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In Problems \(29-30,\) a quantity \(P\) is an exponential function of time \(t .\) Use the given information about the function \(P=P_{0} e^{k t}\) to: (a) Find values for the parameters \(k\) and \(P_{0}\). (b) State the initial quantity and the continuous percent rate of growth or decay. \(P=140\) when \(t=3\) and \(P=100\) when \(t=1\)

Short Answer

Expert verified
\( k = \frac{\ln(1.4)}{2} \), \( P_0 \) is the initial quantity, with a rate based on \( k \).

Step by step solution

01

Write the exponential equations

We start by writing the exponential function formula given for both scenarios: when \( t = 3 \), \( P = 140 \), and when \( t = 1 \), \( P = 100 \).For \( t = 3 \):\[ 140 = P_0 e^{3k} \]For \( t = 1 \):\[ 100 = P_0 e^{k} \]
02

Eliminate \(P_0\) by dividing the equations

Divide the equation for \( t = 3 \) by the equation for \( t = 1 \):\[ \frac{140}{100} = \frac{P_0 e^{3k}}{P_0 e^{k}} \]This simplifies to:\[ 1.4 = e^{2k} \]
03

Solve for \(k\)

Take the natural logarithm on both sides of the equation \( 1.4 = e^{2k} \):\[ \ln(1.4) = 2k \]Thus, solving for \(k\):\[ k = \frac{\ln(1.4)}{2} \]
04

Solve for \(P_0\)

Substitute \( k \) back into one of the original equations. We'll use the equation for \( t = 1 \):\[ 100 = P_0 e^{k} \]\[ P_0 = \frac{100}{e^{k}} \]
05

Calculate initial quantity and continuous rate

From Step 4, we know that \( P_0 = \) 100 divided by \( e^{k} \). Substitute \( k = \frac{\ln(1.4)}{2} \) to find the initial quantity:\( P_0 = \frac{100}{e^{\frac{\ln(1.4)}{2}}} \)The initial quantity \( P_0 \) and \( k \) will also tell us the continuous percent rate. Converting \( k \) to percentage:\( \text{Rate} = k \times 100 \% \). Calculate using \( k = \frac{\ln(1.4)}{2} \).

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Key Concepts

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

Continuous Growth Rate
The continuous growth rate is a key part of understanding how exponential functions behave over time. This rate, often denoted by the symbol \( k \), tells us how quickly a quantity increases or decreases continuously. In the context of exponential growth or decay, \( k \) influences how steeply or gently the curve of the function rises or falls.

To determine \( k \), we utilize given data points within the exponential function \( P = P_0 e^{kt} \). Through solving the equation, \( k \) is extracted by using logarithmic transformations, as demonstrated in the solution steps. Specifically, by taking the logarithm of the ratio of the two quantities and dividing by the time span between them, we can solve for \( k \).

Once \( k \) is known, converting it into a percentage helps give a more intuitive sense of the growth or decay rate. This conversion is achieved by multiplying \( k \) by 100 to express it as a percent. This percentage represents the continuous percent rate of growth or decay, making it very helpful in real-world interpretations where understanding the rate in annual or continuous terms is crucial.
Initial Quantity
The initial quantity, denoted as \( P_0 \), represents the starting value of the quantity in question before it undergoes any exponential growth or decay. Understanding this parameter is essential, as it serves as the baseline from which all exponential changes occur.

To find \( P_0 \), apply the initial condition when time \( t = 0 \). In calculations, it's common to rearrange the exponential function after determining \( k \), allowing the extraction of \( P_0 \) using known values of \( P \) at a specific time. As seen in the original problem's solution, substituting \( k \) back into the equation gives \( P_0 \) through the equation \( P_0 = \frac{P}{e^{kt}} \). This relation allows us to effectively isolate \( P_0 \) and determine its value.

The initial quantity sets the stage for understanding how large or small a system is at the start, offering crucial context for assessing future growth or decay. In practical terms, knowing \( P_0 \) helps predict and plan for future needs or changes based on the calculated growth or decay behavior.
Exponential Function
An exponential function is a mathematical expression used to model situations where quantities grow or shrink at a rate proportional to their current value. The general form is \( P = P_0 e^{kt} \), where \( P \) is the quantity at time \( t \), \( P_0 \) is the initial quantity, and \( k \) is the continuous growth rate.

Characteristics of exponential functions include:
  • The effect of \( t \) is exponential, meaning every time \( t \) changes, \( P \) multiplies by a factor related to base \( e \) and the rate \( k \).
  • If \( k > 0 \), the function models exponential growth; if \( k < 0 \), it models exponential decay.
  • Exponential growth functions rise rapidly, showcasing what is often called compound growth.
These functions are widely used because they accurately describe processes like population growth, radioactive decay, and interest calculations. Using them is beneficial when dealing with any scenario where the rate of change is proportional to the current state of the system.

Understanding exponential functions allows for better prediction, planning, and optimization in various fields such as finance, natural sciences, and engineering.

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Most popular questions from this chapter

Hydroelectric power is electric power generated by the force of moving water. The table shows the annual percent change in hydroelectric power consumption by the US industrial sector. $$\begin{array}{c|c|c|c|c|c} \hline \text { Year } & 2005 & 2006 & 2007 & 2008 & 2009 \\ \hline \text { \% growth over previous yr } & -1.9 & -10 & -45.4 & 5.1 & 11 \\ \hline \end{array}$$ (a) According to the US Department of Energy, the US industrial sector consumed about 29 trillion BTUs of hydroelectric power in \(2006 .\) Approximately how much hydroelectric power (in trillion BTUs) did the US consume in \(2007 ?\) In \(2005 ?\) (b) Graph the points showing the annual US consumption of hydroelectric power, in trillion BTUs, for the years 2004 to \(2009 .\) Label the scales on the horizontal and vertical axes. (c) According to this data, when did the largest yearly decrease, in trillion BTUs, in the US consumption of hydroelectric power occur? What was this decrease?

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Write the functions in Problems \(21-24\) in the form \(P=P_{0} a^{t}\) Which represent exponential growth and which represent exponential decay? $$P=2 e^{-0.5 t}$$

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Concern biodiesel, a fuel derived from renewable resources such as food crops, algae, and animal oils. The table shows the percent growth over the previous year in US biodiesel consumption. $$\begin{array}{c|c|c|c|c|c|c|c} \hline \text { Year } & 2003 & 2004 & 2005 & 2006 & 2007 & 2008 & 2009 \\ \hline \text { * growth } & -12.5 & 92.9 & 237 & 186.6 & 37.2 & -11.7 & 7.3 \\ \hline \end{array}$$ (a) According to the US Department of Energy, the US consumed 91 million gallons of biodiesel in 2005 Approximately how much biodiesel (in millions of gallons) did the US consume in \(2006 ?\) In \(2007 ?\) (b) Graph the points showing the annual US consumption of biodiesel, in millions of gallons of biodiesel, for the years 2005 to \(2009 .\) Label the scales on the horizontal and vertical axes.

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