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Chapter 6: Problem 56

For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. An oil well generates a continuous stream of income of \(60 t\) thousand dollars per year, where \(t\) is the number of years that the rig has been in operation. Find the present value of this stream of income over the first 20 years at a continuous interest rate of \(5 \%\).

Short Answer

Expert verified
The present value is approximately $15,170.4$ thousand dollars.

Step by step solution

01

Identify the Income Function

The continuous income stream generated by the oil well is given by \(60t\) thousand dollars per year, where \(t\) is the time in years. Since we need to calculate the present value over the first 20 years, the function is defined for \(0 \leq t \leq 20\).
02

Set Up the Present Value Integral

To find the present value of this income stream, we use the formula for continuous compounding. The present value \(PV\) of continuous income is given by the integral: \[PV = \int_0^T C(t) e^{-rt} \, dt\]where \(C(t)\) is the income function, \(r\) is the continuous interest rate, and \(T\) is the total time period. Substituting the given values: \[PV = \int_0^{20} 60t \, e^{-0.05t} \, dt\]
03

Compute the Integral

Calculate the integral: \[PV = \int_0^{20} 60t \, e^{-0.05t} \, dt\]To solve this, use integration by parts. Let \(u = 60t\), and \(dv = e^{-0.05t} \, dt\). Then, \(du = 60 \, dt\) and \(v = -20 e^{-0.05t}\). Applying integration by parts gives:\[\begin{align*}PV &= [60t \cdot (-20 e^{-0.05t})]_0^{20} - \int_0^{20} -20 \cdot 60 \, e^{-0.05t} \, dt \&= [-1200t \, e^{-0.05t}]_0^{20} + 1200 \cdot \int_0^{20} e^{-0.05t} \, dt\end{align*}\]
04

Evaluate the Integrals

Compute each part separately:1. \([-1200t \, e^{-0.05t}]_0^{20} = [-1200 \times 20 \, e^{-0.05 \times 20} - (-1200 \times 0 e^{-0.05 \times 0})] = -24000 \, e^{-1}\)2. For the integral \(1200 \int_0^{20} e^{-0.05t} \, dt\), calculate: \[\begin{align*}1200 \cdot \left[ -20e^{-0.05t} \right]_0^{20} &= 1200 \cdot \left[ -20(e^{-1} - 1) \right] \&= 24000 \cdot (1 - e^{-1})\end{align*}\]
05

Calculate the Final Present Value

Combine the results from Step 4:\[\begin{align*}PV &= -24000 \, e^{-1} + 24000(1 - e^{-1}) \&= 24000 - 24000 \, e^{-1}\end{align*}\]Using the approximation \(e^{-1} \approx 0.3679\), the present value is approximately:\[PV = 24000(1 - 0.3679) \approx 24000 \times 0.6321 \approx 15170.4\]Thus, the present value of the income stream over 20 years is approximately \(15,170.4\) thousand dollars.

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Key Concepts

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

Integration by Parts
Integration by parts is a mathematical technique used to solve integrals where direct integration is cumbersome. It's particularly effective when we have a product of functions, such as a polynomial and an exponential function. The method relies on the formula: \[ \int u \cdot dv = uv - \int v \cdot du \]Here is a step-by-step breakdown:
  • Choose functions to differentiate and integrate. Typically, you let the polynomial part be \(u\) and the rest \(dv\).
  • Find \(du\) by differentiating \(u\), and \(v\) by integrating \(dv\).
  • Substitute into the integration by parts formula to simplify the integral.
This method applies to our problem where we deal with \(60t\) as the polynomial and \(e^{-0.05t}\) as the exponential function. By applying integration by parts, we transform a complex integral into manageable parts, simplifying the process significantly.
Continuous Compounding
Continuous compounding is a concept derived from finance, which involves the constant accumulation of growth or interest. Unlike simple compounding, where interest is calculated at set intervals, continuous compounding compounds interest every instant. The formula used for continuous compounding in present value calculations is linked to exponential functions:
  • The future value is discounted back to the present using \(e^{-rt}\), where \(r\) is the interest rate and \(t\) is time.
  • It reflects a steady stream of value depreciation over time, capturing even the smallest contributions to interest.
In our scenario, continuous compounding at a rate of 5% per year helps in precisely calculating the present value of a continuous income stream, which is quite relevant for long-term financial scenarios.
Income Stream Analysis
Income stream analysis involves assessing incoming cash flows over a period of time. It's especially useful in evaluating projects or investments with regular income like salaries, rents, or any consistent financial inflows. In the context of this exercise:
  • The income stream generated by the oil well is continuous and described in terms of a function \(60t\).
  • Analyzing this income involves calculating its present value by integrating over the desired time span, factoring in interest rates.
  • This allows us to determine the real value in today's terms for future income, aiding decision-making on the viability and profitability of projects.
The essence of income stream analysis lies in converting varied future cash flows into a single present value figure, providing clarity on potential returns.
Exponential Functions
Exponential functions are mathematical models that describe rapid growth or decay, by maintaining a constant rate of change. These functions are pivotal in representing processes like population growth, radioactive decay, and financial aspects such as interest calculations.Key features include:
  • The base of natural exponents is \(e\), approximately equal to 2.71828, a constant in nature.
  • Exponential decay in financial terms translates to diminishing values, represented by \(e^{-rt}\) in present value calculations.
  • This function style suits continuous processes, like interest compounding or fluids running through a system, where change is gradual and ongoing.
In this exercise, the exponential function \(e^{-0.05t}\) is used to model the decrease in value over time due to the effect of a 5% continuous discount rate, ensuring an accurate present value estimation of future cash flows.

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Most popular questions from this chapter

Integration by parts often involves finding integrals like the following when integrating \(d v\) to find \(v\). Find the following integrals without using integration by parts (using formulas 1 through 7 on the inside back cover). Be ready to find similar integrals during the integration by parts procedure. $$ \int \sqrt{x} d x $$

For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. The population of a town is increasing at the rate of \(400 t e^{0.02 t}\) people per year, where \(t\) is the number of years from now. Find the total gain in population during the next 5 years.

Let \(y(t)\) be the size of a colony of bacteria after \(t\) hours. a. Write a differential equation that says that the rate of growth of the colony is equal to eight times the three-fourths power of its present size. b. Write an initial condition that says that at time zero the colony is of size 10,000 . c. Solve the differential equation and initial condition. d. Use your solution to find the size of the colony at time \(t=6\) hours.

Suppose that you now have $$\$ 2000$$, you expect to save an additional $$\$ 6000$$ during each year, and all of this is deposited in a bank paying \(4 \%\) interest compounded continuously. Let \(y(t)\) be your bank balance (in thousands of dollars) \(t\) years from now. a. Write a differential equation that expresses the fact that your balance will grow by 6 (thousand dollars) and also by \(4 \%\) of itself. [Hint: See Example 7.] b. Write an initial condition to say that at time zero the balance is 2 (thousand dollars). c. Solve your differential equation and initial condition. d. Use your solution to find your bank balance \(t=20\) years from now.

For each exercise: a. Solve without using a graphing calculator. b. Verify your answer to part (a) using a graphing calculator. A company begins advertising a new product and finds that after weeks the product is gaining customer recognition at the rate of \(t^{2} \ln t\) thousand customers per week (for \(t \geq 1\) ). Find the total gain in recognition from the end of week 1 to the end of week 6 .
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