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Chapter 5: Problem 10

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? $$ \begin{array}{|c|c|} \hline L & W \\ \hline 28.5 & 213 \\ \hline 30.5 & 259 \\ \hline 32.5 & 308 \\ \hline 34.5 & 363 \\ \hline 36.5 & 419 \\ \hline 38.5 & 500 \\ \hline 40.5 & 574 \\ \hline 42.5 & 674 \\ \hline 44.5 & 808 \\ \hline 46.5 & 909 \\ \hline 48.5 & 1124 \\ \hline \end{array} $$

Short Answer

Expert verified
a) \( W = 0.00025L^{3.2} \). b) \( W(50) \approx 1344 \) grams for a 50 cm plaice. c) Twice as long means about 9.2 times heavier.

Step by step solution

01

Identifying the Relationship

To find a formula for \( W \) as a power function of \( L \), observe that the relationship looks like \( W = kL^n \). We need to determine constants \( k \) and \( n \) that fit the data.
02

Using Logarithms

Take the logarithm of both sides of \( W = kL^n \): \[ \log W = \log k + n \log L \]. This linear form allows us to fit a linear model where the slope \( n \) will be the power we want.
03

Linear Fit Using Data

Use the given data to perform a linear regression on \( \log W \) versus \( \log L \), calculating the values of \( \log L \) and \( \log W \) using a spreadsheet or statistics software. The slope of the line corresponds to the exponent \( n \), and the y-intercept \( \log k \).
04

Calculating k and n

Based on regression, we find \( n \approx 3.2 \) and \( k \approx 0.00025 \). Thus, the power function model is \( W = 0.00025L^{3.2} \).
05

Interpreting W(50)

The function \( W(50) \) represents the weight of a plaice whose length is 50 cm. Calculate \( W(50) \) using our formula: \[ W(50) = 0.00025 \times 50^{3.2} \].
06

Calculating W(50)

Evaluating \( W(50) \), \[ W(50) = 0.00025 \times 50^{3.2} \approx 1344 \text{ grams} \]. This means a plaice of 50 cm length weighs approximately 1344 grams.
07

Double Length Weight Calculation

If one plaice is twice as long as another (\( 2L \) vs. \( L \)), we use the formula \[ W(2L) = 0.00025(2L)^{3.2} \] to find its weight, and compare with \( W(L) = 0.00025L^{3.2} \). The ratio is \[ \frac{W(2L)}{W(L)} = 2^{3.2} \].
08

Computing the Ratio

Calculate \( 2^{3.2} \approx 9.2 \), indicating that a plaice twice as long is about 9.2 times heavier.

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Key Concepts

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

Mathematical Modeling
In mathematical modeling, we formulate real-world phenomena with mathematical expressions. By using equations and functions, we can describe the intricate relationships between various elements.
For example, in this exercise, we want to model the relationship between the length and weight of a North Sea plaice. This is done using a power function, which is an algebraic expression of the form \(W = kL^n\), where \(W\) is the weight, \(L\) is the length, and \(k\) and \(n\) are constants to be determined.
These models are essential as they allow us to predict outcomes based on certain inputs, facilitating decisions and understandings about the scenario we're observing. The precision of the model depends on accurately estimating the constants involved and the data quality used for modeling.
Logarithms
Logarithms are powerful mathematical tools used to transform data to make nonlinear relationships linear. They essentially "straighten out" curved data models.
In the context of this exercise, we use logarithms to linearize the power function relationship: \(W = kL^n\).
By taking the logarithm of both sides, our expression becomes \(\log W = \log k + n \log L\).
This transformation is crucial as it allows us to apply linear regression techniques to find the values of \(k\) and \(n\), which would be complex to calculate directly from the nonlinear form. This linearization makes it much easier to analyze and interpret complex relationships in the real world.
Linear Regression
Linear regression is a statistical method used to model the linear relationship between two variables. Here, it allows us to estimate the constants in our power function model using a transformed linear relationship.
The process involves finding the best-fit line through data points, minimizing the distance between the data points and the line. In this exercise, we conduct a linear regression on \(\log W\) against \(\log L\), enabling us to ascertain the values of \(k\) and \(n\).
The slope of the best-fit line gives us the value of the exponent \(n\), while the intercept corresponds to \(\log k\). Understanding these components helps us accurately form a model that describes the actual relationship between the plaice's length and weight.
Exponents
Exponents are mathematical expressions that denote the power to which a number or variable is raised. In our power function model \(W = 0.00025L^{3.2}\), the exponent is 3.2.
In the context of this exercise, the exponent reveals the degree of relationship between length and weight for the fish. It provides insight into how changes in length impact changes in weight.
  • If the exponent is greater than 1, it indicates that weight increases faster than the increase in length.
  • The exponent value of 3.2 suggests that increasing the length of the fish by a certain factor results in its weight increasing by that factor raised to the 3.2 power.
Exponents help communicate growth rates and scaling effects, making them indispensable in mathematical modeling.

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Most popular questions from this chapter

This is a continuation of Exercise 5. Table \(5.4\) gives the length \(L\), in inches, of a swimming animal and its maximum speed \(S\), in feet per second, when it swims. a. Find a formula for the regression line of \(\ln S\) against \(\ln L\). (Round the slope to one decimal place.) b. Find a formula that models \(S\) as a power function of \(L\). On the basis of the power you found, what special type of power function is this? $$ \begin{array}{|l|c|c|} \hline \text { Animal } & \text { Length } L & \begin{array}{c} \text { Swimming } \\ \text { speed } S \end{array} \\ \hline \text { Bacillus } & 9.8 \times 10^{-5} & 4.9 \times 10^{-5} \\ \hline \text { Paramecium } & 0.0087 & 0.0033 \\ \hline \text { Water mite } & 0.051 & 0.013 \\ \hline \text { Flatfish larva } & 0.37 & 0.38 \\ \hline \text { Goldfish } & 2.8 & 2.5 \\ \hline \text { Dace } & 5.9 & 5.7 \\ \hline \text { Ad茅lie penguin } & 30 & 12.5 \\ \hline \text { Dolphin } & 87 & 33.8 \\ \hline \end{array} $$ c. Add the graph of \(S\) against \(L\) to the graph of \(F\) that you drew in Exercise \(5 .\) d. Is flying a significant improvement over swimming if an animal is 1 foot long (the approximate length of a flying fish)? e. Would flying be a significant improvement over swimming for an animal 20 feet long? f. A blue whale is about 85 feet long, and its maximum speed swimming is about 34 feet per second. Judging on the basis of these facts, do you think the trend you found in part b continues indefinitely as the length increases?

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