Question

a cup of coffee contains 100 mg of caffeine, which leaves the body at a continuous rate of 31% per hour.
(a) write a formula for the amount, a mg, of caffeine in the body t hours after drinking a cup of coffee.
a =
(b) use the graph of a (shown below) to estimate the half-life of caffeine to the nearest hour.
half-life ≈ 2 hours
(graph: vertical axis with 25, 50, 75, 100; horizontal t-axis from 0 to 4; curve starting at (0,100) decreasing)
(c) use logarithms to find the half-life of caffeine to two decimal places.
half-life = hours

Explanation:

Part (a)

Step1: Recall continuous decay formula

The formula for continuous decay is \( A = P e^{rt} \), where \( P \) is the initial amount, \( r \) is the continuous decay rate, and \( t \) is time.
Here, \( P = 100 \) mg, \( r = - 0.31 \) (negative because it's a decay rate).

Step2: Substitute values into formula

Substitute \( P = 100 \) and \( r=- 0.31 \) into the formula: \( A = 100e^{-0.31t} \)

The half - life is the time it takes for the amount of caffeine to reduce to half of its initial amount. The initial amount is 100 mg, so half - life is the time when \( A = 50 \) mg. From the graph, we can see that when \( A = 50 \) mg, the value of \( t \) (time) is approximately 2 hours (by looking at the horizontal line \( A = 50 \) and finding the corresponding \( t \) - value on the curve).

Step1: Set up the equation for half - life

We know that the formula for the amount of caffeine is \( A = 100e^{-0.31t} \). For half - life, \( A=\frac{100}{2}=50 \) mg. So we set up the equation:
\( 50=100e^{-0.31t} \)

Step2: Divide both sides by 100

Divide both sides of the equation \( 50 = 100e^{-0.31t} \) by 100:
\( \frac{50}{100}=e^{-0.31t} \)
\( 0.5=e^{-0.31t} \)

Step3: Take the natural logarithm of both sides

Take the natural logarithm (ln) of both sides of the equation \( 0.5 = e^{-0.31t} \). Recall that \( \ln(e^{x})=x \). So we have:
\( \ln(0.5)=\ln(e^{-0.31t}) \)
\( \ln(0.5)=- 0.31t \)

Step4: Solve for \( t \)

We can solve for \( t \) by dividing both sides of the equation \( \ln(0.5)=-0.31t \) by \( - 0.31 \):
\( t=\frac{\ln(0.5)}{-0.31} \)
We know that \( \ln(0.5)\approx - 0.6931 \), so:
\( t=\frac{- 0.6931}{-0.31}\approx2.24 \)

Answer:

\( A = 100e^{-0.31t} \)

Part (b)