后勤职能

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increases without bound.  This is reasonable for some situations; however, for populations there is usually some type of upper bound.  This can be caused by limitations on food, space or other scarce resources.  The effect of this limiting upper bound is a curve that grows exponentially at first and then slows down and hardly grows at all.  This type of growth is called logistic growth . What are some other situations in which logistic growth would be an appropriate model?
::对于某些情况,这是合理的;但是,对于人口来说,通常有某种上限,这可能由于对粮食、空间或其他稀缺资源的限制而造成,这种限制上限的效果是曲线起先成倍增长,然后放慢速度,几乎没有增长。这种增长被称为后勤增长。在哪些其他情况下,后勤增长将是一个适当的模式?

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Logistic Functions
::后勤职能

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Logistic growth can be described with a logistic equation . The logistic equation is of the form:  $\begin{align*}f(x)=\frac{c}{1+a \cdot b^x}\end{align*}$
::后勤增长可以用后勤等式描述,后勤等式为:f(x)=c1+abx。

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The letters $\begin{align*}a, b\end{align*}$  and $\begin{align*}c\end{align*}$  are constants that can be changed to match the situation being modeled.  You will have to solve for $\begin{align*}a\end{align*}$  and  $\begin{align*}b\end{align*}$  with the information that is given to you in each problem. The constant $\begin{align*}c\end{align*}$  is particularly important because it is the limit to growth.  This is also known as the carrying capacity .  
::a、b和c 字母是常数,可以更改以与正在模拟的情况相匹配。您需要用在每一个问题中给您的信息来解决a和b。 c 常数特别重要,因为它是增长的极限。 这也被称为承载能力。

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The following logistic function has a carrying capacity of 2 which can be directly observed from its graph. 
::以下后勤职能的承载能力为2人,可从其图表中直接看出。

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$\begin{align*}f(x)=\frac{2}{1+0.1^x}\end{align*}$
:xx)=21+0.1x

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An important note about the logistic function is that it has an inflection point .  From the previous graph you can observe that at the point (0, 1) the graph transitions from curving up (concave up) to curving down (concave down).  This change in curvature will be studied more in calculus, but for now it is important to know that the inflection point occurs halfway between the carrying capacity and the  $\begin{align*}x\end{align*}$ axis.
::后勤功能的一个重要说明是它有一个交叉点。 从上一个图表中,你可以观察到,在(0, 1)点,图形从弯曲(弯曲)向弯曲(弯曲)向弯曲(弯曲)向弯曲(弯曲)向弯曲(弯曲)向弯曲(弯曲)的过渡。这种曲曲的改变将更多地用微积分来研究,但现在重要的是要知道,这个交叉点是在承载能力和X轴之间的中间。

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Examples
::实例

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Example 1
::例1

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Earlier, you were asked what situations the logistic model is appropriate for. The logistic model is appropriate whenever the total count has an upper limit and the initial growth is exponential.  Examples are the spread of rumors and disease in a limited population and the growth of bacteria or human population when resources are limited. 
::早些时候,有人问到后勤模式适合什么情况。当总计算有上限且初始增长指数指数增长时,后勤模式是合适的。例如谣言和疾病在有限的人口中蔓延,以及在资源有限时细菌或人类人口增长。

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Example 2
::例2

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A rumor is spreading at a school that has a total student population of 1200.  Four people know the rumor when it starts and three days later three hundred people know the rumor.  About how many people at the school know the rumor by the fourth day? 
::4人知道谣言开始的时候,3天后,300人知道谣言。 学校中有多少人到第四天才知道谣言?

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In a limited population, the count of people who know a rumor is an example of a situation that can be modeled using the logistic function.  The population is 1200 so this will be the carrying capacity. 
::在有限的人口中,了解谣言的人数是使用后勤功能进行模拟的一个实例。人口为1200人,这就是承载能力。

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Identifying information:  $\begin{align*}c=1200; (0, 4); (3,300)\end{align*}$ .  First, use the point (0, 4) to solve for $\begin{align*}a\end{align*}$ .
::识别信息: c=1200; (0) 4; (3 300). 首先, 使用点 (0, 4) 来解答 a 。

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$$\begin{align*}\frac{1200}{1+a \cdot b^0} &= 4\\ \frac{1200}{1+a} &=4\\ \frac{1200}{4} &=1+a\\ a &= 299\end{align*}$$
::12001+ab0=412001+a=412004=1+aa=299

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Next, use the point (3, 300) to solve for $\begin{align*}b\end{align*}$ .
::接下来,使用点( 3, 300) 解答 b 。

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$$\begin{align*}\frac{1200}{1+299 \cdot b^3} &= 300\\ 4 &= 1+299 \ b^3\\ \frac{3}{299} &= b^3\\ 0.21568 & \approx b\end{align*}$$
::12001+2999b3=3004=1+299 b33299=b30.21568b

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The modeling equation at $\begin{align*}x=4\end{align*}$ :
::x=4 的建模方程式:

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$\begin{align*}f(x)=\frac{1200}{1+299 \cdot 0.21568^x} \rightarrow f(4) \approx 729 \ people\end{align*}$
::f(x) =2001+299_0.21568x_f(4)____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

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A similar growth pattern will exist with any kind of infectious disease that spreads quickly and can only infect a person or animal once.
::任何传染性疾病迅速蔓延,只能一次感染一个人或动物,这种疾病也会出现类似的增长模式。

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Example 3
::例3

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A special kind of algae is grown in giant clear plastic tanks and can be harvested to make biofuel.  The algae are given plenty of food, water and sunlight to grow rapidly and the only limiting resource is space in the tank.  The algae are harvested when 95% of the tank is full leaving the tank 5% full of algae to reproduce and refill the tank.  Currently the time between harvests is twenty days and the payoff is 90% harvest.  Would you recommend a more optimal harvest schedule? 
::一种特殊的藻类在巨大的清晰塑料罐中生长,可以收获以生产生物燃料。藻类有充足的食物、水和阳光,可以迅速生长,唯一的限制资源是罐中的空间。当95%的罐体满载5%的藻类时,藻类就会收获,以再生和填充罐体。目前,收获期为20天,收益期为90%。您是否推荐一个更理想的收成时间表 ?

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Identify known quantities and model the growth of the algae. 
::查明已知数量并模拟藻类的生长。

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Known quantities: $\begin{align*}(0, 0.05); (20, 0.95); c=1 \ or \ 100\%\end{align*}$
::已知数量: (0,0.005); (20,0.95); c=1或100%

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$$\begin{align*}0.05 &= \frac{1}{1+a \cdot b^0}\\ 1+a &= \frac{1}{0.05}\\ a &= 19\\ 0.95 &= \frac{1}{1+19 \cdot b^{20}}\\ 1+19 \cdot b^{20} &= \frac{1}{0.95}\\ b^{20} &= \frac{\left(\frac{1}{0.95}-1\right)}{19}\\ b & \approx 0.74495\end{align*}$$
::0.05=11+ab01+a=10.05a=190.95=11+19_b201+19_b20=10.95b20=(10.95-1)19b=0.74495

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The model for the algae growth is:
::藻类生长的模式是:

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$\begin{align*}f(x)=\frac{1}{1+19 \cdot (0.74495)^x}\end{align*}$
::f(x) = 11+19( 0. 74495) x

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The question asks about optimal harvest schedule.  Currently the harvest is 90% per 20 day or a unit rate of 4.5% per day.  If you shorten the time between harvests where the algae are growing the most efficiently, then potentially this unit rate might be higher.  Suppose you leave 15% of the algae in the tank and harvest when it reaches 85%.  How much time will that take to yield 70%?
::问题询问了最佳收成时间表。 目前,收成为每20天90%或每天4.5 % 。 如果您缩短了藻类生长效率最高的收获间隔时间, 那么这个单位速度可能会更高 。 假设你把15%的藻类留在水槽里, 当它达到85% , 收获时间会有多长 。 需要多少时间才能收成70% ?

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$$\begin{align*}0.15 &=\frac{1}{1+19 \cdot (0.74495)^x}\\ x_1 & \approx 4.10897\\ 0.85 &= \frac{1}{1+19 \cdot (0.74495)^x}\\ x_2 & \approx 15.8914\end{align*}$$
::0.15=11+19(0.744995xxx14.108970.85=11+19(0.744995xx215.8914)

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$$\begin{align*}{x}_{2}-{x}_{1}\approx 15.8914-4.10897\approx 11.78\end{align*}$$
::15.8914-4.10897-11.78

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It takes about 12 days for the batches to yield 70% harvest which is a unit rate of about 6% per day.  This is a significant increase in efficiency.  A harvest schedule that maximizes the time where the logistic curve is steepest creates the fastest overall algae growth.
::批量需要大约12天才能收获70%的收成,即每天大约6%的单位收成率。这是效率的显著提高。 丰收时间表将后勤曲线最陡峭的时间最大化,从而创造出最快的全面藻类生长速度。

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Example 4
::例4

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Determine the logistic model given $\begin{align*}c=12\end{align*}$  and the points (0, 9) and (1, 11).
::确定给定的后勤模式C=12以及点(0,9)和点(1,11)。

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The two points give two equations, and the logistic model has two variables. Use these points to solve for  $\begin{align*}a\end{align*}$  and  $\begin{align*}b\end{align*}$ .
::两点给出了两个方程式, 后勤模式有两个变量。 用这些点来解答 a 和 b 。

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$$\begin{align*}9 &= \frac{12}{1+a \cdot b^0}\\ 1+a &= \frac{12}{9}\\ a &= \frac{1}{3}\\ 11 &= \frac{12}{1+\left(\frac{1}{3}\right) \cdot b^1}\\ 1+ \left(\frac{1}{3}\right) \cdot b &= \frac{12}{11}\\ b &= 0.\overline{27} = \frac{3}{11}\end{align*}$$
::9=121+ab01+a=129a=1311=121+(13)\b11+(13)\b11+(13)\b=1211b=0。 27=311

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Thus the approximate model is:
::因此,近似模式是:

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$\begin{align*}f(x)=\frac{12}{1+\left(\frac{1}{3}\right) \cdot (\frac{3}{11})^x}\end{align*}$
:xx)=121+(13)__(311)x

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Example 5
::例5

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Determine the logistic model given $\begin{align*}c=7\end{align*}$  and the points (0, 2) and (3, 5).
::确定给定的C=7的后勤模式以及点数(0,2)和点数(3,5)。

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The two points give two equations, and the logistic model has two variables. Use these two points to solve for  $\begin{align*}a\end{align*}$  and  $\begin{align*}b\end{align*}$ .
::这两点给出了两个方程式, 后勤模式有两个变量。 用这两个点解决 a 和 b 。

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$$\begin{align*}2 &= \frac{7}{1+a}\\ 1+a &= \frac{7}{2}\\ a &= 2.5\\ 5 &= \frac{7}{1+(2.5) \cdot b^3}\\ 1+(2.5) \cdot b^3 &= \frac {7}{5}\\ b^3& = 0.16\\ b &\approx 0.5429\end{align*}$$
::2=71+a1+a=72a=2.55=71+(2.5)_b31+(2.5)_b3=75b3=0.16b=0.16b=0.5429

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Thus the approximate model is:
::因此,近似模式是:

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$\begin{align*}f(x)=\frac{7}{1+(2.5) \cdot (0.5429)^x}\end{align*}$
:x) = 71+(2.5) = (0.5429)x

Summary

播放段落 Logistic growth is a type of growth that starts exponentially but slows down and hardly grows at all due to limiting factors such as scarce resources. ::物流增长是一种快速增长,但因资源稀缺等限制因素,增长速度放慢,几乎没有增长。 播放段落 Logistic equation has the form: $\begin{align*}f(x) = \frac{c}{1 + a \cdot b^{x}}\end{align*}$   播放段落 The constant $\begin{align*}c\end{align*}$  represents the carrying capacity or limit to growth. ::常数c 代表承载能力或增长限制。 ::后勤方程式的窗体为: f(x)=c1+abx 播放段落 The logistic function has an inflection point, where the graph transitions from curving up (concave up) to curving down (concave down). ::后勤功能有一个透视点,图形从弯曲向上(弯曲向上)向弯曲(弯曲向下)转变。 播放段落 The inflection point occurs halfway between the carrying capacity and the axis. ::穿孔点在承载能力与轴之间的中间点。

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Review
::回顾

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For 1-5, determine the logistic model given the carrying capacity and two points.
::1-5,根据承载能力和两点确定后勤模式。

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1. $\begin{align*}c=12; (0, 5); (1, 7)\end{align*}$
::1. c=12; (0, 5); (1, 7)

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2.  $\begin{align*}c=200; (0, 150); (5, 180)\end{align*}$
::2. c=200; (0,150); (5,180)

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3.  $\begin{align*}c=1500; (0, 150); (10, 1000)\end{align*}$
::3. c=1500; (0,150); (10,1000)

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4.  $\begin{align*}c=1000000; (0, 100000); (-40, 20000)\end{align*}$
::4. c=100000; (0 100 000); (- 40 20000)

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5.  $\begin{align*}c= 30000000; (-60, 10000); (0, 8000000)\end{align*}$
::5. c=30000; (- 60, 100,00); (0,800,000)

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For 6-8, use the logistic function $\begin{align*}f(x)=\frac{32}{1+3e^{-x}}\end{align*}$ .
::6-8使用后勤功能f(x)=321+3e-x。

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6. What is the carrying capacity of the function?
::6. 该职能的承载能力是什么?

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7. What is the $\begin{align*}y\end{align*}$ -intercept of the function?
::7. 函数的Y 界面是什么?

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8. Use your answers to 6 and 7 along with at least two points on the graph to make a sketch of the function.
::8. 使用您对 6 和 7 的回答以及图上至少两点来绘制函数的草图。

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For 9-11, use the logistic function $\begin{align*}g(x)=\frac{25}{1+4 \cdot 0.2^x}\end{align*}$ .
::9-11时,使用后勤功能g(x)=251+40.2x。

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9. What is the carrying capacity of the function?
::9. 该职能的承载能力是什么?

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10. What is the $\begin{align*}y\end{align*}$ -intercept of the function?
::10. 函数的Y 界面是什么?

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11. Use your answers to 9 and 10 along with at least two points on the graph to make a sketch of the function.
::11. 利用对9和10的答复以及图上至少两点来绘制函数的草图。

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For 12-14, use the logistic function $\begin{align*}h(x)=\frac{4}{1+2 \cdot 0.68^x}\end{align*}$ .
::对于12-14,使用后勤功能h(x)=41+20.68x。

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12. What is the carrying capacity of the function?
::12. 该职能的承载能力是什么?

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13. What is the $\begin{align*}y\end{align*}$ -intercept of the function?
::13. 函数的Y 界面是什么?

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14. Use your answers to 12 and 13 along with at least two points on the graph to make a sketch of the function.
::14. 使用您对12和13的答复以及图上至少两点来绘制函数的草图。

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15. Give an example of a logistic function that is decreasing (models decay). In general, how can you tell from the equation if the logistic function is increasing or decreasing?
::15. 请举一个正在减少的后勤功能(模式衰减)的例子,一般而言,如何从方程式中看出后勤功能在增加还是减少?

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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 。