/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 14 The retrovirus HIV can be transm... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 14

The retrovirus HIV can be transmitted from mother to child during breastfeeding. But under conditions of poverty or poor hygiene, the alternative to breastfeeding carries its own risk for infant mortality. These risks differ in that the risk of HIV transmission is the same regardless of the age of the child, whereas the risk of mortality due to artificial feeding is high for newborns but decreases with the age of the child. \({ }^{50}\) Figure \(5.85\) illustrates the additional risk of death due to artificial feeding and due to HIV transmission from breastfeeding. Here the lighter curve is additional risk of death from artificial feeding, and the darker curve is additional risk of death from breastfeeding. For the purposes of this exercise, assume that the additional risk of death from artificial feeding is \(A=5 \times 0.785^{t}\) and that the additional risk from breastfeeding is \(B=2\), where \(t\) is the child's age in months. a. At what age is the additional risk of HIV transmission from breastfeeding the same as the additional risk from artificial feeding? b. For which ages of the child does breastfeeding carry the smaller additional risk? c. What would be the optimal plan for feeding a child to minimize additional risk of death? d. Write a formula for a piecewise-defined function \(R\) of \(t\) giving the additional risk of death under the optimal plan from part c.

Short Answer

Expert verified
a. Approximately 12.57 months. b. Ages greater than 12.57 months. c. Artificial feeding until 12.57 months, then breastfeeding. d. \( R(t) = \begin{cases} 5 \times 0.785^t, & t \leq 12.57 \\ 2, & t > 12.57 \end{cases} \)

Step by step solution

01

Equate Risks for Part (a)

To find when the additional risk from breastfeeding (B) equals the additional risk from artificial feeding (A), equate the two expressions: \[ 5 \times 0.785^t = 2 \]. This equation allows us to solve for the age \( t \) at which the risks are equal.
02

Solve for Age in Part (a)

Divide both sides of the equation by 5: \[ 0.785^t = \frac{2}{5} \]. Take the logarithm of both sides to solve for \( t \): \[ t = \frac{\log{\frac{2}{5}}}{\log{0.785}} \]. Using a calculator, \( t \approx 12.569 \), meaning the risks are equal at approximately 12.57 months.
03

Analyze Risks for Part (b)

Breastfeeding carries a smaller additional risk when \( B \lt A \). Thus, solve \( 2 \lt 5 \times 0.785^t \). Dividing both sides gives \( 0.4 \lt 0.785^t \), and taking the logarithm of both sides yields \( t \gt \frac{\log{0.4}}{\log{0.785}} \). Calculating gives \( t \gt 12.569 \). Therefore, for ages greater than approximately 12.57 months, breastfeeding carries less risk.
04

Optimal Feeding Plan for Part (c)

The optimal feeding plan to minimize risk would be to use artificial feeding for ages 0 through approximately 12.57 months, as the risk is lower than breastfeeding risk during this period. After 12.57 months, breastfeeding should be used as it carries a lower risk.
05

Define the Piecewise Function for Part (d)

Create the piecewise function \( R(t) \) representing the risk under the optimal plan: \[ R(t) = \begin{cases} 5 \times 0.785^t, & t \leq 12.57 \ 2, & t > 12.57 \end{cases} \]. This function indicates using artificial feeding before 12.57 months and breastfeeding thereafter.

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Key Concepts

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

HIV transmission risk
Understanding the risk of HIV transmission is critical, especially when considering feeding practices for infants. HIV, primarily known to spread through specific bodily fluids, can be transmitted from a mother to her child during breastfeeding. While breastfeeding offers numerous benefits, including essential nutrients and immune support, it introduces the potential risk of HIV if the mother is infected.

This risk of HIV transmission remains constant regardless of the child鈥檚 age. In contrast, alternatives like artificial feeding present a risk of mortality that varies with age. The challenge therefore lies in minimizing the overall risk by balancing these two factors. Families, especially in areas with different socioeconomic conditions, must make informed decisions based on the health profiles and access to resources, to prioritize the safety of their infants.
piecewise functions
Piecewise functions are mathematical tools that allow us to describe a function which behaves differently over various segments of its domain. They are especially useful in situations like our exercise, where different actions are optimal based on specific conditions or intervals.

In the context of feeding practices, piecewise functions help us represent the additional risk of death due to feeding methods as a function of age. By defining piecewise functions, we can distinctly mark the transition between using artificial feeding and breastfeeding based on age-related risk assessments. For instance, in the given problem, artificial feeding was optimal until approximately 12.57 months, after which breastfeeding became the safer option.

The piecewise function can be represented as:
  • For ages up to 12.57 months: The risk is modeled by the exponential decay function corresponding to artificial feeding.
  • For ages greater than 12.57 months: The risk is constant and corresponds to breastfeeding.
This format allows us to effectively capture the decision-making process in a structured, mathematical way.
exponential decay functions
Exponential decay functions describe situations where a quantity decreases at a rate proportional to its current value. This concept is applied to model changes in risk over time.

In our context, the additional risk of death from artificial feeding diminishes as the child ages. The function is defined as:
  • \[ A = 5 \times 0.785^t \]
Here, the base (0.785) reflects the rate at which the risk decreases over time \( t \) (which represents the child's age in months).

Exponential decay functions are pivotal for illustrating diminishing risks because they effectively capture the nature of situations where the initial stages have higher risk, which tapers off over time. This behavior is why the risk from artificial feeding drops dramatically in initial months but reduces its rate of decline as the age increases. Students learning to harness exponential functions can better model real-world scenarios where quick or persistent changes in risk or other quantities occur.

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Most popular questions from this chapter

The power \(P\) required for level flight by an airplane is a function of the speed \(u\) of flight. Consideration of drag on the plane yields the model $$ P=\frac{u^{3}}{a}+\frac{b}{u}. $$ Here \(a\) and \(b\) are constants that depend on the characteristics of the airplane. This model may also be applied to the flight of a bird such as the budgerigar (a type of parakeet), where we take \(a=7800\) and \(b=600\). Here the flight speed \(u\) is measured in kilometers per hour, and the power \(P\) is the rate of oxygen consumption in cubic centimeters per gram per hour. \({ }^{68}\) a. Make a graph of \(P\) as a function of \(u\) for the budgerigar. Include flight speeds between 25 and 45 kilometers per hour. b. Calculate \(P(39)\) and explain what your answer means in practical terms. c. At what flight speed is the required power minimized?

The rate of growth \(G\), in thousands of dollars per year, in sales of a certain product is a function of the current sales level \(s\), in thousands of dollars, and the model uses a quadratic function: $$ G=1.2 s-0.3 s^{2} . $$ The model is valid up to a sales level of 4 thousand dollars. a. Draw a graph of \(G\) versus \(s\). b. Express using functional notation the rate of growth in sales at a sales level of \(\$ 2260\), and then estimate that value. c. At what sales level is the rate of growth in sales maximized?

Assume that a long horizontal pipe connects the bottom of a reservoir with a drainage area. Cox's formula provides a way of determining the velocity \(v\) of the water flowing through the pipe: $$ \frac{H d}{L}=\frac{4 v^{2}+5 v-2}{1200} . $$ Here \(H\) is the depth of the reservoir in feet, \(d\) is the pipe diameter in inches, \(L\) is the length of the pipe in feet, and the velocity \(v\) of the water is in feet per second. (See Figure \(5.103\) on the following page.) a. Graph the quadratic function \(4 v^{2}+5 v-2\) using a horizontal span from 0 to 10 . b. Judging on the basis of Cox's formula, is it possible to have a velocity of \(0.25\) foot per second? c. Find the velocity of the water in the pipe if its diameter is 4 inches, its length is 1000 feet, and the reservoir is 50 feet deep. d. If the water velocity is too high, there will be erosion problems. Assuming that the pipe length is 1000 feet and the reservoir is 50 feet deep, determine the largest pipe diameter that will ensure that the water velocity does not exceed 10 feet per second.

The following table shows carbon monoxide emissions \(M,{ }^{60}\) in millions of metric tons, \(t\) years after 1940 . $$ \begin{array}{|c|c|} \hline t & M \\ \hline 0 & 82.6 \\ \hline 10 & 87.6 \\ \hline 20 & 89.7 \\ \hline 30 & 101.4 \\ \hline 40 & 79.6 \\ \hline 41 & 77.4 \\ \hline 42 & 72.4 \\ \hline 43 & 74.5 \\ \hline 44 & 71.8 \\ \hline 45 & 68.7 \\ \hline \end{array} $$ a. Plot the data points. b. Use quadratic regression to model the data. c. Add the plot of the quadratic model to your data plot. d. According to your model, when were carbon monoxide emissions at a maximum?

By comparing the surface area of a sphere with its volume and assuming that air resistance is proportional to the square of velocity, it is possible to make a heuristic argument to support the following premise: For similarly shaped objects, terminal velocity varies in proportion to the square root of length. Expressed in a formula, this is $$ T=k L^{0.5}, $$ where \(L\) is length, \(T\) is terminal velocity, and \(k\) is a constant that depends on shape, among other things. This relation can be used to help explain why small mammals easily survive falls that would seriously injure or kill a human. a. A 6-foot man is 36 times as long as a 2-inch mouse (neglecting the tail). How does the terminal velocity of a man compare with that of a mouse? b. If the 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of the 2 -inch mouse? c. Neglecting the tail, a squirrel is about 7 inches long. Again assuming that a 6-foot man has a terminal velocity of 120 miles per hour, what is the terminal velocity of a squirrel?
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