Problem 11
Question
A cup of coffee contains \(100 \mathrm{mg}\) of caffeine, which leaves the body at a continuous rate of \(17 \%\) per hour. (a) Write a formula for the amount, \(A \mathrm{mg}\), of caffeine in the body \(t\) hours after drinking a cup of coffee. (b) Graph the function from part (a). Use the graph to estimate the half-life of caffeine. (c) Use logarithms to find the half-life of caffeine.
Step-by-Step Solution
Verified Answer
(a) Formula: \( A(t) = 100 e^{-0.186t} \). (b) Graph shows half-life \( \approx 4 \) hours. (c) Algebraic half-life \( \approx 3.73 \) hours.
1Step 1: Establish the Exponential Decay Formula
The situation describes an exponential decay, where caffeine leaves the body at a continuous rate each hour. The formula for exponential decay is given by: \[ A(t) = A_0 e^{-kt} \] where \( A_0 \) is the initial amount of caffeine (100 mg in this case), \( k \) is the decay constant, and \( t \) is time in hours.
2Step 2: Determine the Decay Constant
We know that caffeine decreases by 17% per hour. To find the decay constant \( k \), we set up the equation for continuous decay: \[ 0.83 = e^{-k \cdot 1} \]Taking the natural logarithm of both sides, we get:\[ -k = \ln(0.83) \]\[ k = -\ln(0.83) \approx 0.186 \]
3Step 3: Write the Formula for Caffeine Amount
Using the initial amount \( A_0 = 100 \) mg and the decay constant \( k = 0.186 \), the formula for the caffeine amount is:\[ A(t) = 100 e^{-0.186t} \]
4Step 4: Plot the Decay Function
To graph the function, plot \( A(t) = 100 e^{-0.186t} \) over a reasonable range of \( t \), say from 0 to 10 hours. The graph should show a decreasing exponential curve starting at 100 mg and getting closer to zero as time increases.
5Step 5: Estimate the Half-Life from the Graph
The half-life is the time at which the caffeine amount is reduced to half its initial amount, i.e., 50 mg. Using the graph plotted in Step 4, estimate the time \( t \) at which \( A(t) = 50 \). Visually, this occurs around \( t \approx 4 \) hours.
6Step 6: Solve for Half-Life Using Logarithms
To find the half-life algebraically, set \( A(t) = 50 \) and solve for \( t \):\[ 50 = 100 e^{-0.186t} \]Divide both sides by 100:\[ 0.5 = e^{-0.186t} \]Take the natural logarithm:\[ \ln(0.5) = -0.186t \]Solve for \( t \):\[ t = \frac{\ln(0.5)}{-0.186} \approx 3.73 \text{ hours} \]
Key Concepts
Half-lifeDecay ConstantLogarithmic Functions
Half-life
The half-life of a substance is the time it takes for the amount of that substance in a system to reduce to half of its initial value. In our example, the caffeine from a cup of coffee decreases by a set percentage every hour.
This type of situation is modeled by exponential decay. Hence, we use the exponential decay formula:- Formula: \[ A(t) = A_0 e^{-kt} \] Where: \( A_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time. For caffeine, we want to find when \( A(t) = 50 \) mg if we start with \( A_0 = 100 \) mg.
This is the definition of the half-life. Visually, in a graph of this exponential function, you would find this point where the curve decreases to half of its original height.
The algebraic method involves solving the equation \( 50 = 100 e^{-0.186t} \), which lets us exactly calculate the half-life time.
This type of situation is modeled by exponential decay. Hence, we use the exponential decay formula:- Formula: \[ A(t) = A_0 e^{-kt} \] Where: \( A_0 \) is the initial amount, \( k \) is the decay constant, and \( t \) is time. For caffeine, we want to find when \( A(t) = 50 \) mg if we start with \( A_0 = 100 \) mg.
This is the definition of the half-life. Visually, in a graph of this exponential function, you would find this point where the curve decreases to half of its original height.
The algebraic method involves solving the equation \( 50 = 100 e^{-0.186t} \), which lets us exactly calculate the half-life time.
Decay Constant
The decay constant \( k \) is a crucial part of any exponential decay function. It dictates how quickly the substance decreases over time. In this exercise, caffeine leaves the body at a rate of 17% per hour, which we express in the decay constant formula.
To determine \( k \), we follow these steps:- Recognize that after 1 hour, 83% of caffeine remains, implying a decay of 17%. - Set up the equation: \[ 0.83 = e^{-k \cdot 1} \] - Solve for \( k \) using logarithms: \[ -k = \ln(0.83) \] \[ k = -\ln(0.83) \approx 0.186 \] With \( k \), we have a complete function to predict the caffeine content over time: \( A(t) = 100 e^{-0.186t} \).
This function shows an exponential decrease consistent with the biological elimination of caffeine from the body.
To determine \( k \), we follow these steps:- Recognize that after 1 hour, 83% of caffeine remains, implying a decay of 17%. - Set up the equation: \[ 0.83 = e^{-k \cdot 1} \] - Solve for \( k \) using logarithms: \[ -k = \ln(0.83) \] \[ k = -\ln(0.83) \approx 0.186 \] With \( k \), we have a complete function to predict the caffeine content over time: \( A(t) = 100 e^{-0.186t} \).
This function shows an exponential decrease consistent with the biological elimination of caffeine from the body.
Logarithmic Functions
Logarithmic functions allow us to solve exponential equations, particularly useful for finding unknown variables like time in half-life problems. They are the inverse functions of exponentials, and they unravel complex exponential expressions.Here's how logarithms help solve for caffeine's half-life:- When we have the equation \( 50 = 100 e^{-0.186t} \), we isolate \( 0.5 = e^{-0.186t} \) by division. - To eliminate the exponential, take the natural logarithm of both sides: \[ \ln(0.5) = -0.186t \] - This linearizes our equation, allowing us to solve for \( t \): \[ t = \frac{\ln(0.5)}{-0.186} \approx 3.73 \text{ hours} \] Logarithms are invaluable for finding precise solutions where graphical estimation may fall short.
They provide a robust mathematical tool to cross-validate visual estimations of exponential functions.
They provide a robust mathematical tool to cross-validate visual estimations of exponential functions.
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