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Chapter 5: Problem 15

Many science fiction movies feature animals such as ants, spiders, or apes growing to monstrous sizes and threatening defenseless Earthlings. (Of course, they are in the end defeated by the hero and heroine.) Biologists use power functions as a rough guide to relate body weight and cross-sectional area of limbs to length or height. Generally, weight is thought to be proportional to the cube of length, whereas cross-sectional area of limbs is proportional to the square of length. Suppose an ant, having been exposed to "radiation," is enlarged to 500 times its normal length. (Such an event can occur only in Hollywood fantasy. Radiation is utterly incapable of causing such a reaction.) a. By how much will its weight be increased? b. By how much will the cross-sectional area of its legs be increased? c. Pressure on a limb is weight divided by crosssectional area. By how much has the pressure on a leg of the giant ant increased? What do you think is likely to happen to the unfortunate ant? \({ }^{15}\)

Short Answer

Expert verified
a) Weight increases by 125,000,000 times. b) Area increases by 250,000 times. c) Pressure increases by 500 times, likely causing limb collapse.

Step by step solution

01

Understanding the problem

We need to establish the relationships between an ant's length, weight, and cross-sectional area of limbs. Weight is proportional to the cube of the length (\(weight \propto length^3\)), while the cross-sectional area is proportional to the square of the length (\(area \propto length^2\)). We must determine the factor by which these increase when the ant's length is scaled by a factor of 500.
02

Calculate weight increase

If the length is increased 500 times, the weight will increase by the cube of this factor. Thus, the increase in weight is given by\(500^3 = 125,000,000\). So, the weight increases by a factor of 125,000,000.
03

Calculate cross-sectional area increase

The cross-sectional area will increase by the square of the factor by which the length increases. Therefore, it will increase by\(500^2 = 250,000\). So, the area increases by a factor of 250,000.
04

Calculate change in pressure

Pressure is defined as weight divided by area. The original pressure is proportional to \(\frac{weight_{original}}{area_{original}}\). The new pressure is proportional to \(\frac{weight_{new}}{area_{new}} = \frac{500^3}{500^2} = 500\),meaning the pressure increases by a factor of 500.
05

Effects on the ant

Given that the pressure on the legs increases by a factor of 500, this dramatic increase is unsustainable structurally. Likely, the ant's limbs would collapse under its own increased weight, rendering it immobile or crushed.

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Key Concepts

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

Cross-Sectional Area
In biology, the concept of cross-sectional area is often used to refer to the area of a cut-through section of any part of an organism's body, like a limb. Imagine slicing a carrot. The circle you see - that's the cross-section. Now, in the context of this ant, we are trying to understand how much the legs' supporting area increases when the length of the ant increases.

The cross-sectional area of the ant's limbs is proportional to the square of its length. This means that if the length of the ant is multiplied by a factor, the cross-sectional area will be multiplied by the square of that factor.
For example, when the ant's length increases by 500 times:
  • The new cross-sectional area is calculated by multiplying the old area by 500 squared, or 250,000.
  • This means the legs' ability to support increases, but not nearly as much as the overall weight.
This imbalance plays a significant role in determining whether this oversized ant can bear its own weight.
Body Weight
Body weight is a biological parameter that can be modeled using power functions, which help relate changes in size to volume.

For our ant, biologists suggest that weight is proportional to the cube of its body length. In mathematical terms, this means if an ant's length is multiplied by a factor, its weight will increase by the cube of that factor.
Here's what happened with the giant ant:
  • Initial length: 1 unit (just for simplicity)
  • New length: 500 units
  • Weight increase: calculated by (500)^3 = 125,000,000.
So, when the length goes up 500 times, the weight balloons to 125 million times its initial weight. Now think of how an object this heavy can be supported! That's the problem Hollywood misses when enlarging creatures.
Length Proportionality
In biological modeling, especially for scaling purposes seen in movies, length proportionality is crucial. Imagine your height tripled overnight. How would the rest of your body react?

This is a quite simplified explanation. In biological equations, different attributes of an organism, like weight and cross-sectional area, are related to powers of length. Length proportionality helps predict what might happen when size changes.
For our ant problem:
  • Cross-sectional area and weight are related to length but with different exponents: 2 (square) and 3 (cube) respectively.
  • This understanding highlights that some properties grow faster than others as size scales up.
Comprehending proportionality helps explain why the ant couldn't logically survive its Hollywood-induced enlargement.
Biological Modeling
Biological modeling uses mathematics to represent biological processes, such as growth patterns for creatures or plants. By using power functions, these models help predict how different biological parameters affect each other.

In our example of the ant, such models demonstrate:
  • How limb areas and total body weight scale with size.
  • Potential consequences when parameters are disproportionate.
A key point in biological modeling is that as organisms grow, not all parts scale equally. For our ant, while the body length increases 500 times, the implications for weight and support (cross-sectional area) increase at different rates, revealing potential issues.
Understanding biological modeling gives us insight into the problem's practical challenges and why an ant-sized to giant proportions in the real world is a fantasy.

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Most popular questions from this chapter

The headway \(h\) is the average time between vehicles. On a highway carrying an average of 500 vehicles per hour, the probability \(P\) that the headway is at least \(t\) seconds is given \(^{45}\) by $$ P=0.87^{t} . $$ a. What is the limiting value of \(P\) ? Explain what this means in practical terms. b. The headway \(h\) can be calculated as the quotient of the spacing \(f\), in feet, which is the average distance between vehicles, and the average speed \(v\), in feet per second, of traffic. Thus the probability that spacing is at least \(f\) feet is the same as the probability that the headway is at least \(f / v\) seconds. Use function composition to find a formula for the probability \(Q\) that the spacing is at least \(f\) feet. Note: Your formula will involve both \(f\) and \(v\). c. If the average speed is 88 feet per second \((60\) miles per hour), what is the probability that the spacing between two vehicles is at least 40 feet?

Gray wolves recolonized in the Upper Peninsula of Michigan beginning in 1990 . Their population has been documented as shown in the accompanying table. \({ }^{49}\) a. Explain why one would expect an exponential model to be appropriate for these data. b. Find an exponential model for the data given. $$ \begin{array}{|c|c|} \hline \text { Year } & \text { Wolves } \\ \hline 1990 & 6 \\ \hline 1991 & 17 \\ \hline 1992 & 21 \\ \hline 1993 & 30 \\ \hline 1994 & 57 \\ \hline 1995 & 80 \\ \hline 1996 & 116 \\ \hline 1997 & 112 \\ \hline 1998 & 140 \\ \hline 1999 & 174 \\ \hline 2000 & 216 \\ \hline \end{array} $$ c. Graph the data and the exponential model. Would it be better to use a piecewise-defined function? d. Find an exponential model for 1990 through 1996 and another for 1997 through \(2000 .\) e. Write a formula for the number of wolves as a piecewise-defined function using the two exponential models. Is the combined model a better fit?

When a car skids to a stop, the length \(L\), in feet, of the skid marks is related to the speed \(S\), in miles per hour, of the car by the power function \(L=\frac{1}{30 h} S^{2}\). Here the constant \(h\) is the friction coefficient, which depends on the road surface. \({ }^{11}\) For dry concrete pavement, the value of \(h\) is about \(0.85\). a. If a driver going 55 miles per hour on dry concrete jams on the brakes and skids to a stop, how long will the skid marks be? b. A policeman investigating an accident on dry concrete pavement finds skid marks 230 feet long. The speed limit in the area is 60 miles per hour. Is the driver in danger of getting a speeding ticket? c. This part of the problem applies to any road surface, so the value of \(h\) is not known. Suppose you are driving at 60 miles per hour but, because of approaching darkness, you wish to slow to a speed that will cut your emergency stopping distance in half. What should your new speed be? (Hint: You should use the homogeneity property of power functions here. By what factor should you change your speed to ensure that \(L\) changes by a factor of \(0.5\) ?)

If you invest \(P\) dollars (the present value of your investment) in a fund that pays an interest rate of \(r\), as a decimal, compounded yearly, then after \(t\) years your investment will have a value \(F\) dollars, which is known as the future value. The discount rate \(D\) for such an investment is given by $$ D=\frac{1}{(1+r)^{t}}, $$ where \(t\) is the life, in years, of the investment. The present value of an investment is the product of the future value and the discount rate. Find a formula that gives the present value in terms of the future value, the interest rate, and the life of the investment.

The accompanying table shows the relationship between the length \(L\), in centimeters, and the weight \(W\), in grams, of the North Sea plaice (a type of flatfish). \({ }^{32}\) a. Find a formula that models \(W\) as a power function of \(L\). (Round the power to one decimal place.) b. Explain in practical terms what \(W(50)\) means, and then calculate that value. c. If one plaice were twice as long as another, how much heavier than the other should it be? $$ \begin{array}{|c|c|} \hline L & W \\ \hline 28.5 & 213 \\ \hline 30.5 & 259 \\ \hline 32.5 & 308 \\ \hline 34.5 & 363 \\ \hline 36.5 & 419 \\ \hline 38.5 & 500 \\ \hline 40.5 & 574 \\ \hline 42.5 & 674 \\ \hline 44.5 & 808 \\ \hline 46.5 & 909 \\ \hline 48.5 & 1124 \\ \hline \end{array} $$
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