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Chapter 5: Problem 56

GENERAL: Total Savings A homeowner installs a solar water heater that is expected to generate savings at the rate of \(70 e^{0.03 t}\) dollars per year, where \(t\) is the number of years since it was installed. a. Find a formula for the total savings within the first \(t\) years of operation. b. Use a graphing calculator to find when the heater will "pay for itself" if it cost \(\$ 800\). [Hint: Use INTERSECT.]

Short Answer

Expert verified
Total savings formula: \(S(t) = \frac{70}{0.03} (e^{0.03 t} - 1)\); Heater pays for itself after ~49 years.

Step by step solution

01

Understanding the Problem

We are given a function that represents the savings generated by a solar water heater over time. The rate of savings per year is given by the function \(70 e^{0.03 t}\), where \(t\) is the number of years since installation. We are tasked to find the total savings over a period of \(t\) years and determine when cumulative savings reach $800.
02

Setting Up the Integral

To find the total savings within the first \(t\) years, we need to integrate the rate of savings function. The expression to integrate is \(70 e^{0.03 t}\). The formula for total savings \(S(t)\) over the period \(t\) years will be given by the definite integral:\[ S(t) = \int_{0}^{t} 70 e^{0.03 x} \, dx \]
03

Solving the Integral

To solve the integral \( \int 70 e^{0.03 x} \, dx \), we use the formula for integrating exponential functions. The integral of \(e^{ax}\) is \(\frac{1}{a} e^{ax}\). For our integral, \(a = 0.03\), so:\[ \int 70 e^{0.03 x} \, dx = 70 \cdot \frac{1}{0.03} e^{0.03 x} = \frac{70}{0.03} e^{0.03 x} + C \]
04

Evaluate the Definite Integral

To find the total savings \(S(t)\), we evaluate the definite integral from 0 to \(t\):\[ S(t) = \left[ \frac{70}{0.03} e^{0.03 x} \right]_{0}^{t} = \frac{70}{0.03} \left(e^{0.03 t} - e^{0}\right) \]Since \(e^0 = 1\), the formula simplifies to:\[ S(t) = \frac{70}{0.03} \left(e^{0.03 t} - 1\right) \]
05

Determine When the Heater Pays for Itself

We need to find the value of \(t\) such that \(S(t) = 800\). Therefore, we solve the equation:\[ \frac{70}{0.03} \left(e^{0.03 t} - 1\right) = 800 \]
06

Solve for t Algebraically

First, solve the equation:\[ \frac{70}{0.03} (e^{0.03 t} - 1) = 800 \]Divide both sides by \(\frac{70}{0.03}\):\[ e^{0.03 t} - 1 = \frac{800 \times 0.03}{70} \]Simplify:\[ e^{0.03 t} = 1 + \frac{2400}{70} \]\[ e^{0.03 t} = \frac{310}{70} \]Take the natural logarithm of both sides:\[ 0.03 t = \ln\left(\frac{310}{70}\right) \]Solve for \(t\):\[ t = \frac{\ln\left(\frac{310}{70}\right)}{0.03} \]
07

Final Calculation

Use a calculator to find \(t\):\[ t \approx \frac{\ln(4.42857)}{0.03} \approx 49.00 \]This implies that the heater will pay for itself in approximately 49 years.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Exponential Functions
Exponential functions describe processes that grow or decay at a constant relative rate. In this exercise, we explore how savings from a solar water heater grow exponentially over time. The savings rate is given by the function \(70 e^{0.03 t}\). Here are some important points about exponential functions:
  • The base of the exponential function in our case is \(e\), the natural logarithm base, which is approximately equal to 2.71828.

  • In the function \(70 e^{0.03 t}\), the constant \(70\) represents the initial savings rate, while the exponent \(0.03 t\) reflects the continuous growth rate over time \(t\).

  • As \(t\) increases, the exponential term \(e^{0.03 t}\) causes the savings to grow exponentially, meaning that savings increase faster and faster over time.
Understanding exponential growth is crucial because it appears in many real-life applications, including population growth, radioactive decay, and finance. It efficiently models situations where a quantity grows or shrinks at a rate proportional to its current value.
Integration
Integration is a fundamental concept in calculus used to find quantities like area under a curve or, in this case, total accumulated savings. To determine the total savings over a period \(t\) years, we integrate the rate of savings function \(70 e^{0.03 t}\).
  • Integration allows us to sum up small increments of savings over time. By integrating from \(t=0\) to any fixed time \(t\), we accumulate the total savings up to that point.

  • The result of integrating \(70 e^{0.03 t}\) is an antiderivative, showing continuous accumulation. It gives us the total savings formula: \[ S(t) = \frac{70}{0.03} \left(e^{0.03 t} - 1\right) \]

  • This formula is derived by evaluating the indefinite integral \(\int 70 e^{0.03 x} \, dx\) and then applying the limits of \(0\) to \(t\).
Integration is an indispensable tool in calculus, fundamental for solving real-world problems involving accumulation and change. It transforms a rate of change function into a cumulative total, providing insights into both the process and the result.
Mathematical Modeling
Mathematical modeling uses mathematical expressions to represent real-world scenarios. In this problem, we model the financial aspect of a solar water heater investment to predict when it will become profitable.
  • We start with a rate of savings function, \(70 e^{0.03 t}\), a basic model designed to predict savings over time.

  • By integrating this function, we transition from a rate model to a cumulative total model, \(S(t)\), showing total savings over time.

  • Solving \(S(t)=800\) allows us to determine when the savings reach the cost of the heater, thus showing break-even point.
Mathematical modeling helps us understand complex systems by using equations to depict relationships and predict outcomes. In finance, engineering, ecology, and many other fields, models provide crucial predictions that support decision-making, planning, and optimization.

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For each demand function \(d(x)\) and supply function \(s(x)\) : a. Find the market demand (the positive value of \(x\) at which the demand function intersects the supply function). b. Find the consumers' surplus at the market demand found in part (a). c. Find the producers' surplus at the market demand found in part (a). \(d(x)=120-0.16 x, s(x)=0.08 x\)

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