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Chapter 15: Problem 63

Recall that the compound interest formula for annual compounding is $$ A(P, r, t)=P(1+r)^{t} $$ where \(A\) is the future value of an investment of \(P\) dollars after \(t\) years at an interest rate of \(r\). a. Calculate \(\frac{\partial A}{\partial P}, \frac{\partial A}{\partial r}\), and \(\frac{\partial A}{\partial t}\), all evaluated at \((100,0.10,10)\). (Round your answers to two decimal places.) Interpret your answers. b. What does the function \(\left.\frac{\partial A}{\partial P}\right|_{(100,0.10, t)}\) of \(t\) tell about your investment?

Short Answer

Expert verified
The partial derivatives of the compound interest formula at (100, 0.10, 10) are as follows: \(\frac{\partial A}{\partial P}|_{(100,0.10,10)} = 2.59\) \(\frac{\partial A}{\partial r}|_{(100,0.10,10)} = 17157.48\) \(\frac{\partial A}{\partial t}|_{(100,0.10,10)} = 1933.02\) Interpretation: 1. A derivative of 2.59 means that for each additional dollar invested, the future value of the investment will increase by approximately $2.59. 2. A derivative of 17157.48 means that for each 1% increase in interest rate (0.01), the future value of the investment will increase by approximately $171.57. 3. A derivative of 1933.02 means that for each additional year the investment is held, the future value of the investment will increase by approximately $1933.02. The function \(\left.\frac{\partial A}{\partial P}\right|_{(100,0.10, t)}\) gives the increase in the future value of the investment for each additional dollar invested at any given time t. Analyzing this function over time helps us understand how our investment grows with respect to the initial investment at different points in time and, consequently, decide when to invest more or withdraw money.

Step by step solution

01

Calculate \(\frac{\partial A}{\partial P}\)

To find the partial derivative with respect to P, we treat r and t as constants. \[ \frac{\partial A}{\partial P} = (1+r)^t \]
02

Calculate \(\frac{\partial A}{\partial r}\)

To find the partial derivative with respect to r, we treat P and t as constants. \[ \frac{\partial A}{\partial r} = P\cdot t\cdot(1+r)^{t-1} \]
03

Calculate \(\frac{\partial A}{\partial t}\)

To find the partial derivative with respect to t, we treat P and r as constants. Using the chain rule, \[ \frac{\partial A}{\partial t} = P(1+r)^t \ln(1+r) \] #Step 2: Evaluate Partial Derivatives at (100,0.10,10)#
04

Evaluate \(\frac{\partial A}{\partial P}\) at (100,0.10,10)

Plug in the values into the expression we found in Step 1 for \(\frac{\partial A}{\partial P}\) and evaluate. \[ \frac{\partial A}{\partial P}\bigg|_{(100,0.10,10)} = (1+0.10)^{10} = 2.59 \]
05

Evaluate \(\frac{\partial A}{\partial r}\) at (100,0.10,10)

Plug in the values into the expression we found in Step 1 for \(\frac{\partial A}{\partial r}\) and evaluate. \[ \frac{\partial A}{\partial r}\bigg|_{(100,0.10,10)} = 100\cdot 10\cdot(1+0.10)^{10-1} = 17157.48 \]
06

Evaluate \(\frac{\partial A}{\partial t}\) at (100,0.10,10)

Plug in the values into the expression we found in Step 1 for \(\frac{\partial A}{\partial t}\) and evaluate. \[ \frac{\partial A}{\partial t}\bigg|_{(100,0.10,10)} = 100(1+0.10)^{10}\ln(1+0.10) = 1933.02 \] #Step 3: Interpret the derivatives evaluated at (100,0.10,10)#
07

Interpret \(\frac{\partial A}{\partial P}\)

A derivative of 2.59 means that for each additional dollar invested, the future value of the investment will increase by approximately $2.59.
08

Interpret \(\frac{\partial A}{\partial r}\)

A derivative of 17157.48 means that for each 1% increase in interest rate (0.01), the future value of the investment will increase by approximately $171.57.
09

Interpret \(\frac{\partial A}{\partial t}\)

A derivative of 1933.02 means that for each additional year the investment is held, the future value of the investment will increase by approximately $1933.02. #Step 4: Analyze function \(\left.\frac{\partial A}{\partial P}\right|_{(100,0.10, t)}\)
10

Function interpretation

The function \(\left.\frac{\partial A}{\partial P}\right|_{(100,0.10, t)}\) gives the increase in the future value of the investment for each additional dollar invested at any given time t. By analyzing this function over time, we can understand how our investment grows with respect to the initial investment at different points in time and, consequently, decide when to invest more or withdraw money.

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Key Concepts

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

Compound Interest
Compound interest is a powerful financial concept that involves earning "interest on interest." With compound interest, your investment grows not only on the initial principal but also on the accumulated interest over time. The compound interest formula for annual compounding is \( A(P, r, t) = P(1+r)^t \), where:
  • \( P \) is the initial principal or the amount of money initially invested.
  • \( r \) is the annual interest rate expressed as a decimal.
  • \( t \) is the time period in years.
  • \( A \) is the future value of the investment after \( t \) years.
Understanding how compound interest works can help you maximize your financial growth. By reinvesting your interest, you can significantly increase your returns over time. This is why it's essential to consider the frequency and rate of compounding throughout your investment journey.
Investment Growth
Investment growth refers to the increase in value of an investment over time, influenced by factors like the interest rate, initial principal, and investment duration. To accurately quantify how these factors impact growth, partial derivatives can be utilized.

The derivative \( \frac{\partial A}{\partial P} \) represents how the investment's future value changes with each additional dollar invested. Evaluating this derivative can guide investment decisions by showing potential returns on additional investments.

Similarly, \( \frac{\partial A}{\partial r} \) highlights how changes in the interest rate affect the investment's value. Small changes in \( r \) can lead to significant differences in future returns, emphasizing the importance of securing favorable rates.

Lastly, \( \frac{\partial A}{\partial t} \) demonstrates the effect of time on investment growth. The longer you hold an investment, the more significant the compounding effect becomes, underscoring the benefits of long-term investing.
Calculus Applications
Calculus, with its rich control over change, finds a significant application in financial modeling, especially in understanding investment returns. Partial derivatives, a vital tool in multivariable calculus, allow us to isolate the effect of individual variables on a function.

In the context of compound interest, the use of partial derivatives helps in:
  • Assessing the increase in investment value for an additional dollar (\( \frac{\partial A}{\partial P} \)).
  • Understanding the sensitivity of investment growth to changes in the interest rate (\( \frac{\partial A}{\partial r} \)).
  • Evaluating the impact of time on the overall growth of the investment (\( \frac{\partial A}{\partial t} \)).
The precision offered by calculus allows investors to fine-tune their strategies, reacting optimally to changes in market conditions. By applying these derivatives, investors can make informed decisions about where, when, and how much to invest, thereby optimizing their financial portfolios.

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