Problem 66 - Algebra and Trigonometry Real Mathematics, Real People

Question

The depreciation \(D\) (in millions of dollars) of the WD-40 Company assets from 2009 through 2013 can be approximated by the function $$D(t)=1.9 \sqrt{t+3.7}$$,where \(t=0\) represents 2009.(a) Describe the transformation of the parent function \(f(t)=\sqrt{t}\). (b) Use a graphing utility to graph the model over the interval \(0 \leq t \leq 4\). (c) According to the model, in what year will the depreciation of WD-40 assets be approximately 6 million dollars? (d) Rewrite the function so that \(t=0\) represents 2011 . Explain how you got your answer.

Step-by-Step Solution

Answer

The function \(D(t) = 1.9 \sqrt{t+3.7}\) is a transformation of the parent function \(f(t) = \sqrt{t}\) which involves a 3.7-units shift to the left and a vertical stretch by a factor of 1.9. Graphing this function demonstrates how the depreciation of WD-40 Company assets changes over time. In 2013, the depreciation will be approximately \6 million dollars. Upon shifting the function to denote 2011 as the start year, the new function would be \(D(t) = 1.9 \sqrt{t+1.7}\)

Step 1: Transformation of the Function

The given function \(D(t) = 1.9 \sqrt{t+3.7}\) is a transformation of the parent function \(f(t) = \sqrt{t}\). Here, the term 3.7 inside the square root shifts the function 3.7 units to the left and the factor 1.9 multiplied with the square root function stretches the function vertically by a factor of 1.9.

Step 2: Graphing the Function

To graph the function, input \(D(t) = 1.9 \sqrt{t+3.7}\) into the graphing utility and set the range for \(t\) from 0 to 4. The resulting graph shows how the depreciation \(D\) changes over time \(t\) from 2009 (t = 0) to 2013 (t = 4).

Step 3: Finding the Depreciation Year

To find the year when the depreciation will be approximately 6 million dollars, set \(D(t) = 6\) and solve for \(t\). Upon solving: \(t = (\frac{6}{1.9})^2 - 3.7\) which gives \(t\) around 3.54 years. This means, considering \(t=0\) as 2009, the depreciation will be approximately \6 million dollars around mid 2013.

Step 4: Shifting the Function

To rewrite the function so that \(t=0\) represents the year 2011, \(t\) has to be shifted by 2 years to the right. This can be done by replacing \(t\) in the initial function by \(t-2\), giving us the new function: \(D(t) = 1.9 \sqrt{t-2+3.7} = 1.9 \sqrt{t+1.7}\)

Key Concepts

Parent Function Transformation

In mathematics, transformations allow us to tweak functions to better match real-life scenarios. The parent function in this exercise is \( f(t) = \sqrt{t} \). This is a basic square root function, whose graph starts at the origin and increases slowly. However, for the WD-40 depreciation function given by \( D(t) = 1.9 \sqrt{t+3.7} \), we make two key transformations to the parent function.

• First, the term \( t+3.7 \) inside the square root shifts the graph horizontally. Specifically, it moves the graph 3.7 units to the left. This is because adding a positive number inside the function effectively shifts it towards the negative axis.
• Next, the coefficient 1.9 outside the square root affects the graph's vertical stretch. It stretches the graph upwards by a factor of 1.9, making it taller. This means for every unit of input, the output is amplified by 1.9 times the original value from the parent function.

Understanding these transformations helps in visualizing how the basic function adapts to meet the specifics of asset depreciation.

Graphing Utility

Graphing utilities are powerful tools that help visualize mathematical functions. These can be software programs like Desmos, graphing calculators, or even online graphing sites. By inputting a function into a graphing utility, students can see how the function behaves over a specific interval.
To graph \( D(t) = 1.9 \sqrt{t+3.7} \), set the interval for \( t \) from 0 to 4, corresponding to the years from 2009 to 2013. Once inputted, the tool will draw the curve showing how asset depreciation progresses over these years.

• The curve will display a gradual increase, illustrating the relationship between time and the depreciation amount.
• This visualization makes it easier to predict or estimate values based on the graph, such as determining the approximate year when depreciation might reach a certain level.

Graphing utilities simplify the complex task of plotting functions by hand and offer a clearer insight into mathematical models.

Approximate Depreciation Year

To determine when the depreciation of assets reaches about 6 million dollars, we solve for \(t\) in \( D(t) = 6 \). Setting the equation, \( 6 = 1.9 \sqrt{t + 3.7} \), solve for \(t\):

• Divide both sides by 1.9: \( 6/1.9 = \sqrt{t+3.7} \).
• Square the result to eliminate the square root: \( (6/1.9)^2 = t + 3.7 \).
• Subtract 3.7 to solve for \(t\): \( t = (6/1.9)^2 - 3.7 \).

After simplification, \( t \approx 3.54 \). Given that \( t=0 \) represents 2009, adding 3.54 years suggests we reach around mid-2013.
This calculation estimates when the asset would depreciate to that value, aiding businesses in financial forecasting.

Function Rewriting

Rewriting a function often involves changing its reference point, such as shifting the base year. To have \( t = 0 \) represent 2011 instead of 2009, adjust the function accordingly.
Given \( D(t) = 1.9 \sqrt{t+3.7} \), we need a 2-year shift to the right.
Replace \( t \) with \( t-2 \):

• This change reflects the new starting year.
• Our equation becomes \( D(t) = 1.9 \sqrt{t - 2 + 3.7} \).
• Simplify to \( D(t) = 1.9 \sqrt{t+1.7} \).

Now, \( t=0 \) aligns with 2011, maintaining the same depreciation behavior but adjusting the timeline for new analyses or comparisons.