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Chapter 5: Problem 4
Answer True or False. You do not need a calculator for these exercises. Rather, use the fact that e is approximately 2.7 $$e^{2}<4$$
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Radiocarbon Dating: Because rubidium-87 decays so slowly, the technique of rubidium-strontium dating is generally considered effective only for objects older than 10 million years. In contrast, archeologists and geologists rely on the radiocarbon dating method in assigning ages ranging from 500 to 50,000 years. Two types of carbon occur naturally in our environment: carbon-12, which is nonradioactive, and carbon-14, which has a half-life of 5730 years. All living plant and animal tissue contains both types of carbon, always in the same ratio. (The ratio is one part carbon- 14 to \(10^{12}\) parts carbon-12.) As long as the plant or animal is living, this ratio is maintained. When the organism dies, however, no new carbon-14 is absorbed, and the amount of carbon-14 begins to decrease exponentially. since the amount of carbon-14 decreases exponentially, it follows that the level of radioactivity also must decrease exponentially. The formula describing this situation is $$\mathcal{N}=\mathcal{N}_{0} e^{k T}$$ where \(T\) is the age of the sample, \(\mathcal{N}\) is the present level of radioactivity (in units of disintegrations per hour per gram of carbon), and \(\mathcal{N}_{0}\) is the level of radioactivity \(T\) years ago, when the organism was alive. Given that the half-life of carbon-14 is 5730 years and that \(\mathcal{N}_{0}=920\) disintegrations per hour per gram, show that the age \(T\) of a sample is given by $$T=\frac{5730 \ln (\mathcal{N} / 920)}{\ln (1 / 2)}$$
(a) Use a graphing utility to estimate the root(s) of the equation to the nearest one-tenth (as in Example 6). (b) Solve the given equation algebraically by first rewriting it in logarithmic form. Give two forms for each answer: an exact expression and a calculator approximation rounded to three decimal places. Check to see that each result is consistent with the graphical estimate obtained in part (a). $$e^{3 x^{2}}=112$$
Solve the inequality \(\log _{10}\left(x^{2}-6 x-6\right)>0.\)
Exercises \(55-60\) introduce a model for population growth that takes into account limitations on food and the environment. This is the logistic growth model, named and studied by the nineteenth century Belgian mathematician and sociologist Pierre Verhulst. (The word "logistic" has Latin and Greek origins meaning "calculation" and "skilled in calculation," respectively. However, that is not why Verhulst named the curve as he did. See Exercise 56 for more about this.) In the logistic model that we "I study, the initial population growth resembles exponential growth. But then, at some point owing perhaps to food or space limitations, the growth slows down and eventually levels off, and the population approaches an equilibrium level. The basic equation that we'll use for logis- tic growth is where \(\mathcal{N}\) is the population at time \(t, P\) is the equilibrium population (or the upper limit for population), and a and b are positive constants. $$\mathcal{N}=\frac{P}{1+a e^{-b t}}$$ (Continuation of Exercise 55 ) The author's ideas for this exercise are based on Professor Bonnie Shulman's article "Math-Alive! Using Original Sources to Teach Mathematics in Social Context," Primus, vol. VIII (March \(1998)\) (a) The function \(\mathcal{N}\) in Exercise 55 expresses population as a function of time. But as pointed out by Professor Shulman, in Verhulst's original work it was the other way around; he expressed time as a function of population. In terms of our notation, we would say that he was studying the function \(\mathcal{N}^{-1}\). Given \(\mathcal{N}(t)=4 /\left(1+8 e^{-t}\right)\) find \(\mathcal{N}^{-1}(t)\) (b) Use a graphing utility to draw the graphs of \(\mathcal{N}, \mathcal{N}^{-1}\), and the line \(y=x\) in the viewing rectangle [-3,8,2] by \([-3,8,2] .\) Use true portions. (Why?) (c) In the viewing rectangle [0,5,1] by \([-3,2,1],\) draw the graphs of \(y=\mathcal{N}^{-1}(t)\) and \(y=\ln t .\) Note that the two graphs have the same general shape and characteristics. In other words, Verhulst's logistic function (our \(\mathcal{N}^{-1}\) ) appears log-like, or logistique, as Verhulst actually named it in French. (For details, both historical and mathematical, see the paper by Professor Shulman cited above.)
The Chernobyl nuclear explosion (in the former Soviet Union, on April 26,1986 ) released large amounts of radioactive substances into the atmosphere. These substances included cesium-137, iodine-131, and strontium-90. Although the radioactive material covered many countries, the actual amount and intensity of the fallout varied greatly from country to country, due to vagaries of the weather and the winds. One area that was particularly hard hit was Lapland, where heavy rainfall occurred just when the Chernobyl cloud was overhead. (a) Many of the pastures in Lapland were contaminated with cesium-137, a radioactive substance with a half- life of 33 years. If the amount of cesium- 137 was found to be ten times the normal level, how long would it take until the level returned to normal? Hint: Let \(\mathcal{N}_{0}\) be the amount that is ten times the normal level. Then you want to find the time when \(\mathcal{N}(t)=\mathcal{N}_{0} / 10\) (b) Follow part (a), but assume that the amount of cesium-137 was 100 times the normal level. Remark: Several days after the explosion, it was reported that the level of cesium- 137 in the air over Sweden was 10,000 times the normal level. Fortunately there was little or no rainfall.
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