Problem 9 - Applied Calculus

Question

An air-freshener starts with 30 grams and evaporates. In each of the following cases, write a formula for the quantity. \(Q\) grams, of air-freshener remaining \(t\) days after the start and sketch a graph of the function. The decrease is: (a) 2 grams a day (b) \(12 \%\) a day

Step-by-Step Solution

Answer

(a) Linear: \(Q(t) = 30 - 2t\). (b) Exponential: \(Q(t) = 30 \times (0.88)^t\).

Step 1: Identify the model type for each case

For part (a), the air freshener decreases by 2 grams each day, which implies a linear decrease over time. For part (b), it decreases by 12% each day, suggesting an exponential decay.

Step 2: Formulate the equation for Case (a)

Since it is a linear model, we use the formula for linear decrease: \(Q(t) = Q_0 - r \times t\), where \(Q_0\) is the initial amount (30 grams) and \(r\) is the rate of decrease (2 grams per day). Thus, \(Q(t) = 30 - 2t\).

Step 3: Sketch the graph for Case (a)

Plot the linear equation \(Q(t) = 30 - 2t\) on a graph. It starts at 30 grams when \(t = 0\) and decreases linearly. The slope is \(-2\), indicating a decrease of 2 grams per day until it reaches zero.

Step 4: Formulate the equation for Case (b)

This case follows an exponential decay model. Use the formula \(Q(t) = Q_0 \times (1 - r)^t\), where \(Q_0\) is 30 grams, and \(r\) is 0.12 (12%). So, the equation becomes \(Q(t) = 30 \times (0.88)^t\).

Step 5: Sketch the graph for Case (b)

Plot the exponential decay equation \(Q(t) = 30 \times (0.88)^t\). It starts at 30 grams and decreases exponentially, approaching zero without ever actually reaching it.

Key Concepts

Linear decrease

Linear decrease is a very straightforward concept in mathematics. In this case, the value of something decreases by the same amount every time period, such as each day. For the air freshener problem, linear decrease means that every day, 2 grams of the air freshener evaporates until none is left.

This can be described using a linear equation. The general formula for a linear decrease is:

• \(Q(t) = Q_0 - r \times t\)

Where:

• \(Q(t)\) is the quantity remaining after \(t\) days,
• \(Q_0\) is the initial quantity, which is 30 grams for our problem, and
• \(r\) is the rate of decrease, which is 2 grams per day.

So, in the given exercise, the formula becomes \(Q(t) = 30 - 2t\). This simply means that if the starting amount is 30 grams, it will decrease by 2 grams each day.

Exponential function

An exponential function, unlike a linear function, changes at a rate proportional to its current value. This creates a curve on a graph, which can rise or fall sharply depending on the nature of the change. In our air freshener case, we are dealing with exponential decay.

This happens when the air freshener loses a fixed percentage of its amount each day. The formula for this type of decay is:

• \(Q(t) = Q_0 \times (1 - r)^t\)

Where:

• \(Q(t)\) is the quantity remaining after \(t\) days,
• \(Q_0\) is the initial quantity, which is 30 grams here, and
• \(r\) is the rate of the decrease expressed as a decimal. For 12%, it is 0.12.

Hence, the formula in the exercise is \(Q(t) = 30 \times (0.88)^t\). This reflects that every day, the air freshener retains 88% of what it was the day before, steadily lessening in a non-linear fashion.

Graphing functions

Graphing functions helps us visually interpret how quantities change over time. With the air freshener example, plotting the graphs for both the linear decrease and the exponential decay reveals their distinct characteristics.

For the linear decrease, the graph will be a straight line. It begins at 30 grams when \(t = 0\) and should extend downward with a slope of \(-2\). This shows the steady reduction by 2 grams per day.

In contrast, the graph of the exponential decay starts similarly at 30 grams but forms a curve that gradually flattens out. This reflects how the air freshener doesn’t lose a constant amount each day, but a constant percentage, showing a rapid decline initially that slows over time.

Visualizing these graphs helps understand not just the final outcome but also how the rate of change can affect the overall progression of deformation or depletion. By examining these graphs, one quickly learns that linear and exponential dynamics differ significantly in how changes accumulate.