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Chapter 4: Problem 99

Bornstein and Bornstein found in a study that the average walking speed \(v\) of a person living in a city of population \(p\), in thousands, is given by $$ v(p)=0.37 \ln p+0.05 $$ where \(v\) is in feet per second. \({ }^{12}\) a) The population of Seattle is \(531,000 .\) What is the average walking speed of a person living in Seattle? b) The population of New York is \(7,900,000\). What is the average walking speed of a person living in New York? c) Find \(v^{\prime}(p)\). d) Interpret \(v^{\prime}(p)\) found in part (c).

Short Answer

Expert verified
a) 2.37 ft/s, b) 3.37 ft/s, c) \( v'(p) = \frac{0.37}{p} \), d) Represents the rate of change of walking speed with respect to population.

Step by step solution

01

Understand the Given Function

The average walking speed of a person living in a city with population \( p \) (in thousands) is given by the function: \[ v(p) = 0.37 \ln p + 0.05 \]where \( v \) is in feet per second. We will use this function to solve the problems.
02

Convert Population to Thousands for Part (a)

The population of Seattle is \( 531,000 \). Convert this to thousands: \( p = 531 \).
03

Calculate the Walking Speed for Seattle

Substitute \( p = 531 \) into the function \( v(p) \):\[ v(531) = 0.37 \ln(531) + 0.05 \]First, calculate \( \ln(531) \). Next, multiply by 0.37 and add 0.05.
04

Compute \( \ln(531) \)

Using a calculator, \( \ln(531) \approx 6.275 \). Now substitute this value back in:\[ v(531) = 0.37 \times 6.275 + 0.05 \]\[ v(531) \approx 2.32175 + 0.05 \]\[ v(531) \approx 2.37175 \] feet per second
05

Convert Population to Thousands for Part (b)

The population of New York is \( 7,900,000 \). Convert this to thousands: \( p = 7900 \).
06

Calculate the Walking Speed for New York

Substitute \( p = 7900 \) into the function \( v(p) \):\[ v(7900) = 0.37 \ln(7900) + 0.05 \]First, calculate \( \ln(7900) \). Next, multiply by 0.37 and add 0.05.
07

Compute \( \ln(7900) \)

Using a calculator, \( \ln(7900) \approx 8.977 \). Now substitute this value back in:\[ v(7900) = 0.37 \times 8.977 + 0.05 \]\[ v(7900) \approx 3.32149 + 0.05 \]\[ v(7900) \approx 3.37149 \] feet per second
08

Differentiate the Function for Part (c)

To find \( v'(p) \), differentiate \( v(p) = 0.37 \ln(p) + 0.05 \):\[ v'(p) = \frac{d}{dp}(0.37 \ln(p) + 0.05) \]Using the derivative rule for \( \ln(p) \), we get:\[ v'(p) = 0.37 \times \frac{1}{p} \]\[ v'(p) = \frac{0.37}{p} \]
09

Interpret the Derivative for Part (d)

The derivative \( v'(p) \) represents the rate of change of the average walking speed with respect to the population size (in thousands). Specifically, \( \frac{0.37}{p} \) indicates how much the walking speed changes as the population changes.

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Key Concepts

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

Differentiation
Differentiation is a fundamental concept in calculus. It involves finding the derivative of a function, which measures how the function's output changes as its input changes. For the function \( v(p) = 0.37 \ln p + 0.05 \), we differentiate with respect to \( p \) to find the rate at which the average walking speed changes as the population changes. The differentiation of a natural logarithm function involves using the rule \( \frac{d}{dp} \ln(p) = \frac{1}{p} \). Applying this, we find that \( v'(p) = 0.37 \times \frac{1}{p} \), which simplifies to \( v'(p) = \frac{0.37}{p} \). This derivative provides insight into the behavior of the function across different population sizes.
Natural Logarithm
A natural logarithm is a logarithm with the base \( e \), where \( e \) is an irrational and transcendental number approximately equal to 2.71828. It is denoted as \( \ln \). In the context of this exercise, the natural logarithm helps model real-world phenomena such as population growth or decay. For the problem at hand, the function \( v(p) = 0.37 \ln p + 0.05 \) uses the natural logarithm to relate population size \( p \) to average walking speed \( v \). Understanding and calculating the natural logarithm is crucial for solving the exercise, as it permits a nonlinear relationship that more accurately reflects real-world situations.
Population Analysis
Population analysis involves studying how certain variables, such as average walking speed, change with population size. In the given problem, Bornstein and Bornstein have provided a function that links the average walking speed to the population of a city. By converting the population to thousands (e.g., Seattle's 531,000 becomes \( p = 531 \)), students can use the function \( v(p) = 0.37 \ln p + 0.05 \) to find the average walking speed. This form of analysis is invaluable in urban planning, public health, and other fields where understanding human behavior at different population scales is essential.
Rate of Change
The rate of change of a function describes how quickly the function's output changes relative to changes in its input. It is given by the derivative. For the function \( v(p) = 0.37 \ln p + 0.05 \), the rate of change of average walking speed with respect to population size is found by differentiating the function. The derivative, \( v'(p) = \frac{0.37}{p} \), shows how the average walking speed changes as the population increases or decreases. This rate of change is particularly useful for interpreting and predicting trends in data, providing insights that static values alone cannot offer.

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Most popular questions from this chapter

The population of the United Kingdom in July 2002 was estimated to be \(59.8\) million. At that time, it was estimated that the population was growing according to the differential equation $$ \frac{d P}{d t}=0.0021 P $$ where \(t\) is the number of years, after July \(2002 .{ }^{20}\) a) Find the function that satisfies the initial-value problem. b) Estimate the population of the United Kingdom in 2015 . c) After what period of time will the population be double that in \(2002 ?\)

Double Declining-Balance Depreciation. An office machine is purchased for \(\$ 5200 .\) Under certain assumptions, its salvage value \(V\) depreciates according to a method called double declining balance, basically \(80 \%\) each year, and is given by $$ V(t)=\$ 5200(0.80)^{\ell} $$ where \(t\) is the time, in years. a) Find \(V^{\prime}(t)\). b) Interpret the meaning of \(\mathbf{V}^{\prime}(t)\).

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The cumulated oil distribution of a french fry fried at \(185^{\circ}\) for 1 min is approximated by the function $$ G(x)=\frac{100}{1+43.3 e^{-0.0425 x}} $$ where \(x\) is the distance from the surface of the potato, in \(\mu \mathrm{m}\left(1 \mu \mathrm{m}=10^{-6} \mathrm{~m}\right)\), and \(G(x)\) is the percent of oil the potato absorbed within a distance of \(x\) from the surface. For example, \(G(100)=21.1\) indicates that \(21.1 \%\) of the absorbed oil is in the first \(100 \mu \mathrm{m}\) of the potato. \({ }^{21}\) a) Use the formula to compute the percent of oil within \(100 \mu \mathrm{m}\) from the surface, \(150 \mu \mathrm{m}\) from the surface, and \(300 \mu \mathrm{m}\) from the surface. b) Compute the derivative \(G^{\prime}(x)\). c) Find the point of inflection for the function \(G(x)\) d) Sketch the graph of the function.
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