Section: 指数增长和衰减 | 7.5 指数增长和衰减

Section outline

When the rate of change of the amount of a substance, or a population, is proportional to the amount present at any time, we say that this substance or population is going through either a decay or a growth, depending on the sign of the constant of proportionality. Do you know how to write a differential equation that expresses this condition? This kind of growth or decay, common in nature and in the business world, is called exponential growth or and is characterized by rapid change.
::当物质或人口的数量变化速度与任何时间的当前数量成正比时,我们说这种物质或人口正在经历衰减或增长,这取决于相称性不变的标志。你知道如何写出表达这一条件的差别方程式吗?这种在性质和商业界中常见的成长或衰变被称为指数增长,或以快速变化为特征。

If the solution to the differential equation that governs the quantity of your favorite things is $\begin{aligned} y = 5 \cdot {0.3}^{t} \end{aligned}$ , is this a growth or decay solution? Can you write the differential equation for which $\begin{aligned} y = 5 \cdot {0.3}^{t} \end{aligned}$ is the solution?
::如果制约您最喜爱事物数量的差异方程式的解决方案是 y= 50. 3t, 这是增长或衰变的解决方案吗? 您能写出以y= 50. 3t为解决方案的差别方程式吗 ?

Exponential Growth and Decay
::指数增长和衰减

In this section, we will look at a formulation of and decay, and some applications that are modeled by exponential growth or decay.
::在本节中,我们将研究一种成形和衰变,以及一些以指数增长或衰变为模型的应用。

The concept of exponential growth or decay arises as the solution to the problem that the rate of change of a quantity, $\begin{aligned} y (t) \end{aligned}$ , with respect to time, $\begin{aligned} t \end{aligned}$ , varies directly as the quantity. The mathematical formulation of this differential equation and general solution can be summarized as follows:
::指数增长或衰减的概念是用来解决一个问题,即一个数量(y(t))相对于时间(t)的变化速度随着数量而直接变化。

Given the differential equation: $\begin{aligned} \frac{d y}{d t} = k y \end{aligned}$ , where $\begin{aligned} k \end{aligned}$ is a constant.
::考虑到差别方程: dydt=ky, K是常数。

Then, $\begin{aligned} y = C e^{k t} \end{aligned}$ is a solution to the differential equation with $\begin{aligned} y = C \end{aligned}$ at $\begin{aligned} t = 0 \end{aligned}$ .
::然后,y=Cekt是y=C t=0的差别方程的解决方案。

If $\begin{aligned} k > 0 \end{aligned}$ : The function $\begin{aligned} y \end{aligned}$ represents exponential growth (increasing values).
::如果 k>0 : 函数 y 表示指数增长( 增加值) 。

If $\begin{aligned} k < 0 \end{aligned}$ : The function $\begin{aligned} y \end{aligned}$ represents exponential decay (decreasing values).
::如果 k< 0 : 函数 y 表示指数衰减( 衰减值) 。

The above formulation for $\begin{aligned} y \end{aligned}$ is expressed with a base $\begin{aligned} b = e \end{aligned}$ . The general formulation would be $\begin{aligned} y = C b^{\frac{k t}{\ln b}} \end{aligned}$ , where $\begin{aligned} b \end{aligned}$ is any base such that $\begin{aligned} b > 0 \end{aligned}$ and $\begin{aligned} b \neq 1 \end{aligned}$ .

::y的上述制剂用 b=e 基表示。一般制剂为 y= Cbktlnb, b 是 b>0 和 b=1 的基数。

The statement above comes from the solution to the differential equation:
::以上陈述来自差别等式的解决办法:

$\begin{aligned} \frac{d y}{d t} = k y \end{aligned}$
::迪迪特=基

Separating variables,
::分离变量,

$\begin{aligned} \frac{d y}{y} = k d t \end{aligned}$
::dyy=kdt dyy=kdt dyy=kdt dyy=kdt dyy=kdt dyy=kdt dyy=kdt dyy=kdt dyy=kdt

and integrating both sides,
::并使双方一体化,

$\begin{aligned} \int \frac{d y}{y} = \int k d t \end{aligned}$ ,
::迪伊・卡达特,

gives us
::给了我们

$\begin{aligned} \ln y & = k t + C \\ y & = e^{k t + C} \\ = e^{k t} e^{C} \\ = C e^{k t} \dots The general constant C is used as a replacement for e^{C} . \end{aligned}$

::Iny=kt+Cy=ekt+C=ekteC=Cekt * 通用常数 C 用于替换 eC 。

Applications of Exponential Growth and Decay: Radioactive Decay
::指数增长和衰变的应用:放射性衰变

In physics, radioactive decay is a process in which an unstable atomic nucleus loses energy by emitting radiation in the form of electromagnetic radiation (like gamma rays) or particles (such as beta and alpha particles). During this process, the nucleus will continue to decay, in a chain of decays, until a new stable nucleus is reached (called an isotope). Physicists measure the rate of decay by the time it takes a sample to lose half of its nuclei due to radioactive decay. Initially, as the nuclei begins to decay, the rate starts very fast and furious, but it slows down over time as more and more of the available nuclei have decayed. The figure below shows a typical radioactive decay of a nucleus. As you can see, the graph has the shape of an exponential function with $\begin{aligned} k < 0 \end{aligned}$ .
::在物理学中,放射性衰变是一个过程,不稳定原子核通过以电磁辐射(如伽马射线)或粒子(如乙型和甲型粒子)的形式发射辐射而失去能量。在这一过程中,核在衰变链中将继续衰变,直到达到一个新的稳定核(称为同位素)为止。物理学家在样品取而代之以因放射性衰变而失去半核时测量了衰变速度。最初,当核开始衰变时,该速度开始非常快,而且非常暴躁,但随着现有核的逐渐衰变,该速度会放慢。下图显示核的典型放射性衰变。如你所见,该图的形状是以 k < 0 为单位的指数函数。

The equation that is used to model radioactive decay is $\begin{aligned} y = C e^{k t} \end{aligned}$ . To find an expression for the half-life of an isotope, use the definition of half-life as the time it takes for a sample to lose half of its nuclei. If there is an initial mass $\begin{aligned} y = C \end{aligned}$ (measured in grams) at $\begin{aligned} t = 0 \end{aligned}$ , then at some later time $\begin{aligned} t = t_{\frac{1}{2}}, y \end{aligned}$ , will become half the initial amount, or $\begin{aligned} \frac{C}{2} \end{aligned}$ . The relationship is expressed in the decay model as
::用于模拟放射性衰变的方程式是 y=Cekt。要找到同位素半衰期的表达式, 使用半衰期的定义作为样本失去一半核核所需的时间。如果初始质量 y=C (以克计) 等于 t=0, 那么在以后某个时候 t=t12,y 将变成初始数量的一半, 或 C2 。衰变模型将显示这种关系。

$\begin{aligned} \frac{C}{2} = C e^{k t_{\frac{1}{2}}} \end{aligned}$ , or
::C2=Cekt12,或

$\begin{aligned} \frac{1}{2} = e^{k t_{\frac{1}{2}}} \end{aligned}$
::12=ekt12

Solve for the half-life $\begin{aligned} t_{\frac{1}{2}} \end{aligned}$ , by taking the natural logarithm of both sides,
::解决半衰期T12, 通过采取双方的自然对数,

$\begin{aligned} \ln \frac{1}{2} & = \ln e^{k t_{\frac{1}{2}}} \\ - \ln 2 & = k t_{\frac{1}{2}} \\ t_{\frac{1}{2}} & = - \frac{\ln 2}{k} \end{aligned}$

::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}... {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}

The half-life is then,
::那么,半衰期是,

$\begin{aligned} t_{\frac{1}{2}} = \frac{- \ln 2}{k} = \frac{- 0.693}{k} \end{aligned}$ .
::{\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}12号2k0.693k {\fn黑体\fs22\bord1\shad0\3aHBE\4aH00\fscx67\fscy66\2cHFFFFFF\3cH808080}2k

This is a famous expression in physics for measuring the half-life of a substance if the decay constant $\begin{aligned} k \end{aligned}$ is known. It can also be used to compute $\begin{aligned} k \end{aligned}$ if the half-life $\begin{aligned} t_{\frac{1}{2}} \end{aligned}$ is known.
::这是物理学中的一个著名表述,用于测量物质半衰期,如果已知衰变常数 k 。如果已知半衰期 t12 ,也可以用来计算 k。

Say that a radioactive sample contains 2 grams of nobelium. If you know that the half-life of nobelium is 25 seconds, how much will remain after 3 minutes?
::如果你知道的半衰期是25秒,3分钟后还剩多少?

Before we compute the mass of nobelium after 3 minutes, we need to first know its decay rate $\begin{aligned} k \end{aligned}$ . Using the half-life formula,
::在我们计算3分钟后的质量之前我们需要首先知道它的衰变速率 k。使用半衰期公式,

$\begin{aligned} t_{\frac{1}{2}} & = \frac{- \ln 2}{k} \\ k & = \frac{- \ln 2}{t_{\frac{1}{2}}} \\ = \frac{- \ln 2}{25} \\ = - 0.028 s e c^{- 1} \end{aligned}$

::========================================================================================================================================= ====================================================================

So the decay rate is $\begin{aligned} k = - 0.028 / s e c \end{aligned}$ . The common unit for the decay rate is the Becquerel (Bq): 1 Bq is equivalent to 1 decay per sec.
::衰变率是 k0.028/sec。衰变率的共同单位是 Becquerel (Bq): 1 Bq 等于每秒一个衰变。

To calculate the mass after 3 minutes (180 seconds), use the radioactive decay formula:
::为计算3分钟(180秒)后的质量,使用放射性衰变公式:

$\begin{aligned} y & = C e^{k t} \\ = 2 e^{(- 0.028) (180)} \\ = 0.013 grams \end{aligned}$

::y=Cekt=2e(- 0.028)(180)=0.013克

So after 3 minutes, the mass of the isotope is approximately 0.013 grams.
::3分钟后,同位素质量约为0.013克。

Applications of Exponential Growth and Decay: Population Growth
::指数增长和衰变的应用:人口增长

The same formula $\begin{aligned} y = C e^{k t} \end{aligned}$ can be used for population growth, except that $\begin{aligned} k > 0 \end{aligned}$ for an increasing function.
::同一公式 y= Cekt 可用于人口增长,但K>0 用于不断增长的函数除外。

A certain population of bacteria increases continuously at a rate that is proportional to its present number. The initial population of the bacterial culture is 140; it jumped to 720 bacteria in 4 hours. How many bacteria will be present in 10 hours? How long will it take the population to double?
::某些细菌的数量以与其现有数量成正比的速度持续增长。最初的细菌文化人口为140个,4小时内跳跃到720个细菌。 10小时内会出现多少细菌? 人口需要多长时间才能翻倍?

From reading the first sentence in the problem, we learn that the number of bacteria is increasing exponentially. Therefore, the exponential growth formula $\begin{aligned} y = C e^{k t} \end{aligned}$ is the correct model to use.
::从读到问题的第一句,我们了解到细菌的数量正在成倍增长。因此,指数增长公式y=Cekt是正确的使用模式。

As in the previous problem , first find $\begin{aligned} k \end{aligned}$ , the growth rate:
::与前一个问题一样,首先发现k,增长率:

$\begin{aligned} y & = C e^{k t} \\ 720 & = 140 e^{k (4)} \\ \ln \frac{720}{140} & = k (4) \\ 1.6376 & = k (4) \\ 0.409 & = k \end{aligned}$

::y= Cekt720= 140ek(4)ln @ 720140=k(4)1. 6376=k(4)0. 409=k

Therefore, $\begin{aligned} k = 0.409 \end{aligned}$ .
::因此,k=0.409。

Using the value of $\begin{aligned} k \end{aligned}$ , we can determine how many bacteria will be present after 10 hours.
::使用k值,我们可以确定10小时后会有多少细菌存在。

$\begin{aligned} y & = C e^{k t} \\ = 140 e^{(0.409) (10)} \\ = 8364 bacteria \end{aligned}$

::y=Cekt=140e(0.409(10))=8364细菌

To find the time required for the population to double means to find the time at which $\begin{aligned} y = 2 C \end{aligned}$ .
::找到人口需要的时间翻倍,以找到y=2C的时间。

$\begin{aligned} y & = C e^{k t} \\ 2 C & = C e^{k t} \\ 2 & = e^{k t} \\ \ln 2 & = k t \end{aligned}$

::y= Cekt2C= Cekt2=ektln @ @ @%2=kt

Solving for $\begin{aligned} t \end{aligned}$ ,
::解决 t,

$\begin{aligned} t & = \frac{\ln 2}{k} \\ = \frac{\ln 2}{0.409} \\ = 1.7 hours \end{aligned}$

::2k=20.409=1.7小时

After about 1.7 hours (102 minutes), the population of bacteria will double in number.
::在大约1.7小时(102分钟)之后,细菌的数量将翻一番。

Applications of Exponential Growth and Decay: Continuous Compound Interest
::指数增长和衰减的应用:连续复利

Investors and bankers depend on compound interest to increase their investment. Traditionally, banks added interest after certain periods of time, such as a month or a year, and the phrase was “the interest is being compounded monthly or yearly.” With the advent of computers, the compunding could be done daily or even more often. Our exponential model represents continuous, or instantaneous, compounding, and it is a good model of current banking practices. Our model states that
::投资者和银行家依赖复合利息来增加投资。传统上,银行在某个时期(如一月或一年)之后增加利息是“利息是每月或每年的复合 ” 。随着计算机的出现,分解可以每天进行,甚至更经常地进行。我们的指数模型代表连续或即时的复合,是当前银行业务的良好模式。我们的模型指出,随着计算机的出现,分解可以每天进行,甚至更经常地进行。我们的指数模型代表连续或即时的组合,并且是当前银行业务的良好模式。

$\begin{aligned} A = P e^{r t}, \end{aligned}$

::A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=佩尔特,A=P=Pert,A=Pert,A=P=Pert,A=Pert,A=P=Pert,A=P=Pert,A=P=P=Pert,A=P=Pert,A=P=P=Pert,A=P=P=Pert,A=P=P=P=Pert,A=P=P=P=P=Pert,A=Pert,A=P.

where $\begin{aligned} P \end{aligned}$ is the initial investment (present value) and $\begin{aligned} A \end{aligned}$ is the future value of the investment after time $\begin{aligned} t \end{aligned}$ at an interest rate of $\begin{aligned} r \end{aligned}$ . The interest rate $\begin{aligned} r \end{aligned}$ is usually given in percentage per year. The rate must be converted to a decimal number, and $\begin{aligned} t \end{aligned}$ must be expressed in years.
::P是初始投资(现值),A是时间t之后投资的未来价值,利率为 r。利率r通常以年百分比表示。利率必须转换为小数数, t必须用年表示。

Say that an investor invests an amount of $10,000 and discovers that its value has doubled in 5 years. What is the annual interest rate that this investment is earning? How long will it take the invested money to triple?
::说一个投资者投资了10 000美元,发现其价值在5年内翻了一番。这一投资的年利率是多少? 投资要花多久才能达到三倍?

To find the interest rate, use the exponential growth model for continuously compounded interest,
::为了找到利率, 使用指数增长模式持续复合利息,

$\begin{aligned} A & = P e^{r t} \\ 20, 000 & = 10, 000 e^{r (5)} \\ 2 & = e^{5 r} \end{aligned}$

::A=Pert20 000=10 000er(5)2=e5r

Thus,
::因此,

$\begin{aligned} r & = \frac{\ln 2}{5} \\ = 0.139 \\ r & = 13.9 % \end{aligned}$

::25=0.0.139r=13.9%

The investment has grown at a rate of 13.9% per year.
::投资以每年13.9%的速度增长。

To find out how long it will take the invested money to triple, use the growth model again:
::要知道投资资金要花多久才能翻三番,

$\begin{aligned} A & = P e^{r t} \\ 30, 000 & = 10, 000 e^{(0.139) t} \\ 3 & = e^{0.139 t} \\ \ln 3 & = 0.139 t \\ t & = \frac{\ln 3}{0.139} \\ = 7.9 years \end{aligned}$

::A=Pert30,000=10,000e(0.139)t3=e0.139 tln3=0.139 tttt=ln30.139=7.9年

Applications of Exponential Growth and Decay: Other Models
::指数增长和衰减的应用:其他模式

Not all exponential growth and decay models are of the form $\begin{aligned} y = C e^{k t} \end{aligned}$ . Some use a base other than the natural base $\begin{aligned} e \end{aligned}$ ; some are solutions to differential equations other than $\begin{aligned} \frac{d y}{d x} = k y \end{aligned}$ . Actually, in everyday life most are constructed from empirical data and regression techniques.
::并不是所有指数增长和衰变模型都以y=Cekt的形式出现。有些人使用除自然基数e以外的基数; 有些是丁氧=ky以外的差异方程式的解决方案。事实上,在日常生活中,大多数都是用实验数据和回归技术构建的。

Suppose that in the business world the demand function for the price of a motorcycle is described by the formula
::假设在商业界,对摩托车价格的需求功能由公式来描述

$\begin{aligned} p = 12, 400 - \frac{11, 000}{2.2 + e^{- 0.0003 x}} \end{aligned}$
::p=12 400-11,0002.2+e-000003x

where $\begin{aligned} p \end{aligned}$ is the motorcycle price per unit and $\begin{aligned} x \end{aligned}$ is the number of units produced. If the business is interested in basing the price of its unit on the number that it is projecting to sell, this formula becomes very helpful.
::P是单位摩托车价格,x是生产单位数量。如果企业有兴趣将单位价格以其预计销售的数量为基础,这一公式将非常有用。

If a motorcycle factory is projecting to sell 7000 units in one month, what price should the factory set on each motorcycle?
::如果一家摩托车厂在一个月内打算出售7000个单位,那么每辆摩托车上工厂应该买多少价钱?

$\begin{aligned} p & = 12, 400 - \frac{11, 000}{2.2 + e^{0.0003 x}} \\ = 12, 400 - \frac{11, 000}{2.2 + e^{0.0003 (7000)}} \\ = 12, 400 - \frac{11, 000}{2.2 + 0.122} \\ = 7, 663. \end{aligned}$

::p=12,400-11,0002.2+0.0003x=12,400-11,0002.2+0.0003x=12,400-11,0002.2+0.0003+0.0003(700)=12,400-11,0002.2+0.122=7,663。

Thus the factory’s base price for each motorcycle should be set at $7,663.
::因此,工厂每辆摩托车的基本价格应定为7 663美元。

Examples
::实例

Example 1
::例1

Earlier, you were asked if the solution to a growth/decay differential equation, $\begin{aligned} y = 5 \cdot {0.3}^{t} \end{aligned}$ , is a growth or decay solution. Can you write the differential equation for which $\begin{aligned} y = 5 \cdot {0.3}^{t} \end{aligned}$ is the solution?
::早些时候, 有人询问您, 成长/ 衰减差异方程式的解决方案, y= 50. 3t, 是成长或衰变的解决方案。您能否写成以y= 50. 3t为解决方案的差别方程式 ?

$\begin{aligned} y = 5 \cdot {0.3}^{t} \end{aligned}$ is an exponential function of the kind $\begin{aligned} f (x) = b^{x} \end{aligned}$ , where $\begin{aligned} b = 0.3 \end{aligned}$ . As $\begin{aligned} t \end{aligned}$ increases, $\begin{aligned} y = 5 \cdot {0.3}^{t} \end{aligned}$ decreases. The function represents exponential decay. We can also write the equation in the following form: $\begin{aligned} y = 5 \cdot {0.3}^{t} = 5 \cdot (e^{\ln 0.3})^{t} = 5 \cdot e^{- 1.2 t} \end{aligned}$ . It is more clear that this is a decreasing function.
::y=5=0.3t 是 f(x) = bx 的指数函数, b=0.3。随着 t 的增加, y= 5= 0.3t 递减。该函数代表指数衰减。我们也可以以下列形式写入方程式: y= 5= 5=0.3t= 5( eln=0.3)t= 5e- 1. 2t。更明显的是, 这是一个递减函数。

Since $\begin{aligned} y = 5 \cdot {0.3}^{t} \end{aligned}$ is the solution to a differential equation where the rate of change of the function is proportional to the function, we can write
::y=50.3t 是函数变化率与函数成比例的差别方程的解决方案, 因此我们可以写入

$\begin{aligned} \frac{d}{d x} (5 \cdot {0.3}^{t}) & = k (5 \cdot {0.3}^{t}) \\ 5 \cdot \ln 0.3 \cdot ({0.3}^{t}) & = k (5 \cdot {0.3}^{t}) \\ k & = \ln 0.3 = - 1.2 \end{aligned}$

::dx( 50. 3t) =k( 50. 3t) 5 = 0. 3 (0. 3t) = k( 50. 0. 3t) k= ln 0. 3. 1.2

The differential equation is: $\begin{aligned} \frac{d y}{d x} = - 1.2 y \end{aligned}$
::差别方程是: dydx==1.2y

Example 2
::例2

A medical researcher is studying the spread of the flu virus through a certain campus during the winter months. The model for the spread is described by
::一位医学研究人员正在研究流感病毒在冬季几个月通过某个校园在某个校园中传播的问题。

$\begin{aligned} P = \frac{4500}{1 + 4499 e^{- 0.8 x}}, x \geq 0 \end{aligned}$
::P=450001+4499ee-0.8xxx_0

where $\begin{aligned} P \end{aligned}$ represents the total number of infected students and $\begin{aligned} x \end{aligned}$ is the time, measured in days. How many students will be infected in the next week (7 days)? How long it will take until 1000 students become infected with the flu virus?
::P代表受感染学生的总数,x是按天数衡量的时间。下周(7天)有多少学生将受到感染? 要多久才能有1000名学生感染流感病毒?

Use the model $\begin{aligned} P = \frac{4500}{1 + 4499 e^{- 0.8 x}} \end{aligned}$ with $\begin{aligned} x = 7 \end{aligned}$ to determine the number of students infected in the next week.
::使用P=45001+4499e-0.8xxxxxx=7的模型确定下周受感染的学生人数。

$\begin{aligned} P & = \frac{4500}{1 + 4499 e^{- 0.8 x}} \\ = \frac{4500}{1 + 4499 e^{- 0.8 (7)}} \\ = \frac{4500}{1 + 4499 (0.004)} \\ = 255. \end{aligned}$

::P=45001+4499e-0.8x=45001+4499e-0.8(7)=45001+4499(0.004)=255。

According to the model, 255 students will become infected with the flu virus.
::根据该模式,255名学生将感染流感病毒。

Use the model $\begin{aligned} P = \frac{4500}{1 + 4499 e^{- 0.8 x}} \end{aligned}$ and solve for $\begin{aligned} x \end{aligned}$ in terms of $\begin{aligned} P \end{aligned}$ . $\begin{aligned} P = \frac{4500}{1 + 4499 e^{- 0.8 x}} \end{aligned}$ .
::使用P=45001+4499e-0.8x的模型,并用P=45001+4499e-0.8x的模型解决x。

Cross-multiplying,
::交叉倍增,

$\begin{aligned} P (1 + 4499 e^{- 0.8 x}) & = 4500 \\ 1 + 4499 e^{- 0.8 x} & = \frac{4500}{P} \\ 4499 e^{- 0.8 x} & = \frac{4500}{P} - 1 \\ = \frac{4500 - P}{P} \\ e^{- 0.8 x} & = \frac{4500 - P}{4499 P} . \end{aligned}$

::P(1+4499e-0.8x)=45001+4499e-0.8x=4500P4499e-0.8x=4500P-1=4500P-1=4500-PPE-0.8x=4500-PPE-0.8x=4500-P4499P。

Taking the natural log of both sides,
::以双方的自然记录,

$\begin{aligned} - 0.8 x & = \ln [\frac{4500 - P}{4499 P}] \\ x & = \ln [\frac{4500 - P}{4499 P}] \div (- 0.8) . \end{aligned}$

::-0.8x=ln[4500-P4499 P]x=ln[4500-P4499 P](-0.8)。

Substituting for $\begin{aligned} P = 1000 \end{aligned}$ ,
::P=1000,替代P=1000,

$\begin{aligned} x = 9 days \end{aligned}$ .
::x=9天。

So the flu virus will spread to 1000 students in 9 days.
::流感病毒将在9天内传播到1000名学生。

Review
::回顾

In 1990, the population of the USA was 249 million. Assume that the annual growth rate is 1.8%.

According to this model, what was the population in the year 2000?
::根据这一模式,2000年的人口是多少?

According to this model, in which year will the population reach 1 billion?
::根据这一模式,人口将在哪一年达到10亿?

::1990年,美国人口为2.49亿,假设年增长率为1.8 % 。根据这个模式,2000年的人口是多少?根据这个模式,人口将在哪一年达到10亿?

Prove that if a quantity $\begin{aligned} A \end{aligned}$ is exponentially growing and if $\begin{aligned} A_{1} \end{aligned}$ is the value at $\begin{aligned} t_{1} \end{aligned}$ and $\begin{aligned} A_{2} \end{aligned}$ at time $\begin{aligned} t_{2} \end{aligned}$ , then the growth rate will be given by $\begin{aligned} k = \frac{1}{t_{1} - t_{2}} \ln (\frac{A_{1}}{A_{2}}) \end{aligned}$ .
::证明如果一个A数量指数增长,如果A1是t1和A2在t2时间的值,则以k=1t1-t2ln(A1A2)表示增长率。

Newton’s Law of Cooling states that the rate of cooling of object with respect to the temperature of its surroundings is proportional to the difference in temperature between the object and the surroundings.

Write the differential equation that expresses Newton’s Law of Cooling. Hint: Write the equation using the dependent variable $\begin{aligned} D = T (t) - T_{r} \end{aligned}$ where $\begin{aligned} T_{r} \end{aligned}$ is the room temperature (the surroundings)
::写入表达牛顿冷却定律的差别方程。提示 : 使用附属变量 D=T( t) - Tr 写方程, Tr 是房间温度( 周围环境) 。

Show that the formula $\begin{aligned} T (t) = (T_{0} - T_{r}) e^{- k t} + T_{r} \end{aligned}$ satisfies the equation, where $\begin{aligned} T_{0} \end{aligned}$ is the initial temperature of the object at $\begin{aligned} t = 0 \end{aligned}$ and $\begin{aligned} k \end{aligned}$ is a constant that is unique for the measuring instrument (the thermometer) called the time constant.
::显示公式T(t)=(T0-Tr)e-kt+Tr 满足方程式,其中T0是天体在 t=0 的初始温度, k是测量仪器(温度计)称为时间常数的唯一常数。

::牛顿的《冷却法》规定,物体的冷却速度与其周围温度的差值成正比。写下表达牛顿冷却法的差别方程式。提示: 使用附属变量 D=T( t)- Tr 写方程式, Tr 是房间温度( 周围) 显示公式 T( t) = ( T0- Tr) = ( T0- Tr) e- kt+Tr ) 满足方程式, T0 是天体在 t=0 和 k 的初始温度, k 是测量仪( 温度计) 唯一需要的时间常数。

Suppose a liter of juice at $\begin{aligned} 23^{\circ} C \end{aligned}$ is placed in the refrigerator to cool. If the temperature of the refrigerator is kept at $\begin{aligned} 11^{\circ} C \end{aligned}$ and $\begin{aligned} k = 0.417 \end{aligned}$ , use Newton’s law of cooling to find the temperature of the juice after 3 minutes?
::如果冰箱温度保持在11°C和K=0.417时,用牛顿冷却法则在3分钟后找到果汁温度的话?

Referring to problem #3, if it takes an object 320 seconds to cool from $\begin{aligned} 40^{\circ} C \end{aligned}$ above room temperature to $\begin{aligned} 22^{\circ} C \end{aligned}$ above room temperature, how long will it take to cool another $\begin{aligned} 10^{\circ} C \end{aligned}$ after it reaches $\begin{aligned} 22^{\circ} C \end{aligned}$ above room temperature?
::提及问题3:如果一个物体需要320秒从40°C高于房间温度到22°C高于房间温度冷却到22°C,那么在达到22°C高于房间温度后再冷却10°C需要多长时间?

Polonium-210 is a radioactive isotope with half-life of 140 days. If a sample has a mass of 10 grams, how much will remain after 10 weeks?
::210是放射性同位素,半衰期为140天,如果样品质量为10克,10周后还剩多少?

A Logarithmic Model: In the physics of acoustics, there is a relationship between the subjective sensation of loudness and the physically measured intensity of sound. This relationship is called the sound level $\begin{aligned} β \end{aligned}$ . It is specified on a logarithmic scale and measured with units of decibels (dB). The sound level $\begin{aligned} β \end{aligned}$ (in decibels, dB) of any sound is defined in terms of its intensity $\begin{aligned} I \end{aligned}$ (measured in watts per square meter, $\begin{aligned} \frac{W}{m^{2}} \end{aligned}$ , in the SI-mks unit system): $\begin{aligned} β = 10 \log \frac{I}{I_{0}} = 10 \log \frac{I}{10^{- 12}} \end{aligned}$ , where $\begin{aligned} I_{0} = 10^{- 12} \frac{W}{m^{2}} \end{aligned}$ is a standard threshold of human hearing at 1000 Hz. A $\begin{aligned} β \end{aligned}$ of 0 dB $\begin{aligned} (I = I_{0}) \end{aligned}$ is the threshold of hearing, i.e. the minimum sound that can be heard by humans. The sound level 120 dB is considered the threshold of pain for the human ear.

If a decibel meter registered 130 dB at a heavy metal rock concert, what is the intensity $\begin{aligned} I \end{aligned}$ of this sound level?
::如果在重金属摇滚音乐会上分贝仪注册了130 dB, 那么音量的强度是多少?

What is the sound level (in dB) of a sound whose intensity is $\begin{aligned} 2.0 \times 10^{- 6} \frac{W}{m^{2}} \end{aligned}$ ?
::强度为2.0x10-6Wm2 的音频( dB) 音频水平是多少?

::对数模型:在声学物理学中,声响的主观感知与物理测测声音强度之间存在某种关系。此关系被称为声级。它在对数尺度上指定, 并用除颤器单位( dB) 进行测量。音音的音量水平 β( 在除颤器中, dB) 以其强度I( 以每平方米瓦特测量, 在 SI- mks 单元系统中以 Wm2 测量) : @ 10log II0= 10logI10-12, 其中I0=10-12Wm2 是1,000赫兹的人听力标准门槛。 0 dB ( I=I0) 是听觉的门槛, 即人类能听到的最起码的声音。音量 120 dB 被认为是人类耳痛的门槛。如果在重金属岩石音乐会上登记了130 dB, 这个音强度是多少? 声音的音级是多少( 在 dB ) ?

Referring to problem #7, if a single mosquito 10 meters away from a person makes a sound that is barely heard by the person (threshold 0 dB), what will be the sound level of 1000 mosquitoes at the same distance?
::提到问题7, 如果一只蚊子距离一个人10米远, 发出一个人几乎听不到的声音(高度0 dB), 同一距离1000蚊子的声音会达到什么程度?

Referring to problem #7, if a noisy machine at a factory produces a sound level of 90 dB at a certain distance, what is the combined sound level when an identical machine is placed beside it?
::提到问题7:如果工厂的噪音机器在一定距离内产生90 dB的音量,那么将同一机器放在它旁边时,声音的组合水平是多少?

At the tail end of a news report, you hear that the fact that your town has been growing at a yearly rate of 5% for the last 6 years means its population will, at this rate, be 50,000 people in 2 years. What was the population 3 years ago?
::在一个新闻报道的尾端,你听到,在过去6年里,你的城镇以5%的年增长率增长,这意味着其人口在两年内将达到5万人。3年前的人口是多少?

The mass of a radioactive substance is currently 20 grams, but was 25 grams two weeks ago. What is the half-life of the substance?
::放射性物质的质量目前为20克,但两周前为25克。该物质的半衰期是多少?

A motorcycle is worth $10,000, and loses value at a rate of 6.5% per year, what will it be worth in 5 years?
::摩托车价值10 000美元,以每年6.5%的速度损失价值,五年内价值多少?

A Logarithmic Model: On the Richter scale, the magnitude, $\begin{aligned} R \end{aligned}$ , of an earthquake of intensity $\begin{aligned} I \end{aligned}$ is given by $\begin{aligned} R = \log \frac{I}{I_{0}} \end{aligned}$ , where $\begin{aligned} I_{0} = 1 \end{aligned}$ is the reference intensity. Intensity is a measure of the wave energy of an earthquake. For two earthquakes with Richter magnitudes of 4 and 7.9, what are there respective intensities?
::逻辑模型 : 在 Richter 规模上, R 是 R=log II0 给出的强度I型地震的大小 R, 参考强度为 I0=1 。强度是地震波能量的量度。对于两次 Richter 规模为 4 和 7. 9 级的地震来说, 强度是多少?

On a high school campus of 2000 students, a student returns from a break with a contagious flu virus. The spread of the virus is modeled by the logistic growth function
$\begin{aligned} y = \frac{2000}{1 + 1999 e^{- 0.7 t}} for t \geq 0 \end{aligned}$
where $\begin{aligned} y \end{aligned}$ is the total number of students infected after $\begin{aligned} t \end{aligned}$ days. The school will cancel classes when 10% or more of the students are infected. After how many days will the high school cancel classes?
::在2000年学生的高中校园里,学生从与传染性流感病毒的分离中返回。病毒的传播以后勤增长功能y=20001+1999e-0.7t为模范,这里是24天后受感染的学生总数。当10%或以上学生感染时,学校将取消课程。在几天后,高中将取消多少天?

The amount of money, $\begin{aligned} A \end{aligned}$ , accumulated in an account that earns compound interest is given by the formula $\begin{aligned} A = P {[1 + \frac{r}{n}]}^{n t} \end{aligned}$ , where $\begin{aligned} P \end{aligned}$ is the initial deposit, $\begin{aligned} r \end{aligned}$ is the interest rate, $\begin{aligned} n \end{aligned}$ is the number of compounding periods per year, and $\begin{aligned} t \end{aligned}$ is the number of years. Show that as $\begin{aligned} n \to \infty \end{aligned}$ , the accumulated earnings is given by the continuous compound interest formula $\begin{aligned} A = P e^{r t} \end{aligned}$ .
::货币 A , 累积在赚取复合利息的账户中的金额,由公式A=P[1+rn]nt给出,其中P为初始存款,r为利率,r为利率,n为年复利期数, t为年数。显示,作为n,累计收益由连续复利公式A=Pert给出。

Review (Answers)
::回顾(答复)

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。

Section outline

Exponential Growth and Decay ::指数增长和衰减

Applications of Exponential Growth and Decay: Radioactive Decay ::指数增长和衰变的应用:放射性衰变

Applications of Exponential Growth and Decay: Population Growth ::指数增长和衰变的应用:人口增长

Applications of Exponential Growth and Decay: Continuous Compound Interest ::指数增长和衰减的应用:连续复利

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Review ::回顾

Review (Answers) ::回顾(答复)

Exponential Growth and Decay
::指数增长和衰减

Applications of Exponential Growth and Decay: Radioactive Decay
::指数增长和衰变的应用:放射性衰变

Applications of Exponential Growth and Decay: Population Growth
::指数增长和衰变的应用:人口增长

Applications of Exponential Growth and Decay: Continuous Compound Interest
::指数增长和衰减的应用:连续复利

Examples
::实例

Example 1
::例1

Example 2
::例2

Review
::回顾

Review (Answers)
::回顾(答复)