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Chapter 7: Problem 32

Future Value The value \(F(P, r)\) of an investment of \(P\) dollars after 2 years in an account with annual percentage yield \(100 r \%\) is given by the function \(F(P, r)=P(1+r)^{2}\) dollars. a. Write the first partial derivatives of \(F\). b. Write each of the second partial derivative formulas and interpret them for \(P=10,000\) and \(r=0.09 .\)

Short Answer

Expert verified
First partial derivatives: \( (1+r)^2 \) and \( 2P(1+r) \). Second partials at \( P=10,000 \) and \( r=0.09 \): \(0\), \(20,000\), and \(2.18\).

Step by step solution

01

Find the partial derivative with respect to P

To find the partial derivative of \(F(P, r) = P(1+r)^2\) with respect to \(P\), treat \(r\) as a constant. So the derivative is simply \((1+r)^2\) because all terms involving \(P\) can be considered as constants during differentiation:\[\frac{\partial F}{\partial P} = (1+r)^2\]
02

Find the partial derivative with respect to r

To find the partial derivative of \(F(P, r) = P(1+r)^2\) with respect to \(r\), treat \(P\) as a constant and differentiate with respect to \(r\). Consider the chain rule and simplify:\[\frac{\partial F}{\partial r} = P \cdot 2(1+r)\]Thus, the partial derivative with respect to \(r\) is:\[\frac{\partial F}{\partial r} = 2P(1+r)\]
03

Calculate the second partial derivative with respect to P

Differentiate \(\frac{\partial F}{\partial P} = (1+r)^2\) again with respect to \(P\). Since \( (1+r)^2\) is independent of \(P\), its derivative is zero:\[\frac{\partial^2 F}{\partial P^2} = 0\]
04

Calculate the second partial derivative with respect to r

Differentiate \(\frac{\partial F}{\partial r} = 2P(1+r)\) again with respect to \(r\). Using the power rule, this becomes:\[\frac{\partial^2 F}{\partial r^2} = 2P\cdot 1 = 2P\]
05

Calculate the mixed partial derivative (with respect to P and r)

For the mixed partial derivative, differentiate \(\frac{\partial F}{\partial P} = (1+r)^2\) with respect to \(r\) again:\[\frac{\partial^2 F}{\partial r \partial P} = 2(1+r)\]Similarly, differentiating \(\frac{\partial F}{\partial r} = 2P(1+r)\) with respect to \(P\):\[\frac{\partial^2 F}{\partial P \partial r} = 2(1+r)\]
06

Evaluate the second partial derivatives at P=10,000 and r=0.09

Now evaluate the second partial derivatives:1. \(\frac{\partial^2 F}{\partial P^2} = 0\) (no dependence on \(P\) or \(r\)).2. \(\frac{\partial^2 F}{\partial r^2} = 2P = 2 \times 10,000 = 20,000\).3. \(\frac{\partial^2 F}{\partial r \partial P} = 2(1+0.09) = 2 \times 1.09 = 2.18\).4. \(\frac{\partial^2 F}{\partial P \partial r} = 2.18\) (by Clairaut's theorem).
07

Interpret the results

1. \(\frac{\partial^2 F}{\partial P^2} = 0\) indicates linear behavior with respect to \(P\).2. \(\frac{\partial^2 F}{\partial r^2} = 20,000\) implies that the investment's curvature with respect to \(r\) is increasing and significant at this point.3. \(\frac{\partial^2 F}{\partial r \partial P} = 2.18\) implies how sensitive the rate growth effect is, given a small change in \(P\), and vice versa for \(\frac{\partial^2 F}{\partial P \partial r}\).

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Key Concepts

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

Partial Derivatives
In calculus, partial derivatives are a fundamental concept especially useful when dealing with functions of multiple variables. When we calculate a partial derivative, we differentiate with respect to one variable while treating all other variables as constants. This allows us to understand how the function changes as we vary one of the inputs.

Given the investment value function, \(F(P, r) = P(1+r)^2\), the function depends on both the principal \(P\) and the annual interest rate \(r\).
  • The partial derivative of \(F\) with respect to \(P\) is \(\frac{\partial F}{\partial P} = (1+r)^2\), indicating how the future value changes as the initial investment \(P\) changes while keeping \(r\) constant.
  • The partial derivative with respect to \(r\) is \(\frac{\partial F}{\partial r} = 2P(1+r)\), showing the change in future value as the interest rate \(r\) changes while \(P\) remains constant.
Investment Value Function
The investment value function \(F(P, r) = P(1+r)^2\) models the future value of an investment after two years based on two variables: the principal \(P\) and the interest rate \(r\). Understanding this function helps in analyzing how varying these parameters can impact investment returns.

  • Increasing \(P\) linearly increases \(F\), because \(F\) directly scales with \(P\).
  • An increase in \(r\) not only increases \(F\) due to a larger multiplier on \(P\), but this change is amplified because the interest acts exponentially via \((1+r)^2\).
This exponential growth reflects the compounding effect of interest, a crucial concept in investment and finance.
Second Partial Derivatives
The second partial derivatives provide valuable insights into the curvature of the investment value function and can help gauge the sensitivity of the investment to changes in \(P\) and \(r\).

  • \(\frac{\partial^2 F}{\partial P^2} = 0\): This indicates the relationship between future value and principal \(P\) is linear - no additional curvature beyond linearity.
  • \(\frac{\partial^2 F}{\partial r^2} = 2P\): This shows a positive curvature with respect to \(r\); the investment value grows at an increasing rate as \(r\) increases, due to the compounding effect.
  • Mixed Partial Derivative \(\frac{\partial^2 F}{\partial r \partial P} = \frac{\partial^2 F}{\partial P \partial r} = 2(1+r)\): The consistency of mixed partial derivatives, as per Clairaut's theorem, suggests that the influence of \(r\) on \(P\) and vice versa is symmetric, which helps in understanding the model sensitivity.
These derivatives allow for more precise financial planning, by evaluating how different factors can affect the growth of investments.

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Most popular questions from this chapter

For Activities 17 through \(22,\) write the first and second partial derivatives. $$ j(x, y)=y^{2} \ln x $$

For Activities 9 through \(16,\) write formulas for the indicated partial derivatives for each of the multivariable functions. \(m(t, s)=s \ln t+3.75 s+14.96\) a. \(m_{t}\) b. \(m_{s}\) c. \(\left.m_{s}\right|_{t=3}\)

Competitive Sales \(\quad\) Two vending machines sit side by side in a college dorm. One machine sells Coke products, and the other sells Pepsi products. Daily sales of Coke products, based on the prices of the products in the two machines, can be modeled as $$ \begin{aligned} S(c, p)=& 196.42 p-50.2 c^{2}+9.6 c \\ &+66.4-1.04 c p \text { cans } \end{aligned} $$ when Coke products cost \(c\) dollars and Pepsi products cost \(p\) dollars. a. Calculate the rate of change of the sale of Coke products with respect to the price of Coke products when Coke products cost \(\$ 0.75\) and Pepsi products cost \(\$ 1.25\) b. Calculate the rate of change of the sale of Coke products with respect to the price of Pepsi products when Coke products cost \(\$ 0.75\) and Pepsi products \(\operatorname{cost} \$ 1.25\) c. Calculate and interpret the two rates of change \(\left.\frac{\partial S}{\partial c}\right|_{(c, p)=(1.30,1.20)}\) and \(\left.\frac{\partial S}{\partial p}\right|_{(c, p)=(1.30,1.20)}\)

Future Value The value \(F(r, t)\) of an investment of 1 million dollars with an annual yield of \(100 r \%\) is given by the function \(F(r, t)=(1+r)^{t}\) million dollars. a. Write the partial derivative \(\frac{\partial F}{\partial t}\). What are the units on \(\frac{\partial F}{\partial t}\) ? b. Write the partial derivative \(\frac{\partial F}{\partial r}\). What are the units on \(\frac{\partial F}{\partial r}\) ? c. How quickly will the value of the investment be changing with respect to time 5 years after the investment is made if the investment yields \(15 \%\) annually? d. Illustrate the answer to part \(c\).

For Activities 17 through \(22,\) write the first and second partial derivatives. $$ h(x, y)=e^{2 x-3 y} $$
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