Problem 91 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The concentration (in milligrams/cubic centimeter) of a certain drug in a patient's body \(t\) hr after injection is given by $$ C(t)=\frac{t^{2}}{2 t^{3}+1} \quad(0 \leq t \leq 4) $$ When is the concentration of the drug increasing, and when is it decreasing?

Step-by-Step Solution

Answer

The concentration of the drug is increasing when \(0 < t < 1\) hours and decreasing when \(1 < t \leq 4\) hours.

Step 1: Find the derivative of the concentration function

Given the function: $$ C(t)=\frac{t^{2}}{2 t^{3}+1} \quad(0 \leq t \leq 4) $$ We need to find its derivative, C'(t), with respect to time t. Using the quotient rule, where \(u(t) = t^{2}\) and \(v(t) = 2t^{3} + 1\), we get: $$ C'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{[v(t)]^2} $$ Now we must find \(u'(t)\) and \(v'(t)\) by taking the derivatives of u(t) and v(t) with respect to t: $$ u'(t) = \frac{d(t^2)}{dt} = 2t $$ $$ v'(t) = \frac{d(2t^3 + 1)}{dt} = 6t^2 $$ Now, plug these derivatives back into the quotient rule formula for C'(t): $$ C'(t) = \frac{(2t)(2t^3 + 1) - (t^2)(6t^2)}{ (2t^3 + 1)^2} $$

Step 2: Simplify the derivative

Now, we simplify the expression for C'(t): $$ C'(t) = \frac{4t^4 + 2t - 6t^4}{(2t^3 + 1)^2} $$ $$ C'(t) = \frac{-2t^4 + 2t}{(2t^3 + 1)^2} $$ We can factor out the common factor (2t) from the numerator: $$ C'(t) = \frac{2t(-t^3 + 1)}{(2t^3 + 1)^2} $$

Step 3: Analyze the derivative to determine when the concentration is increasing or decreasing

Now, we find the intervals where C'(t) > 0 (increasing) and C'(t) < 0 (decreasing): $$ C'(t) > 0 \implies 2t(-t^3 + 1) > 0 $$ The function is increasing when t > 0 and -t^3 + 1 > 0, which implies t < 1 by solving for t in the inequality. So, the concentration is increasing when \(0 < t < 1\). Similarly, $$ C'(t) < 0 \implies 2t(-t^3 + 1) < 0 $$ The function is decreasing when t > 0 and -t^3 + 1 < 0, which implies t > 1 by solving for t in the inequality. So, the concentration is decreasing when \(1 < t \leq 4\). Therefore, the concentration of the drug is increasing when \(0 < t < 1\) hours and decreasing when \(1 < t \leq 4\) hours.

Key Concepts

Calculus in Pharmacokinetics

Calculus plays an instrumental role in studying and predicting the concentration of drugs within the human body, a crucial aspect of pharmacokinetics. One might wonder why the math of change and motion, calculus, is so vital in determining the fate of a drug administered to a patient. The answer lies in our body's dynamic nature and how it processes substances over time.

Drug concentration, as described by mathematical models, can help predict how long a medication will be effective and determine the appropriate dosing schedules. The concentration of a drug over time typically follows a curve that requires differential calculus to analyze. As in the given exercise, by differentiating the concentration function with respect to time, we can determine when the drug levels are increasing or decreasing, indicating the periods of absorption and elimination respectively. This understanding is critical for clinicians to design an appropriate dosage regimen to ensure the drug is delivered at the right time, in the right amount, for optimal therapeutic effect.

Quotient Rule Differentiation

Differentiation is a fundamental concept in calculus used to find the rate at which something changes. When functions are divided by one another, the quotient rule comes into play for differentiation. It's an efficient way to tackle complex problems in which the division of two functions is involved, such as the concentration function in our original exercise.

The quotient rule states that for two differentiable functions, u(t) and v(t), the derivative of their quotient u(t)/v(t) is given by \
\[\frac{{d}}{{dt}}\left(\frac{{u}}{{v}}\right) = \frac{{u'(t) \cdot v(t) - u(t) \cdot v'(t)}}{{[v(t)]^2}}\]
The quotient rule is a reliable tool used in pharmacokinetics for assessing how drug concentrations change over time. By applying this rule, as shown in the exercise solution, we obtain the first derivative of the concentration function, which informs us about the rate and direction (increase or decrease) of change in the drug's concentration.

Analyzing Function's Increasing and Decreasing Intervals

Determining when a function is increasing or decreasing is essential in various fields, including economics, physics, and, as illustrated in our exercise, pharmacokinetics. This analysis is based on the sign of the function's first derivative.

When the derivative of a function is positive over an interval, it indicates that the function is increasing on that interval. Conversely, if the derivative is negative, the function is decreasing. By setting the derivative equal to zero, we can find critical points, which could represent peaks, valleys, or inflection points in the function. In the context of the drug concentration problem, finding where the function C'(t) changes its sign reveals when the concentration of the drug stops increasing and starts decreasing, and vice versa. This fine detail provides valuable insight into the drug's absorption and metabolism, which are critical for ensuring the safety and efficacy of pharmacological treatments.