Problem 44 - Single Variable Calculus

Question

Air Pollution The amount of nitrogen dioxide, a brown gas that impairs breathing, that is present in the atmosphere on a certain day in May in the city of Long Beach is approximated by $$ A(t)=\frac{136}{1+0.25(t-4.5)^{2}}+28 \quad 0 \leq t \leq 11 $$ where \(A(t)\) is measured in pollutant standard index (PSI) and \(t\) is measured in hours with \(t=0\) corresponding to 7 A.M. When is the PSI increasing, and when is it decreasing? At what time is the PSI highest, and what is its value at that time?

Step-by-Step Solution

Answer

The PSI is increasing from 7 A.M. (\(t=0\)) to 11:30 A.M. (\(t=4.5\)) and decreasing from 11:30 A.M. to 6 P.M. (\(t=11\)). The highest PSI value is 164, which occurs at 11:30 A.M. (\(t=4.5\)).

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

To find \(A'(t)\), we will apply the Quotient Rule, which states that if we have a function in the form of \(\frac{f(t)}{g(t)}\), then its derivative can be found using \(\frac{f'(t)g(t)-f(t)g'(t)}{g(t)^2}\). In this case, \(f(t) = 136\) and \(g(t) = 1+0.25(t-4.5)^{2}\). First, let's calculate the derivatives of \(f(t)\) and \(g(t)\): \[ f'(t) = 0 \\ g'(t) = 0.25\cdot2(t-4.5)=0.5(t-4.5) \] Next, we can use the Quotient Rule to find \(A'(t)\): \[ A'(t) = \frac{0 \cdot (1+0.25(t-4.5)^{2}) - 136 \cdot 0.5(t-4.5)}{(1+0.25(t-4.5)^{2})^2} \\ A'(t) = \frac{-68(t-4.5)}{(1+0.25(t-4.5)^{2})^2} \]

Step 2: Analyze the sign of the first derivative

Now that we have the first derivative of A(t), we need to analyze its sign. We have: \[ A'(t) = \frac{-68(t-4.5)}{(1+0.25(t-4.5)^{2})^2} \] As we can see, the denominator is always positive, since it is squared. Therefore, the sign of \(A'(t)\) is determined by the numerator, \(-68(t-4.5)\): - If \(t<4.5\), then \(t-4.5<0\) and \(A'(t)>0\), meaning that PSI is increasing. - If \(t>4.5\), then \(t-4.5>0\) and \(A'(t)<0\), meaning that PSI is decreasing. So, the PSI is increasing from \(0 \leq t < 4.5\) hours and decreasing from \(4.5 < t \leq 11\) hours.

Step 3: Determine the time with the highest PSI value

From our analysis of \(A'(t)\), we know that the function has a critical point at \(t=4.5\). As we showed, the PSI increases before this point and decreases after it, which means that this is indeed the time when the PSI reaches its highest value. To find the maximum value of PSI, plug \(t=4.5\) into the original function: \[ A(4.5) = \frac{136}{1+0.25(4.5-4.5)^{2}}+28 = 136+28 = 164 \text{ PSI} \] Therefore, the PSI is highest at \(t = 4.5\) hours, or \(11:30\) A.M., and the value of PSI at that time is 164.

Key Concepts

Quotient Rule

The Quotient Rule is a fundamental technique in calculus used to find the derivative of a quotient of two functions. If you have a function expressed as \( \frac{f(t)}{g(t)} \), the quotient rule tells you how to differentiate it. The formula is given by:

• \( \frac{f'(t)g(t)-f(t)g'(t)}{g(t)^2} \)

This rule requires that you know both the functions \( f(t) \) and \( g(t) \), as well as their derivatives, \( f'(t) \) and \( g'(t) \).
In the provided problem, to find the derivative of \( A(t) = \frac{136}{1+0.25(t-4.5)^{2}}+28 \), we separate the constant \( 28 \) since it does not affect the derivative. The focus is on differentiating \( \frac{136}{1+0.25(t-4.5)^2} \).
Here, \( f(t) = 136 \) and \( g(t) = 1+0.25(t-4.5)^2 \). To apply the quotient rule, we compute \( f'(t) = 0 \) because the derivative of a constant is zero, and \( g'(t) = 0.5(t-4.5) \), which stems from the derivative of a quadratic term.
With these components, you apply the Quotient Rule to get \( A'(t) = \frac{-68(t-4.5)}{(1+0.25(t-4.5)^{2})^2} \). This result helps us move on to understanding how the function behaves by analyzing the sign of this derivative.

First Derivative

The first derivative of a function provides critical insights into the behavior of the function. It tells us the rate at which the function's value is changing with respect to its input variable.
In this particular problem, the first derivative \( A'(t) = \frac{-68(t-4.5)}{(1+0.25(t-4.5)^2)^2} \) helps us understand whether the PSI levels are increasing or decreasing over time.

Understanding the Sign of the First Derivative

We focus on the numerator, \( -68(t-4.5) \), because the denominator \( (1+0.25(t-4.5)^2)^2 \) is positive for all real \( t \) due to the squaring of a non-negative term.

• If \( t < 4.5 \), then \( t-4.5 < 0 \), making the product \(-68(t-4.5)\) positive. Therefore, \( A'(t) > 0 \) and the PSI is increasing.
• If \( t > 4.5 \), then \( t-4.5 > 0 \), making the product \(-68(t-4.5)\) negative. Hence, \( A'(t) < 0 \) and the PSI is decreasing.

This analysis explains that from \( 0 \leq t < 4.5 \) hours, the PSI is increasing, and from \( 4.5 < t \leq 11 \) hours, the PSI is decreasing.

Critical Points

Critical points in a function occur where its derivative equals zero or is undefined. These points are significant because they can indicate local maxima, minima, or points of inflection.
To find critical points, we set \( A'(t) = 0 \) or consider where the derivative does not exist. For the function \( A(t) \), the derivative \( A'(t) = \frac{-68(t-4.5)}{(1+0.25(t-4.5)^2)^2} \) equals zero when the numerator is zero, i.e., where \( -68(t-4.5) = 0 \).

Solving for Critical Points

From the equation above, \( -68(t-4.5) = 0 \) simplifies to \( t-4.5 = 0 \), giving a critical point at \( t = 4.5 \).
This point is a candidate for a local maximum or minimum. Given the sign change of \( A'(t) \) around \( t = 4.5 \) (from positive to negative), this critical point is actually a local maximum.
By plugging \( t = 4.5 \) back into the original function, we find the PSI at this critical time to be \( A(4.5) = 164 \). Hence, at 11:30 A.M., the PSI reaches its peak value of 164, confirming this as the time and value of maximum pollution.