Problem 66 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

A study on formaldehyde levels in 900 homes indicates that emissions of various chemicals can decrease over time. The formaldehyde level (parts per million) in an average home in the study is given by $$ f(t)=\frac{0.055 t+0.26}{t+2} \quad(0 \leq t \leq 12) $$ where \(t\) is the age of the house in years. How fast is the formaldehyde level of the average house dropping when it is new? At the beginning of its fourth year?

Step-by-Step Solution

Answer

The formaldehyde level in an average house is dropping by 0.0375 parts per million per year when it is new and by \(\frac{1}{240}\) parts per million per year at the beginning of its fourth year.

Step 1: Find the derivative of the function, \(f'(t)\)

To find the derivative of the given function, \(f(t) = \frac{0.055t + 0.26}{t + 2}\), we'll use the quotient rule, which states: \[ \frac{d}{dt} \frac{u}{v} = \frac{vu' - uv'}{v^2} \] Here, \(u = 0.055t + 0.26\) and \(v = t + 2\). So, let's find their derivatives: \[ u' = \frac{d}{dt}(0.055t + 0.26) = 0.055 \\ v' = \frac{d}{dt}(t + 2) = 1 \] Now, use the quotient rule to find the derivative of \(f\): \[ f'(t) = \frac{ (t + 2)(0.055) - (0.055t + 0.26)(1) }{(t + 2)^2} \]

Step 2: Simplify the derivative

Now, let's simplify the derivative, \(f'(t)\): \[ f'(t) = \frac{(0.055t + 0.11) - (0.055t + 0.26)}{(t + 2)^2} = \frac{-0.15}{(t + 2)^2} \]

Step 3: Evaluate the derivative when the house is new

To find how fast the formaldehyde level is dropping when the house is new, we'll evaluate the derivative at \(t = 0\): \[ f'(0) = \frac{-0.15}{(0 + 2)^2} = \frac{-0.15}{4} = -0.0375 \] So, the formaldehyde level in an average house is dropping by 0.0375 parts per million per year when it is new.

Step 4: Evaluate the derivative at the beginning of the fourth year

To find how fast the formaldehyde level is dropping at the beginning of the fourth year, we'll evaluate the derivative at \(t = 4\): \[ f'(4) = \frac{-0.15}{(4 + 2)^2} = \frac{-0.15}{36} = -\frac{1}{240} \] So, the formaldehyde level in an average house is dropping by \(\frac{1}{240}\) parts per million per year at the beginning of its fourth year.

Key Concepts

Formaldehyde Levels

Formaldehyde is a common indoor air pollutant found in homes, often released from building materials and household products. Over time, the concentration of formaldehyde in a home can decrease due to the natural off-gassing of materials. In the given exercise, the function models how formaldehyde levels decrease over time in a home.The function provided is \( f(t) = \frac{0.055t + 0.26}{t + 2} \),
where \( t \) represents the age of the house in years. Studying how quickly the level of formaldehyde drops is useful to understand indoor air quality improvement over time.

• This function helps predict and manage health risks associated with prolonged exposure to formaldehyde.
• By calculating the rate of change, we gain insights into how fast indoor air quality might improve as a house ages.

Understanding these changes over time is important for both health and environmental considerations.

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. This rule simplifies the process of finding the rate of change for such functions.Given two functions, \( u(t) \) and \( v(t) \), the derivative of their quotient is given by:\[\frac{d}{dt} \frac{u}{v} = \frac{vu' - uv'}{v^2}\]For the function in the exercise, we have:

• \( u(t) = 0.055t + 0.26 \) with \( u' = 0.055 \)
• \( v(t) = t + 2 \) with \( v' = 1 \)

Applying the quotient rule helps to find how quickly the formaldehyde level is decreasing in a more systematic and accurate manner.

• This rule is very helpful when dealing with real-world applications where changes in the ratio of two quantities are involved.
• It ensures continuity and accuracy in deriving complex functions where both the numerator and denominator are subject to change.

Derivatives of Functions

Derivatives measure the rate at which a quantity changes with respect to another. In the context of this exercise, the derivative of the function \( f(t) \) indicates how the formaldehyde level changes as the house gets older.Finding the derivative \( f'(t) \) involves:

• Determining \( u'(t) \) and \( v'(t) \), the derivatives of the numerator and the denominator.
• Using these derivatives in the quotient rule formula to find \( f'(t) \).

In practical terms:

• Evaluating \( f'(0) \) tells us how fast formaldehyde is decreasing when the house is new, showing a rate of \(-0.0375\) parts per million per year.
• Calculating \( f'(4) \) shows how quickly the levels drop at the start of the fourth year, giving a rate of \(-\frac{1}{240}\) parts per million per year.

Understanding derivatives is crucial because:

• They help predict future values based on the current rate of change.
• These calculations guide decisions related to health standards and safety regulations regarding air quality over time.