Problem 10 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Plasma penicillin concentration is $$P(t)=5 e^{-0.3 t}-5 e^{-0.4 t}$$ \(t\) hours after ingestion of a penicillin pill into the stomach. A small amount of the drug diffuses into tissue and the tissue concentration, \(C(t)\), is $$ C(t)=-e^{-0.3 t}+0.5 e^{-0.4 t}+0.5 e^{-0.2 t} \quad \mu \mathrm{g} / \mathrm{ml} $$ a. Use your technology (calculator or computer) to find the time at which the concentration of the drug in tissue is maximum and the value of \(C\) at that time. b. Compute \(C^{\prime}(t)\) and solve for \(t\) in \(C^{\prime}(t)=0\). This is really bad, for you must solve for \(t\) in $$ 0.3 e^{-0.3 t}-0.2 e^{-0.4 t}-0.1 e^{-0.2 t}=0 $$ Try this: $$ \text { Let } \quad Z=e^{-0.1 t} \quad \text { then solve } \quad 0.3 Z^{3}-0.2 Z^{4}-0.1 Z^{2}=0 . $$ c. Solve for the possible values of \(Z\). Remember that \(Z=e^{-0.1 t}\) and solve for \(t\) if possible using the possible values of \(Z\). d. Which value of \(t\) solves our problem?

Step-by-Step Solution

Answer

The maximum concentration occurs at \( t = \frac{\ln(0.5)}{-0.1} \).

Step 1: Compute the derivative C'(t)

The function for tissue concentration is given by \( C(t) = -e^{-0.3 t} + 0.5 e^{-0.4 t} + 0.5e^{-0.2 t} \). To find the derivative \( C'(t) \), apply the derivative rules for exponential functions:\[C'(t) = 0.3e^{-0.3t} - 0.2e^{-0.4t} - 0.1e^{-0.2t}\]

Step 2: Set C'(t) to zero

Set the derivative \( C'(t) \) to zero to find the critical points:\[0.3e^{-0.3t} - 0.2e^{-0.4t} - 0.1e^{-0.2t} = 0\]

Step 3: Introduce substitution Z=e^{-0.1t}

Let \( Z = e^{-0.1t} \). This converts the equation to:\[0.3Z^3 - 0.2Z^4 - 0.1Z^2 = 0\]

Step 4: Factor the equation

Factor the equation \( 0.3Z^3 - 0.2Z^4 - 0.1Z^2 = 0 \) by factoring out \( Z^2 \):\[Z^2(0.3Z - 0.2Z^2 - 0.1) = 0\]This simplifies to:\[Z^2 = 0 \quad \text{or} \quad 0.3Z - 0.2Z^2 - 0.1 = 0\]

Step 5: Solve each factor separately

The solution \( Z^2 = 0 \) gives \( Z = 0 \), which is impossible since \( Z = e^{-0.1t} \).Focus on the second equation:\[0.3Z - 0.2Z^2 - 0.1 = 0\]Rearrange and solve the quadratic equation:\[0.2Z^2 - 0.3Z + 0.1 = 0 \]

Step 6: Use the quadratic formula

Apply the quadratic formula \( Z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to \( 0.2Z^2 - 0.3Z + 0.1 = 0 \) where \( a = 0.2 \), \( b = -0.3 \), \( c = 0.1 \):\[Z = \frac{0.3 \pm \sqrt{(-0.3)^2 - 4 \times 0.2 \times 0.1}}{2 \times 0.2} = \frac{0.3 \pm \sqrt{0.01}}{0.4}\]This results in two possible roots for \( Z \).

Step 7: Solve for t using Z

Calculate \( Z = \frac{0.3 \pm 0.1}{0.4} \), yielding \( Z_1 = 1 \) and \( Z_2 = 0.5 \).Since \( Z = e^{-0.1t} \), solve for \( t \):- \( Z_1 = 1 \) gives \( e^{-0.1t} = 1 \) leading to \( t = 0 \)- \( Z_2 = 0.5 \) gives \( e^{-0.1t} = 0.5 \). Solve: \(-0.1t = \ln(0.5) \), hence \( t = \frac{\ln(0.5)}{-0.1} \)

Key Concepts

Exponential Functions

Exponential functions are a cornerstone in calculus, especially when dealing with biological processes like drug concentration decay. An exponential function can be represented in the form \( f(t) = ae^{bt} \), where \( a \) is the initial amount, \( b \) is the rate of growth or decay, and \( e \) is the base of the natural logarithm, approximately equal to 2.718.

Exponential decay occurs when \( b \) is negative, leading to a decrease over time.

Exponential growth happens when \( b \) is positive, indicating an increase.

The given problem involves two exponential expressions showing how the concentrations of penicillin change over time: in plasma and tissue. Negative coefficients in the exponents illustrate a decay, common in processes such as drug absorption or radiation decay.
This decay is crucial for understanding how long a drug remains effective in the body. By interpreting these functions, students can predict when the concentration of a drug peaks and starts to decrease, which is essential knowledge for dosing regimes in medicine.

Critical Points

Critical points occur where the derivative of a function is zero or undefined. In biological contexts, such as finding the maximum concentration of a drug within tissues, identifying critical points helps determine key values.

The derivative \( C'(t) \) tells us the rate of change at any given moment.

Setting \( C'(t) = 0 \) indicates where the concentration neither increases nor decreases, hinting at potential maximums or minimums.

The function for tissue concentration, given as a series of exponential terms, uses a derivative to find these critical points. By substituting values into the derivative and solving the equation \( 0.3e^{-0.3t} - 0.2e^{-0.4t} - 0.1e^{-0.2t} = 0 \), we locate the moments when the concentration shifts from increasing to decreasing. This involves changing variables (e.g., \( Z = e^{-0.1t} \)) to simplify solving the equation.
Finding critical points is essential for optimizing dosages and ensuring that drugs are effective without overdosing.

Quadratic Equations

Quadratic equations play a vital role in simplifying complex calculus problems to make them solvable, especially when searching for critical points in exponential functions. A quadratic equation takes the standard form \( ax^2 + bx + c = 0 \).

The solution to a quadratic equation involves using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).

In this context, \( Z \, \) a substitute for \( e^{-0.1t} \, \) leads to a quadratic form: \( 0.2Z^2 - 0.3Z + 0.1 = 0 \).

Solving this quadratic equation by applying the formula gives potential values of \( Z \), which can then be used to find \( t \) by substituting back. For instance, \( Z = 0.5 \) leads to \( t = \frac{\ln(0.5)}{-0.1} \).
Understanding and solving quadratic equations are critical skills that empower us to handle diverse and often complex biological data.