Problem 85 - Precalculus Mathematics for Calculus

Question

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) The maximum concentration is 2.5 mg/L at \( t = 1 \). (b) The concentration approaches zero over time. (c) It takes slightly over 0.02 hours for the concentration to drop below 0.3 mg/L.

Step 1: Find the Maximum Concentration

To find the maximum concentration, we need to determine the critical points of the function \( c(t) = \frac{5t}{t^2 + 1} \). First, find the derivative \( c'(t) \) using the quotient rule: \( c'(t) = \frac{(t^2 + 1)(5) - (5t)(2t)}{(t^2 + 1)^2} = \frac{5 - 5t^2}{(t^2 + 1)^2} \). Set the derivative equal to zero to find critical points: \( 5 - 5t^2 = 0 \) which gives \( t^2 = 1 \). Therefore, \( t = 1 \) (since \( t \geq 0 \)).

Step 2: Verify Maximum Concentration

Evaluate the concentration at \( t = 1 \): \( c(1) = \frac{5 \times 1}{1^2 + 1} = \frac{5}{2} = 2.5 \ \mathrm{mg/L} \). Check the behavior of the function around \( t = 1 \) or use the second derivative to confirm that it is a maximum. Since \( c'(t) \) changes from positive to negative, \( t = 1 \) is a local maximum.

Step 3: Long-term Behavior Analysis

To determine the long-term behavior of \( c(t) \), find the limit of \( c(t) \) as \( t \) approaches infinity. The limit is: \( \lim_{t \to \infty} \frac{5t}{t^2 + 1} \). Simplifying, \( \lim_{t \to \infty} \frac{5}{t} = 0 \). Hence, the concentration approaches zero as \( t \to \infty \).

Step 4: Determine Time for Concentration to Drop

Solve \( \frac{5t}{t^2 + 1} < 0.3 \) to find when \( t \) drops below 0.3. Set \( 5t < 0.3(t^2 + 1) \), giving the equation \( t^2 - \frac{50}{3}t + 1 > 0 \). Use the quadratic formula: \( t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1, b = \frac{50}{3}, c = -1 \). Solve for \( t \) to find critical points where \( t \approx 0.0201 \) and check \( f(t) < 0.3 \) for \( t > 0.0201 \).

Key Concepts

Quotient Rule

The Quotient Rule is a technique in calculus used to find the derivative of a function that is the ratio of two differentiable functions. Here, the concentration function given is \( c(t) = \frac{5t}{t^2 + 1} \). It is crucial to utilize the Quotient Rule to derive its derivative, which helps in finding the rate of change of the concentration with respect to time.
Here's a quick refresher on the Quotient Rule: If you have two functions \( u(t) \) and \( v(t) \), with \( u(t) = 5t \) and \( v(t) = t^2 + 1 \) in this case, then the derivative \( \frac{d}{dt}\left( \frac{u}{v} \right) \) is given by:

• \( \frac{v(t)u'(t) - u(t)v'(t)}{v(t)^2} \).

Applying the rule in our context involves:

• The derivative of the numerator \( u'(t) = 5 \).
• The derivative of the denominator \( v'(t) = 2t \).
• Thus, \( c'(t) = \frac{(t^2 + 1)(5) - (5t)(2t)}{(t^2 + 1)^2} \).

This derivative will be key in finding where the concentration reaches its maximum (critical points).

Derivative

Derivatives are fundamental in calculus as they measure how a function changes as its inputs change. In the context of this drug concentration function, the derivative \( c'(t) \) informs us about the behavior of drug concentration over time.
Understanding \( c'(t) = \frac{5 - 5t^2}{(t^2 + 1)^2} \):

• The sign of \( c'(t) \) helps us determine where the function is increasing or decreasing.
• Setting \( c'(t) = 0 \) finds us the critical points where the function might have a maxima, minima, or a point of inflection.

Using simple algebra, set \( 5 - 5t^2 = 0 \) to yield \( t^2 = 1 \) and since \( t \) must be non-negative, we take \( t = 1 \). This tells us that the changes in concentration (increase or decrease) might hit a pause at \( t = 1 \) possibly indicating a maximum.

Critical Points

Critical points are where the derivative of a function is either zero or undefined, often hinting at a peak or valley in the graph. Here, we see how these critical points relate to the drug concentration's highest point in the bloodstream.
With the calculated derivative \( c'(t) \), set \( c'(t) = 0 \) to identify potential critical points:

• The critical point found is \( t = 1 \).
• Evaluating \( c(t) \) gives \( c(1) = \frac{5}{2} = 2.5 \text{ mg/L} \).
• Examine the sign change in \( c'(t) \) around \( t = 1 \) to confirm it is indeed a local maximum. \( c'(t) \) shifts from positive to negative, confirming the peak.

This point allows us to identify the maximum concentration achieved after administering the drug.

Long-term Behavior

Long-term behavior of a function involves analyzing what happens as the variable approaches infinity. In drug concentration studies, it shows how the concentration changes over a prolonged time.
In this problem, consider the behavior of function \( c(t) \) as \( t \) approaches infinity:

• The expression \( \lim_{t \to \infty} \frac{5t}{t^2 + 1} \) determines this behavior.
• Simplifying, it becomes \( \lim_{t \to \infty} \frac{5}{t} \).

The denominator \( t^2 + 1 \) grows much faster than \( 5t \), causing the limit to approach zero. Therefore, over time, the drug concentration diminishes and asymptotically approaches zero as time goes on indefinitely. This information is critical in understanding when the drug ceases to have a notable effect in the bloodstream.