/*! 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 55 The number of cable telephone su... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 55

The number of cable telephone subscribers stood at \(3.2\) million at the beginning of \(2004(t=0)\). For the next \(5 \mathrm{yr}\), the number was projected to grow at the rate of $$ R(t)=3.36(t+1)^{0.05} \quad(0 \leq t \leq 5) $$ million subscribers/year. If the projection held true, how many cable telephone subscribers were there at the beginning of \(2008(t=4)\) ?

Short Answer

Expert verified
There were approximately 5.38 million cable telephone subscribers at the beginning of 2008.

Step by step solution

01

Write down the given information

Initial number of subscribers: 3.2 million Growth rate function, R(t): \(3.36(t+1)^{0.05}\) million subscribers/year
02

Set up the integral

To find the total growth in subscribers between 2004 (t=0) and 2008 (t=4), we need to integrate the growth rate function, R(t), over that time interval: $$ \int_{0}^{4} 3.36(t+1)^{0.05} dt $$
03

Integrate the function

Integrate the function with respect to t: $$ \int 3.36(t+1)^{0.05} dt = \frac{3.36}{0.05+1}(t+1)^{0.05+1}+C $$
04

Calculate the definite integral between the limits 0 and 4

Now, we will evaluate the integral from 0 to 4: $$ \left[\frac{3.36}{1.05}(t+1)^{1.05}\right]_{0}^{4} $$ Compute the value at the upper limit (t = 4): $$ \frac{3.36}{1.05}(4 + 1)^{1.05} = \frac{3.36}{1.05}(5)^{1.05} \approx 5.38 $$ Compute the value at the lower limit (t = 0): $$ \frac{3.36}{1.05}(0 + 1)^{1.05} = \frac{3.36}{1.05}(1)^{1.05} \approx 3.20 $$ Subtract the lower limit from the upper limit: $$ 5.38 - 3.20 \approx 2.18 $$ This means that the total growth in subscribers between 2004 and 2008 is approximately 2.18 million.
05

Add the growth to the initial subscriber count

To find the number of subscribers at the beginning of 2008 (t=4), add the total growth to the initial subscriber count: $$ 3.2 \, \text{million} + 2.18 \, \text{million} \approx 5.38 \, \text{million} $$ Thus, there were approximately 5.38 million cable telephone subscribers at the beginning of 2008.

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.

Definite Integral
Understanding the concept of a definite integral is like grasping how to measure the total accumulation of something over a period of time or space. In applied mathematics, it's a crucial tool for quantifying this accumulation in a precise way. Imagine you're filling up a swimming pool and want to know exactly how much water you've added over several hours鈥攖hat's where the definite integral comes into play.

A definite integral, in its essence, is a calculation of the net area under a curve within specific boundaries on a graph. This is integral (pun intended!) in applications ranging from physics to engineering, where it's often necessary to add up continuous quantities. In our cable telephone subscribers example, we utilized a definite integral to find the total increase in subscribers from the beginning of 2004 to the start of 2008.
Growth Rate Function
A growth rate function, symbolized as R(t) in our scenario, represents how quickly something is changing over time. In the context of our problem, it measures the change in the number of cable telephone subscribers as a function of time, t. This kind of function is pivotal in mathematical modeling because it provides a dynamic view of changes, rather than a static snapshot.

The growth rate isn't always consistent; it can vary depending on various factors, much like the acceleration of a car changes with the pressure on the gas pedal. The specific function we dealt with, R(t)=3.36(t+1)^{0.05}, interestingly, indicates that the rate of growth itself is changing, albeit slightly, as time goes on due to the exponent 0.05. This nuanced detail helps reflect more realistic scenarios where growth doesn't follow a straight-line pattern.
Mathematical Modeling
Mathematical modeling is akin to crafting a miniature version of the real world using the language of mathematics so we can study and predict complex systems in a simplified form. It involves creating equations and functions that mimic real-life processes, like the increase of subscribers, weather patterns, or the trajectory of a spaceship.

In our example, we constructed a model using the growth rate function to predict the number of cable telephone subscribers over a five-year period. This model's strength lies in its ability to transform abstract numbers into a digestible narrative about how a situation might unfold. By integrating the growth rate function, we essentially completed the story of our subscriber growth from chapter 2004 to chapter 2008, giving us a rumored ending of approximately 5.38 million subscribers.

From budget forecasts to epidemiological studies, mathematical modeling helps experts in various fields make informed decisions based on projected outcomes. When combined with other tools like the definite integral, it becomes a potent method for analyzing changes over time.

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

The White House wants to cut the gasoline usage from 140 billion gallons per year in 2007 to 128 billion gallons per year in 2017 . But estimates by the Department of Energy's Energy Information Agency suggest that won't happen. In fact, the agency's projection of gasoline usage from the beginning of 2007 until the beginning of 2017 is given by $$ A(t)=0.014 t^{2}+1.93 t+140 \quad(0 \leq t \leq 10) $$ where \(A(t)\) is measured in billions of gallons/year and \(t\) is in years, with \(t=0\) corresponding to 2007 . a. According to the agency's projection, what will be gasoline consumption at the beginning of \(2017 ?\) b. What will be the average consumption/year over the period from the beginning of 2007 until the beginning of 2017 ?

Based on a preliminary report by a geological survey team, it is estimated that a newly discovered oil field can be expected to produce oil at the rate of $$ R(t)=\frac{600 t^{2}}{t^{3}+32}+5 \quad(0 \leq t \leq 20) $$ thousand barrels/year, \(t\) yr after production begins. Find the amount of oil that the field can be expected to yield during the first 5 yr of production, assuming that the projection holds true.

Find the amount of an annuity if $$\$ 400 /$$ month is paid into it for a period of \(20 \mathrm{yr}\), earning interest at the rate of \(6 \% / y\) ear compounded continuously.

Because of the increasingly important role played by coal as a viable alternative energy source, the production of coal has been growing at the rate of $$ 3.5 e^{0.05 \mathrm{y}} $$ billion metric tons/year, \(t\) yr from 1980 (which corresponds to \(t=0\) ). Had it not been for the energy crisis, the rate of production of coal since 1980 might have been only $$ 3.5 e^{0.011} $$ billion metric tons/year, \(t\) yr from 1980 . Determine how much additional coal was produced between 1980 and the end of the century as an alternate energy source.

Sketch the graph and find the area of the region bounded below by the graph of each function and above by the \(x\) -axis from \(x=a\) to \(x=b\). $$f(x)=-x e^{-x^{2}} ; a=0, b=1$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Statistics

Read Explanation

Mechanics Maths

Read Explanation

Decision Maths

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

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.