Problem 109 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The level of ozone, an invisible gas that irritates and impairs breathing, present in the atmosphere on a certain May day in the city of Riverside was approximated by $$ A(t)=1.0974 t^{3}-0.0915 t^{4} \quad(0 \leq t \leq 11) $$ where \(A(t)\) is measured in pollutant standard index (PSI) and \(t\) is measured in hours, with \(t=0\) corresponding to 7 a.m. Use the second derivative test to show that the function \(A\) has a relative maximum at approximately \(t=9\). Interpret your results.

Step-by-Step Solution

Answer

The function \(A(t) = 1.0974t^3 - 0.0915t^4\) represents the ozone level in the atmosphere. Using the second derivative test, we found that there is a relative maximum at approximately \(t=9\), which corresponds to 4 p.m. This implies that the ozone level is highest at around 4 p.m., making the air quality worse for people with sensitive respiratory systems.

Step 1: Find the first derivative of A(t)

To identify any potential relative maximum points, we need to find the first derivative of the function \(A(t)\). We have given the function: \( A(t) = 1.0974t^3 - 0.0915t^4 \) Now, let's find the first derivative of \(A(t)\), denoted by \(A'(t)\): \[ A'(t) = \frac{d}{dt}(1.0974t^3 - 0.0915t^4) = 1.0974(3)t^2 - 0.0915(4)t^3 \] After simplifying, we get: \[ A'(t) = 3.2922t^2 - 0.366t^3 \]

Step 2: Find critical points

Next, we need to find the critical points by setting the first derivative equal to zero, \[ 3.2922t^2 - 0.366t^3 = 0 \] To solve for the critical points, factor out \(t^2\): \[ t^2(3.2922 - 0.366t) = 0 \] This equation has two solutions: \(t = 0\) and \(t = \frac{3.2922}{0.366} \approx 8.996\), which is approximately \(t=9\).

Step 3: Find the second derivative of A(t)

To apply the second derivative test, we need to find the second derivative of \(A(t)\), denoted by \(A''(t)\): \[ A''(t) = \frac{d^2}{dt^2}(3.2922t^2 - 0.366t^3) = 6.5844t - 0.366(3)t^2 \] After simplifying, we get: \[ A''(t) = 6.5844t - 1.098t^2 \]

Step 4: Evaluate the second derivative at the critical point t = 9

We will evaluate the second derivative at the critical point \(t=9\) to determine if it is a relative maximum: \[ A''(9) = 6.5844(9) - 1.098(9^2) \] \[ A''(9) \approx -88.398 \] Since \(A''(9) < 0\), we can conclude that there is a relative maximum at \(t = 9\).

Step 5: Interpret the results

Based on the second derivative test, the function \(A(t)\) reaches a relative maximum at approximately \(t=9\), which corresponds to 4 p.m. This means the level of ozone is at its highest in the atmosphere around 4 p.m. on this particular day in the city of Riverside. Therefore, the air quality is worse at this time, and people with sensitive respiratory systems should be cautious when outdoors.

Key Concepts

Second Derivative Test

The second derivative test is a mathematical tool that helps determine whether a function has a relative maximum or minimum at a given point. The idea is that if you have already identified a critical point (a point where the first derivative is zero), the second derivative can tell you the nature of that critical point.

• If the second derivative is positive at that point, the function has a relative minimum there.
• If the second derivative is negative, the function has a relative maximum.
• If the second derivative is zero, the test is inconclusive.

Applying the second derivative test involves these steps:

• Finding the second derivative of the function by differentiating the first derivative.
• Evaluating this second derivative at the critical points.

In the exercise considered, we saw that the second derivative at the critical point around 9 hours is negative, indicating a relative maximum. This means the function's rate of change is decreasing at this point, confirming that the ozone level peaks around 4 p.m.

Critical Points

Critical points are points on the graph of a function where its derivative is zero or undefined. These points are important because they indicate possible locations of relative maxima, minima, or saddle points.
To find critical points, set the function's first derivative to zero and solve for the variable. This tells you where the slope of the tangent line to the function is zero, indicating potential peak or trough points. Critical points also need to lie within the domain of the function you're investigating:

• If the first derivative changes from positive to negative at a critical point, you have a relative maximum.
• If it changes from negative to positive, you have a relative minimum.

In our ozone problem, setting the first derivative equal to zero helped find that one critical point was around 9 hours. Knowing how to identify these points correctly enables other tests, such as the second derivative test, to interpret the behavior of the function further.

Polynomial Functions

Polynomial functions are algebraic expressions that involve sums and products of variables and coefficients, with the variable raised to non-negative integer powers. They are used extensively in both pure and applied mathematics because they are easy to differentiate and integrate.
A polynomial in one variable can be written as:\[ P(x) = a_n x^n + a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 \]where \(a_n, a_{n-1}, \ldots, a_0\) are constants and \(x\) is the variable.
Key characteristics of polynomial functions include:

• They are smooth and continuous, meaning they have no breaks or jumps.
• The degree of the polynomial, given by the highest power of the variable, often influences the number of turning points and the general shape of the graph.
• Lower degrees have fewer turning points; for instance, a quadratic (degree 2) has at most one turning point.

The exercise about ozone levels involved a cubic polynomial, which could potentially have up to two turning points. Knowing this guides our analysis as we locate critical points and use the second derivative test to determine maximums or minimums.