Problem 72 - Single Variable Calculus

Question

Air Pollution According to the South Coast Air Quality Management district, the level of nitrogen dioxide, a brown gas that impairs breathing, that is present in the atmosphere between 7 A.M. and 2 P.M. on a certain May day in downtown Los Angeles is approximated by $$ I(t)=0.03 t^{3}(t-7)^{4}+60.2 \quad 0 \leq t \leq 7 $$ where \(I(t)\) is measured in pollutant standard index (PSI) and \(t\) is measured in hours, with \(t=0\) corresponding to 7 A.M. Determine the time of day when the PSI is the lowest and when it is the highest.

Step-by-Step Solution

Answer

The minimum PSI occurs at \(t=3\) hours (10 A.M.) with a value of approximately 48.2 PSI, while the maximum PSI occurs at both \(t=0\) hours (7 A.M.) and \(t=7\) hours (2 P.M.) with a value of 60.2 PSI.

Step 1: Identify the given function

We are given the function: \[ I(t)=0.03t^{3}(t-7)^{4}+60.2 \quad 0 \leq t \leq 7 \] Here, \(I(t)\) represents the PSI at time \(t\) (in hours), where \(t=0\) corresponds to 7 A.M.

Step 2: Find the derivative of I(t)

To find the critical points, we need to compute the derivative of \(I(t)\) with respect to \(t\). We need to apply the product rule and chain rule for differentiation. Using the chain rule, we have: \[ \begin{aligned} \dfrac{d I(t)}{dt} &= 0.03\left(\frac{d}{dt}(t^3)\right)(t-7)^4 + 0.03t^3\left(\frac{d}{dt}((t-7)^4)\right) \\ &= 0.03(3t^2)(t-7)^4 + 0.03t^3(4(t-7)^3\cdot1) \\ &= 0.09t^2(t-7)^4 + 0.12t^3(t-7)^3 \end{aligned} \]

Step 3: Find the critical points

Set the derivative equal to zero and solve for \(t\) to find the critical points: \[ 0.09t^2(t-7)^4 + 0.12t^3(t-7)^3 = 0 \] We can factor out a common term \(0.03t^2(t-7)^3\): \[ 0.03t^2(t-7)^3(3(t-7)+4t) = 0 \] From the factored equation, we can see the critical points \(t\) are when \(t = 0\), \(t=7\), and when \(3(t-7)+4t = 0\), which simplifies to: \[ 3t - 21 + 4t = 0 \quad \Longrightarrow \quad 7t = 21 \quad \Longrightarrow \quad t=3 \] Thus, our critical points are \(t=0\), \(t=3\), and \(t=7\).

Step 4: Find the value of I(t) at the critical points and endpoints

Evaluate the function at the critical points and the endpoints of the interval [0, 7]: \[ I(0) = 0.03(0)^3(0-7)^4 + 60.2 = 60.2 \\ I(3) = 0.03(3)^3(3-7)^4 + 60.2 \approx 48.2 \\ I(7) = 0.03(7)^3(7-7)^4 + 60.2 = 60.2 \]

Step 5: Determine the time when PSI is lowest and highest

Now, we can see the values of \(I(t)\) at the critical points: \(I(0) = 60.2\) PSI \(I(3) \approx 48.2\) PSI \(I(7) = 60.2\) PSI From these results, we can conclude that the minimum PSI occurs at \(t=3\) hours (which corresponds to 10 A.M.), with a value of approximately 48.2 PSI. The maximum PSI occurs at both \(t=0\) hours (7 A.M.) and \(t=7\) hours (2 P.M.) with a value of 60.2 PSI.

Key Concepts

Understanding Critical Points

Critical points are locations on a graph where the derivative equals zero or doesn't exist. They often indicate where a function's graph might change direction, such as highs, lows, or points of inflection.
Critical Points are vital in calculus because they help identify where a function reaches its highest or lowest values.

To find critical points:

• Take the derivative of the function.
• Set the derivative equal to zero and solve for the variable.
• Check if the derivative is undefined, which can also indicate a critical point.

In our exercise, we set the derivative of the nitrogen dioxide function to zero. Solving it gives us potential critical points during the day.

Applying the Chain Rule

The chain rule in calculus is essential for finding derivatives of composite functions. It states that to differentiate a composite function, you take the derivative of the outer function and multiply it by the derivative of the inner function.
The formula is: \( \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) \)

In our exercise, when differentiating \((t-7)^4\), the chain rule helps compute how this inner function changes over time.The key idea is to work from the outside in:

• Differentiate the outer function, keeping the inside the same.
• Multiply the result by the derivative of the inside function.

By using the chain rule correctly, we can find crucial parts of the derivative needed to locate critical points.

Using the Product Rule

The product rule is used when differentiating products of two functions. If you have a function that is a product, like \( u(t) \cdot v(t) \), you need this rule.
The product rule formula is:\( \frac{d}{dt}[u(t) \cdot v(t)] = u'(t)v(t) + u(t)v'(t) \)

In our problem, we see the function \(0.03t^3(t-7)^4\) where both parts need differentiation. The product rule helps by guiding us how to differentiate each part in turn and combine these results.

By applying the product rule, you maintain accuracy when differentiating complex products, such as polynomial and composite functions.

Characteristics of Polynomial Functions

Polynomial functions are expressions consisting of variables and coefficients, formed using addition, subtraction, multiplication, and non-negative integer exponents.
Polynomial functions have several important characteristics:

• They are smooth and continuous.
• Their graphs are easy to analyze with calculus tools like derivatives.
• They exhibit predictable patterns, such as turning points based on their degree.

In our task, \(I(t)\) is a polynomial function due to terms like \(t^3\) and \((t-7)^4\). These terms allow us to use differentiation techniques to find critical points.
Understanding polynomial behavior aids in locating maximum and minimum points, explaining the daily variation in pollution levels.