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Section 7.7 Projects for Chapter 7

Project 7.3 Solving cubics: Part I

In this project, we solve cubic equations of the form

\begin{equation*} x^3 + mx = n \end{equation*}

Note that there is no quadratic term. This special form was first solved by the Italian mathematicians Scipione del Ferro and Niccolò Fontana Tartaglia early in the sixteenth century. Tartaglia revealed the secret to solving the special cubic equation in a poem. He first found values \(u\) and \(v\) to satisfy the system

\begin{gather*} u-v=n\\ uv=\left(\frac{m}{3}\right)^3 \end{gather*}
  1. We will use Tartaglia's method to solve

    \begin{equation*} x^3 + 6x = 7 \end{equation*}

    What are the values of \(m\) and \(n\text{?}\)

  2. Substitute the values of \(m\) and \(n\) into Tartaglia's system, then use substitution to solve for \(u\) and \(v\text{.}\) You should find two possible solutions.

  3. For each solution of the system, compute \(x=\sqrt[3]{u}- \sqrt[3]{v} \text{.}\) You should get the same value of \(x\) for each \((u, v)\text{.}\)

  4. Check that your value for \(x\) is a solution of \(x^3 + 6x = 7\text{.}\)

Project 7.4 Solving cubics: Part II

Tartaglia's method always works to solve the special cubic equation, even when \(u\) and \(v\) are not convenient values. We will show why in this project.

  1. Expand the expression \((a - b)^3 + 3ab(a - b)\) and complete the identity.

    \begin{equation*} (a-b)^3+3ab(a-b)=\underline{\hspace{6.818181818181818em}} \end{equation*}
  2. Your answer to part (a) is actually Tartaglia's special cubic in disguise. Substitute \(x = a - b\text{,}\) \(m = 3ab\text{,}\) and \(n = a^3 - b^3\) to see this. Therefore, if we can find numbers \(a\) and \(b\) that satisfy

    \begin{align*} 3ab \amp = m\\ a^3 - b^3 \amp = n \end{align*}

    then the solution to Tartaglia's cubic is \(x = a - b\text{.}\)

  3. Compare the system in part (b) to the system from Project 7.3,

    \begin{gather*} u - v = n\\ uv=\left(\frac{m}{3}\right)^3 \end{gather*}

    to show that \(u = a^3\) and \(v = b^3\text{.}\)

  4. Use your answer to part (c) to show that Tartaglia's value, \(x=\sqrt[3]{u}- \sqrt[3]{v} \text{,}\) is a solution of \(x^3 + mx = n\text{.}\)

Project 7.5 Solving cubics: Part III

Use Tartaglia's method to solve the equation

\begin{equation} x^3 + 3x = 2\label{equation-cubic}\tag{7.1} \end{equation}

by carrying out the following steps.

  1. Identify the values of \(m\) and \(n\) from (7.1) and write two equations for \(u\) and \(v\text{.}\)

  2. Solve for values of \(u\) and \(v\text{.}\) You will need to use the quadratic formula.

  3. Take the positive values of \(u\) and \(v\text{.}\) Write the solution \(x=\sqrt[3]{u}- \sqrt[3]{v} \text{.}\) Do not try to simply the radical expression; instead, use your calculator to check the solution numerically.

Project 7.6 Solving cubics: Part IV

We can solve any cubic equation by first using a substitution to put the equation in Tartaglia's special form.

  1. Consider the equation \(X^3 + bX^2 + cX + d = 0\text{.}\) Make the substitution \(X = x - \dfrac{b}{3}\text{,}\) and expand the left side of the equation.

  2. What is the coefficient of \(x^2\) in the resulting equation? What are the values of \(m\) and \(n\text{?}\)

  3. If you solve the special form in part (a) for \(x\text{,}\) how can you find the value of \(X\) that solves the original equation?

Project 7.7 Duration of eclipse

The time, \(T\text{,}\) it takes for the Moon to eclipse the Sun totally is given (in minutes) by the formula

\begin{equation*} T=\frac{1}{v}\left(\frac{rD}{R}-d \right) \end{equation*}

where \(d\) is the diameter of the Moon, \(D\) is the diameter of the Sun, \(r\) is the distance from the Earth to the Moon, \(R\) is the distance from the Earth to the Sun, and \(v\) is the speed of the Moon.

  1. Solve the formula for \(v\) in terms of the other variables.

  2. It takes \(2.68\) minutes for the Moon to eclipse the Sun. Calculate the speed of the Moon given the following values:

    \begin{align*} d = 3.48 \times 10^3 \text{ km}\amp\amp D = 1.41 \times 10^6 \text{ km}\\ r = 3.82 \times 10^5 \text{ km} \amp\amp R = 1.48 \times 10^8 \text{ km} \end{align*}
Project 7.8 Optimal traffic flow

The stopping distance, \(s\text{,}\) for a car traveling at speed \(v\) meters per second is given (in meters) by

\begin{equation*} s=vT+\frac{v^2}{2a} \end{equation*}

where \(T\) is the reaction time of the driver and \(a\) is the average deceleration as the car brakes. Suppose that all the cars on a crowded motorway maintain the appropriate spacing determined by the stopping distance for their speed. What speed allows the maximum flow of cars along the road per unit time? Using the formula \(\text{time} = \dfrac{\text{distance}}{\text{speed}}\text{,}\) we see that the time interval, \(t\text{,}\) between cars is

\begin{equation*} t=\frac{s}{v}+\frac{L}{v} \end{equation*}

where \(L\) is the length of the car. To achieve the maximum flow of cars, we would like \(t\) to be as small as possible. (Source: Bolton, 1974)

  1. Substitute the expression for \(s\) into the formula for \(t\text{,}\) then simplify.

  2. A typical reaction time is \(T = 0.7\) seconds, a typical car length is \(L=5\) meters, and \(a = 7.5\) meters per second squared. With these values, graph \(t\) as a function of \(v\) in the window

    \begin{align*} {\text{Xmin}} \amp = 0 \amp\amp {\text{Xmax}} = 20\\ {\text{Ymin}} \amp = 0 \amp\amp {\text{Ymax}} = 3 \end{align*}
  3. To one decimal place, what value of \(v\) gives the minimum value of \(t\text{?}\) Convert your answer to miles per hour.

Project 7.9 Effective population of endangered species

Many endangered species have fewer than \(1000\) individuals left. To preserve the species, captive breeding programs must maintain a certain effective population, \(N\text{,}\) given by

\begin{equation*} N=\frac{4FM}{F+M} \end{equation*}

where \(F\) is the number of breeding females and \(M\) the number of breeding males. (Source: Chapman and Reiss, 1992)

  1. What is the effective population if there are equal numbers of breeding males and females?

  2. In 1972, a breeding program for Speke's gazelle was established with just three female gazelle. Graph the effective population, \(N\text{,}\) as a function of the number of males.

  3. What is the largest effective population that can be created with three females? How many males are needed to achieve the maximum value?

  4. With three females, for what value of \(M\) is \(N = M\text{?}\)

  5. The breeding program for Speke's gazelle began with only one male. What was the effective population?

Project 7.10 Biological half-life

When a drug or chemical is injected into a patient, biological processes begin removing that substance. If no more of the substance in introduced, the body removes a fixed fraction of the substance each hour. The amount of substance remaining in the body at time \(t\) is an exponential decay function, so there is a biological half-life to the substance denoted by \(T_b\text{.}\) If the substance is a radioisotope, it undergoes radioactive decay and so has a physical half-life as well, denoted \(T_p\text{.}\)

The effective half-life, denoted by \(T_e\text{,}\) is related to the biological and physical half-lives by the equation

\begin{equation*} \frac{1}{T_e}=\frac{1}{T_b}+\frac{1}{T_p} \end{equation*}

The radioisotope \(^{131}\text{I}\) is used as a label for the human serum albumin. The physical half-life of \(^{131}\text{I}\) is \(8\) days. (Source: Pope, 1989)

  1. If \(^{131}\text{I}\) is cleared from the body with a half-life of \(21\) days, what is the effective half-life of \(^{131}\text{I}\text{?}\)

  2. The biological half-life of a substance varies considerably from person to person. If the biological half-life of \(^{131}\text{I}\) is \(x\) days, what is the effective half life?

  3. Let \(f (x)\) represent the effective half-life of \(^{131}\text{I}\) when the biological half-life is \(x\) days. Graph \(y = f (x)\text{.}\)

  4. What would the biological half-life of \(^{131}\text{I}\) need to be to produce an effective half-life of \(6\) days? Label the corresponding point on your graph.

  5. For what possible biological half-lives of \(^{131}\text{I}\) will the effective half-life be less than \(4\) days?

Project 7.11 Rate of eating

Animals spend most of their time hunting or foraging for food to keep themselves alive. Knowing the rate at which an animal (or population of animals) eats can help us determine its metabolic rate or its impact on its habitat. The rate of eating is proportional to the availability of food in the area, but it has an upper limit imposed by mechanical considerations, such as how long it takes the animal to capture and ingest its prey. (Source: Burton, 1998)

  1. Sketch a graph of eating rate as a function of quantity of available food. This will be a qualitative graph only; you do not have enough information to put scales on the axes.

  2. Suppose that the rate at which an animal catches its prey is proportional to the number of prey available, or \(r_c = ax\text{,}\) where \(a\) is a constant and \(x\) is the number of available prey. The rate at which it handles and eats the prey is constant, \(r_h = b\text{.}\) Write expressions for \(T_c\) and \(T_h\text{,}\) the times for catching and handling \(N\) prey.

  3. Show that the rate of food consumption is given by

    \begin{equation*} y=\dfrac{abx}{b+ax}=\frac{bx}{b/a+x} \end{equation*}

    Hint: \(y=\dfrac{N}{T} \text{,}\) where \(N\) is the number of prey consumed in the interval \(T\text{,}\) where \(T=T_c+T_h\text{.}\)

  4. In a study of ladybirds, it was discovered that larvae in their second stage of development consumed aphids at a rate \(y_2\) aphids per day, given by

    \begin{equation*} y_2=\frac{20x}{x+16} \end{equation*}

    where \(x\) is the number of aphids available. Larvae in the third stage ate at rate \(y_3\text{,}\) given by

    \begin{equation*} y_3=\frac{90x}{x+79} \end{equation*}

    Graph both of these functions on the domain \(0\le x\le 140\text{.}\)

  5. What is the maximum rate at which ladybird larvae in each stage of development can consume aphids?

Project 7.12 Buoyancy

A person will float in fresh water if his or her density is less than or equal to \(1\) kilogram per liter, the density of water. (Density is given by the formula \(\text{density} = \dfrac{\text{weight}}{\text{volume}}\text{.}\)) Suppose a swimmer weighs \(50 + F\) kilograms, where \(F\) is the amount of fat her body contains. (Source: Burton, 1998)

  1. Calculate the volume of her nonfat body mass if its density is \(1.1\) kilograms per liter.

  2. Calculate the volume of the fat if its density is \(0.901\) kilograms per liter.

  3. The swimmer’s lungs hold \(2.6\) liters of air. Write an expression for the total volume of her body, including the air in her lungs.

  4. Write an expression for the density of the swimmer's body.

  5. Write an equation for the amount of fat needed for the swimmer to float in fresh water.

  6. Solve your equation. What percent of the swimmer's weight is fat?

  7. Suppose the swimmer's lungs can hold \(4.6\) liters of air. What percent body fat does she need to be buoyant?