Section: 与指数函数和对数函数建模 | 5.8 与指数函数和对数函数建模-interactive

Section outline

Lesson Objectives

Write logarithmic and functions to model real-world problems.
::写入对数和函数以模拟现实世界的问题。

Find the asymptote and intercept of a logarithmic function .
::查找对数函数的无时点和截取功能。

Introduction: The Tournament
::导言:锦标赛

Basketball Tournament

Every spring, the best teams in college basketball compete in a tournament to crown a champion. Half the teams are eliminated each round until there is only one team left at the end of six rounds. Can you write a function to represent the number of teams left in the tournament based on the round?
::每年春天,大学篮球最佳球队都参加锦标赛,以赢得冠军。每轮淘汰一半球队,直到六轮比赛结束时只剩下一个球队。你能写一个函数来代表比赛中剩下的球队数量吗?

Activity 1: The Tournament Continued
::活动1:继续比赛

T wo pieces of information are needed to write the exponential equation $\begin{aligned} y = a b^{x} : \end{aligned}$ the starting amount, $\begin{aligned} a, \end{aligned}$ and the growth factor , $\begin{aligned} b . \end{aligned}$ In the scenario presented in the introduction , the growth factor is $\begin{aligned} \frac{1}{2}, \end{aligned}$ but the starting amount is not given. Since you know that after 6 rounds, there is 1 team left, this means that (6,1) is a point on the function. S ubstitute this into the equation to solve for $\begin{aligned} a . \end{aligned}$
::写成指数方程式 y=abx 需要两块信息: 起始数, a 和增长系数, b。在导言中介绍的假设情景中, 增长系数为 12 , 但没有给出起始数。既然你知道在六轮后, 只剩下一个团队, 这意味着函数上的一个点( 6, 1) 。将其替换为方程式, 以解答一个。

However, what if you didn't have $\begin{aligned} a or b ? \end{aligned}$
::但是,如果你没有a或b呢?

Example
::示例示例示例示例

Lee's parents opened a savings account with $10,000 for college when he was 10. Every year the account earns interest, which is compounded annually. Lee is now 13, and there is $12,597.12 in the account. How much money will be in the account when Lee is 18?
::李的父母在他10岁时为大学开了一个10 000美元的储蓄账户,每年该账户都赚取利息,每年的利息会变本加厉。李现在13岁,账户中有12 597.12美元。李18岁时,该账户里会有多少钱?

To solve this problem, you need to find the growth factor since you have the starting amount. In year 0 there was $10,000 in the account, and in year 3 there was $12,597.12 in the account resulting in the points $\begin{aligned} (0, 10000) \end{aligned}$ and $\begin{aligned} (3, 12597.12) . \end{aligned}$ From these points, you can see that $\begin{aligned} a = 10, 000 \end{aligned}$ because the starting amount, when $\begin{aligned} x \end{aligned}$ is 0, is 10,000. U se the point $\begin{aligned} (3, 12597.12) \end{aligned}$ to find the growth factor by substituting it into the standard exponential equation for $\begin{aligned} x \end{aligned}$ and $\begin{aligned} y \end{aligned}$ respectively.
::要解决这个问题, 您需要找到增长系数, 因为您有起始金额。在0年, 账户中有10,000美元, 在3年, 账户中有12, 597. 12美元, 导致点数( 1100美元) 和 3, 12597. 12 美元。从这些点上, 您可以看到 a= 10,000, 因为起始金额, 当 x为 0 时, 是 10,000 。使用点 ( 3, 125.97. 12) 来找到增长系数, 将其分别替换成 x 和 y 的标准指数方程式。

$\begin{aligned} 12597.12 = 10, 000 \cdot b^{3} \end{aligned}$

::12597.12=10 000b3

Our first step in solving for $\begin{aligned} b \end{aligned}$ is to divide both sides by 10,000.
::我们解决b的第一步是将双方除以1万

$\begin{aligned} 12597.12 & = 10000 \cdot b^{3} \\ \div 10000 & \div 10000 \\ 1.259712 & = b^{3} \end{aligned}$

::12597.12=10000b31000010000=1000001.259712=b3

Next, find the cube root of both sides.
::接下来,找到两边的立方根

$\begin{aligned} \sqrt[3]{1.259712} & = \sqrt[3]{b^{3}} \\ 1.08 = b \end{aligned}$

::1.2597123=b331.08=b

Since the growth factor is 1.08, the interest rate in 8% and you also know the equation in standard form :
::由于增长系数为1.08,利率为8%,您还知道标准方程式的方程式:

$\begin{aligned} y = 10, 000 \cdot {1.08}^{x} \end{aligned}$

::y=10 000 = 1.08x

We can use this to find the value of the account when Lee is 18, which is 8 years after the account was opened.
::我们可以用这个来找到李十八岁时的账户价值, 也就是开户8年后的李十八岁。

$\begin{aligned} y = 10, 000 \cdot {1.08}^{8} \\ y \approx 18, 509.30 \end{aligned}$

::y=10 000 1.088y 18 509.30

Answer: Lee will have approximately $18,509.30 in the account when he turns 18.
::回答:Lee18岁时,账户里大约有18 509.30美元。

Activity 2: Newton's Law of Cooling
::活动2:牛顿冷却法

In the early 1700s, Isaac Newton discovered that the rate of change of the temperature of an object is proportional to the difference in the temperatures between the object and its surroundings. Newton used the formula below to express this relationship.
::1700年代初期,Isaac Newton发现一个物体的温度变化速度与该物体与其周围环境之间的温度差异成正比。牛顿使用下面的公式来表达这种关系。

$\begin{aligned} T_{t} = T_{env} + {(T_{0} - T_{env}) e}^{- k t} \end{aligned}$

::Tt=Tenv+(T0-Tenv)e-kt

$\begin{aligned} T_{t} \end{aligned}$ is the temperature after time $\begin{aligned} t \end{aligned}$
::Tt 是时间后的温度 t

$\begin{aligned} T_{e n v} \end{aligned}$ is the temperature of the environment
::Tenv 是环境的温度

$\begin{aligned} T_{0} \end{aligned}$ is the initial temperature of the object
::T0 是对象初始温度

$\begin{aligned} t \end{aligned}$ is the time
::t 是时间

$\begin{aligned} k \end{aligned}$ is a constant determined by the material the object is made of
::k 是一个常数,由物体的制成物质决定。

This formula is regularly used by forensic scientists to estimate the time a murder was committed.
::法医科学家经常使用这一公式来估计谋杀的发生时间。

Example
::示例示例示例示例

Rearranging this formula for the time will tell you how long it will take for an object to reach a certain temperature. Some items of food need to be refrigerated to prevent spoiling. Solving for time can be used to produce the amount of time it will take for an item to spoil. Solve Newton's law of cooling for time.
::重新排列此公式的时间将告诉您一个对象要达到一定温度需要多长时间。有些食物需要冷冻以防止变质。解冻时间可以用来生成变质时间。解冻时间可以用来生成变质时间。解冻时间是牛顿的冷却时间规则。

B egin by subtracting $\begin{aligned} T_{e n v} \end{aligned}$ on both sides of the equation.
::开始在方程两侧减去 Tenv 。

$\begin{aligned} T_{t} & = T_{e n v} + {(T_{0} - T_{e n v}) e}^{- k t} \\ - T_{e n v} & - T_{e n v} \\ T_{t} - T_{e n v} & = {(T_{0} - T_{e n v}) e}^{- k t} \end{aligned}$

::Tt=Tenv+(T0-Tenv)e-kt-Tenv-Tenv-TenvTT-Tenv=(T0-Tenv)e-kt

Next, divide both sides of the equations by $\begin{aligned} (T_{0} - T_{e n v}) . \end{aligned}$
::接下来,将方程的两边除以( T0- Tenv ) 。

$\begin{aligned} T_{t} - T_{e n v} & = {(T_{0} - T_{e n v}) e}^{- k t} \\ \div (T_{0} - T_{e n v}) & \div (T_{0} - T_{e n v}) \\ \frac{T_{t} - T_{e n v}}{T_{0} - T_{e n v}} & = e^{- k t} \end{aligned}$

::T- Tenv= (T0- Tenv) e- kt (T0- Tenv) (T0- Tenv) Tt- TenvT0- Tenv= e- kt

Take the natural log of both sides to cancel the $\begin{aligned} e . \end{aligned}$
::使用双方的自然日志取消e。

$\begin{aligned} \ln (\frac{T_{t} - T_{e n v}}{T_{0} - T_{e n v}}) & = \ln e^{- k t} \\ \ln (\frac{T_{t} - T_{e n v}}{T_{0} - T_{e n v}}) & = - k t \end{aligned}$

::内 (Tt- TenvT0- Tenv) = lne- ktln (Tt- TenvT0- Tenv) kt

Finally, divide both sides by $\begin{aligned} - k . \end{aligned}$
::最后,将双方一分为二。

$\begin{aligned} \ln (\frac{T_{t} - T_{e n v}}{T_{0} - T_{e n v}}) & = - k t \\ \div - k & \div - k \\ \ln (\frac{T_{t} - T_{e n v}}{T_{0} - T_{e n v}}) \div - k & = t \end{aligned}$

:Tt-TenvT0-Tenv) @ktkkkln(Tt-TenvT0-Tenv) @k=t

Answer: $\begin{aligned} t = \ln (\frac{T_{t} - T_{e n v}}{T_{0} - T_{e n v}}) \div - k \end{aligned}$
::答复:t=ln(Tt-TenvT0-Tenv)k

Answer the questions below to practice rearranging formulas using logarithms and natural logarithms.
::回答下列问题:使用对数和自然对数来实施重新排列公式。

Discussion Question : How can a logarithmic function be created from an exponential function if you only have either the starting amount or the growth factor and a point?
::讨论问题:如果只有起始数或增长系数和点,如何从指数函数中创建对数函数?

Debate this among others in your class, or online in the CK-12 Math Cafe'. What are the differences in process between your idea(s) and those of others? Why do you think yours i s right?
::在你们班级或CK-12数学咖啡厅网上辩论这个问题。你们的想法和别人的想法之间有什么不同?你们为什么认为你们的想法是对的?

Activity 3: Population Growth
::活动3:人口增长

Example
::示例示例示例示例

A town had a population of 12,500 in 1970, and a population of 26,900 was reported in the 2015 census. Write an exponential function to estimate population as a function of time, in years, since 1970 based on the model below:
::1970年,一个城镇人口为12,500人,2015年人口普查报告人口为26,900人。

$\begin{aligned} P = P_{0} \cdot e^{r t} \end{aligned}$

::P=P0ERT = P0

$\begin{aligned} P \end{aligned}$   is the population
::P 是人口

$\begin{aligned} P_{0} \end{aligned}$   is the initial population
::P0是初始人口

$\begin{aligned} e \end{aligned}$   is the exponential constant 2.718...
::e 是指数常数2.718...

$\begin{aligned} r \end{aligned}$   is the growth rate
::r 是增长率

$\begin{aligned} t \end{aligned}$   is time in years
::t 是年中的时间

Since 1970 is the starting year, you know that $\begin{aligned} P_{0} = 12, 500. \end{aligned}$ Additionally, the year 2015 is 45 years after 1970, so the point $\begin{aligned} (45, 26900) \end{aligned}$ lies on the graph of the function. Plug all of this information into the equation above to find the growth rate.
::1970年是起始年,你知道P0=12,500。此外,2015年是1970年后的45年,因此这个点(45,26900)就在函数图上。将所有这些信息都插入以上方程以找到增长率。

$\begin{aligned} 26, 900 = 12, 500 \cdot e^{r \times 45} \end{aligned}$

::26,900=12,500ER×45

Our first step in solving for the growth rate, r, is to divide the 12,500 to the other side of the equation.
::我们解决增长率问题的第一步,r,是将12,500分到等式的另一边。

$\begin{aligned} 26, 900 & = 12, 500 \cdot e^{r \times 45} \\ \div 12, 500 & \div 12, 500 \\ 2.152 & = e^{r \times 45} \end{aligned}$

::26 900 = 12,500___________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________

The next step is to cancel the exponent . Since the base is $\begin{aligned} e, \end{aligned}$ use $\begin{aligned} \log_{e}, \end{aligned}$ which is also referred to as the natural log $\begin{aligned} \ln . \end{aligned}$
::下一步是取消指数。由于基数是 e, 请使用loge, 也就是自然对数In。

$\begin{aligned} \ln (2.152) = \ln (e^{r \times 45}) \end{aligned}$

::In ( 2.152) =ln ( er×45)

The natural log will cancel the base $\begin{aligned} e, \end{aligned}$ resulting in the following:
::自然日志将取消基数e, 其结果如下:

$\begin{aligned} \ln (2.152) = r \times 45 \end{aligned}$

::IN( 2.152) =rx45

Although you can use a calculator to estimate the value of $\begin{aligned} \ln 2.152 \approx 0.7664, \end{aligned}$ wait until the end to keep the answer as accurate as possible. Approximating before you finish simplifying might thro w off the accuracy of your calculations.
::尽管您可以使用计算器来估算 In2.1520.7664 的值, 但要等到最后才能尽可能准确地回答问题。完成简化前的接近可能会降低计算准确性。

$\begin{aligned} \ln (2.152) & = r \times 45 \\ \div 45 & \div 45 \\ \frac{\ln (2.152)}{45} & = r \\ 0.017 \approx r \end{aligned}$

::在( 2.152) =rx45\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\

The population grows at a rate of approximately 1.7% a year.
::人口的年增长率约为1.7%。

Answer: $\begin{aligned} P (t) = 12, 500 \cdot e^{0.017 t} \end{aligned}$
::答复:P(t)=12 500e0.017t

Discussion Question : Develop a general strategy for increasing or decreasing the starting year while keeping the starting population equal to 12,500? How would other transformations affect the graph in context?
::讨论问题:制定一项总战略,增加或减少起始年数,同时将起始人口保持在12 500人的水平上?其他变化如何影响图表的上下文?

Activity 4: Oil Spill
::活动4:漏油

Fitting to data is a useful tool when modeling data. Use the models to make predictions.
::适合数据是模拟数据时有用的工具。使用模型进行预测。

Example
::示例示例示例示例

Scientists are gathering data about an oil spill in the ocean. They have taken measurements of the square mileage of the spill at various times and want to create a model that can be used to predict how much damage will be caused to the environment as a result of the spill.
::科学家们正在收集海洋石油泄漏的数据,他们在不同的时间对泄漏平方英里进行了测量,并想建立一个模型,用来预测溢漏将对环境造成多大损害。

Use the interactive below to create an exponential function model for the data and use it to predict the square mile s covered by the spill in one year.
::使用下面的交互式数据来创建数据指数函数模型,并用它来预测一年后溢出所覆盖的平方英里。

INTERACTIVE

Exponential Functions from Real World Data

Try to guess the function that fits the data points.
::尝试猜测符合数据点的函数。

You have 5 attempts before you are given the next question.
::在回答下一个问题之前,你有5次尝试。

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Please check your internet connection and try again.

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Yes

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Discussion Question : How accurate was your model? How do you account for any error? Do you think it affects the results whether or not you assumed the growth was continuous ?
::讨论问题:模型的准确性如何?你如何解释错误?你是否认为它会影响结果,而不论你是否认为增长是连续的?

Wrap-Up: Review Questions
::总结:审查问题

Summary
::摘要

An exponential function has the form $\begin{aligned} f (x) = a b^{x}, \end{aligned}$ with $\begin{aligned} a \end{aligned}$ being the starting amount and $\begin{aligned} b \end{aligned}$ the growth factor.
::指数函数有 f(x) =abx 的表单,以起始数为单位,b为生长系数。

$\begin{aligned} P = P_{0} e^{r t} \end{aligned}$ can be used to model population growth where $\begin{aligned} P_{0} \end{aligned}$ is the initial population, $\begin{aligned} r \end{aligned}$ is the growth rate, and $\begin{aligned} t \end{aligned}$ is the time.
::P=P0ert可用于模拟人口增长,其中P0为初始人口,r为增长率,t为时间。

$\begin{aligned} l o g_{b} (x^{a}) = a \log_{b} x \end{aligned}$
::logb(xa) = alogb =x

$\begin{aligned} \ln x^{n} = n \ln x \end{aligned}$
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}伊尼诺 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}伊尼诺 {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}伊尼诺

Section outline

Introduction: The Tournament ::导言:锦标赛

Activity 1: The Tournament Continued ::活动1:继续比赛

Activity 2: Newton's Law of Cooling ::活动2:牛顿冷却法

Activity 3: Population Growth ::活动3:人口增长

Activity 4: Oil Spill ::活动4:漏油

Wrap-Up: Review Questions ::总结:审查问题

Summary ::摘要

Introduction: The Tournament
::导言:锦标赛

Activity 1: The Tournament Continued
::活动1:继续比赛

Activity 2: Newton's Law of Cooling
::活动2:牛顿冷却法

Activity 3: Population Growth
::活动3:人口增长

Activity 4: Oil Spill
::活动4:漏油

Wrap-Up: Review Questions
::总结:审查问题

Summary
::摘要