指数增长和衰减函数

播放段落

Lesson Objectives
::经验教训目标

播放段落
• Understand the characteristics of the general exponential growth function , where $\begin{align*}r\end{align*}$  is growth rate .
  ::理解一般指数增长函数的特性,其中r为增长率。
播放段落
• Solve problems involving interest and .
  ::解决涉及利息和... .
播放段落
• Derive and use the formula for the sum of a finite geometric series  to solve problems.
  ::产生并使用限定几何序列总和的公式来解决问题。
播放段落
• Write exponential growth and decay functions given  an appropriate context.
  
  ::以适当的环境写入指数增长和衰变函数。

---

播放段落

Introduction: Compound Interest
::导言:复合利息

播放段落

The principle of a loan is the original loaned amount. Interest is a percentage of the principle that is added as a cost to pay for the privilege of borrowing the money. Compound interest occurs when interest is added to principle and further interest is then  calculated  on previous interest as well as the principle.
::贷款原则是原始贷款额,利息是作为借款权支付费用而增加的原则的一个百分比,如果将利息列入原则,然后根据先前的利息和原则计算进一步利息,则产生复利。

播放段落

Use the interactive below to re view how compound interest works.
::利用以下互动方式审查复合利息如何运作。

 

 

 

---

播放段落

Activity 1: Compound Interest Continued
::活动1:复利

播放段落

The formula for compound interest can be written as  $\begin{align*}A={P\left(1+\frac{r}{n}\right)}^{nt}.\end{align*}$  The variables are used to represent the following:
::复利公式可以用 A=P(1+rn)nt 写入。 变量用于代表以下内容:

播放段落
• $\begin{align*}A\end{align*}$  is the total amount owed
  ::A是欠款总额。
播放段落
• $\begin{align*}P\end{align*}$  is the principal, which is the starting amount
  ::P是本金,即起始金额
播放段落
• $\begin{align*}r\end{align*}$  is the interest rate
  ::r 是利率
播放段落
• $\begin{align*}n\end{align*}$  is the number of times the  principal is compounded in a year
  ::n 是本本金在一年内复数的倍数。
播放段落
• $\begin{align*}t\end{align*}$  is the number of years
  ::t 是年数

播放段落

The compound interest formula comes from the exponential growth formula. Exponential growth can be expressed as a percent of the starting amount. The function  $\begin{align*}f(x)=2^x\end{align*}$  represents a quantity that repeatedly doubles. Another way to think of doubling is an increase of 100%.  This  can be rewritten as follows:
::复合利息公式来自指数增长公式。 指数增长可以表示为起始量的%。 函数 f( x) = 2x 代表一个多次翻倍的数量。 另一种设想翻倍的方法是增加100%。 这可以重写如下:

播放段落

  $\begin{align*}f(x) = (200\%)^x = (100\% + 100\%)^x = (1+100\%)^x\end{align*}$
::f(x) = (200%)x = (100 100%)x = (1+100%)x

播放段落

E verything  is being taken and added to itself, resulting in the general exponential growth equation :  $\begin{align*}f(x) = a(1+r)^x\end{align*}$  where $\begin{align*}a\end{align*}$  is the starting amount and $\begin{align*}r\end{align*}$  is the growth rate, written as a decimal.
::所有东西都被取出并加到自己身上,从而形成总的指数增长方程式:f(x)=a(1+r)x,其中a是起始数,r是增长率,以小数数表示。

播放段落

Example
::示例示例示例示例

播放段落

A population of 10 mice  increases by 300% every month.
::10只小鼠每月增加300%。

播放段落

a. Write a function model th at  expresses the number of mice in population $\begin{align*}P\end{align*}$  as a function of $\begin{align*}x\end{align*}$  months.  
::a. 写一个功能模型,表示P群中小鼠数的乘以x个月。

播放段落

This problem highlights a challenge when it comes to wording. If a sentence reads "increases by 300%," this means that the growth rate is 300%, and the quantity is quadrupling. K eep everything and add 300%. The common ratio is 4 or 400%. The growth rate  is the rate at which an amount increases; the common ratio  is the rate at which an amount is multiplied. In this case, the starting population is  $\begin{align*}a=10,\end{align*}$  and the  growth rate is   $\begin{align*}r=300\%\end{align*}$  or 3. Substituting this into the exponential growth formula will give the following:
::这个问题凸显了在措辞上的一个挑战。 如果一句话是“ 增加300 % ” , 这意味着增长率是300 %, 数量是四倍。 保持一切并增加300 % 。 共同比率是 4 或 400 % 。 增长率是 数额增加的速率; 共同比率是 数额乘以的速率。 在这种情况下, 起始人口是 = 10, 增长率是 r= 300 % 或 3 。 将这一增长率转换成指数增长公式, 将作出如下说明:

播放段落

$$\begin{align*}P(x)=10(1+3)^x \:\:\text{ or }\:\: P(x)=10(4)^x\end{align*}$$
::P(x)=10(1+3)x或P(x)=10(4)x

播放段落

Answer:   $\begin{align*}P(x)=10(4)^x\end{align*}$
::答复:P(x)=10(4)x

播放段落

A graph of this function is displayed below.
::此函数的图形显示在下面。

播放段落

As portrayed in the graph of this function above, the population of 10 is increasing exponentially. Every month, add triple the amount of mice, which is the same as quadrupling the total.
::正如上文关于此函数的图表所示,10人的人口正以指数速度增长。 每月,每增加3倍于小鼠的数量,这与总数翻了2倍相同。

播放段落

b. How many mice will there be after six months?
::b. 六个月后将有多少小鼠?

播放段落

To determine the population of mice after 6 months, set up and solve the equation:
::确定六个月后的小鼠数量,设置和解决方程式:

播放段落

$$\begin{align*}P(x)&=10(1+3)^6 \\ &= 10(4)^6\\ &= 10\times4,096\\ &=40,960\end{align*}$$
::P(x)=10(1+3)6=10(4)6=10x4,096=40,960

播放段落

Answer:  There will be 40,960 mice after six months.
::答复:六个月后将有40 960只小鼠。

 

 

播放段落

Discussion Question :
::讨论问题:

播放段落

Suppose you deposit $50 into a savings account. The account grows at a 1.2% annual interest rate, compounded monthly. Assuming you do not deposit or withdraw money from the account, how long will it take for there to be $70 in the account ?
::假设你把50美元存入储蓄账户。该账户以1.2%的年利率增长,每月复计。假设你没有从账户中存款或提款,账户中需要70美元多久?

 

 

   

---

播放段落

Activity 2:  Paying off a Loan
::活动2:偿还贷款

播放段落

Compound interest is used in nearly all  scenarios where money is borrowed. However, as you pay off your loan, the interest is applied to smaller and smaller amounts. 
::在借款的几乎所有情况下,都使用复利,但是,在偿还贷款时,利息适用于较小数额的金额。

播放段落

Example
::示例示例示例示例

播放段落

You  take out  a 30-year mortgage on a house for $350,000 at a monthly interest rate of 0.4% per month. How much will you pay per month?
::每月利率为0.4%,每月利率为0.4%。每月支付多少钱?

播放段落

F irst consider the idea that you pay nothing until the 30 year period is over, at which point you pay the full amount. If this was the case, you would owe over $1,000,000 for the house, including interest. Paying  this much for a $350,000 house seems pretty ridiculous, but it goes to show how quickly interest can build. The reason this seems like a high number is because the interest is only charged to the amount left on the loan as you pay it off. To determine y our monthly mortgage payment, you will need to take it month by month . U se the variable  $\begin{align*}m\end{align*}$  to represent the monthly payment, $\begin{align*}p\end{align*}$  to represent the principal loan amount, and $\begin{align*}r\end{align*}$  to represent the monthly interest. 
::首先考虑你直到30年期限结束才支付任何款项的想法, 在那个时候你支付全额。 如果是这样的话, 你将会为房子欠一亿多美元, 包括利息。 支付这笔钱给一个350 000美元的房子似乎相当可笑, 但这说明利息能迅速积累。 原因似乎很大, 这是因为利息只按贷款的剩余金额来支付。 为了确定每月按揭的支付额, 您需要按月支付。 使用变量m来表示每月的支付额, p 代表本金贷款金额, r 代表每月的利息 。

Pay Period	Amount Due	Amount Due Simplified
Starting Amount Owed	$\begin{align*}p\end{align*}$	$\begin{align*}p\end{align*}$
Amount O wed A fter Month 1	$\begin{align*}p(1+r) - m\end{align*}$	$\begin{align*}p(1+r) - m\end{align*}$
Amount Owed After Month 2	$\begin{align*}(p(1 + r) - m)(1 + r) - m\end{align*}$	$\begin{align*}p(1 + r)² - m(1 + r) - m\end{align*}$
Amount Owed After Month 3	$\begin{align*}(p(1 + r) - m)(1 + r) - m)(1 + r) - m\end{align*}$	$\begin{align*}p(1 + r)³ - m(1 + r)² - m(1 + r) - m\end{align*}$

播放段落

Looking at the simplified amount owed, t here is   a pattern emerging. This pattern can be extended to write an expression representing the amount owed at the end of the loan:
::从简化的欠款数额来看,出现了一种模式。这种模式可以扩大,以写出贷款结束时所欠数额的表示:

播放段落

$$\begin{align*}p(1 + r)^n - m(1 + r)^{n-1} - m(1 + r)^{n-2} - ... - m(1 + r)^{0}\end{align*}$$
:1+r)n-m(1+r)n-m(1+r)n-1-m(1+r)n-2-...-m(1+r)0

播放段落

Since you will owe $0 at the end of the loan, you can use the expression above to write the following equation:
::如果您在贷款结束时欠了0美元,您可以用上面的表达式写出以下方程式:

播放段落

$$\begin{align*}p(1 + r)^n - m(1 + r)^{n-1} - m(1 + r)^{n-2} - ... - m(1 + r)^{0} = 0\end{align*}$$
:1+r)n-m(1+r)n-m(1+r)n-1-m(1+r)n-2-...-m(1+r)0=0

播放段落

A dd each term with an  $\begin{align*}m\end{align*}$   coefficient to the right side of the equation:
::在方程右侧加上每个词,加上一个 m 系数:

播放段落

$$\begin{align*}p(1 + r)^n = m(1 + r)^{n-1} + m(1 + r)^{n-2} + ... + m(1 + r)^{0}\end{align*}$$
::p(1+r)n=m(1+r)n-1+m(1+r)n-2+...+m(1+r)0

播放段落

F actor out an  $\begin{align*}m\end{align*}$  from each term on the right side of the equation .
::在方程的右侧从每个学期中计出一米。

播放段落

$$\begin{align*}p(1 + r)^n = m[(1 + r)^{n-1} + (1 + r)^{n-2} + ... + (1 + r)^{0}]\end{align*}$$
::p(1+r)n=m[(1+r)n-1+(1+r)n-2+...+(1+r)0]

播放段落

Finally, solve for the monthly payment by dividing the expression in the brackets to both sides of the equation.
::最后,通过将括号中的表达式除以方程的两侧,解决每月付款问题。

播放段落

$$\begin{align*}\frac{p(1 + i)^n}{(1 + i)^{n-1} + (1 + i)^{n-2} + ... + (1 + i)^{0}} = m\end{align*}$$
:1+一)n(1+一)n(1+一)n-1+(1+一)n-2+...+(1+一)0=m

播放段落

While this does represent a formula that can be used to find the monthly payment, the denominator will take a long time to calculate. The expression in the denominator represents a geometric sequence .
::虽然它确实代表一种公式,可以用来寻找月度付款, 分母则需要很长的时间来计算。 分母中的表达式代表几何序列 。

播放段落

Ideally, this formula to be easy to calculate. To accomplish this, write an expression to represent the sum of a geometric sequence. Recall from  that a geometric  series can be written as  terms of a geometric sequence : 
::理想情况下, 这个公式容易计算 。 要完成此任务, 请写一个表达式来表示几何序列的总和 。 回顾几何序列可以写成几何序列的条件 :

播放段落

$$\begin{align*}S_n=a_1+a_1r+a_1r^2+a_1r^3+ \ldots +a_1r^{n-2}+a_1r^{n-1}\end{align*}$$
::Sn=a1+a1r+a1r+a1r2+a1r3+...+a1rn-2+1

播放段落

Now, factor out $\begin{align*}a_1\end{align*}$  to get   $\begin{align*}a_1(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}).\end{align*}$   By isolating what is in the parenthesis and multiplying  it by   $\begin{align*}(1-r)\end{align*}$   as shown below, the sum can be simplified:
::乘以 a1 以获得 a1 (1+r2+r3+...+rn-2+rn- 1) 。 通过将括号中的内容分离并乘以下文所示的乘数(1-r),总和可以简化:

播放段落

$$\begin{align*}&(1-r)S_n=a_1 (1-r)(1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}) \\ &= a_1 (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= a_1 (1+r+r^2+r^3+ \ldots +r^{n-2}+r^{n-1}-r-r^2-r^3-r^4- \ldots -r^{n-1}-r^n) \\ &= a_1 (1-r^n)\end{align*}$$
:1-r)Sn=a1(1+r+r2+r3+...+rn2+rn1)=a1(1+r+r2+r2+r3+...+r2+r3+...+r2+r2+r-rn-r2-r3-r4-...-rn-r4_r-rn)=a1(1+r+r+r2+r3+...+r2+r2+r_r2_r3-r4_...-rn-rn_rn)=a1(1+r+r+r+r2+r2+r3+r3+...-r4-r___rn)=a1(1-rn)

播放段落

M ultiplying $\begin{align*}(1+r^2+r^3+ \ldots +r^{n-2}+r^{n-1})\end{align*}$  by   $\begin{align*}1-r\end{align*}$   resulted in the middle terms being  canceled, leaving only $\begin{align*}(1-r^n).\end{align*}$  However, both sides of the equation must be multiplied by  $\begin{align*}(1-r)\end{align*}$   to preserve the equation. The final step is to solve for  $\begin{align*}S_n\end{align*}$  by dividing  $\begin{align*}(1-r)\end{align*}$  on both sides of the equation. The sum of a finite geometric series can be written as follows:
::乘以 1+r2+r3+...+rn2+rn_1+1+1+1+r_rn_r+1+1+1+r2+rn_rr+1+1+r2+rn_rn_r1+1+1+r2+r2+1+r2+rn_rn_r_l+1+1+r2+r2+1+r2+r2+r2+1+r2+r2+r2+r_r_l+1+1+1+1+r2+r2+r2+rn_r_l+1+1+1+r2+rn_r2+r_r_r_r_l+1+r2+1+1+r2+r2+r2+1+r2+r2+r2+r2+1+r2+r2+1+1+r2+r2+r2+r_1+r1+r_1+r1+r_1+r_1+1+r1+r2+r1+r2+r1+r2+r1+r1+r2+r1+r1+r2+r2+r1+r1+r1+r1+r1+r1+r2+r2+r1+r2+r1+r2+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r2+r1+r1+r2+r2+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r2+r1+r1+r2+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+r1+

播放段落

$$\begin{align*}S_n=\frac{a_1(1-r^n)}{1-r}\end{align*}$$
::Sn=a1( 1- rn) 1-r

播放段落
• $\begin{align*}S_n\end{align*}$  is the sum of the finite geometric series.
  ::Sn是有限的几何序列的总和。
播放段落
• $\begin{align*}a_1\end{align*}$  is the first term in the series. 
  ::a1 是系列的第一个学期。
播放段落
• $\begin{align*}n\end{align*}$  is the  number of  terms in the series.
  ::n 是序列中的术语数。
播放段落
• $\begin{align*}r\end{align*}$   is the common ratio .
  ::r 是共同比率。

播放段落

U se this equation to rewrite or monthly mortgage payment formula above. The geometric sequence that needs to be rewritten is  $\begin{align*}(1 + i)^{n-1} + (1 + i)^{n-2} + ... + (1 + i)^{0}.\end{align*}$  The value of  $\begin{align*}a_{1}\end{align*}$  is 1, the value of $\begin{align*}r\end{align*}$  is  $\begin{align*}i,\end{align*}$  and the value of $\begin{align*}n\end{align*}$  is $\begin{align*}n.\end{align*}$
::使用此方程式重写或按月按揭支付公式。 需要重写的几何序列是(1+i)n-1+(1+i)n-2+...+(1+i)0。 a1 值为 1, r值为 i, n值为 n。

播放段落

$$\begin{align*}(1 + i)^{n-1} + (1 + i)^{n-2} + ... + (1 + i)^{0}&=\frac{1(1-(1+i)^n)}{1-(1+i)}\\\\ &=\frac{(1+i)^n-1}{(1+i)-1}\\\\ &=\frac{(1+i)^n-1}{i}\end{align*}$$
:1+1)n-n-1+1+(1+1)n-2+...+(1+1)0=1-1-1-1-1-1(1+1)n+1=(1+1)n-1-1-1(1+1)n-1=(1+1)n-1=(1+1)n-1=(1+1)n-1=(1+1)n-1-i

播放段落

This substitution will look as follows :
::这一替代如下:

播放段落

$$\begin{align*}m = \frac{p(1 + i)^n}{(1 + i)^{n-1} + (1 + i)^{n-2} + ... + (1 + i)^{0}} = \frac{p(1 + i)^n}{\frac{(1+i)^n-1}{i}}\end{align*}$$
::m=p(1+一)n(1+一)n-1+(1+一)n-2+...+(1+一)0=p(1+一)n(1+一)n-1i

播放段落

From here, simplify  by multiplying $\begin{align*}p(1 + i)^n\end{align*}$  by the reciprocal of  $\begin{align*}\frac{(1+i)^n-1}{i}:\end{align*}$  
::从这里,将p(1+i)n乘以(1+i)n-1i的对等性,简化为p(1+i)n:

播放段落

$$\begin{align*}m &= \frac{p(1 + i)^n}{1} \cdot{\frac{i}{(1+i)^n-1}} = \frac{i \cdot p(1 + i)^n}{(1+i)^n-1}\end{align*}$$
::mp(1+一)n1i(1+一)n-1=ip(1+一)n(1+一)n-1

播放段落

N ow solve for the monthly payment given the following:
::现在根据以下条件解决月度付款 :

播放段落
• $\begin{align*}p=350,000\end{align*}$  
  ::p=350 000美元
播放段落
• $\begin{align*}i=0.004\end{align*}$  
  ::i=0.004
播放段落
• $\begin{align*}n=12\cdot30=360\end{align*}$  
  ::n=12=30=360

播放段落

$$\begin{align*}m &= \frac{0.004 \cdot 350000(1 + 0.004)^{360}}{(1+0.004)^{360}-1}=1,836.33\end{align*}$$
::m=0.004350000(1+0.0004360(1+0.0004360-1=1,836.33)

播放段落

Answer:  The monthly mortgage payment will be $1,836.33.
::答复:每月按揭付款为1 836.33美元。

播放段落

Answer the questions  below to practice finding the sum of a geometric series.
::回答下述问题,以便从实践中找出几何序列之和。

 

 

 

播放段落

Discussion Question : How can the formula for the sum of a geometric series be useful when working with the exponential function $\begin{align*}f(x) = 3^x?\end{align*}$  
::讨论问题:在与指数函数f(x)=3x合作时,几何序列总和的公式如何有用?

 

---

播放段落

Activity 3: Euler's Number
::活动3:EULER编号

播放段落

In the previous activity,  an example was presented of how  interest could be compounded. This process is important because there are many examples of exponential growth in nature that compound. Continuous growth can be seen in  b acteria,  populations, and many more examples .  However, these examples do not compound yearly or quarterly; they grow continuously.  Continuous growth means that an object is constantly compounding.  The growth is growing. In activity 1,  you explored how the function  $\begin{align*}f(x)=2^x\end{align*}$   grows as a  pattern of doubling. However, while this function doubles an amount, the doubled quantity is not doubling until after each iteration. Use the interactive below to explore this visually.
::在先前的活动中,有人举了一个例子,说明利息如何会变本加厉。这个过程很重要,因为许多例子表明自然会变本加厉。细菌、人口和更多例子都可以看到持续的增长。但是,这些例子并不每年或每季度复增;它们会持续增长。持续的增长意味着一个物体在不断增殖。增长正在增长。在活动1中,你探讨了函数f(x)=2x如何作为翻一番的模式增长。然而,虽然这个函数使数量翻倍一倍,但每次迭代后数量不会翻一番。使用下文的交互作用来直观地探索这一点。

 

 

播放段落

When a quantity is growing continuously, the growth of the quantity is growing. One way to think of this relationship is that the interest is earning interest. Swiss mathematician Jacob Bernoulli first studied continuous growth in the late 1600s. Bernoulli wanted to better understand what was happening during compound interest so he focused on what would happen to 1 single dollar as it was compounded at an interest rate of 100%. Recall from above  that  interest rate of 100% means that every year, the dollar quantity will double. Bernoulli was especially focused on what would happen to the dollar over the first year.
::当一个数量持续增长时,数量的增长在不断增长。这种关系中的一种思考方式是利息是赚取利息。瑞士数学家雅各布·伯努利在1600年代末期首次研究持续增长。伯努利想更好地了解在复合利息期间发生了什么。伯努利想更好地了解在复合利息期间发生的情况,因此他集中关注一美元将会发生什么,因为美元以100%的利率复杂化。回顾一下超过100%的利率意味着每年美元数量将翻一番。伯努利尤其关注第一年美元将发生什么。

播放段落

Use the interactive below to visually explore what happens as the dollar is compounded at varying frequencies: annually, bi-annually, tri-annually, etc.
::利用下面的交互式互动,对美元在不同频率上复合时发生的情况进行视觉探索:每年、每半年、每三年等等。

 

 

播放段落

Bernoulli saw that the value of the dollar increased as the frequency. However, the value of the dollar increased toward a certain number, which he was never able to determine. It was not until the early 1700s that another Swiss mathematician, Leonhard Euler, discovered this value to be an irrational constant with the value 2.71828... Euler used the letter $\begin{align*}e\end{align*}$  to refer to the constant, and it is now known as Euler's Number.
::Bernoulli发现美元值随着频率增加而增加。 但是,美元值上升到一个他永远无法确定的数字。 直到1700年代初期,另一位瑞士数学家Leonhard Euler才发现这一值是一个非理性的常数,值为2.71828。 尤勒用这封信来表示常数,现在它被称为Euler数字。

播放段落

Recall that continuous growth happens constantly, which represents compounding with infinite frequency. U se the function  $\begin{align*}f(x) = 2.71828...^x \end{align*}$   or  $\begin{align*}f(x) = e^x\end{align*}$  to model continuous growth. The value of $\begin{align*}x\end{align*}$  represents the number of doubling periods, and i n the context of continuous growth, $\begin{align*}x\end{align*}$  represents the number of years. This function assumes a starting value of 1. If, for example, $3 was compounded continuously, the function  $\begin{align*}f(x) = 3e^x\end{align*}$  could be used. Imagine the $1 above undergoing continuous growth 3 times.  This function can be transformed  in a variety of ways. Changing the starting amount will stretch or shrink the function vertically . Multiplying the input, or $\begin{align*}x,\end{align*}$  by a value to cause a horizontal stretch or shrink. This multiplication is performed to alter the rate of doubling. For example, if the interest rate is 50% instead of 100%, the function will increase at a slower rate.  This can be accomplished by multiplying the input value by 0.5 or 50%. he total amount of continuous growth can be modeled as a function of time using the  following function:
::提醒注意连续增长会不断发生, 这代表着无限的频率。 使用函数 f( x) = 2. 71828... x 或 f( x) =ex 来模拟连续增长。 x 值代表双倍期数, 在连续增长的情况下, x 代表年数。 这个函数假定起始值为 1 。 如果, 例如, 3 美元连续增加, 函数 f( x) = 3ex 可以使用。 想象上面的 1 美元 正在连续增长 3 次 。 这个函数可以以多种方式转换 。 改变起始值将垂直拉伸或缩缩缩缩 函数 。 将输入或 x 乘以一个数值导致水平拉伸或缩缩缩 。 此倍数是用来改变翻倍率 。 例如, 如果利率为 50% 而不是 100% , 则函数会以较慢的速度增长 。 这一点可以通过将输入值乘以 0. 5 或 50% 来完成 。 持续增长的总量可以用以下函数作为时间函数的模型 :

播放段落

$\begin{align*}f(t) = a \cdot e^{\:rt}\end{align*}$  
::f(t) =a-iert

播放段落
• $\begin{align*}f(t)\end{align*}$  is the amount of total growth.
  :t) 是总增长额。
播放段落
• $\begin{align*}a\end{align*}$   is the starting amount.
  ::a 是起始数额。
播放段落
• $\begin{align*}e\end{align*}$   is the constant 2.71828...
  ::e 是常数2.71828...
播放段落
• $\begin{align*}r\end{align*}$   is the rate of growth
  ::r 是 r 的增长率
播放段落
• $\begin{align*}t\end{align*}$   is the number of periods, often years.
  ::t 是周期数, 通常是年数 。

播放段落

Example
::示例示例示例示例

播放段落

A $2,500 loan earns interest continuously at a rate of 6% interest.
::2 500美元的贷款以6%的利率持续赚取利息。

播放段落

a.  Write a function to model the total amount owed on the loan as a function of time, in years.
::a. 写入一项函数,将贷款所欠总额按年时间函数来模拟。

播放段落

Since the interest rate is 6%, the  $\begin{align*}r\end{align*}$  value will be 0.06 and will horizontally stretch the parent continuous growth function.  The   $\begin{align*}a\end{align*}$  value is $2,500 because this is the starting amount of the loan. Imagine  Ta king $1 and continuously growing it 2,500 times simultaneously.
::由于利率为6%,r值将为0.06,并将水平延伸父/母连续增长功能。值为2,500美元,因为这是贷款的起始金额。想象一下用1美元并同时持续增长2,500倍。

播放段落

Answer:   $\begin{align*}f(t) = 2,500{e}^{0.06t}\end{align*}$
::答复:f(t)=2 500e0.06t

播放段落

b. How much will be owed on the loan in 12 years ?
::b. 12年内贷款将欠多少?

播放段落

To find the amount owed on the loan in 12 years, find $\begin{align*}f(12).\end{align*}$
::为了找到12年内所欠贷款的数额,请参看f(12)。

播放段落

$\begin{align*}f(12) &= 2,500{e}^{0.06\times 12}\\ &= 2,500{e}^{0.72}\\ &= 5,136.08302... \end{align*}$
::f(12)=2 500e0.06x12=2 500e0.72=5 36.08302...

播放段落

Answer: $5,136.08 
::答复:5 136.08美元

播放段落

Use the interactive below to explore  how various transformations of the parent function   $\begin{align*}f(x) = e^x\end{align*}$  will look.
::使用下面的交互效果来探索父函数 f( x) =ex 的各种变换将如何外观 。

 

 

 

 

播放段落

Discussion Question : How can you explain the answers to the previous 2 questions?
::讨论问题:你如何解释对前两个问题的答复?

  

---

播放段落

Extension: Continuous Interest
::延展期: 连续利息

播放段落

Use the interactive below to explore further.
::利用下面的交互方式进一步探讨。

 

 

 

---

播放段落

Activity 4: Exponential Decay
::活动4:灾害衰减

播放段落

As you saw in the previous section, Exponential Functions, behaves the same as exponential growth, with the exception that the common ratio is between 0 and 1. With a slight change to the exponential growth formula, the exponential decay formula can be written:   $\begin{align*}f(x) = a(1-r)^x\end{align*}$  where $\begin{align*}a\end{align*}$  is the starting amount and $\begin{align*}r\end{align*}$  is the rate of decay, written as a decimal.
::如上一节“指数函数”所示,其行为与指数增长相同,但共同比率介于0和1之间。 如果指数增长公式稍有变化,指数衰变公式可以写成:f(x)=a(1-r)x,其中以起始数为起始数,r为衰变率,以小数记为小数。

播放段落

Example
::示例示例示例示例

播放段落

When businesses calculate the value of their assets, they consider depreciation. Depreciation is the decrease in the value of an object over time. Brandon runs a construction company and bought a new excavator for $60,000, which depreciates 20% per year.
::当企业计算其资产的价值时,它们会考虑折旧。折旧是指一个物体的价值随时间推移而下降。布兰登经营一家建筑公司,并用60 000美元购买了一台新的挖土机,每年贬值20%。

播放段落

a.  Write a function  to model this scenario?
::a. 写一个函数来模拟这一假设情景?

播放段落

The rate of decay is the rate at which an amount decreases. In this case, the rate of decay is 20%. Substituting this into the exponential decay formula will give the following:
::衰变率是数量下降的速率。 在这种情况下,衰变率为20%。 将其转换成指数衰变公式可以提供以下信息:

播放段落

$$\begin{align*}f(x)&=60,000(1-0.20)^x\\ &=60,000(0.8)^x\end{align*}$$
::f(x) = 60,000(1-0.20)x = 60,000(0.8)x

播放段落

Answer:  $\begin{align*}f(x)=60,000(0.8)^x\end{align*}$  
::答复:f(x)=60 000(0.8)x

播放段落

Since the excavator loses 20% of its value every year, this means that it retains 80% of its value. The graph of this model can be seen below:
::由于挖掘机每年损失其价值的20%,这意味着它保留其价值的80%。

播放段落

b. How much will the excavator be worth in 5 years?
::b. 挖掘机五年内价值多少?

播放段落

To find the value in five years, reduce the value by 20% five times, which can be done finding $\begin{align*}f(5).\end{align*}$  
::为了在五年内找到价值,将价值减少20%5倍,可以找到f(5)。

播放段落

$\begin{align*}f(x)&=60,000 \cdot {0.8}^{(5)}\\ &=60,000 \cdot 0.32768\\ &=19,660.8\end{align*}$  
::f(x)=6000.8(5)=6000.32768=19660.8

播放段落

Answer:   $\begin{align*}$19,660.80\end{align*}$
::答复:19 660.80美元

播放段落

In  the section  Modeling With Quadratic Functions,  a piecewise function was used to represent the height of a bouncing ball. However, this could also be represented using an exponential function since the ball only bounces to a percentage of its previous height. Use the interactive below to explore this further.
::在具有二次函数建模的一节中,用一个片断函数表示弹跳球的高度。但是,也可以用指数函数表示,因为球只弹出到其先前高度的某个百分比。用下面的交互函数来进一步探讨这一点。

 

 

 

 

   

---

播放段落

Wrap-Up: Review Questions
::总结:审查问题