Problem 8 - Calculus with Concepts in Calculus

Question

The concentration of a drug in the blood stream \(t\) seconds after injection into a muscle is given by $$ y=c\left(e^{-b t}-e^{-a t}\right) \quad \text { for } t \geq 0 $$ where \(a, b\), and \(c\) are positive constants with \(a>b\). a. Find the time at which the concentration is maximum. b. Find the time at which the concentration of the drug in the blood stream is decreasing most rapidly.

Step-by-Step Solution

Answer

Maximum concentration at \(t = \frac{\ln\left(\frac{b}{a}\right)}{b-a}\); most rapid decrease at \(t = \frac{\ln\left(\frac{b^2}{a^2}\right)}{b-a}\).

Step 1: Find the derivative of the concentration function

The concentration function is given by \[y = c(e^{-bt} - e^{-at})\]To find the derivative, apply the differentiation rules to each exponential term:1. The derivative of \(e^{-bt}\) with respect to \(t\) is \(-b e^{-bt}\)2. The derivative of \(e^{-at}\) with respect to \(t\) is \(-a e^{-at}\)Therefore:\[y' = c(-b e^{-bt} + a e^{-at})\]

Step 2: Find critical points for maximum concentration

To find when the concentration is maximum, set the derivative \(y'\) to zero.\[0 = c(-b e^{-bt} + a e^{-at})\]Solving for \(t\):\[b e^{-bt} = a e^{-at} \\frac{b}{a} = \frac{e^{-at}}{e^{-bt}} \\frac{b}{a} = e^{(b-a)t}\]Taking the natural log on both sides:\[\ln\left(\frac{b}{a}\right) = (b-a)t \t = \frac{\ln\left(\frac{b}{a}\right)}{b-a}\]

Step 3: Calculate the second derivative for concavity

To confirm that this point is a maximum, calculate the second derivative:\[y'' = c(b^2 e^{-bt} - a^2 e^{-at})\]Substitute \(t = \frac{\ln\left(\frac{b}{a}\right)}{b-a}\) into \(y''\):The second derivative should satisfy \(y''(t) < 0\) ensuring it is a maximum.

Step 4: Find the time of most rapid decrease

The rate of decrease is given by the negative of the derivative. To find where it is maximum, find the critical points of the second derivative:\[0 = y'' \0 = c(b^2 e^{-bt} - a^2 e^{-at})\]Solving gives:\[b^2 e^{-bt} = a^2 e^{-at} \\frac{b^2}{a^2} = e^{(b-a)t} \t = \frac{\ln\left(\frac{b^2}{a^2}\right)}{b-a}\]

Key Concepts

Differentiation

Differentiation is a core concept in calculus that deals with rates of change. In simpler terms, it's about figuring out how a function behaves as you make small changes to its input. For a given function, the derivative helps us find out how quickly the function's value is changing at any point.
In the context of our exercise, the drug concentration function is modeled as:

• \(y = c(e^{-bt} - e^{-at})\)

When we differentiate this function, we apply differentiation rules to each term. This involves:

• The derivative of \(e^{-bt}\) with respect to \(t\) being \(-b e^{-bt}\), reflecting the decrease rate dictated by the constant \(b\).
• The derivative of \(e^{-at}\) with respect to \(t\) being \(-a e^{-at}\), showing the faster decrease owing to a larger exponent \(a\).

After differentiation, we acquire the derivative:

• \(y' = c(-b e^{-bt} + a e^{-at})\)

Here, the derivative indicates how the concentration changes over time, crucial for both determining peaks and the most rapid decrease in concentration.

Exponential Functions

Exponential functions are a vital tool in various mathematical modeling scenarios due to their ability to represent rapidly changing phenomena. They are expressed generally in the form \(f(t) = e^{kt}\), where \(t\) is the variable, and \(k\) governs the rate of increase or decrease.
In the drug concentration function:

• \(y = c(e^{-bt} - e^{-at})\)

The terms \(e^{-bt}\) and \(e^{-at}\) represent exponential decay. Because \(a > b\), \(e^{-at}\) decreases faster than \(e^{-bt}\), which plays a crucial role in altering the behavior of the overall function over time.
Exponential functions are unique due to their constant proportionate rate of change:

• As time \(t\) increases, the function’s rate at any point is directly proportional to its current value.

This means exponential functions are extraordinarily useful in scenarios like population growth, radioactive decay, and here, drug concentration in the bloodstream after injection.

Critical Points

Critical points are those values of the input where the derivative of a function is zero or undefined. These points are crucial for pinpointing maximum or minimum values of a function, indicating peaks or troughs in its graph.
For the drug concentration function, the task involves determining when the blood concentration is at its maximum. This implies finding where the first derivative, or rate of change of concentration, crosses zero:

• Setting \(y' = c(-b e^{-bt} + a e^{-at}) = 0\) gives us the critical point

Solving the equation:

• \[b e^{-bt} = a e^{-at}\]
• \[\frac{b}{a} = e^{(b-a)t}\]
• Solve for \(t\): \[t = \frac{\ln\left(\frac{b}{a}\right)}{b-a}\]

These calculations indicate when the concentration reaches its peak (maximum), and a second derivative test confirms it's indeed a maximum by showing the curve bends downwards at this point.