Question

ch 8 closure

1. a team of doctors reports in a medical journal that a drug is metabolized (broken down in the blood stream)

at the rate $y = 150(0.997)^t$, where $t$ is the time in minutes and $y$ is the amount of the drug remaining in the
bloodstream in ppm (parts per million). interpret the meaning of 150 and 0.997 in the context of this
situation. does more than half the drug remain after 4 hours?

1. the price of milk has been steadily increasing 3% per year. if the cost of a gallon is now $4.

a. what will it cost in 10 years?

b. what did it cost 5 years?

Explanation:

Problem 3

Step 1: Interpret 150

When \( t = 0 \) (initial time, \( t = 0 \) minutes), we substitute \( t = 0 \) into the equation \( y = 150(0.997)^t \). Using the property \( a^0 = 1 \) for any non - zero \( a \), we get \( y=150\times(0.997)^0 = 150\times1=150 \). So, 150 represents the initial amount of the drug in the bloodstream (at \( t = 0 \) minutes) in ppm.

Step 2: Interpret 0.997

The general form of an exponential decay function is \( y = a(b)^t \), where \( a \) is the initial amount and \( b \) is the decay factor. If we rewrite \( b \) as \( 1 - r \), where \( r \) is the rate of decay, then for \( y = 150(0.997)^t \), \( 0.997=1 - 0.003 \). This means that the amount of the drug remaining in the bloodstream is \( 99.7\% \) of the amount from the previous minute, or the drug is decaying (being metabolized) at a rate of \( 0.3\% \) per minute.

Step 3: Check drug after 4 hours

First, convert 4 hours to minutes. Since 1 hour = 60 minutes, 4 hours \(=4\times60 = 240\) minutes. Now, substitute \( t = 240 \) into the equation \( y = 150(0.997)^{240} \). Calculate \( (0.997)^{240}\approx e^{240\times\ln(0.997)}\) (using the formula \( a^x=e^{x\ln(a)} \)). \(\ln(0.997)\approx - 0.0030045\), so \( 240\times\ln(0.997)\approx240\times(- 0.0030045)=-0.72108\), and \( e^{-0.72108}\approx0.485 \). Then \( y = 150\times0.485 = 72.75 \). Half of the initial amount is \( \frac{150}{2}=75 \). Since \( 72.75<75 \), less than half the drug remains after 4 hours.

Step 1: Identify the formula

The formula for exponential growth is \( A = P(1 + r)^t \), where \( P \) is the initial amount, \( r \) is the annual growth rate (as a decimal), \( t \) is the number of years, and \( A \) is the amount after \( t \) years. Here, \( P=\$4 \), \( r = 0.03 \) (since 3% \(=0.03\)), and \( t = 10 \) years.

Step 2: Substitute values into the formula

Substitute \( P = 4 \), \( r=0.03 \), and \( t = 10 \) into \( A = P(1 + r)^t \). We get \( A=4\times(1 + 0.03)^{10}=4\times(1.03)^{10} \).

Step 3: Calculate \( (1.03)^{10} \)

Using a calculator, \( (1.03)^{10}\approx1.343916379 \). Then \( A = 4\times1.343916379\approx5.375665516 \).

Step 1: Identify the formula for exponential growth (solving for past value)

We use the formula \( P=\frac{A}{(1 + r)^t} \), where \( A \) is the current amount, \( r \) is the annual growth rate, and \( t \) is the number of years in the past. Here, \( A = 4 \), \( r=0.03 \), and \( t = 5 \) years.

Step 2: Substitute values into the formula

Substitute \( A = 4 \), \( r = 0.03 \), and \( t = 5 \) into \( P=\frac{A}{(1 + r)^t} \). We get \( P=\frac{4}{(1 + 0.03)^{5}}=\frac{4}{(1.03)^{5}} \).

Step 3: Calculate \( (1.03)^{5} \)

Using a calculator, \( (1.03)^{5}\approx1.159274074 \). Then \( P=\frac{4}{1.159274074}\approx3.450580596 \).

Answer:

• 150: Initial amount of drug (at \( t = 0 \) minutes) in bloodstream, 150 ppm.
• 0.997: Decay factor, drug is 99.7% of previous minute's amount (decays at 0.3% per minute).
• After 4 hours, less than half the drug remains.

Problem 4a