课程： 8.3 使用指数增长和衰减模型

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使用指数增长和衰减模型

The half-life of an isotope of barium is about 10 years. The half-life of a substance is the amount of time it takes for half of that substance to decay. If a nuclear scientist starts with 200 grams of barium, how many grams will remain after 100 years?
::同位素的半衰期约为10年左右。一种物质的半衰期是该物质一半衰减所需的时间。如果核科学家从200克开始,那么100年后还剩下多少克?

Growth and Decay Factors
::增长和衰减因素

When a real-life quantity increases by a percentage over a period of time, the final amount can be modeled by the equation : $\begin{align*}A=P(1+r)^t\end{align*}$ , where $\begin{align*}A\end{align*}$ is the final amount, $\begin{align*}P\end{align*}$ is the initial amount, $\begin{align*}r\end{align*}$ is the rate (or percentage), and $\begin{align*}t\end{align*}$ is the time (in years). $\begin{align*}1+r\end{align*}$ is known as the growth factor .
::当一个实际生命量在一段时间内以百分比增长时,最终数量可以方程式模拟:A=P(1+r)t,其中A为最终数量,P为初始数量,r为比例(或百分比),t为时间(年)。 1+r称为增长系数。

Conversely, a real-life quantity can decrease by a percentage over a period of time. The final amount can be modeled by the equation: $\begin{align*}A=P(1-r)^t\end{align*}$ , where $\begin{align*}1-r\end{align*}$ is the decay factor .
::反之,一个实际生命量可在一段时间内减少一个百分比。最终数量可以通过公式来模拟:A=P(1-r)t,其中1-r是衰变系数。

Let's solve the following problems.
::让我们解决以下的问题。

The population of Coleman, Texas grows at a 2% rate annually. If the population in 2000 was 5981, what was the population is 2010? Round up to the nearest person.
::得克萨斯州科尔曼的人口年增长率为2%,如果2000年的人口为5981人,那么2010年的人口是多少?

First, set up an equation using the growth factor. $\begin{align*}r=0.02, t=10,\end{align*}$ and $\begin{align*}P=5981\end{align*}$
::首先,使用增长系数(r=0.02,t=10)和P=5981来设置方程式

\begin{aligned} A & = 5981 (1 + 0.02)^{10} \\ = 5981 (1.02)^{10} \\ = 7291 people \end{aligned}

$\begin{align*}A & =5981(1+0.02)^{10} \\ & =5981(1.02)^{10} \\ & =7291 \ \text{people}\end{align*}$
::A=5981(1+0.02)10=5981(1.02)10=7291人

You deposit $1000 into a savings account that pays 2.5% annual interest. Find the balance after 3 years if the interest rate is compounded a) annually, b) monthly, and c) daily.
::您将 1000 美元存入储蓄账户 , 储蓄账户支付 2.5% 的年息。如果年利率 (a) 、 (b) 月、 (c) 日 , 请在 3 年后找到余额。

For part a, we will use $\begin{align*}A=1000(1.025)^3=1008.18\end{align*}$ .
::a部分将使用A=1000(1.025)3=1008.18。

But, to determine the amount if it is compounded in amounts other than yearly, we need to alter the equation. For compound interest, the equation is $\begin{align*}A=P \left(1+ \frac{r}{n}\right)^{nt}\end{align*}$ , where $\begin{align*}n\end{align*}$ is the number of times the interest is compounded within a year. For part b, $\begin{align*}n=12\end{align*}$ .
::但是,要确定是否以年度以外的金额加在一起的金额,我们需要改变方程。对于复合利息, 方程是 A=P(1+rn)n, 其中 n 是利息在一年之内加在一起的倍数。 b, n=12 部分。

\begin{aligned} A & = 1000 {(1 + \frac{0.025}{12})}^{12 \cdot 3} \\ = 1000 (1.002)^{36} \\ = 1077.80 \end{aligned}

$\begin{align*}A & =1000 \left(1+ \frac{0.025}{12}\right)^{12 \cdot 3} \\ & =1000(1.002)^{36} \\ & =1077.80\end{align*}$
::A=1000(1+0.02512)12_3=1000(1.002)36=1077.80

In part c, $\begin{align*}n=365\end{align*}$ .
::部分c, n=365。

\begin{aligned} A & = 1000 {(1 + \frac{0.025}{365})}^{365 \cdot 3} \\ = 1000 (1.000068)^{1095} \\ = 1077.88 \end{aligned}

$\begin{align*}A & =1000 \left(1+ \frac{0.025}{365}\right)^{365 \cdot 3} \\ & =1000(1.000068)^{1095} \\ & =1077.88\end{align*}$
::A=1000(1+0.0253653653653=1000(1.000068)1095=1077.88)

You buy a new car for $35,000. If the value of the car decreases by 12% each year, what will the value of the car be in 5 years?
::你买一辆新车要三万五千美元如果汽车每年减少12% 5年后车的价值会是多少?

This is a decay function because the value decreases .
::这是一个衰变函数,因为值会下降。

\begin{aligned} A & = 35000 (1 - 0.12)^{5} \\ = 35000 (0.88)^{5} \\ = 18470.62 \end{aligned}

$\begin{align*}A & =35000 (1-0.12)^5 \\ & =35000(0.88)^{5} \\ & =18470.62\end{align*}$
::A=35000(1-0.12.5)5=35000(0.885=18470.62)

The car would be worth $18,470.62 after five years.
::五年后车价18 470.62美元

Examples
::实例

Example 1
::例1

Earlier, you were asked to find the number of grams of barium that will remain after 100 years if you start with 200 grams and the half-life of this barium isotope is 10 years.
::早些时候,有人要求你找到在100年后会留下的克数量如果你从200克开始, 同位素的半衰期是10年。

This is an example of , so we can once again use the exponential form $\begin{align*}f(x)=a \cdot b^{x-h}+k\end{align*}$ . In this case, a = 200, the starting amount; b is 1/2, the rate of decay; x-h = 100/10 = 10, and k = 0.
::这是一个示例, 所以我们可以再次使用指数表 f( x) = abx-h+k。在此情况下, a = 200, 起始数; b 是 1/2, 衰变速度; x-h = 100/ 10 = 10, k = 0 。

\begin{aligned} P = 200 \cdot {\frac{1}{2}}^{10} \\ = 200 \cdot \frac{1}{1024} = 0.195 \end{aligned}

$\begin{align*}P = 200 \cdot {\frac{1}{2}}^{10}\\ = 200 \cdot {\frac{1}{1024}} = 0.195\end{align*}$
::P=2001210=20011024=0.195

Therefore , 0.195 grams of the barium still remain 100 years later.
::因此,100年后,的0.195克仍保留在100年之后。

Example 2
::例2

Tommy bought a truck 7 years ago that is now worth $12,348. If the value of his truck decreased 14% each year, how much did he buy it for? Round to the nearest dollar.
::Tommy7年前买了一辆卡车,现在价值12 348美元,如果他的卡车价值每年下降14%,他买多少?

Tommy needs to use the formula $\begin{align*}A = P(1-r)^t\end{align*}$ and solve for $\begin{align*}P\end{align*}$ .
::汤米需要使用 A=P(1-r)t 公式并解决 P 。

\begin{aligned} 12348 & = P (1 - 0.14)^{7} \\ 12348 & = P (0.86)^{7} Tommy's truck was originally $ 35, 490. \\ \frac{12348}{(0.86)^{7}} & = P \approx 35490 \end{aligned}

$\begin{align*}12348 &= P(1-0.14)^7 \\ 12348 &= P(0.86)^7 \qquad \qquad \text{Tommy's truck was originally} \ \$35,490. \\ \frac{12348}{(0.86)^7} &= P \approx 35490\end{align*}$
::托米的卡车最初为35,490.12348(0.866)7 = P35490

Example 3
::例3

The Wetakayomoola credit card company charges an Annual Percentage Rate (APR) of 21.99%, compounded monthly. If you have a balance of $2000 on the card, what would the balance be after 4 years (assuming you do not make any payments)? If you pay $200 a month to the card, how long would it take you to pay it off? You may need to make a table to help you with the second question.
::Wetakayomoola信用卡公司每月收取21.99%的年百分比(APR),复数每月。如果卡上有2 000美元的余额,四年后余额会是多少(假设你没有支付任何款项)?如果你每月向卡支付200美元,你需要花多久才能付清?你可能需要做一张表格来帮助你解决第二个问题。

You need to use the formula $\begin{align*}A=P \left(1+ \frac{r}{n}\right)^{nt}\end{align*}$ , where $\begin{align*}n=12\end{align*}$ because the interest is compounded monthly.
::您需要使用公式 A=P(1+rn)n, n=12 因为利息是每月复合的。

\begin{aligned} A & = 2000 {(1 + \frac{0.2199}{12})}^{12 \cdot 4} \\ = 2000 (1018325)^{48} \\ = 4781.65 \end{aligned}

$\begin{align*}A & =2000 \left(1+ \frac{0.2199}{12}\right)^{12 \cdot 4} \\ & =2000(1018325)^{48} \\ & =4781.65\end{align*}$
::A=2000(1+0.219912)124=2000(1018325)48=4781.65

To determine how long it will take you to pay off the balance, you need to find how much interest is compounded in one month, subtract $200, and repeat. A table might be helpful. For each month after the first, we will use the equation, $\begin{align*}B=R \left(1+ \frac{0.2199}{12}\right)^{12 \cdot (\frac{1}{12})} = R(1.018325)\end{align*}$ , where $\begin{align*}B\end{align*}$ is the current balance and $\begin{align*}R\end{align*}$ is the remaining balance from the previous month. For example, in month 2, the balance (including interest) would be $\begin{align*}B=1800 \left(1+ \frac{0.2199}{12}\right)^{12 \cdot \left(\frac{1}{12}\right)}= 1800 \cdot 1.08325=1832.99\end{align*}$ .
::要确定清结余额需要多长时间, 您需要找到一个月的利息加起来多少, 减去200美元, 重复。表格可能有用。对于第一个月之后的每个月, 我们将使用公式 B=R(1+0. 219912)12( 112) = R (1.0 18325), B 是当前余额, R 是前一个月的余额。例如, 在第2个月, 余额( 包括利息) 为 B= 1800(1+0. 219912) 12( 112) = 18001.08325 = 1832.99 。

Month	1	2	3	4	5	6	7	8	9	10	11
Balance	2000	1832.99	1662.91	1489.72	1313.35	930.09	790.87	640.06	476.69	299.73	108.03
Payment	200	200.00	200.00	200.00	200.00	200.00	200.00	200.00	200.00	200.00	108.03
Remain	$1800	1632.99	1462.91	1289.72	913.35	730.09	590.87	440.06	276.69	99.73	0

It is going to take you 11 months to pay off the balance and you are going to pay 108.03 in interest, making your total payment $2108.03.
::你们要花11个月的时间来还清余额你们要付108.03的利息总共支付2108.03美元

Example 4
::例4

As the altitude increases, the atmospheric pressure (the pressure of the air around you) decreases. For every 1000 feet up, the atmospheric pressure decreases about 4%. The atmospheric pressure at sea level is 101.3. If you are on top of Hevenly Mountain at Lake Tahoe (elevation about 10,000 feet) what is the atmospheric pressure?
::随着高度升高,大气压力(你周围的空气压力)下降。每上升1000英尺,大气压力下降约4%。海平面的大气压力为101.3。如果你在Tahoe湖的Hevenly山顶(高度约10,000英尺),那么大气压力是什么?

The equation will be $\begin{align*}A=101,325(1-0.04)^{100}=1709.39\end{align*}$ . The decay factor is only raised to the power of 100 because for every 1000 feet the pressure decreased. Therefore, $\begin{align*}10,000 \div 1000=100\end{align*}$ . Atmospheric pressure is what you don’t feel when you are at a higher altitude and can make you feel light-headed. The picture below demonstrates the atmospheric pressure on a plastic bottle. The bottle was sealed at 14,000 feet elevation (1), and then the resulting pressure at 9,000 feet (2) and 1,000 feet (3). The lower the elevation, the higher the atmospheric pressure, thus the bottle was crushed at 1,000 feet.
::方程为 A= 101 325 (1- -0.04100 = 1709.39). 衰变系数仅提高到100, 因为每1000英尺压力降下。因此, 10 000 1000 = 100 。大气压力是高海拔时你感受不到的, 可以让你感觉到光头。下面的图片显示了塑料瓶的大气压力。瓶子密封在14 000英尺高一级(1) , 然后产生的压力在9 000英尺(2) 和 1 000英尺(3) 。高度越低, 大气压力越高, 瓶子被压压在1 000英尺。

Review
::回顾

Use an exponential growth or exponential decay function to model the following scenarios and answer the questions.
::使用指数增长或指数衰减函数来模拟以下情景并回答问题。

Sonya’s salary increases at a rate of 4% per year. Her starting salary is $45,000. What is her annual salary, to the nearest $100, after 8 years of service?
::索尼亚的年工资增长率为4%。她的起始工资为45,000美元。服务8年后,她的年工资是多少,最接近的100美元?

The value of Sam’s car depreciates at a rate of 8% per year. The initial value was $22,000. What will his car be worth after 12 years to the nearest dollar?
::萨姆的汽车年折旧率为8%。最初价值为22,000美元。他的汽车在12年到最接近的美元之后值多少?

Rebecca is training for a marathon. Her weekly long run is currently 5 miles. If she increase her mileage each week by 10%, will she complete a 20 mile training run within 15 weeks?
::Rebecca正在接受马拉松训练。她每周的长程是5英里。如果她每周的里程增加10%,她是否会在15周内完成20英里的训练?

An investment grows at a rate of 6% per year. How much, to the nearest $100, should Noel invest if he wants to have $100,000 at the end of 20 years?
::投资以每年6%的速度增长。如果诺埃尔想要在20年结束时获得10万美元,那么他应该投资多少,再到最近的100美元呢?

Charlie purchases a 7 year old used $\begin{align*}RV\end{align*}$ for $54,000. If the rate of depreciation was 13% per year during those 7 years, how much was the $\begin{align*}RV\end{align*}$ worth when it was new? Give your answer to the nearest one thousand dollars.
::Charlie用54,000美元的房车购买一个7岁的房车。如果这7年的折旧率为每年13%,那么新房车的价值是多少?回答最接近的1000美元。

The value of homes in a neighborhood increase in value an average of 3% per year. What will a home purchased for $180,000 be worth in 25 years to the nearest one thousand dollars?
::邻里住宅的价值每年平均增长3%。以180,000美元购买的房屋在25年内价值多少? 最接近1,000美元?

The population of a community is decreasing at a rate of 2% per year. The current population is 152,000. How many people lived in the town 5 years ago?
::一个社区的人口正在以每年2%的速度下降,目前的人口为152,000人,5年前有多少人居住在该镇?

The value of a particular piece of land worth $40,000 is increasing at a rate of 1.5% per year. Assuming the rate of appreciation continues, how long will the owner need to wait to sell the land if he hopes to get $50,000 for it? Give your answer to the nearest year.
::价值40,000美元的一块土地的价值正在以每年1.5%的速度增长。假设升值率还在继续,如果所有者希望得到50,000美元的土地,需要等待多久才能出售土地? 请回答最近的一年。

For problems 9-15, use the formula for compound interest: $\begin{align*}A=P \left(1+ \frac{r}{n}\right)^{nt}\end{align*}$ .
::对于问题9-15,使用复利公式:A=P(1+rn)nt。

If $12,000 is invested at 4% annual interest compounded monthly, how much will the investment be worth in 10 years? Give your answer to the nearest dollar.
::如果12 000美元投资为每月4%的年息复利,那么10年的投资值是多少?给最近的美元回答。

If $8,000 is invested at 5% annual interest compounded semiannually, how much will the investment be worth in 6 years? Give your answer to the nearest dollar.
::如果每半年投资8 000美元,年利率为5%,再加上每半年投资一次,那么6年的投资价值是多少?回答最接近的美元。

If $20,000 is invested at 6% annual interested compounded quarterly, how much will the investment be worth in 12 years. Give your answer to the nearest dollar.
::如果投资20 000美元,按每年6%有兴趣的复合季度计算,12年内投资价值是多少。请回答最接近的美元。

If $5,000 is invested at 8% annual interest compounded quarterly, how much will the investment be worth in 15 years? Give your answer to the nearest dollar.
::如果以8%的年利率投资5 000美元,每季复计,15年的投资价值是多少?回答最接近的美元。

How much of an initial investment is required to insure an accumulated amount of at least $25,000 at the end of 8 years at an annual interest rate of 3.75% compounded monthly? Give your answer to the nearest one hundred dollars.
::最初投资需要多少资金才能在8年结束时以每月3.75%的年利率保证累计金额至少为25 000美元,每月3.75%的复利? 回答最接近的100美元。

How much of an initial investment is required to insure an accumulated amount of at least $10,000 at the end of 5 years at an annual interest rate of 5% compounded quarterly? Give your answer to the nearest one hundred dollars.
::最初投资需要多少资金才能保证在5年结束时以5%的年利率按季度复合利率支付至少10 000美元的累计金额?回答最接近的100美元。

Your initial investment of $20,000 doubles after 10 years. If the bank compounds interest quarterly, what is your interest rate?
::10年后,你最初投资了20,000美元的双倍。如果银行每季增加利息,你的利率是多少?

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。

章节大纲

常规

使用指数增长和衰减模型

Growth and Decay Factors ::增长和衰减因素

Examples ::实例

Example 1 ::例1

Example 2 ::例2

Example 3 ::例3

Example 4 ::例4

Review ::回顾

Review (Answers) ::回顾(答复)