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Chapter 6: Problem 3

Deaths from heart disease: Tables \(\mathrm{A}\) and \(\mathrm{B}\) show the deaths per 100,000 caused by heart disease in the United States for males and females aged 55 to 64 years. The function \(H_{m}\) gives deaths per 100,000 for males, and \(H_{f}\) gives deaths per 100,000 for females. a. Approximate the value of \(\frac{d H_{s}}{d r}\) in 1980 using the average rate of change from 1980 to 1985 . b. Explain the meaning in practical terms of the number you calculated in part a. You should, among other things, tell what the sign means. $$ \begin{array}{|c|c|} \hline t=\text { year } & H_{m}=\text { deaths per } 100,000 \\ \hline 1970 & 987.2 \\ \hline 1980 & 746.8 \\ \hline 1985 & 651.9 \\ \hline 1990 & 537.3 \\ \hline 1991 & 520.8 \\ \hline \end{array} $$ Table A: Heart disease deaths per 100,000 for males aged 55 to 64 years c. Use your answer from part a to estimate the heart disease death rate for males aged 55 to 64 years in \(1983 .\) 1\. Approximate the value of \(\frac{d H_{f}}{d t}\) for 1980 using the average rate of change from 1980 to \(1985 .\) e. Explain what your calculations from parts a and d tell you about comparing heart disease deaths for men and women in 1980 . $$ \begin{array}{|c|c|} \hline t=\text { year } & H_{f}=\text { deaths per } 100,000 \\ \hline 1970 & 351.6 \\ \hline 1980 & 272.1 \\ \hline 1985 & 250.3 \\ \hline 1990 & 215.7 \\ \hline 1991 & 210.0 \\ \hline \end{array} $$

Short Answer

Expert verified
Death rates decreased faster for males than females from 1980 to 1985.

Step by step solution

01

Calculate the Average Rate of Change for Males (1980-1985)

To approximate \( \frac{d H_{m}}{d t} \) in 1980, we use the average rate of change between 1980 and 1985. This is calculated as: \[ \frac{H_{m}(1985) - H_{m}(1980)}{1985 - 1980} = \frac{651.9 - 746.8}{5} \]. Solving this gives: \[ \frac{-94.9}{5} = -18.98 \] deaths per 100,000 per year.
02

Explain the Practical Meaning of the Calculated Rate

The value \( -18.98 \) means that, on average, the number of deaths per 100,000 males aged 55 to 64 years decreased by 18.98 each year from 1980 to 1985. The negative sign indicates a decrease in the death rate.
03

Estimate the Heart Disease Death Rate for Males in 1983

Using the rate of change \(-18.98\) from 1980, we estimate the death rate in 1983 as follows:\[ H_{m}(1983) = H_{m}(1980) + 3 \times (-18.98) \]\[ = 746.8 - 56.94 = 689.86 \] deaths per 100,000.
04

Calculate the Average Rate of Change for Females (1980-1985)

To approximate \( \frac{d H_{f}}{d t} \) in 1980, we use the rate of change between 1980 and 1985 as: \[ \frac{H_{f}(1985) - H_{f}(1980)}{1985 - 1980} = \frac{250.3 - 272.1}{5} \]. Solving gives: \[ \frac{-21.8}{5} = -4.36 \] deaths per 100,000 per year.
05

Compare the Death Rates for Males and Females in 1980

In 1980, the death rate from heart disease among males decreased more sharply than among females, with a rate of \(-18.98\) compared to \(-4.36\). This indicates a faster reduction in heart disease deaths among males in that period.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heart Disease Mortality
Heart disease mortality refers to the frequency of deaths caused by heart disease within a specified population or group. In our context, it's represented by the number of deaths per 100,000 individuals. Monitoring this is significant because heart disease is one of the leading causes of death globally. By analyzing data鈥攍ike the tables provided for males and females aged 55-64 years鈥攚e can identify trends over time.
For example, the number of deaths per 100,000 for males decreased from 987.2 in 1970 to 520.8 in 1991. Similarly, female death rates declined from 351.6 in 1970 to 210.0 in 1991.
  • This data shows the progression and impacts of medical advancements.
  • It provides a basis for assessing health policy effectiveness.
  • It alerts public health initiatives to areas needing attention.
Understanding mortality trends is essential for targeting interventions and improving national health outcomes.
Average Rate of Change
The average rate of change is a concept used to determine how a quantity changes over a specified interval. In the context of heart disease mortality, it measures how the number of deaths per 100,000 people in a particular demographic changes over time.
To find the average rate of change for males between 1980 and 1985, we calculated \[\frac{H_m(1985) - H_m(1980)}{1985 - 1980} = \frac{651.9 - 746.8}{5} = -18.98\] deaths per 100,000 per year. This means that, every year, the mortality rate decreased by 18.98.
  • A negative result shows a decline in death rates, a positive result would indicate an increase.
  • This metric can be used for short-term analysis to determine whether public health initiatives have been successful.
  • It serves as a building block for more complex calculations in mathematical modeling.
Understanding the average rate of change provides a snapshot of how quickly or slowly things are evolving in public health contexts.
Gender Comparison
When analyzing heart disease mortality, comparing genders reveals how different populations are affected. For the period from 1980 to 1985, both males and females saw declines in death rates, but the rates of decline differed.
The average rate of change
- For males: \[-18.98\] deaths per 100,000 per year - For females: \[-4.36\] deaths per 100,000 per year
  • This shows that male heart disease mortality rates fell faster than those of females over the same period.
  • Gender comparison highlights areas where health interventions might need to be more targeted.
  • It can also reflect broader socio-economic or lifestyle differences between genders impacting health outcomes differently.
Insights from comparing these trends enable more nuanced strategies for reducing heart disease mortality.
Mathematical Modeling
Mathematical modeling in the context of heart disease mortality involves representing real-world health situations using mathematical terms. It helps us predict trends and assess the impacts of different factors on health outcomes. By using models, we can:
- Predict future heart disease mortality rates based on past data.
- Evaluate the effectiveness of public health interventions over time.
  • For example, estimating the mortality rate for males in 1983 involves using the calculated rate of change from earlier years.
  • The calculation: \[H_m(1983) = 746.8 + 3 \times (-18.98) = 689.86\] deaths per 100,000, provides an estimated value, useful for planning.
Mathematical models are tools that help policymakers and healthcare professionals make informed decisions. They offer a quantitative basis for evaluating hypotheses, such as the impact of lifestyle changes or the introduction of new treatments.

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Most popular questions from this chapter

Fishing for sardines: This is a continuation of Example 6.10. If we take into account an annual fish harvest of \(F\) million tons of fish, then the equation of change for Pacific sardines becomes $$ \frac{d N}{d t}=0.338 N\left(1-\frac{N}{2.4}\right)-F . $$ a. Suppose that there are currently \(1.8\) million tons of Pacific sardines off the California coast and that you are in charge of the commercial fishing fleet. It is your goal to leave the Pacific population of sardines as you found it. That is, you wish to set the fishing level \(F\) so that the biomass of Pacific sardines remains stable. What value of \(F\) will accomplish this? (Hint: You want to choose \(F\) so that the current biomass level of \(1.8\) million tons is an equilibrium solution.) b. For the remainder of this exercise, take the value of \(F\) to be \(0.1\) million tons per year. That is, assume the catch is 100,000 tons per year. i. Make a graph of \(\frac{d N}{d t}\) versus \(N\), and use it to find the equilibrium solutions. ii. For what values of \(N\) will the biomass be increasing? For what values will it be decreasing? iii. On the same graph, sketch all equilibrium solutions and the graphs of \(N\) versus \(t\) for each of the initial populations \(N(0)=0.3\) million tons, \(N(0)=1.0\) million tons, and \(N(0)=2.3\) million tons. \(\rightarrow\) iv. Explain in practical terms what the picture you made in part iii tells you. Include in your explanation the significance of the equilibrium solutions.

The spread of AIDS: The table on the following page shows the cumulative number \(N=N(t)\) of AIDS cases in the United States that have been reported to the Centers for Disease Control and Prevention by the end of the year given. (The source for these data, the U.S. Centers for Disease Control and Prevention in Atlanta, cautions that they are subject to retrospective change.) a. What does \(\frac{d N}{d t}\) mean in practical terms? b. From 1986 to 1992 was \(\frac{d N}{d t}\) ever negative? $$ \begin{array}{|c|c|} \hline t=\text { year } & N=\text { total cases reported } \\ \hline 1986 & 28,711 \\ \hline 1987 & 49,799 \\ \hline 1988 & 80,518 \\ \hline 1989 & 114,113 \\ \hline 1990 & 155,766 \\ \hline 1991 & 199,467 \\ \hline 1992 & 244,939 \\ \hline \end{array} $$

Making up a story about a car trip: You begin from home on a car trip. Initially your velocity is a small positive number. Shortly after you leave, velocity decreases momentarily to zero. Then it increases rapidly to a large positive number and remains constant for this part of the trip. After a time, velocity decreases to zero and then changes to a large negative number. a. Make a graph of velocity for this trip. b. Discuss your distance from home during this driving event, and make a graph. c. Make up a driving story that matches this description.

Logistic growth with a threshold: Most species have a survival threshold level, and populations of fewer individuals than the threshold cannot sustain themselves. If the carrying capacity is \(K\) and the threshold level is \(S\), then the logistic equation of change for the population \(N=N(t)\) is $$ \frac{d N}{d t}=-r N\left(1-\frac{N}{S}\right)\left(1-\frac{N}{K}\right) . $$ For Pacific sardines, we may use \(K=2.4\) million tons and \(r=0.338\) per year, as in Example 6.10. Suppose we also know that the survival threshold level for the sardines is \(S=0.8\) million tons. a. Write the equation of change for Pacific sardines under these conditions. b. Make a graph of \(\frac{d N}{d t}\) versus \(N\) and use it to find the equilibrium solutions. How do the equilibrium solutions correspond with \(S\) and \(K\) ? c. For what values of \(N\) is the graph of \(N\) versus \(t\) increasing, and for what values is it decreasing? d. Explain what can be expected to happen to a population of \(0.7\) million tons of sardines. e. At what population level will the population be growing at its fastest?

The rock with a formula: If from ground level we toss a rock upward with a velocity of 30 feet per second, we can use elementary physics to show that the height in feet of the rock above the ground \(t\) seconds after the toss is given by \(S=30 t-16 t^{2}\). a. Use your calculator to plot the graph of \(S\) versus \(t\). b. How high does the rock go? c. When does it strike the ground? d. Sketch the graph of the velocity of the rock versus time.
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