Problem 78
Question
Compound interest. If \(\$ 1000\) is invested at interest rate \(r\) compounded annually, in 3 yr it will grow to an amount A given by (see Section R.1) $$ A=\$ 1000(1+r)^{3} $$ a) Find the rate of change, \(d A / d r\) b) Interpret the meaning of \(d A / d r\).
Step-by-Step Solution
Verified Answer
\(dA/dr = 3000(1 + r)^2\); it shows how the amount changes as the interest rate changes.
1Step 1: Identify the Problem Type
We are dealing with an exercise that includes finding the derivative of a function related to compound interest. Specifically, we want to find the rate of change of the amount with respect to the interest rate.
2Step 2: Write Down the Formula
The amount after 3 years, given an investment of \( \$1000 \) and an interest rate \( r \), is given by the formula:\[ A = 1000(1 + r)^3 \].
3Step 3: Differentiate the Formula with Respect to r
To find the rate of change, we need to differentiate the function \( A = 1000(1 + r)^3 \) with respect to \( r \). Use the chain rule for differentiation:\[ \frac{dA}{dr} = 1000 \cdot 3(1 + r)^2 \cdot (1) \].
4Step 4: Simplify the Derivative
Simplify the expression obtained from differentiation:\[ \frac{dA}{dr} = 3000(1 + r)^2 \].
5Step 5: Interpret the Derivative
The derivative \( \frac{dA}{dr} = 3000(1 + r)^2 \) represents the rate of change of the amount \( A \) with respect to the interest rate \( r \). This tells us how much the total amount increases for a small increase in \( r \).
Key Concepts
DifferentiationRate of ChangeMathematical Interpretation
Differentiation
Differentiation in calculus is all about finding how things change. When dealing with functions, differentiation helps you determine how a small change in an input affects the output. In the case of our exercise, the input is the interest rate \(r\), and the output is the total amount \(A\) after 3 years. Differentiation lets us find the derivative \( \frac{dA}{dr} \), which shows us exactly how the amount \(A\) changes when \(r\) changes slightly.
To differentiate the function \(A = 1000(1 + r)^3\) with respect to \(r\), you apply the chain rule. The chain rule is a crucial technique in calculus that allows you to differentiate composite functions. Here, the chain rule was used to see the change in the formula as a whole in response to a change in \(r\). The result is:
To differentiate the function \(A = 1000(1 + r)^3\) with respect to \(r\), you apply the chain rule. The chain rule is a crucial technique in calculus that allows you to differentiate composite functions. Here, the chain rule was used to see the change in the formula as a whole in response to a change in \(r\). The result is:
- The derivative \( \frac{dA}{dr} = 3000(1 + r)^2 \).
Rate of Change
Rate of change is a concept that tells us how one quantity changes in relation to another. In our scenario, it informs us how the amount \(A\) increases as the interest rate \(r\) incrementally rises. This is distinct from calculating the exact new value of \(A\), focusing instead on the variation's pace.
For our compound interest problem, the rate of change is given by the derivative \( \frac{dA}{dr} \). This derivative, \(3000(1 + r)^2\), expresses how swiftly the future amount \(A\) accumulates in response to a rate boost. Simply put, it's a snapshot of the formula's responsiveness.
The larger the derivative value, the faster \(A\) will grow as \(r\) rises. Conversely, a smaller value means the future amount grows more slowly with similar increases. Understanding this rate of change can be vital in financial contexts like investment, where predicting future returns based on slight interest changes is key.
For our compound interest problem, the rate of change is given by the derivative \( \frac{dA}{dr} \). This derivative, \(3000(1 + r)^2\), expresses how swiftly the future amount \(A\) accumulates in response to a rate boost. Simply put, it's a snapshot of the formula's responsiveness.
The larger the derivative value, the faster \(A\) will grow as \(r\) rises. Conversely, a smaller value means the future amount grows more slowly with similar increases. Understanding this rate of change can be vital in financial contexts like investment, where predicting future returns based on slight interest changes is key.
Mathematical Interpretation
Mathematical interpretation involves translating the formulas and symbols into real-world meanings. For this exercise, it means understanding what the derivative \( \frac{dA}{dr} = 3000(1 + r)^2 \) practically tells us.
This function is a mathematical expression of the financial concept: how compound interest sensitivity alters with changes in interest rate \(r\). When \(r\) increases by a bit, \(A\) increases significantly more when \(r\) is near higher values. This is because the term \((1+r)^2\) gets larger with increasing \(r\).
So, when the interest rate is higher, every incremental increase in \(r\) leads to a larger bump in \(A\). This can be crucial information for investors who are trying to understand how slight changes in interest rates affect their long-term gains.
Understanding this interpretation helps make smarter financial decisions, reinforcing how closely linked mathematics and real-world applications are.
This function is a mathematical expression of the financial concept: how compound interest sensitivity alters with changes in interest rate \(r\). When \(r\) increases by a bit, \(A\) increases significantly more when \(r\) is near higher values. This is because the term \((1+r)^2\) gets larger with increasing \(r\).
So, when the interest rate is higher, every incremental increase in \(r\) leads to a larger bump in \(A\). This can be crucial information for investors who are trying to understand how slight changes in interest rates affect their long-term gains.
Understanding this interpretation helps make smarter financial decisions, reinforcing how closely linked mathematics and real-world applications are.
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Problem 78
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