Problem 37
Question
If \(V\) dollars is the present value of an ordinary annuity of equal payments of \(\$ 100\) per year for \(t\) years at an interest rate of \(100 i\) percent per year, then $$ V=100\left[\frac{1-(1+i)^{-t}}{i}\right] $$ (a) Find the instantaneous rate of change of \(V\) per unit change in \(i\) if \(t\) remains fixed at 8. (b) Use the result of part (a) to find the approximate change in the present value if the interest rate changes from \(6 \%\) to \(7 \%\) and the time remains fixed at 8 years. (c) Find the instantaneous rate of change of \(V\) per unit change in \(t\) if \(i\) remains fixed at \(0.06\). (d) Use the result of part (c) to find the approximate change in the present value if the time is decreased from 8 to 7 years and the interest rate remains fixed at \(6 \%\).
Step-by-Step Solution
Verified Answer
The instantaneous rate of change of V with respect to i is found using the quotient rule and simplification. The approximate change in V from 6% to 7% interest rate and from 8 to 7 years is calculated using the rate of changes.
1Step 1: Find the expression for V
Given the formula for the present value of an ordinary annuity: \( V = 100 \left[ \frac{1 - (1+i)^{-t}}{i} \right] \)
2Step 2: Find the derivative of V with respect to i
To find the instantaneous rate of change of \( V \) with respect to \( i \) when \( t \) is fixed at 8, we need to compute the partial derivative \( \frac{\partial V}{\partial i} \). Use the quotient rule: \( \frac{\partial}{\partial i} \left( \frac{1 - (1+i)^{-t}}{i} \right) = \frac{(i \cdot \frac{d}{di}[1 - (1+i)^{-t}] - [1 - (1+i)^{-t}] \cdot 1)}{i^2} \)
3Step 3: Compute the derivative inside the numerator
The derivative of \( 1 - (1+i)^{-t} \) with respect to \( i \) is \( t(1+i)^{-t-1} \), therefore: \( \frac{\partial}{\partial i} \left( \frac{1 - (1+i)^{-t}}{i} \right) = \frac{i \cdot t (1+i)^{-t-1} - [1 - (1+i)^{-t}] }{i^2} \)
4Step 4: Simplify the expression
Combine and simplify the expression: \( \frac{i t (1+i)^{-t-1} - 1 + (1+i)^{-t}}{i^2} \) This can be written as: \( \frac{t(1+i)^{-t-1}}{i} - \frac{1 - (1+i)^{-t}}{i^2} \) Therefore, \( \frac{\partial V}{\partial i} = 100 \left( \frac{t(1+i)^{-t-1}}{i} - \frac{1 - (1+i)^{-t}}{i^2} \right) \)
5Step 5: Determine the instantaneous rate of change for i = 0.06
Substitute \( i = 0.06 \) and \( t = 8 \) into the derivative to find the rate of change.
6Step 6: Approximate change in V due to change in i
Using the rate of change found in step 5, approximate the change in \( V \) as \( \Delta V \approx \frac{\partial V}{\partial i} \times \Delta i \) where \( \Delta i = 0.01 \).
7Step 7: Find the derivative of V with respect to t
Compute the partial derivative \( \frac{\partial V}{\partial t} \) while keeping \( i \) fixed at 0.06. Use chain rule: \( \frac{\partial V}{\partial t} = 100 \left( \frac{(1+i)^{-t} \cdot (-\ln(1+i))}{i} \right) \)
8Step 8: Simplify the expression for the derivative
The expression simplifies to: \( \frac{\partial V}{\partial t} = -100 \left( \frac{(1+i)^{-t} \cdot \ln(1+i)}{i} \right) \)
9Step 9: Determine the instantaneous rate of change for t = 8
Substitute \( i = 0.06 \) to find the rate of change of \( V \) with respect to \( t \).
10Step 10: Approximate change in V due to change in t
Using the rate of change found in step 9, approximate the change in \( V \) as \( \Delta V \approx \frac{\partial V}{\partial t} \times \Delta t \) where \( \Delta t = -1 \).
Key Concepts
Instantaneous Rate of ChangePartial DerivativesQuotient RuleApproximate Change in Value
Instantaneous Rate of Change
The instantaneous rate of change measures how a quantity changes at a specific point, rather than over an interval. It is essentially the derivative of a function at a given point.
The exercise asks us to find the instantaneous rate of change of the present value, V, with respect to the interest rate, i, while keeping the number of years, t, constant at 8.
In this case, we need to calculate the partial derivative of V with respect to i, denoted as \( \frac{\partial V}{\partial i} \). This helps understand how a small change in the interest rate impacts the present value.
By using the quotient rule, the instantaneous rate of change for this scenario was derived as:
\( \frac{\partial V}{\partial i} = 100 \left( \frac{t(1+i)^{-t-1}}{i} - \frac{1 - (1+i)^{-t}}{i^2} \right) \)
Substituting \( t = 8 \) and \( i = 0.06 \), allows us to find the specific rate of change at this point.
The exercise asks us to find the instantaneous rate of change of the present value, V, with respect to the interest rate, i, while keeping the number of years, t, constant at 8.
In this case, we need to calculate the partial derivative of V with respect to i, denoted as \( \frac{\partial V}{\partial i} \). This helps understand how a small change in the interest rate impacts the present value.
By using the quotient rule, the instantaneous rate of change for this scenario was derived as:
\( \frac{\partial V}{\partial i} = 100 \left( \frac{t(1+i)^{-t-1}}{i} - \frac{1 - (1+i)^{-t}}{i^2} \right) \)
Substituting \( t = 8 \) and \( i = 0.06 \), allows us to find the specific rate of change at this point.
Partial Derivatives
Partial derivatives are used when dealing with functions of multiple variables. They measure how the function changes with respect to one variable while keeping others constant.
In the exercise, V is a function of both i (interest rate) and t (time in years). To find the partial derivative of V with respect to i, with t fixed, we apply the quotient rule.
Given function:
\( V = 100 \left[ \frac{1 - (1+i)^{-t}}{i} \right] \)
To calculate \( \frac{\partial V}{\partial i} \), we use:
\( \frac{\partial}{\partial i} \left( \frac{1 - (1+i)^{-t}}{i} \right) = \frac{(i \cdot \frac{d}{di}[1 - (1+i)^{-t}] - [1 - (1+i)^{-t}] \cdot 1)}{i^2} \)
Then, simplifying gives us the instantaneous rate of change as discussed in the previous section.
In the exercise, V is a function of both i (interest rate) and t (time in years). To find the partial derivative of V with respect to i, with t fixed, we apply the quotient rule.
Given function:
\( V = 100 \left[ \frac{1 - (1+i)^{-t}}{i} \right] \)
To calculate \( \frac{\partial V}{\partial i} \), we use:
\( \frac{\partial}{\partial i} \left( \frac{1 - (1+i)^{-t}}{i} \right) = \frac{(i \cdot \frac{d}{di}[1 - (1+i)^{-t}] - [1 - (1+i)^{-t}] \cdot 1)}{i^2} \)
Then, simplifying gives us the instantaneous rate of change as discussed in the previous section.
Quotient Rule
The quotient rule is used to find the derivative of a fraction of two functions. If you have a function of the form \( \frac{u(i)}{v(i)} \), where both u and v are functions of i, the quotient rule states:
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
In our exercise, the present value function is a ratio involving interest rate (i). Specifically:
\[ V = 100 \left( \frac{1 - (1+i)^{-t}}{i} \right) \]
Here, we apply the quotient rule to find \( \frac{\partial V}{\partial i} \), where\[ u = 1 - (1+i)^{-t} \]
and\[ v = i \]
The quotient rule helps us manage the complexity of the relationship between V and i in the annuity formula.
\[ \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \]
In our exercise, the present value function is a ratio involving interest rate (i). Specifically:
\[ V = 100 \left( \frac{1 - (1+i)^{-t}}{i} \right) \]
Here, we apply the quotient rule to find \( \frac{\partial V}{\partial i} \), where\[ u = 1 - (1+i)^{-t} \]
and\[ v = i \]
The quotient rule helps us manage the complexity of the relationship between V and i in the annuity formula.
Approximate Change in Value
The approximate change in value is useful when we want to estimate how a small change in one variable affects a function's output.
In the exercise, we used the result of the partial derivative of V with respect to i to find the approximate change in V when the interest rate changes from 6% to 7%.
We apply the differential approximation formula:
\( \Delta V \approx \frac{\partial V}{\partial i} \times \Delta i \)
For our specific values, a change of interest rate by 1% (\( \Delta i = 0.01 \)) results in:
\( \Delta V \approx \frac{\partial V}{\partial i} \times 0.01 \)
Similarly, for the change in years from 8 to 7, the approximate change in V was determined using \( \Delta V \approx \frac{\partial V}{\partial t} \times \Delta t \).
In both cases, this approximation provides a linear estimate of how V changes in response to small variations in i or t.
In the exercise, we used the result of the partial derivative of V with respect to i to find the approximate change in V when the interest rate changes from 6% to 7%.
We apply the differential approximation formula:
\( \Delta V \approx \frac{\partial V}{\partial i} \times \Delta i \)
For our specific values, a change of interest rate by 1% (\( \Delta i = 0.01 \)) results in:
\( \Delta V \approx \frac{\partial V}{\partial i} \times 0.01 \)
Similarly, for the change in years from 8 to 7, the approximate change in V was determined using \( \Delta V \approx \frac{\partial V}{\partial t} \times \Delta t \).
In both cases, this approximation provides a linear estimate of how V changes in response to small variations in i or t.
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