Question

The function $$ D(h)=5 e^{-0.4 h} $$ can be used to find the number of milligrams \(D\) of a certain drug that is in a patient's bloodstream \(h\) hours after the drug has been administered. How many milligrams will be present after 1 hour? After 6 hours?

Short answer

After 1 hour: approximately 3.35 mg; after 6 hours: approximately 0.45 mg.

Step-by-step solution

Understand the Function

The function given is \[D(h)=5 e^{-0.4 h}\] which calculates the number of milligrams of a drug in a patient's bloodstream (D) based on the number of hours (h) passed since administration.

Substitute 1 hour into the function

To determine the amount of the drug present after 1 hour (h = 1), substitute 1 for h in the function: \[D(1)=5 e^{-0.4 \times 1}\]

Calculate the Amount for 1 hour

Simplify the expression \[D(1)=5 e^{-0.4}\] Use a calculator to find the value of e^{-0.4}, which is approximately 0.6703. Therefore, \[D(1)=5 \times 0.6703 \ D(1) \ approximate 3.3515 \]

Substitute 6 hours into the function

Next, calculate the amount of the drug present after 6 hours (h = 6). Substitute 6 for h in the function: \[D(6)=5 e^{-0.4 \times 6}\]

Calculate the Amount for 6 hours

Simplify the expression \[D(6)=5 e^{-2.4}\] Use a calculator to find the value of e^{-2.4}, which is approximately 0.0907. Therefore, \[D(6)=5 \times 0.0907 \ D(6) \ approximate 0.4535 \]

Key concepts

drug concentration over time

When a patient takes a drug, its concentration in the bloodstream decreases over time. This is known as exponential decay. Researchers and doctors often use mathematical functions to model this process. For the given problem, the function used is: \[D(h) = 5e^{-0.4h}\]
Here, \(D(h)\) represents the concentration of the drug in milligrams at any given time \(h\) hours after administration. \[e^{-0.4h}\] is the exponential decay factor which shows how the drug concentration declines as time increases.

exponential decay calculation

Exponential decay occurs when the rate of decrease of a quantity is proportional to its current value. The function \[D(h)=5e^{-0.4h}\] models this behavior as follows:

• \(D(h)\): The remaining drug concentration in milligrams at time \(h\)
• \(5\): Initial amount of the drug administered, in milligrams
• \(e^{-0.4h}\): Decay factor, where \(e\) is Euler's number approximately equal to 2.718, and \(-0.4\) is the decay constant

The decay constant -0.4 indicates how rapidly the drug concentration decreases. After substitution, these calculations involve evaluating the exponent to determine the remaining concentration.

substitution in functions

Substitution is a process in which we replace a variable in an equation with a specific value. For our problem, we need to find the concentration of the drug at different time points. For example:
To find the concentration after 1 hour: \[D(1) = 5e^{-0.4 \times 1}\] Replace \(h\) with 1. Simplifying it, \[D(1) = 5e^{-0.4}\] This requires calculation of \(e^{-0.4}\), which is approximately 0.6703. Thus, \[D(1) = 5 \times 0.6703 \] ~ 3.3515 milligrams.

evaluating functions

Once we've substituted the values and simplified, the next step is evaluating the function. This often requires numerical approximations or a calculator, since exact values of \(e\) are complex. For \(6\) hours: \[D(6)=5e^{-0.4 \times 6}\] Substitute \(h\) with 6: \[D(6)=5e^{-2.4}\] Calculating, \(e^{-2.4}\) is approximately 0.0907. Thus, \[D(6)=5 \times 0.0907 \] ~ 0.4535 milligrams.
Evaluating involves both substitution and simplifying the exponent, leading to a clear understanding of how the drug dissipates over time.