Problem 88 - Calculus Single Variable

Question

The rate of elimination of alcohol from the bloodstream is proportional to the amount \(A\) that is present. That is, $$ \frac{d A}{d t}=-\frac{1}{k} A $$ where \(k\) is a time constant that depends on the drug and the individual. If \(k\) is \(1 / 2\) hour for a certain person, how long will it take for his blood alcohol content to reduce from \(0.12 \%\) to \(0.06 \% ?\)

Step-by-Step Solution

Answer

It takes approximately 0.35 hours for the blood alcohol content to reduce from 0.12% to 0.06%.

Step 1: Understand the Differential Equation

The problem states that the rate of elimination of alcohol, \( \frac{dA}{dt} \), is proportional to the negative of the current amount \( A \). This means it decreases over time, following the differential equation: \( \frac{dA}{dt} = -\frac{1}{k}A \).

Step 2: Identify the Time Constant

The time constant \( k \) is given as \( \frac{1}{2} \) hour for this person. This means that the rate of elimination of alcohol is more significant in a smaller timeframe.

Step 3: Solve the Differential Equation

To solve the equation \( \frac{dA}{dt} = -\frac{2}{1}A \) (since \( k = \frac{1}{2} \)), we integrate to find: \[ A(t) = A_0 e^{-\frac{2}{1} t} \] where \( A_0 \) is the initial amount of alcohol content.

Step 4: Set Initial and Final Conditions

We start with an initial alcohol content \( A_0 = 0.12\% \) and want \( A(t) \) to be \( 0.06\% \). So, we set up the equation: \[ 0.06 = 0.12 e^{-2t} \]

Step 5: Solve for Time \( t \)

Rearrange the equation to solve for \( t \): \[ e^{-2t} = \frac{0.06}{0.12} = 0.5 \]Take the natural logarithm of both sides: \[ -2t = \ln{0.5} \]Solve for \( t \): \[ t = -\frac{\ln{0.5}}{2} \] Using the properties of logarithms, this simplifies to: \[ t = \frac{\ln{2}}{2} \] Calculate this value to find \( t \).

Step 6: Calculate the Final Time

Computing \( t = \frac{\ln{2}}{2} \) gives us approximately 0.3466 hours.

Key Concepts

Exponential Growth and Decay

Exponential growth and decay occur when the rate of change of a quantity is proportional to its current value. In simpler terms, this process describes how quickly something can grow or shrink over time.

Let's break it down a bit further:

• Growth: When a quantity increases over time, and its rate of change is positive. For instance, investment capital growing over time with interest.
• Decay: When a quantity decreases, like the amount of alcohol in the bloodstream in our exercise. This kind of decay is expressed by an equation where the rate is directly proportional to the current amount but is negative, meaning it reduces over time.

The mathematical model for decay in this case is given by the equation \[\frac{dA}{dt} = -\frac{1}{k} A\]This signifies that the rate of change of alcohol content in blood is proportional to the current amount, but its decreasing, hence the negative sign.

In real life, this model applies to a variety of processes, such as radioactive decay, population decline, and cooling of substances.

Separation of Variables

Separation of variables is a method used to solve differential equations where variables are grouped together to simplify the process of finding a solution. It's like neatly organizing your equation to make solving it a breeze! In the equation \[ \frac{dA}{dt} = -\frac{2}{1}A\] we use separation of variables, which means getting "all the \(A\)s on one side and all the \(t\)s on the other" to simplify:

• You rearrange the equation  dx = f(x) \cdot g(y) \ dy  so each function involving a separate variable.
• Integrate both sides separately to find a solution.

By following these steps, we transform our model into something more straightforward: the equation \\[ A(t) = A_0 e^{-\frac{2}{1}t}\]. This formula predicts how quickly the alcohol will decay over time, reducing the original content by employing the natural exponential function.

Separation of variables is commonly applied in practical scenarios, like physics, biology, and chemistry, wherever there's a change over time controlled by differential equations.

Natural Logarithms

Natural logarithms are the logarithms in base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. In our solution, we see the presence of natural logarithms when solving for time \(t\).

• The equation \[ e^{-2t} = 0.5 \] is simplified by applying the natural logarithm to both sides: \[ -2t = \ln{0.5} \]
• This allows us to isolate the variable \(t\) as \( t = -\frac{\ln{0.5}}{2} \)

Using natural logarithms is crucial when dealing with exponential functions. They help to "undo" the exponential functions and make equations easier to solve.

Natural logarithms appear in a wide variety of scientific calculations and allow us to handle exponential growth and decay problems efficiently by converting them into linear ones, which are much simpler to solve.