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Chapter 1: Problem 51

The number of pages in a science journal published every year increased exponentially from 625 pages in 1940 to 14,000 pages in \(1980 .\) Sketch two graphs of this increase: one with linear axes, and one with logarithmic axes.

Short Answer

Expert verified
Exponential model: \( N(t) = 625 \times e^{0.1985t} \). Linear axes: curve increasing from 625 to 14,000 pages. Logarithmic axes: straight line.

Step by step solution

01

Identify the given points

We know the number of pages in the science journal was 625 in 1940 and 14,000 in 1980. These can be represented as points (1940, 625) and (1980, 14,000).
02

Convert years to relative time

To simplify the calculations, set 1940 as the starting point (year 0). Therefore, 1980 corresponds to year 40. The points become (0, 625) and (40, 14,000).
03

Determine the exponential growth model

The general form of the exponential growth model is \[ N(t) = N_0 \times e^{kt} \]where \(N_0\) is the initial quantity (625 pages), \(t\) is the time (in years), and \(k\) is the growth rate. We need to find the value of \(k\).
04

Solve for the growth rate, k

Using the point (40, 14,000):\[ 14,000 = 625 \times e^{40k} \]Divide both sides by 625:\[ 22.4 = e^{40k} \]Take the natural logarithm of both sides:\[ \text{ln}(22.4) = 40k \]Solve for \(k\):\[ k = \frac{\text{ln}(22.4)}{40} \approx 0.1985 \]
05

Write the exponential function

Using the values found, the exponential growth function is:\[ N(t) = 625 \times e^{0.1985t} \]
06

Sketch the graph with linear axes

Plot the exponential function \( N(t) = 625 \times e^{0.1985t} \) on a graph with linear time (years) on the x-axis and the number of pages on the y-axis. The result should show a curve that starts at 625 pages in 1940 and increases steeply to 14,000 pages by 1980.
07

Sketch the graph with logarithmic axes

Plot the natural logarithm of the number of pages (\(\text{ln}(N(t))\)) against time (years) on linear axes. Transform the original function to:\[ \text{ln}(N(t)) = \text{ln}(625) + 0.1985t \]This should result in a straight line starting at \(\text{ln}(625)\) (in 1940) with a slope of 0.1985.

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Key Concepts

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

Linear Transformation
A linear transformation allows us to transform time so the calculations become simpler. For this exercise, we start with two key points: 1940 with 625 pages and 1980 with 14,000 pages. Converting these to relative time points where 1940 is 'year 0', makes 1980 become 'year 40'. This simplifies our calculations and plotting steps.
Logarithmic Transformation
Logarithmic transformation is useful to linearize exponential data. By taking the natural logarithm of the exponential function, we can transform it into a linear form. In this exercise, we transform the function \( N(t) = 625 \times e^{0.1985t} \) into \( \text{ln}(N(t)) = \text{ln}(625) + 0.1985t \). This straight line enables easier graphing and understanding of the growth trend.
Graphing Exponential Functions
Exponential functions grow rapidly. In our example, the function \( N(t) = 625 \times e^{0.1985t} \) illustrates the rapid rise in the number of journal pages from 625 in 1940 to 14,000 in 1980. To visualize this, you plot time (years) on the x-axis and the number of pages on the y-axis. This helps us see the steep curve that characterizes exponential growth.
Natural Logarithm
The natural logarithm (ln) is a special way to handle exponential growth. It enables us to solve for the growth rate (k). In our example, given \( 14,000 = 625 \times e^{40k} \), we take the ln of both sides to obtain \( \text{ln}(14,000) = \text{ln}(625) + 40k \). Solving for k involves isolating k: \( k = \frac{\text{ln}(22.4)}{40} \approx 0.1985 \). This process is crucial for accurate growth modeling.
Growth Rate Estimation
Estimating the growth rate (k) is key for understanding exponential trends. In the exercise, we derived \( k\approx 0.1985 \) by comparing the initial and final values over a 40-year span. This tells us how quickly the journal pages grow each year. Understanding this rate helps in predicting future growth and making informed decisions based on data trends.

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