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Given an initial value of 50 units for parts (a)-(d) below, in each case construct a function that represents \(Q\) as a function of time \(t\). Assume that when \(t\) increases by 1: a. \(Q(t)\) doubles b. \(Q(t)\) increases by \(5 \%\) c. \(Q(t)\) increases by ten units d. \(Q(t)\) is multiplied by 2.5

Short Answer

Expert verified
a. \( Q(t) = 50 \times 2^t \) b. \( Q(t) = 50 \times (1.05)^t \) c. \( Q(t) = 50 + 10t \) d. \( Q(t) = 50 \times (2.5)^t \)

Step by step solution

01

Understand the problem

You need to construct functions that represent the quantity Q as a function of time t for different scenarios given the initial value of 50 units.
02

Define the initial value

You are given an initial value of 50 units for each case. This means at time t=0, Q(0) = 50.
03

Determine the function for case (a)

In this case, Q(t) doubles each time t increases by 1. Hence the function will be an exponential function: \[ Q(t) = 50 \times 2^t \]
04

Determine the function for case (b)

Here, Q(t) increases by 5% each time t increases by 1. Hence the function will be: \[ Q(t) = 50 \times (1.05)^t \]
05

Determine the function for case (c)

In this scenario, Q(t) increases by ten units each time t increases by 1. Hence the function is a linear function: \[ Q(t) = 50 + 10t \]
06

Determine the function for case (d)

Here, Q(t) is multiplied by 2.5 each time t increases by 1. So the function will be: \[ Q(t) = 50 \times (2.5)^t \]

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Key Concepts

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

Initial Value Problem
An initial value problem involves finding a function that satisfies a given relationship, with a specific value at a starting point. In our example, we know the quantity at time zero, which is 50 units.
The initial value helps establish the starting point for our functions. Whether it involves exponential growth or linear increase, the initial value sets the context for how the quantity changes over time.
Consider the initial value problem as a foundation upon which different types of growth (like exponential or linear) build. This initial point is essential for solving any time-dependent function accurately.
Rate of Change
The rate of change describes how a quantity changes over time. In math, it helps us understand how fast or slow something increases or decreases, and it's vital in differentiating between exponential and linear functions.
If the rate of change is constant, meaning the same amount is added each time t increases, you have a linear function. This is evident in case (c), where the quantity increases by ten units per time increment: \( Q(t) = 50 + 10t \).
When the rate of change itself changes, the growth can be exponential. In cases (a), (b), and (d), the quantity grows at different rates. Case (a), where the quantity doubles: \(Q(t) = 50 \times 2^t\). This means the rate of change increases as time progresses. Recognizing this will help you identify different types of functions in various problems.
Time-Dependent Functions
Time-dependent functions describe how a quantity varies over time. These kinds of functions are essential in modeling real-world phenomena.
For example, exponential functions, like those in cases (a), (b), and (d), show how a quantity grows rapidly or decays over time. Formulas like \(Q(t) = 50 \times (1.05)^t\) illustrate how compounding growth works—each increment of time multiplies the quantity by a fixed factor.
In contrast, linear functions, such as the one in case (c), show steady and predictable growth: \(Q(t) = 50 + 10t\). Such functions are easier to predict and model certain types of changes where the quantity increases or decreases at a constant rate.
Understanding how to interpret and construct these functions helps in various subjects, including physics, economics, and ecology, where time-related changes are prevalent.

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Most popular questions from this chapter

(Requires technology to find a best-fit function.) Estimates for world population vary, but the data in the accompanying table are reasonable estimates. $$ \begin{aligned} &\text { World Population }\\\ &\begin{array}{cc} \hline & \text { Total Population } \\ \text { Year } & \text { (millions) } \\ \hline 1800 & 980 \\ 1850 & 1260 \\ 1900 & 1650 \\ 1950 & 2520 \\ 1970 & 3700 \\ 1980 & 4440 \\ 1990 & 5270 \\ 2000 & 6080 \\ 2005 & 6480 \\ \hline \end{array} \end{aligned} $$ a. Enter the data table into a graphing program (you may wish to enter 1800 as 0,1850 as \(50,\) etc. ) or use the data file WORLDPOP in Excel or in graph link form. b. Generate a best-fit exponential function. c. Interpret each term in the function, and specify the domain and range of the function. d. What does your model give for the growth rate? e. Using the graph of your function, estimate the following: i. The world population in \(1750,1920,2025,\) and 2050 ii. The approximate number of years in which world population attained or will attain 1 billion (i.e., 1000 million), 4 billion, and 8 billion f. Estimate the length of time your model predicts it takes for the population to double from 4 billion to 8 billion people.

A swimming pool is initially shocked with chlorine to bring the chlorine concentration to 3 ppm (parts per million). Chlorine dissipates in reaction to bacteria and sun at a rate of about \(15 \%\) per day. Above a chlorine concentration of 2 ppm, swimmers experience burning eyes, and below a concentration of 1 ppm, bacteria and algae start to proliferate in the pool environment. a. Construct an exponential decay function that describes the chlorine concentration (in parts per million) over time. b. Construct a table of values that corresponds to monitoring chlorine concentration for at least a 2-week period. c. How many days will it take for the chlorine to reach a level tolerable for swimmers? How many days before bacteria and algae will start to grow and you will need to add more chlorine? Justify your answers.

Write the equation of each exponential growth function in the form \(y=C a^{x}\) where: a. The initial population is 350 and the growth factor is \(\frac{4}{3}\). b. The initial population is \(5 \cdot 10^{9}\) and the growth factor is \(1.25 .\) c. The initial population is 150 and the population triples during each time period. d. The initial population of 2 quadruples every time period.

Generate quick sketches of each of the following functions, without the aid of technology. $$ \begin{array}{ll} f(x)=4(3.5)^{x} & g(x)=4(0.6)^{x} \\ h(x)=4+3 x & k(x)=4-6 x \end{array} $$ a. As \(x \rightarrow+\infty\), which function(s) approach \(+\infty ?\) b. As \(x \rightarrow+\infty\), which function(s) approach \(0 ?\) c. As \(x \rightarrow-\infty\), which function(s) approach \(-\infty\) ? d. As \(x \rightarrow-\infty\), which function(s) approach 0 ?

Rewrite each expression so that no fraction appears in the exponent and each expression is in the form \(a^{x}\). a. \(3^{w / 4}\) b. \(2^{x / 3}\) c. \(\left(\frac{1}{2}\right)^{x_{4}}\) d. \(\left(\frac{1}{4}\right)^{x / 2}\)

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