3.12 腐蚀指数

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• 选择节指数衰落

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  指数衰落

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  You observe that the glass of water you took out of the microwave dropped from 190 deg to 170 deg in 1 min. In another minute, the temperature has dropped to 152 deg., and after 3 mins, it is down to 136 deg.
  ::你观察到你从微波炉中取出的水杯在1分钟内从190度下降到170度。再过一分钟,温度下降到152度,3分钟后下降到136度。

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  What type of function would allow you to calculate the projected temperature of the water over time?
  ::哪种函数允许您计算一段时间内水的预测温度?

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  Exponential Decay
  ::指数衰落

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  A decreasing quantity can be modeled with an exponential function in much the same way as an increasing (growing) quantity can. This kind of situation is referred to as exponential decay.
  ::减少的数量可以与增加(增加)的数量一样,以指数函数模拟。 这种情况被称为指数衰减。

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  Perhaps the most common example of exponential decay is that of radioactive decay, which refers to the transformation of an atom of one type into an atom of a different type, when the nucleus of the atom loses energy. The rate of radioactive decay is usually measured in terms of “half-life,” or the time it takes for half of the atoms in a sample to decay. For example Carbon-14 is a radioactive isotope that is used in “carbon dating,” a method of determining the age of organic materials. The half-life of Carbon-14 is 5730 years. This means that if we have a sample of Carbon-14, it will take 5730 years for half of the sample to decay. Then it will take another 5730 years for half of the remaining sample to decay, and so on.
  ::也许最常见的指数衰减例子是放射性衰变,它是指当原子核失去能量时,将一种原子的原子转换成另一种原子。放射性衰变的速度通常以“半衰期”或样本中一半原子衰变所需的时间来衡量。例如,碳-14是一种放射性同位素,用于“碳约会”,一种确定有机材料年龄的方法。碳-14的半衰期是5730年。这意味着如果我们有碳-14的样本,那么一半样本的衰变将需要5730年。然后,剩下的一半样本还要花5730年才能衰变,等等。

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  We can model decay using the same form of equation we use to model growth, except that the exponent in the equation is negative: A ( t ) = A ₀ e ^{- kt} .
  ::我们可以用我们用来模拟增长的同一种方程式模拟衰变,但方程式中的引号为负数:A(t) = A0e-kt。

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  Newton’s Law of Cooling is an exponential decay model. The Law of Cooling allows us to determine the temperature of a cooling (or warming) object, based on the temperature of the surroundings and the time since the object entered the surroundings. The general form of the cooling function is T ( x ) = T ₅ + ( T ₀ - T ₅ ) e ^{- kx} , where T ₅ , is the surrounding temperature, T ₀ is the initial temperature, and x represents the time since the object began cooling or warming.
  ::牛顿的《冷却法则》是一种指数衰变模式。 《冷却法则》允许我们根据周围的温度和物体进入周围的时间来确定冷却(或变暖)物体的温度。 冷却功能的一般形式是T(x) = T5 + (T0 - T5) e- kx,其中T5是周围的温度,T0是初始温度,x是天体开始冷却或变暖以来的时间。

  

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  The first graph shows a situation in which an object is cooling. The graph has a horizontal asymptote at y = 70. This tells us that the object is cooling to $\begin{align*}70^{\circ}\end{align*}$ F. The second graph has a horizontal asymptote at y = 70 as well, but in this situation, the object is warming up to $\begin{align*}70^{\circ}\end{align*}$ F. We can use the general form of the function to answer questions about cooling (or warming) situations.
  ::第一个图形显示一个对象在冷却中的情况。 图形在y = 70 上有一个水平淡化状态。 这告诉我们该对象正在冷却到 70°F。 第二个图形在y = 70°F 上有一个水平淡化状态, 但在此情况下, 该对象正在变暖到 70°F 。 我们可以使用该函数的一般形式来回答关于冷却( 或变暖) 情况的问题 。

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  Examples
  ::实例

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  Example 1
  ::例1

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  Earlier, you were asked to find the type of function that would allow you to calculate the projected temperature of water over time.
  ::早些时候,你被要求找到一种功能 来计算一段时间内水的预测温度。

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  The temperature drop of the water over time is an example of exponential decay.
  ::随着时间的推移,水的温度下降是指数衰减的一个例子。

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  To predict the temperature of the water over time, you would use a function like $\begin{align*}T_f = T_i (.8947)^t\end{align*}$ where $\begin{align*}T_f\end{align*}$ and $\begin{align*}T_i\end{align*}$ are starting and ending temperature and $\begin{align*}t\end{align*}$ is time in minutes.
  ::为了预测一段时间内水的温度,您将使用Tf=Ti( 8947)t这样的函数,Tf和Ti在开始和结束温度, t是分钟内的时间。

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  Example 2
  ::例2

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  You have a sample of Carbon-14. How much time will pass before 75% of the original sample remains?
  ::你有碳14的样本。在75%的原始样本留下之前,还要花多少时间?

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  We can use the half-life of 5730 years to determine the value of k :
  ::我们可以使用5730年的半衰期来确定 k:

  $\begin{align*}A(t) = A_0e^{-kt}\end{align*}$
  $\begin{align*}\frac{1} {2} = 1e^{-k\cdot 5730}\end{align*}$	We do not know the value of A 0 , so we use “1” as 100%. (1/2) of the sample remains when t = 5730 years
  $\begin{align*}ln \frac{1} {2} = ln e^{-k\cdot 5730}\end{align*}$	Take the ln of both sides
  $\begin{align*}ln \frac{1} {2} = -5730 klne\end{align*}$	Use the power property of logs
  $\begin{align*}ln \frac{1} {2} = -5730k\end{align*}$	$\begin{align*}ln(e) = 1\end{align*}$
  $\begin{align*}-ln 2 = -5730k\end{align*}$	$\begin{align*}ln(1/2) = ln(2^{-1})=-ln2\end{align*}$
  $\begin{align*}ln 2 = 5730k\end{align*}$
  $\begin{align*}k = \frac{ln2} {5730}\end{align*}$	Isolate k

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  Now we can determine when the amount of Carbon-14 remaining is 75% of the original:
  ::现在,我们可以确定碳14的剩余数量是原量的75%:

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  $\begin{align*}0.75 = 1 e^{\frac{-ln 2} {5730}t}\end{align*}$
  ::0.75=1e-ln25730t

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  $\begin{align*}0.75 = 1 e^{\frac{-ln 2} {5730}t}\end{align*}$
  ::0.75=1e-ln25730t

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  $\begin{align*}ln (0.75) = ln e^{\frac{-ln 2} {5730}t}\end{align*}$
  ::In( 0. 75) = lne - ln25730t

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  $\begin{align*}ln (0.75) = \frac{-ln 2} {5730}t\end{align*}$
  ::在( 0. 75 ) 中 = = = = = = = = = = = = = = = = = ======== = ===================== === ============================================================================================================================================================================================================================================================================================================================================================================================================================================

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  $\begin{align*}t = \frac{5730 ln (0.75)} {-ln 2} \approx 2378\end{align*}$
  ::t = 5730n( 0. 75) - ln2 2378

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  Therefore it would take about 2,378 years for 75% of the original sample to be remaining.
  ::因此,保留原样本的75%需要大约2 378年。

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  Example 3
  ::例3

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  You are baking a casserole in a dish, and the oven is set to $\begin{align*}325^{\circ}\end{align*}$  F. You take the pan out of the oven and put it on a cooling rack in your kitchen which is $\begin{align*}70^{\circ}\end{align*}$  F, and after 10 minutes the pan has cooled to $\begin{align*}300^{\circ}\end{align*}$ F. How long will it take for the pan to cool to 200 ⁰  F?
  ::你正在烤盘中的锅炉,烤箱设在325°F。你把锅子从烤箱里拿出来,放在厨房的冷却柜子上,厨房的冷却架子是70°F,10分钟后锅子冷却到300°F。锅子冷却到2000F需要多长时间?

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  We can use the general form of the equation and the information given in the problem to find the value of k :
  ::我们可以使用方程的一般形式和问题中提供的信息来找到 k:

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  $\begin{align*}T(x) = T_s + (T_0 -T_s) e^{-kx}\end{align*}$
  ::T(x) =Ts+(T0-Ts) e-kx

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  $\begin{align*}T(x) = 70 + (325 - 70)e^{-kx}\end{align*}$
  ::T(x)=70+(325-70)e-kx

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  $\begin{align*}T(x) = 70 + (255)e^{-kx}\end{align*}$
  ::T(x)=70+(255)e-kx

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  $\begin{align*}T(10) = 70 + 255e^{-10k} = 300\end{align*}$
  ::T(10)=70+255e-10k=300

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  $\begin{align*}255e^{-10k} = 230\end{align*}$
  ::255e-10k=230

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  $\begin{align*}e^{-10k} = \frac{230} {255}\end{align*}$
  ::电子-10k=230255

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  $\begin{align*}ln e^{-10k} = ln \left (\frac{230} {255}\right )\end{align*}$
  ::Ine- 10k=ln( 230255)

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  $\begin{align*}-10k = ln \left (\frac{230} {255}\right )\end{align*}$
  ::-10k=ln(230255)

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  $\begin{align*}k = \frac{ln \left (\frac{230} {255}\right )} {-10} \approx 0.0103\end{align*}$
  ::kn=00( 230255) - 10 0.0103

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  Now we can determine the amount of time it takes for the pan to cool to 200 degrees:
  ::现在我们可以确定锅冷到200度需要多少时间:

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  $\begin{align*}T(x) = 70 + (255)e^{-.0103}x\end{align*}$
  ::T(x)=70+(255)e-.0103x

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  $\begin{align*}T(x) = 70 + (255)e^{-.0103}x\end{align*}$
  ::T(x)=70+(255)e-.0103x

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  $\begin{align*}200 = 70 + (255)e^{-.0103}x\end{align*}$
  ::200=70+(255)e -.0103x

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  $\begin{align*}130 = (255)e^{-.0103}x\end{align*}$
  ::130=(255)e-.0103x

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  $\begin{align*}\frac{130} {255} = e^{-.0103}x\end{align*}$
  ::130255=e-0103x

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  $\begin{align*}ln \left (\frac{130} {255}\right ) = ln e^{-.0103}x\end{align*}$
  ::In( 130255) = lne - 0103x

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  $\begin{align*}ln \left (\frac{130} {255}\right ) = -.0103 x\end{align*}$
  ::In(130255)0103x

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  $\begin{align*}x = \frac{ln \left (\frac{130} {255}\right )} {-.0103} \approx 65\end{align*}$
  ::x=00(130255)-.010365

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  Therefore, in the given surroundings, it would take about an hour for the pan to cool to 200 degrees.
  ::因此,在给定的周围环境中, 锅需要大约一个小时才能冷到200度。

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  Example 4
  ::例4

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  When doctors prescribe medicine, they consider how quickly the drug’s effectiveness decreases as time passes in order to calculate the time to administer the next dose. If a drug is only 85% as effective each hour as it was the previous hour, at some point the patient must be given another dose. If the initial dose was 350 mg, how long will it take for the initial dose to reach the minimum level of 83 mg, to the nearest hour?
  ::当医生开药时,他们会考虑药效随着时间的流逝而下降的速度,以便计算下一种剂量的服用时间。 如果药物每小时只有前一个小时的85%有效,那么在某个时候病人必须再服一次剂量。 如果最初剂量为350毫克,那么最初剂量要达到83毫克的最低剂量,到最接近的时段还要花多长时间?

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  Use the general decay formula : A ( t ) = A ₀ r ^t , to solve for t , time in hours.
  ::使用一般衰变公式: A(t) = A0rt, 解答 t, 时间以小时计 。

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  Substituting gives: 83 = 350 (.85) ^t
  ::替代给付 : 83 = 350 (.85)t

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  Dividing both sides by 350 gives: .237 = .85 ^t
  ::双方分出350个给:237 = .85t

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  Using logs: log .237 = log .85 ^t
  ::使用日志: log 237 = log.85t

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  Properties of logs: log .237 = t (log .85)
  ::日志属性:日志 237 = t(log.85)

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  Dividing to isolate t : t = log .237 / log .85
  ::分离 t: t = log 237 / log .85

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  Using a calculator: t = (-.625) / (-.07) = 9 hours (apx.)
  ::使用计算器: t = (- 625) / (- . 07) = 9 小时( px.)

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  For the following examples, use the model below:
  ::对于以下例子,请使用以下模式:

  Exponential Decay Model	$\begin{align*}A_f = A_i (1 - r)^t\end{align*}$
  $\begin{align*}A_f\end{align*}$ = final amount
  $\begin{align*}A_i\end{align*}$ = initial amount
  $\begin{align*}r\end{align*}$ = decay rate (percent as a decimal)
  $\begin{align*}t\end{align*}$ = time
  $\begin{align*}(1 - r)\end{align*}$ = decay factor

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  Example 5
  ::例5

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  If a particular Egyptian artifact originally contained 21 grams of carbon-14, and if carbon-14 decays at a rate modeled by $\begin{align*}A_f = A_i (e)^{-0.000121t}\end{align*}$ how much carbon-14 should be present after 25000 years?
  ::如果某一埃及工艺品最初含有21克碳-14,如果碳-14以Af=Ai(e)-0.00011t的模型速度腐烂,25 000年后碳-14应存在多少?

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  To calculate the remaining carbon-14, start with the model:
  ::为计算剩余的碳-14, 从模型开始 :

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  $\begin{align*}A_f = A_i (e)^{-0.000121t}\end{align*}$
  ::Af=Ai(e)-0000121t

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  $\begin{align*}A_f = 21e^{-0.000121(25000)}\end{align*}$ substitute the given values
  ::Af=21e-00001221( 2500) 替换给定值

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  $\begin{align*}A_f = 21e^{-3.025}\end{align*}$ simplify
  ::Af=21e-3.025 简化

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  $\begin{align*}A_f = 21 \cdot .048557821\end{align*}$ with a calculator
  ::Af=21048557821,带有计算器

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  $\begin{align*}A_f \approx 1.019714\end{align*}$
  ::1979年1月14日,纽约

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  $\begin{align*}\therefore\end{align*}$ there will be approximately 1.019714 grams of carbon-14 left after 25,000 years.
  ::在25000年之后, 将剩下大约1.01974克碳-14。

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  Example 6
  ::例6

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  thorium-234 decays with a half-life of 24 days to protactinium-234. If the decay rate can be calculated by $\begin{align*}t_{\frac{1}{2}} = \frac{ln 2}{r}\end{align*}$ where $\begin{align*}t_{\frac{1}{2}}\end{align*}$ is the half-life of the material and $\begin{align*}r\end{align*}$ is the rate of decay, what is the rate of decay of thorium-234?
  ::-234的衰变,半衰期为24天,而-234的衰变为24天。如果可以用t12=n2r来计算衰变率,那么当该物质的半衰期为t12时,而r是衰变率时,-234的衰变率是多少?

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  To calculate decay rate, begin with the formula given in the problem:
  ::要计算衰减率,首先从问题给出的公式开始:

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  $\begin{align*}t_{\frac{1}{2}} = \frac{ln 2}{r}\end{align*}$
  ::t12=n2r

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  $\begin{align*}24 = \frac{ln 2}{r}\end{align*}$ : substitute the given half-life value
  ::24=ln2r : 替换给定半衰期值

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  $\begin{align*}24 = \frac{.69}{r}\end{align*}$ : substitute the value of $\begin{align*}ln 2\end{align*}$ with a calculator
  ::24=.69r : 用计算器取代 I02 值

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  $\begin{align*}r = \frac{.69}{24}\end{align*}$ : re-arrange to solve for $\begin{align*}r\end{align*}$
  ::r=6924 : 重新排列以解 r

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  $\begin{align*}r \approx .029\end{align*}$
  ::r. 029

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  $\begin{align*}\therefore\end{align*}$ the rate of decay is $\begin{align*}\approx 2.9\%\end{align*}$ per day
  ::衰变率为每天2.9%

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  Example 7
  ::例7

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  If thorium-234 has a decay rate of $\begin{align*}\approx .0289\end{align*}$ as calculated above, and if you begin with a sample of thorium-234 weighing 37 grams, how long will it take before you have only 23 grams?
  ::如果234的衰变速率按照上文计算为0289,如果首先对重37克的234进行取样,那么在只有23克之前需要多长时间?

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  To calculate time, begin with the exponential decay formula:
  ::要计算时间, 从指数衰变公式开始 :

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  $\begin{align*}A_f = A_i (1 - r)^t\end{align*}$
  ::Af=Ai(1-r)t

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  $\begin{align*}23 = 34 (.971)^t\end{align*}$ : substitute the given values
  ::23=34(971t):替换给定值

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  $\begin{align*}.6765 = .971^t\end{align*}$ : divide both sides by 34
  ::6765=971t:将双方除以34

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  $\begin{align*}log .6765 = log .971^t\end{align*}$ : take the log of both sides
  ::log.6765=log.971t:使用两侧的日志

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  $\begin{align*}log .6765 = (t) log .971\end{align*}$ : using $\begin{align*}log x^y = y log x\end{align*}$
  ::log.6765=(t)log.971:使用logxy=ylogx

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  $\begin{align*}\frac{log .6765}{log .971} = t\end{align*}$ : divide both sides by $\begin{align*}log .971\end{align*}$
  ::log.6765log.971=t:将两边除以log.971

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  $\begin{align*}13.28 = t\end{align*}$ : with a calculator
  ::13.28=t:使用计算器

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  $\begin{align*}\therefore\end{align*}$ it will be approximately 13.28 days before the 37 gram sample of thorium-234 decays into a 23 gram sample of thorium-234.
  ::大约在37克的-234样本衰变为23克的-234样本之前13.28天。

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  Review
  ::回顾

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  Are the following equations examples of exponential decay?
  ::以下方程式是指数衰减的例子吗?

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  1. $\begin{align*}y = 0.7(1.1)^t\end{align*}$
     ::y= 0.7(1.1)t
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  2. $\begin{align*}y = 0.95(0.3t)^2\end{align*}$
     ::y=0.95(0.3吨)2
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  3. Write an exponential decay model for a new car valued at $28,000 which depreciates at a rate of 8% per year.
     ::为一辆新车写一个指数衰变模型,价值28 000美元,每年贬值8%。
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  4. Write an exponential decay model for a stock in a new dyeing company is valued at $500.00. Its value decreases by 7% each year.
     ::为一家新染色公司的股票写一个指数衰减模型,价值为500美元。其价值每年减少7%。
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  5. What is the value of a car after 5 years, if you buy it for $12,000 and it depreciates at a rate of 10% annually?
     ::5年后的汽车价值是多少? 如果你买一万2千美元,每年以10%的速度贬值?

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  Identify the initial amount and the decay rate in the exponential function.
  ::识别指数函数中的初始数量和衰变率。

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  6. $\begin{align*}y = 10(1 - 0.2)^t\end{align*}$
     ::y=10( 1- 0.2) t
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  7. $\begin{align*} y = 18(0.11)^t\end{align*}$
     ::y=18(0.11)t
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  8. $\begin{align*}y = 2(\frac{1}{4})^t\end{align*}$
     ::y=2( 14) t

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  Write an exponential function to model the situation:
  ::写入一个指数函数以模拟局势 :

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  9. A $32,000 boat depreciates at a rate of 5.2% every other year.
     ::一艘32 000美元的船沉船,每隔一年以5.2%的速度折旧。
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  10. The population of a small country is 185,000 people and it increases by 1.75% semi-annually.
      ::一个小国的人口为185,000人,每半年增加1.75%。
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  11. The initial investment of $732.00 in a mutual fund is losing value at a rate of 5% every 4mos.
      ::对共同基金的初始投资732.00美元正在以每4分钟5%的利率贬值。
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  12. You have owned a car for 6 years. You purchased the car for $16,000. If it has depreciated at a rate of 6.5% annually, how much is it worth today?
      ::你拥有一辆车已经6年了。你以16,000美元购买了这辆车。如果它以每年6.5%的折旧率贬值,那么它今天值多少钱?

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  An artifact originally had 10 grams of carbon-14 present. The decay model $\begin{align*}A = 10e^{-0.000121t}\end{align*}$ describes the amount of carbon after $\begin{align*}t\end{align*}$ years.
  ::A=10e-0.00011t的衰变模型描述了两年后的碳含量。

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  13. How many grams of carbon-14 will be present in this artifact after 17,000 years?
      ::17000年后 有多少克碳14 将存在于这件工艺品中?
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  14. What is the half-life of carbon-14?
      ::碳-14的半衰期是什么?

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  A certain substance decays at a daily relative rate of .495%. The initial amount of the quantity is 300 mg.
  ::某种物质以每天4.95%的相对速度衰减,初始数量为300毫克。

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  15. Express the quantity as a function of time (in days) and find how much of the quantity will be left after one week.
      ::以时间函数(天数)表示数量,并找出一周后将留下的数量。

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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 。