/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 We found a base \(e\) so that \(... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 5

We found a base \(e\) so that \(E(t)=e^{t}\) has the property that the rate of change of \(E\) at 0 is \(1 .\) Suppose we had searched for a number \(B\) so that the average rate of change of \(E_{B}(t)=B^{t}\) on [0,0.01] is 1: $$m_{0,0.01}=\frac{E_{B}(0.01)-E_{B}(0)}{0.01}=\frac{B^{0.01}-B^{0}}{0.01}=\frac{B^{0.01}-1}{0.01}=1$$ a. Solve the last equation for \(B\). b. Solve for \(B\) in each of the equations: $$\frac{B^{0.001}-1}{0.001}=1 \quad \frac{B^{0.00001}-1}{0.00001}=1 \quad \frac{B^{0.0000001}-1}{0.0000001}=1$$

Short Answer

Expert verified
The solved value of \( B \) is approximately 2.718, aligning with the constant \( e \).

Step by step solution

01

Understanding the problem

We need to find the value of \( B \) for which the expression \( \frac{B^{x}-1}{x}=1 \) holds. This indicates that the average rate of change over the interval is 1, similar to the property for the base \( e \).
02

Expressing the equation

We have the general expression \( \frac{B^{x}-1}{x}=1 \). This can be rearranged into the form \( B^{x} - 1 = x \).
03

Solving for B in terms of x=0.01

Substitute \( x = 0.01 \) into \( B^{x} - 1 = x \) and get \( B^{0.01} = 1.01 \). Therefore, \( B = 1.01^{1/0.01} \). Calculating this gives \( B \approx 2.718 \).
04

Solving for B in terms of x=0.001

Substitute \( x = 0.001 \) into \( B^{x} - 1 = x \) so \( B^{0.001} = 1.001 \). Then \( B = 1.001^{1/0.001} \). Calculating this gives \( B \approx 2.718 \).
05

Solving for B in terms of x=0.00001

Substitute \( x = 0.00001 \) to \( B^{x} - 1 = x \), which gives \( B^{0.00001} = 1.00001 \). Then \( B = 1.00001^{1/0.00001} \). Calculating this gives \( B \approx 2.718 \).
06

Solving for B in terms of x=0.0000001

Substitute \( x = 0.0000001 \) to \( B^{x} - 1 = x \), which gives \( B^{0.0000001} = 1.0000001 \). Then \( B = 1.0000001^{1/0.0000001} \). Calculating this gives \( B \approx 2.718 \).
07

Conclusion

In all cases, as \( x \) approaches 0, \( B \) approaches \( e \), which is approximately 2.718. This shows the base \( B \) behaves as the constant \( e \) when the rate of change is set to 1 for infinitely small intervals.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Exponential Functions
Exponential functions are mathematical expressions where the variable is the exponent. This structure makes them very unique and useful, especially in modeling growth processes, compound interest, and natural phenomena.

Exponential functions take the form \( f(t) = a \cdot b^{t} \), where:
  • \( a \) is a constant multiplier.
  • \( b \) is the base of the function, which determines the rate and nature of growth or decay.
  • \( t \) is the variable, often representing time.
In these functions, as \( t \) increases, the value of \( f(t) \) can change dramatically if \( b \) is greater than 1 (growth) or decrease rapidly if \( b \) is between 0 and 1 (decay). Exponential growth is characterized by its accelerating increase, making it essential in fields like finance and population studies.
Rate of Change
The rate of change in mathematics tells us how one quantity changes in relation to another. In calculus, this concept is often explored through the derivative, which shows the instantaneous rate of change at any given point.

For exponential functions like \( f(t) = e^{t} \), the rate of change can be particularly interesting. The derivative of \( e^{t} \) is itself, \( e^{t} \), meaning the rate of change is precisely the value of the function at any point.

This property is fascinating because it implies that the function grows at a rate proportional to its size. For the problem given, the goal is to determine how different bases \( B \) must be altered so their average rate of change over very small intervals equals 1, similar to the natural base \( e \). Understanding rates of change is crucial in fields such as physics, economics, and engineering.
Euler's Number
Euler's number, denoted as \( e \), is a fundamental constant approximately equal to 2.718. It is named after the Swiss mathematician Leonhard Euler and is essential in mathematics because of its properties and occurrence in various natural phenomena.

One of the unique characteristics of \( e \) is that it serves as the base for the natural exponential function \( e^{t} \), whose rate of change at \( t = 0 \) is exactly 1. This makes \( e \) the only base where the derivative of the function at 0 equals 1, giving it an appealing simplicity in calculus.

Euler's number emerges in many areas such as compound interest calculations, where it helps model continuous growth or decay, and in defining logarithms, with the natural logarithm having \( e \) as its base. Understanding \( e \) and its applications can enrich one's knowledge in calculus and beyond, broadening comprehension of mathematical behaviors and functions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

E. O. Wilson, a pioneer in study of area-species relations on islands, states in Diversity of Life, p 221,: "In more exact language, the number of species increases by the area-species equation, \(S=C A^{z},\) where \(A\) is the area and \(\mathrm{S}\) is the number of species. \(C\) is a constant and \(z\) is a second, biologically interesting constant that depends on the group of organisms (birds, reptiles, grasses). The value of \(z\) also depends on whether the archipelago is close to source ares, as in the case of the Indonesian islands, or very remote, as with Hawaii \(\cdots\) It ranges among faunas and floras around the world from about 0.15 to 0.35 ." Discuss this statement as a potential Mathematical Model.

(a) Compute the centered difference $$\frac{P(a+h)-P(a-h)}{2 h}$$ which is an approximation to \(P^{\prime}(a),\) for \(P(t)=t^{2}\) and compare your answer with \(P^{\prime}(a)\). (b) Compute the centered difference $$\frac{P(a+h)-P(a-h)}{2 h}$$ for \(P(t)=5 t^{2}-3 t+7\) and compare your answer with \(P^{\prime}(a)\).

The Body Mass Index, $$\mathrm{BMI}=\frac{\text { Mass }}{\text { Height }^{2}}$$ was introduced by Adolphe Quetelet, a French mathematician and statistician in \(1869 .\) The Center for Disease Control and Prevention (CDC) notes that BMI is a helpful indicator of overweight and obesity in adults. From simple allometric considerations, BMI3 = Mass/Height \(^{3}\) should be approximately a constant, \(C\). If \(B M I 3=\) Mass/Height \(^{3}=C\) then \(B M I=\) Mass/Height \(^{2}=C\) Height. so that BMI should increase with height. CDC also states that "... women are more likely to have a higher percentage of body fat than men for the same BMI." If a male and a female both have \(\mathrm{BMI}=23\) and are of average height for their sex \((1.77\) meters for males and 1.63 meters for females), then BMI3 for the male \(=\frac{23}{1.77}=13.0\) and BMI3 for the female \(=\frac{23}{1.63}=14.1\) Thus BMI3 is larger for the female than for the male and may indicate a larger percentage of body fat for the female. Shown are four Age, and 50 th percentile Weight, Height data points for boys and for girls. Compute BMI and BMI3 for the four points and plot the sixteen points on a graph. Which of the two indices, BMI or BMI3, remains relatively constant with age? Data are from the Centers for Disease Control and Prevention, http://www.cdc.gov/growthcharts/data/set1clinical/cj41c021.pdf and \(\cdots\) cj41c022.pdf. $$ \begin{array}{|l|rrrr|} \hline \text { Age (Boys) } & 8 & 12 & 16 & 20 \\ \hline \text { Weight (kg) (50 percentile) } & 26 & 41 & 62 & 71 \\ \text { Height (cm) (50 percentile) } & 128 & 149 & 174 & 177 \\ \text { BMI kg/m }^{2} & & & & \\ \text { BMI3 kg/m }^{3} & & & & \\ \hline & & & 12 & 16 & 20 \\ \text { Age (Girls) } & 26 & 41 & 54 & 58 \\ \hline \text { Weight (kg) (50 percentile) } & 128 & 151 & 162 & 163 \\ \text { Height (cm) (50 percentile) } & & & & \\ \text { BMI kg/m }^{2} & & & & \\ \text { BMI3 kg/m }^{3} & & & & \\ & & & & \\ \hline \end{array} $$ We suggest that BMI3 might be more useful than BMI as an index of body fat. Other indices of body fat that have been suggested include \(M / H, M^{1 / 3} / H, H / M^{1 / 3},\) and \(c M^{1.2} / H^{3.3}\). The interested reader should visit the web site cdc.gov/ncedphp/dnpa/bmi and read the references there.

Let \(E(t)=10^{t}\) a. Approximate \(E^{\prime}(0)\) using the centered difference quotient on [-0.0001,0.0001] . b. Use your value for \(E^{\prime}(0)\) and \(E^{\prime}(t)=E^{\prime}(0) E(t)\) to approximate \(E^{\prime}(-1), E^{\prime}(1),\) and \(E^{\prime}(2)\). c. Sketch the graphs of \(E(t)\) and \(E^{\prime}(t)\). d. Repeat a., b., and c. for \(E(t)=8^{t}\).

Let \(S_{2}\) denote the points of the \(X\) -axis that have positive \(x\) -coordinate and \(S_{1}\) denote the points of the \(X\) -axis that do not belong to \(S_{2} .\) Does \(S_{2}\) have a left most point?
See all solutions

Recommended explanations on Biology Textbooks

Control of Gene Expression

Read Explanation

Cells

Read Explanation

Cell Cycle

Read Explanation

Astrobiological Science

Read Explanation

Cellular Energetics

Read Explanation

Genetic Information

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.