4.7 变化率

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  变化率

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  Rates of Change
  ::变化率

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  The of a function that describes real, measurable quantities is often called a rate of change . In that case the slope refers to a change in one quantity "> $\begin{array}{r}(y)\end{array}$ per unit of change in another quantity $\begin{array}{r}(x)\end{array}$ . (This is where the equation $\begin{array}{r}m=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\end{array}$ comes in - remember that $\begin{array}{r}\mathrm{\Delta }y\end{array}$ and $\begin{array}{r}\mathrm{\Delta }x\end{array}$ represent the change in $\begin{array}{r}y\end{array}$ and $\begin{array}{r}x\end{array}$ respectively.)
  ::函数描述真实、可计量的数量,通常称为变化率。在此情况下,斜度是指一个数量单位的变化变化,而另一个数量(x)为一个数量单位的变化。 (这就是公式 myx的输入点 - 记住 y 和 x分别代表y和x的变化。)

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  A candle has a starting length of 10 inches. 30 minutes after lighting it, the length is 7 inches. Determine the rate of change in length of the candle as it burns. Determine how long the candle takes to completely burn to nothing.
  ::蜡烛的起始长度为 10 英寸。 点灯后30 分钟, 长度为 7 英寸。 确定蜡烛燃烧时的长度变化速度 。 确定蜡烛完全烧得一干二净需要多久时间 。

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  First , graph the function to visualize what is happening. There are  2 points to start with: at the moment the candle is lit ( $\begin{array}{r}\text{time}=0\end{array}$ ) the length of the candle is 10 inches, and after 30 minutes ( $\begin{array}{r}\text{time}=30\end{array}$ ) the length is 7 inches. Since the candle length depends on the time, plot time on the horizontal axis , and candle length on the vertical axis .
  ::首先, 图形显示正在发生的事情的函数 。 有两点要开始 : 当蜡烛点燃时( 时间= 0) , 蜡烛长度为 10 英寸, 30 分钟后( 时间= 30 ) , 长度为 7 英寸。 因为蜡烛长度取决于时间, 绘制水平轴的时间, 以及垂直轴的蜡烛长度 。

  

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  The rate of change of the candle’s length is simply the slope of the line. With the  2 points $\begin{array}{r}(x_1,y_1)=(0,10)\end{array}$ and $\begin{array}{r}(x_2,y_2)=(30,7)\end{array}$ , you  can use the familiar version of the slope formula :
  ::蜡烛长度的改变速度只是线上的斜坡。 2点(x1,y1)=(0,10)和2点(x2,y2)=(30,7),您可以使用熟悉的斜坡公式版本:

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  $$\begin{array}{rl}\text{Rate of change}& =\frac{y_2-y_1}{x_2-x_1}\\ & =\frac{(7\text{inches})-(10\text{inches})}{(30\text{minutes})-(0\text{minutes})}\\ & =\frac{-3\text{inches}}{30\text{minutes}}\\ & =-0.1\text{inches per minute}\end{array}$$

  
  ::变化率=y2-y1x2-x1=(7英寸)-(10英寸)(30分钟)-(0分钟)-(0分钟)-(3英寸30分钟)-(0.1英寸/每分钟)

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  Note that the slope is negative. A negative rate of change means that the quantity is decreasing over  time,  in this case the height of the candle is decreasing as it burns down .
  ::注意斜坡是负的。负变化率意味着数量逐渐下降,在此情况下,蜡烛的高度随着燃烧而下降。

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  To find the point when the candle reaches zero length, simply read the $\begin{array}{r}x-\end{array}$ intercept off the graph (100 minutes). U se the rate equation to verify this algebraically:
  ::找到蜡烛长度为零的点,只需读取图中的 X- interblock (100分钟) 。 使用速率方程式来校验这个代数 :

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  $$\begin{array}{rl}\text{Length burned}& =\text{rate}\times \text{time}\\ 10& =0.1\times 100\end{array}$$

  
  ::长度燃烧= ratextime10=0. 1x100

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  Since the candle length was originally 10 inches, this  equation confirms that 100 minutes is the time taken.
  ::由于蜡烛的长度最初为10英寸,这个方程式证实需要100分钟的时间。

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  Determining Rate of Change 
  ::确定变化率

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  The population of fish in a certain lake increased from 370 to 420 over the months of March and April. At what rate is the population increasing?
  ::3月和4月,某个湖泊的鱼类人口从370人增加到420人,人口以何种速度增加?

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  Here , there aren't two obvious points from which to  get $\begin{array}{r}x-\end{array}$ and $\begin{array}{r}y-\end{array}$ coordinates to use in  the slope formula. Instead, use the alternate formula, $\begin{array}{r}m=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\end{array}$ .
  ::这里没有两点可以让 x - 和 y - 坐标用于斜坡公式。 相反, 使用替代公式 m\\ yx 。

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  The change in $\begin{array}{r}y\text{-values}\end{array}$ , or $\begin{array}{r}\mathrm{\Delta }y\end{array}$ , is the change in the number of fish, which is $\begin{array}{r}420-370=50\end{array}$ . The change in $\begin{array}{r}x\text{-values}\end{array}$ , $\begin{array}{r}\mathrm{\Delta }x\end{array}$ , is the amount of time over which this change took place: two months. So $\begin{array}{r}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{50\text{fish}}{2\text{months}}\end{array}$ , or 25 fish per month.
  ::Y值或y值的变化是鱼类数量的变化,即420-370=50。X值的变化是这一变化发生的时间:两个月。因此,y*x=50鱼2个月,或每月25鱼2个月。

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  Interpreting a Graph to Compare Rates of Change
  ::解释变化比较率图表

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  The graph below represents a trip made by a large delivery truck on a particular day. The truck departs from the warehouse, and the distance value represents his distance from his starting location, while time is in hours away from the warehouse. During the day the truck made two deliveries, one taking one hour and the other taking two hours. Identify what is happening in the first 3 stages of the trip (stages A through C).
  ::下图显示一辆大型运货卡车在某一天的旅行。卡车离开仓库,距离值代表他离开起动地点的距离,而时间则在距仓库的几个小时之内。卡车在当天两次交货,一次一小时,另一次两小时。确定前三阶段(A至C阶段)发生的情况。

  

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  The first 3 stages of the trip are:
  ::行程的前三个阶段是:

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  A. The truck sets off and travels 80 miles in 2 hours.
  ::A. 卡车起飞,在2小时内行驶80英里。

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  B. The truck covers no distance for 2 hours.
  ::B. 卡车在2小时之内没有距离。

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  C. The truck covers $\begin{array}{r}(120-80)=40\text{miles}\end{array}$ in 1 hour.
  ::C. 卡车覆盖(120-80)=40英里,每1小时40英里。

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  A :   $\begin{array}{r}\text{Rate of change}=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{80\text{miles}}{2\text{hours}}=40\text{miles per hour}\end{array}$
  ::A:变化率=80英里2小时=40英里/小时

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  In the first stage, the truck travels 80 miles in 2 hours, which means his average speed (the rate of change of distance) was 40 miles per hour.
  ::在第一阶段,卡车在2小时内行驶80英里,这意味着他的平均速度(距离变化速度)是每小时40英里。

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  B : The slope here is 0, so the rate of change is 0 mph. The truck is stationary for one hour. This is the first delivery stop.
  ::B: 这里的斜坡是 0, 所以变化速度是 0 mph。 卡车是一个固定小时。 这是第一个交货站 。

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  C :   $\begin{array}{r}\text{Rate of change}=\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{(120-80)\text{miles}}{(4-3)\text{hours}}=40\text{miles per hour}.\end{array}$ The truck is again traveling at 40 mph.
  ::C:变化速率=120-80英里(4-3)小时=每小时40英里。卡车再次以40英里的速度行驶。

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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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  Continue where we left of in the example about interpreting a graph, by identifying what is happening in the last two stages of the trip (stages D and E).
  ::继续我们在解释图表的例子中留下的位置,确定行程最后两个阶段(D阶段和E阶段)发生的情况。

  

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  The last two stages of the trip are:
  ::行程的最后两个阶段是:

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  D. The truck covers no distance for 1 hour.
  ::D. 卡车在1小时之内没有距离。

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  E. The truck covers -120 miles in 2 hours.
  ::E. 卡车在2小时内覆盖120英里。

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  F ind the slopes:
  ::寻找斜坡 :

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  D. Here the slope is 0, so the rate of change is 0 mph. The truck is stationary for two hours. This is the second delivery stop. At this point the truck is 120 miles from the start position.
  ::D. 这里的斜坡是0,因此变化速度是0英里。卡车固定两个小时,这是第二个交货站。此时卡车离起站位置120英里。

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  E. In the last stage, the truck returns to the warehouse , so distance is graphed as a negative value.
  ::E. 在最后阶段,卡车返回仓库,因此以负值表示距离。

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  $$\begin{array}{rl}\text{Rate of change}& =\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}\\ & =\frac{(0-120)\text{miles}}{(8-6)\text{hours}}\\ & =\frac{-120\text{miles}}{2\text{hours}}\\ & =-60\text{miles per hour.}\end{array}$$

  
  ::变化速率=(0-120)英里(8-6)小时 =(120)英里2小时 =(60)英里/小时

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  The truck is traveling at negative 60 mph.
  ::卡车的车速为负60米

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  Wait – a negative rate of change ? Does that mean that the truck is reversing? Well, probably not. Technically, it's  actually the velocity and not the speed that is negative (velocity is just speed in a specified direction), and a negative velocity simply means that the distance from the starting position is decreasing with time. The truck is driving in the opposite direction – back to where it started from. Since it no longer has 2 heavy loads, it travels faster (60 mph instead of 40 mph), covering the 120 mile return trip in 2 hours. Its speed is 60 mph, and its velocity is -60 mph, because it is traveling in the opposite direction from when it started out.
  ::等待- 负变化率? 这是否意味着卡车正在倒转? 嗯, 也许不是。 从技术上讲,它其实是速度而不是速度是负的( 速度只是指定方向的速度) , 负速度只是意味着从起点的距离随着时间而缩小。 卡车正在向相反的方向行驶 — — 回到从那里出发的。 由于它不再有两重重载,它行驶得更快( 60 mph, 而不是 40mph ) , 覆盖了2小时的120英里返回行程。 它的速度是60mph, 速度是 - 60mph, 它的速度是 - 60mph, 它的速度是 - 60mph, 因为它的行驶方向与它离开时相反。

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  Review 
  ::回顾

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  For 1-6, the graph below is a distance-time graph for Mark’s three and a half mile cycle ride to school. During this ride, he rode on cycle paths but the terrain was hilly. He rode slower up hills and faster down them. He stopped once at a traffic light and at one point he stopped to mend a punctured tire. The graph shows his distance from home at any given time. Identify each section of the graph accordingly.
  ::对于1-6, 下面的图表是Mark的三英里半里乘车上学的距离时间图。 在这次旅程中,他骑着自行车,但地形是丘陵的。 他骑着较慢的山坡,走得更快。 他停在一条交通灯前停过一次,有一次他停下来修补了被刺穿的轮胎。 图表显示了他在任何时候离家的距离。 图表的每个部分都相应标明。

  

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  1. Section A.
     ::A节。
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  2. Section B.
     ::B节。
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  3. Section C.
     ::C节。
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  4. Section D.
     ::D节。
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  5. Section E.
     ::E节。
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  6. Section F.
     ::F节。

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  For 7-12, approximate the slope of each part of Mark's ride.
  ::7 -12, 大约是马克骑车的每一段的斜坡。

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  7. Section A.
     ::A节。
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  8. Section B.
     ::B节。
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  9. Section C.
     ::C节。
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  10. Section D.
      ::D节。
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  11. Section E.
      ::E节。
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  12. Section F.
      ::F节。

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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 。