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Chapter 9: Problem 19

Find the associated exponential decay or growth model. $$ Q=1,000 \text { when } t=0 ; \text { half-life }=1 $$

Short Answer

Expert verified
The associated exponential decay model is \(Q(t) = 1000e^{\ln{\frac{1}{2}}t}\).

Step by step solution

01

Determine the initial quantity Q鈧� when t = 0

From the given information, the initial quantity \(Q_0 = 1000\). As \(t = 0\), we have \(Q(0) = 1000\).
02

Use the half-life information to find the value of k

Given that half-life is 1, which means at \(t = 1\), the quantity \(Q(1)\) becomes half of the initial quantity \(Q_0\). So, we have: $$ Q(1) = \frac{Q_0}{2} $$ Now, substitute the values of \(Q_0\) and the general formula for exponential decay \(Q(t) = Q_0e^{-kt}\), we get: $$ \frac{1000}{2} = 1000e^{-k(1)} $$
03

Solve for the decay constant k

Now, let's solve the equation for \(k\). $$ \frac{1000}{2} = 1000e^{-k} $$ $$ \frac{1}{2} = e^{-k} $$ Take the natural logarithm of both sides: $$ \ln{\frac{1}{2}} = -k $$ So, we find the decay constant \(k\) as: $$ k = -\ln{\frac{1}{2}} $$
04

Write the exponential decay model

Now that we have the value of \(k\), we can write the complete exponential decay model \(Q(t) = Q_0e^{-kt}\): $$ Q(t) = 1000e^{-\left(-\ln{\frac{1}{2}}\right)t} $$ So, the associated exponential decay model is: $$ Q(t) = 1000e^{\ln{\frac{1}{2}}t} $$

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Key Concepts

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

Half-Life
The concept of half-life is crucial in the context of an exponential decay model. Half-life refers to the time it takes for a quantity to reduce to half of its initial value in a decay process. It is an extremely useful measure, especially in fields like physics and chemistry, where it helps to understand how substances decay over time. For example, if you start with 1,000 units of a substance, and the half-life is 1 year, then in one year, only 500 units would remain. Understanding half-life can help predict how a substance will diminish over time, and it allows scientists and researchers to calculate the ultimate lifespan of any decaying object or material.
Decay Constant
The decay constant, often represented by the symbol \(k\), is a key factor in exponential decay models. It helps in understanding the rate at which a substance or quantity decreases over time.To find the decay constant, one usually depends on given data about the half-life of a material. To illustrate, we use the fact that: \[ \frac{1}{2} = e^{-k imes ext{half-life}} \] Taking the natural logarithm of both sides helps isolate \(k\).The negative sign in front of \(k\) in the equation \(Q(t) = Q_0e^{-kt}\) indicates decay, showing the exponential decline in the quantity over time.In simpler terms, the decay constant tells you how "fast" something is decaying. A higher value of \(k\) means faster decay.
Exponential Function
An exponential function is a mathematical expression where a constant base is raised to some power or an exponent. In the context of decay, the function takes the form of \(Q(t) = Q_0e^{-kt}\). Here, \(Q(t)\) represents the quantity of a substance at time \(t\), \(Q_0\) is the initial quantity, \(e\) is the base of the natural logarithm (\(\approx 2.718\)), and \(k\) is the decay constant.Exponential functions are incredibly helpful for describing processes where quantities decrease at rates proportional to their current value. This is why they are extremely relevant in scientific fields dealing with phenomena like radioactive decay, population decline, or cooling of materials. The exponential function allows us to model and predict how a quantity changes over time, providing both clarity and accuracy.

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Most popular questions from this chapter

The half-life of cobalt 60 is 5 years. a. Obtain an exponential decay model for cobalt 60 in the form \(Q(t)=Q_{0} e^{-k t}\). (Round coefficients to three significant digits.) b. Use your model to predict, to the nearest year, the time it takes one third of a sample of cobalt 60 to decay.

Convert the given exponential function to the form indicated. Round all coefficients to four significant digits. $$ f(t)=10(0.987)^{t} ; f(t)=Q_{0} e^{-k t} $$

The following table gives the approximate number of Facebook users at various times since its establishment early in \(2004 .^{12}\) $$ \begin{array}{|r|r|r|r|r|r|r|r|r|} \hline \begin{array}{r} \text { Year } \boldsymbol{t} \\ \text { (since start } \\ \text { of 2004) } \end{array} & 0 & 1 & 2 & 2.5 & 3 & 3.5 & 4 & 4.5 \\ \hline \begin{array}{r} \text { Facebook } \\ \text { Members } \boldsymbol{n} \\ \text { (millions) } \end{array} & 0 & 1 & 5.5 & 7 & 12 & 30 & 58 & 80 \\ \hline \end{array} $$ a. Find a quadratic regression model for these data. (Round coefficients to the nearest whole number.) Graph the model, together with the data. b. Assuming the trend had continued, estimate the number of members at the start of 2010 to the nearest 10 million members. c. Is the quadratic model appropriate for long-term prediction of the number of members? Why?

A member of your study group tells you that, because the following set of data does not suggest a straight line, the data are best modeled by a quadratic. $$ \begin{array}{|c|c|c|c|c|c|} \hline \boldsymbol{x} & 0 & 2 & 4 & 6 & 8 \\ \hline \boldsymbol{y} & 1 & 2 & 1 & 0 & 1 \\ \hline \end{array} $$ Comment on her suggestion.

What happens to the function \(P(t)=\frac{N}{1+A b^{-t}}\) if \(A=0 ?\) If
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