Problem 85 - Algebra and Trigonometry

Question

Drug Concentration \(\mathrm{A}\) drug is administered to a patient, and the concentration of the drug in the bloodstream is monitored. At time \(t \geq 0\) (in hours since giving the drug), the concentration (in \(\mathrm{mg} / \mathrm{L}\) ) is given by $$ c(t)=\frac{5 t}{t^{2}+1} $$ Graph the function \(c\) with a graphing device. (a) What is the highest concentration of drug that is reached in the patient's bloodstream? (b) What happens to the drug concentration after a long period of time? (c) How long does it take for the concentration to drop below 0.3 \(\mathrm{mg} / \mathrm{L} ?\)

Step-by-Step Solution

Answer

(a) 2.5 mg/L, (b) Approaches 0 mg/L, (c) After 16.27 hours.

Step 1: Find the Derivative

To find the maximum concentration, we need to find the derivative of function \(c(t)\). The function given is \( c(t) = \frac{5t}{t^2 + 1} \). Use the quotient rule: \( c'(t) = \frac{(t^2 + 1)(5) - (5t)(2t)}{(t^2 + 1)^2} \). Simplifying the expression gives: \( c'(t) = \frac{5 - 5t^2}{(t^2 + 1)^2} \).

Step 2: Find Critical Points

Set the derivative equal to zero to find critical points: \( \frac{5(1 - t^2)}{(t^2 + 1)^2} = 0 \). Simplifying gives \( 1 - t^2 = 0 \), which results in \( t = \pm1 \). Since time \( t \geq 0 \), we only consider \( t = 1 \).

Step 3: Evaluate the Function at Critical Points

Evaluate \( c(t) \) at the critical point \( t = 1 \): \( c(1) = \frac{5 \times 1}{1^2 + 1} = \frac{5}{2} = 2.5 \). The highest concentration is 2.5 mg/L.

Step 4: Analyze Long-term Behavior

Analyze the limit of \( c(t) \) as \( t \to \infty \): \( \lim_{{t \to \infty}} \frac{5t}{t^2 + 1} = \lim_{{t \to \infty}} \frac{5}{t} = 0 \). This shows the concentration approaches zero over time.

Step 5: Determine Time for Concentration to Drop Below 0.3

Set the function equal to 0.3 mg/L and solve for \( t \): \( \frac{5t}{t^2 + 1} = 0.3 \). Rearranging gives \( 5t = 0.3(t^2 + 1) \) which simplifies to \( 0.3t^2 - 5t + 0.3 = 0 \). Solve this quadratic equation using the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 0.3, b = -5, c = 0.3 \). This results in \( t = 0.061 \) or \( t = 16.27 \). Since concentration is decreasing, \( t = 16.27 \) is where the concentration drops below 0.3 mg/L.

Key Concepts

Derivative

When dealing with functions like the concentration of a drug in the bloodstream over time, the derivative helps us understand how the concentration changes. The derivative of a function gives us the rate of change of the function at any point. It is like a snapshot in time, showing whether the concentration is increasing or decreasing. For the drug concentration function given by \( c(t) = \frac{5t}{t^2 + 1} \), the derivative is calculated using the quotient rule. The quotient rule applies because the function is a ratio of two other functions: \(5t\) and \(t^2 + 1\). Using this rule, the derivative \(c'(t)\) is determined as \( \frac{5 - 5t^2}{(t^2 + 1)^2} \). This derivative tells us how quickly the concentration changes with respect to time \(t\). By setting this derivative to zero, you can find points where the change stops, known as critical points.

Critical Points

Critical points of a function are where the derivative equals zero or is undefined. These points are important because they may indicate maximum or minimum values of the function.For our drug concentration function, critical points occur where the derivative \(c'(t) = \frac{5 - 5t^2}{(t^2 + 1)^2} \) is zero. Solving \(1 - t^2 = 0\) gives \(t = \pm1\). However, since time \(t\) cannot be negative, the only relevant critical point is \(t = 1\).Evaluating the function at this critical point helps determine the maximum concentration in the bloodstream. Substituting \(t = 1\) into the original concentration function \(c(t)\) reveals that the highest concentration is \(2.5\;\mathrm{mg/L}\). At this critical point, the concentration stops increasing and starts to decrease.

Quadratic Equation

The quadratic equation comes into play when we need to solve for time \(t\) in complex situations, such as when determining when the concentration drops below a certain level.A quadratic equation is typically in the form \(ax^2 + bx + c = 0\). In this scenario, once we've rearranged the concentration equation \( \frac{5t}{t^2 + 1} = 0.3 \), it becomes \(0.3t^2 - 5t + 0.3 = 0\). To solve this, we apply the quadratic formula: \[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Where \(a = 0.3\), \(b = -5\), and \(c = 0.3\). By calculating, the solutions for \(t\) are approximately \(t = 0.061\) and \(t = 16.27\).Given the context, where we are interested in the time when the concentration has been decreasing to a certain level, \(t = 16.27\) hours is the time when the concentration first falls below \(0.3\;\mathrm{mg/L}\). Remember, in quadratic equations, it's crucial to consider the context to decide which solution is relevant.

Long-term Behavior

Understanding long-term behavior of a function means determining what happens to the concentration after a long period of time, as \(t\) approaches infinity.For the concentration function \(c(t) = \frac{5t}{t^2 + 1}\), long-term behavior can be studied using limits. We look at what happens to \(c(t)\) as \(t\) becomes very large. This is symbolically expressed as \[ \lim_{{t \to \infty}} \frac{5t}{t^2 + 1} \]By dividing every term by \(t\), we find that the concentration approaches \[ \lim_{{t \to \infty}} \frac{5}{t} = 0 \]This indicates that, over time, the concentration of the drug in the bloodstream decreases and approaches zero. It implies that the drug eventually gets cleared from the bloodstream, reflecting the body's natural process of metabolizing and excreting the drug.