Problem 37 If \(V\) dollars is the present ... [FREE SOLUTION] (2024)

Short Answer

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.

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.

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.

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.