赤道的地方,克值略小于靠近极点的地方,对于高地地区,克值略小于较低高度的地方。在赫尔辛基,克=9.819m/s2,克=9.796m/s2,在洛杉矶,克=9.796m/s2,在墨西哥城,克=9.779m/s2,克=9.779m/s2。用同一图表显示三个城市的钟摆长度。使用公式来找出每个城市的钟摆长度(以米计)是8秒?
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  7. The aspect ratio of a wide-screen TV is 2.39:1.
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     1. Graph the length of the diagonal of a screen as a function of the area of the screen.
        ::作为屏幕区域函数的屏幕对角长度图。
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     2. What is the diagonal of a screen with area $\begin{align*}150 \ in^2\end{align*}$ ?
        ::区域 150 英寸2 的屏幕的对角值是多少?
     
     ::宽屏电视的侧边比为2.39:1. 绘制屏幕对角线的长度,作为屏幕面积的函数。区域 150 英寸2 的屏幕对角线是什么?

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  For 8-10, rationalize the denominator.
  ::8 -10,把分母合理化

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  8. The volume of a soup can is $\begin{align*}452 \ cm^3\end{align*}$ . The height of the can is three times the radius of the base. Find the radius of the base of the cylinder.
     ::汤罐的体积是452厘米3. 罐头的高度是基底半径的三倍。 找到圆筒底部的半径 。
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  9. The volume of a spherical balloon is $\begin{align*}950 \ cm^3\end{align*}$ . Find the radius of the balloon. (Volume of a sphere $\begin{align*}= \frac{4}{3} \pi R^3\end{align*}$ ).
     ::球球球气球的体积为950厘米3. 找到气球的半径。 (球体的体积=431立方厘米)。
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  10. A rectangular picture is 9 inches wide and 12 inches long. The picture has a frame of uniform width. If the combined area of picture and frame is $\begin{align*}180 \ in^2\end{align*}$ , what is the width of the frame?
      ::矩形图象宽9英寸,长12英寸。 图片有统一的宽度框。 如果图像和框架的合并区域是180英寸2, 则框架的宽度是多少?

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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 。