Problem 11 - Applied Calculus

Question

The value, \(V\), of a Tiffany lamp, worth \(\$ 225\) in 1975, increases at \(15 \%\) per year. Its value in dollars \(t\) years after 1975 is given by $$ V=225(1.15)^{t} $$ Find the average value of the lamp over the period \(1975-\) \(2010 .\)

Step-by-Step Solution

Answer

The average value of the lamp from 1975 to 2010 is approximately \( \$3496.30 \).

Step 1: Understand the Problem

We are asked to find the average value of the Tiffany lamp from 1975 to 2010. The value function given is \( V = 225(1.15)^t \).

Step 2: Define the Average Value Formula

The average value of a function \( f(t) \) over a period \([a, b]\) is given by the formula: \( \frac{1}{b-a} \int_{a}^{b} f(t) \, dt \). Here, \( f(t) = 225(1.15)^t \), and we want to find the average over \( a = 0 \) (1975) to \( b = 35 \) (2010).

Step 3: Set Up the Integration for Average Value

In this case, we must evaluate \( \int_{0}^{35} 225(1.15)^t \, dt \) as part of finding the average value. The formula for the average is then \( \frac{1}{35-0} \int_{0}^{35} 225(1.15)^t \, dt \).

Step 4: Find the Antiderivative

The antiderivative of \( (1.15)^t \) is \( \frac{(1.15)^t}{\ln(1.15)} \). Thus, \( \int 225(1.15)^t \, dt = 225 \times \frac{(1.15)^t}{\ln(1.15)} + C \), where \( C \) is the constant of integration.

Step 5: Evaluate the Definite Integral

Now compute \( \int_{0}^{35} 225(1.15)^t \, dt \). Using the antiderivative, this evaluates to: \[ 225 \cdot \left[ \frac{(1.15)^{35}}{\ln(1.15)} - \frac{(1.15)^{0}}{\ln(1.15)} \right] = 225 \cdot \frac{(1.15)^{35} - 1}{\ln(1.15)}. \]

Step 6: Calculate the Average Value

Substitute the definite integral's result into the average formula: \[ \frac{1}{35} \times 225 \cdot \frac{(1.15)^{35} - 1}{\ln(1.15)}. \] Calculate this value using a calculator for approximate results.

Step 7: Compute the Final Result

Finally, perform the calculations to determine the average value of the lamp from 1975 to 2010. Using the numbers, the computation yields approximately \( 3496.30 \).

Key Concepts

Exponential Growth

Exponential growth is when something increases rapidly over time at a constantly growing rate. It occurs when the growth rate of a function is proportional to its current value. In the case of the Tiffany lamp, the value grows by 15% annually. Here’s why exponential growth is significant:

• Rapid Increase: A small initial quantity can grow into a huge amount over time, as seen with the lamp's value.
• Compounding Effect: The increase itself grows, compounding the initial growth, which makes exponential functions powerful for modeling real-world phenomena.

This is different from linear growth, where things increase by a constant amount. For exponential growth, we use functions like \( V = 225(1.15)^t \) to model.These functions help predict future values and understand how things like investments or populations evolve.

Integration

Integration is a fundamental concept in calculus that helps in finding the total or accumulation of quantities. If differentiation is about finding the rate of change, integration is about finding the total change. It can answer questions like "how much space is under this curve?In our exercise, we need integration to determine the average value of the lamp's price increase over time.Here’s what you should know about integration related to exponential functions:

• Antiderivatives: This is finding the function whose derivative would give us our original function. For example, the function whose derivative is \( 225(1.15)^t \) is \( 225 \frac{{(1.15)^t}}{{\ln(1.15)}} \).
• Definite vs. Indefinite: Indefinite integrals give a general form, while definite integrals include limits and give a specific numerical result.

In practice, integration helps us calculate total values over intervals, whether it’s distance, volume, or cost, as seen in this lamp valuation problem.

Definite Integral

The definite integral allows us to calculate the exact total change of a quantity over a specific interval. In our lamp exercise, the definite integral is how we calculate the total value increase from 1975 to 2010.Here's why it's important:

• Specific Range: It helps to find the total change over a fixed period, such as 35 years for the lamp value.
• Evaluating Functions: Using the antiderivative, we can evaluate the function at given limits, 0 and 35 years in this case, to get precise results.

The definite integral uses an antiderivative, and we calculated it as:\[\int_{0}^{35} 225(1.15)^t \, dt = 225 \left[\frac{(1.15)^{35}}{\ln(1.15)} - \frac{1}{\ln(1.15)}\right].\]This process gives us the essential numerical value needed to determine the average value. By evaluating this expression, we can find the total increase in the lamp's value between defined limits. This is a tangible application of how definite integrals help solve real-world problems.