Problem 64 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

Nitrogen dioxide is a brown gas that impairs breathing. The amount of nitrogen dioxide present in the atmosphere on a certain May day in the city of Long Beach measured in PSI (pollutant standard index) at time \(t\). where \(t\) is measured in hours, and \(t=0\) corresponds to 7 a.m., is approximated by $$ \frac{136}{1+0.25(t-4.5)^{2}}+28 \quad(0 \leq t \leq 11) $$ Find the time of the day when the amount of nitrogen diox-

Step-by-Step Solution

Answer

The greatest amount of nitrogen dioxide is present at 11:30 a.m. on that day.

Step 1: Compute the derivatives

We are given the function: \( f(t) = \frac{136}{1+0.25(t-4.5)^{2}}+28 \quad(0 \leq t \leq 11) \) To find the critical points, we will first take the derivative of function f(t) with respect to time t.

Step 2: Apply the quotient rule and simplify

We use the quotient rule to find the derivative: \( f'(t) = \frac{d}{dt} \left( \frac{136}{1+0.25(t-4.5)^{2}} \right) \) The quotient rule states that for \(f(t) = \frac{u(t)}{v(t)}\), the derivative is: \( f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{ v(t)^2 } \) Here, \(u(t)=136\) and \(v(t)=1+0.25(t-4.5)^2\). Taking the derivative w.r.t t, we get: \(u'(t) = 0\) and \(v'(t) = 0.5(t - 4.5)\) Now, using the quotient rule in our context, we have: \( f'(t) = -\frac{(136)(0.5)(t-4.5)}{(1+0.25(t-4.5)^{2})^2} \)

Step 3: Set the first derivative equal to zero

To find the critical points, we must set the first derivative equal to zero and solve for t: \( -\frac{(136)(0.5)(t-4.5)}{(1+0.25(t-4.5)^{2})^2} = 0 \) As the negative sign won't affect the equality, we remove it: \( \frac{(136)(0.5)(t-4.5)}{(1+0.25(t-4.5)^{2})^2} = 0 \)

Step 4: Solve the equation for critical points

Since the denominator won't equal zero, we only need to solve for the numerator: \( (136)(0.5)(t - 4.5) = 0 \) \( (t - 4.5) = 0 \) \( t = 4.5 \) So, there's a critical point at \(t = 4.5\) hours.

Step 5: Analyze the given interval

Since the critical point \(t = 4.5\) is within the given interval (0 ≤ t ≤ 11), we can find the nitrogen dioxide level at this time. Since this is the only critical point, it is the time when the amount of nitrogen dioxide is the greatest.

Step 6: Calculate the time of the day

Now, we need to convert the value of t (4.5) to the time of the day. Since t=0 corresponds to 7 a.m., we add the 4.5 hours to 7 a.m. 7:00 a.m. + 4.5 hours = 11:30 a.m. Thus, the amount of nitrogen dioxide is the greatest at 11:30 a.m. on that day.

Key Concepts

Derivative

A derivative is a fundamental concept used in calculus to determine how a function changes as its input changes. In the exercise, we're looking at how the amount of nitrogen dioxide changes over time, which we can express by finding the derivative of the given function. The process of differentiation provides us with a new function that shows the rate of change at any given point.

• The original function given is: \(f(t) = \frac{136}{1+0.25(t-4.5)^{2}}+28\).
• By differentiating, we discover at which point in time the amount of nitrogen dioxide reaches its maximum level over the interval \(0 \leq t \leq 11\).

Understanding derivatives helps in motion analysis, optimization problems, and data trends. In this context, it helps pinpoint when air quality is at its worst for the day.

Critical Points

Critical points in calculus refer to points on a curve where the derivative is zero or undefined. These are significant because they can indicate maximum or minimum points or points of inflection. In our problem, finding where the derivative equals zero informs us about possible extrema of nitrogen dioxide levels.

• We find the critical points by setting the derivative \(f'(t) = -\frac{(136)(0.5)(t-4.5)}{(1+0.25(t-4.5)^{2})^2} = 0\).
• This simplifies to find \(t = 4.5\) as the critical point within the given interval.

This step is essential because it reveals when the pollutant reaches its peak concentration. Evaluating these points within the constraints provides insights into optimal times for exposure reduction.

Quotient Rule

The quotient rule is a technique for finding the derivative of a function that is the quotient of two other functions. It is especially useful when you encounter more complex fractional forms like in our nitrogen dioxide function.

• The quotient rule formula is: \(f'(t) = \frac{u'(t)v(t) - u(t)v'(t)}{v(t)^2}\), where functions \(u(t)\) and \(v(t)\) are differentiable.
• In our example, \(u(t) = 136\) and \(v(t) = 1+0.25(t-4.5)^2\).

Applying the quotient rule simplifies the process of differentiation for complex rational functions, allowing us to compute derivatives that indicate slopes or rates of change effectively.

Pollutant Standard Index

The Pollutant Standard Index (PSI) is a numerical value used to communicate the level of air pollution at a particular time. It helps understand the quality of air and its potential effects on health, especially when considering pollutants like nitrogen dioxide.

• Nitrogen dioxide affects respiratory functions and is a major concern in areas with heavy traffic or industrial activity.
• The PSI provides a way to easily communicate information about air quality and necessary measures for public safety.

In this exercise, calculating when the PSI is highest helps identify the most critical times of the day for exposure. This understanding allows for better planning and response to air quality issues.