Problem 6
Question
Bank Account The future value of a certain savings account with no activity besides compounding of interest is modeled as $$ F(t)=1500\left(1.0407^{t}\right) \text { dollars } $$ where \(t\) is the number of years since \(\$ 1500\) was invested. a. Numerically estimate to the nearest cent the rate of change of the future value when \(t=10 .\) b. Calculate the percentage rate of change of the future value when \(t=10\).
Step-by-Step Solution
Verified Answer
The rate of change at \( t = 10 \) is approximately \$88.71, and the percentage rate is approximately 3.98\%.
1Step 1: Differentiate the Function
To find the rate of change of the future value, we need to differentiate the function \( F(t) = 1500 \times 1.0407^t \) with respect to \( t \). The derivative of \( F(t) = a \times b^t \) is \( F'(t) = a \times \ln(b) \times b^t \). Here, \( a = 1500 \) and \( b = 1.0407 \). Thus, \( F'(t) = 1500 \times \ln(1.0407) \times 1.0407^t \).
2Step 2: Substitute t = 10
To find the numerical rate of change at \( t = 10 \), substitute \( t = 10 \) into the derivative. This gives \( F'(10) = 1500 \times \ln(1.0407) \times 1.0407^{10} \).
3Step 3: Calculate Numerical Rate of Change
Compute \( \ln(1.0407) \), and the exponential term \( 1.0407^{10} \). Next, calculate \( F'(10) \) using these values. Use a calculator for precise computation: \( \ln(1.0407) \approx 0.0398 \) and \( 1.0407^{10} \approx 1.4863 \). Then, \( F'(10) \approx 1500 \times 0.0398 \times 1.4863 \approx 88.71 \). Thus, the rate of change is approximately \$88.71 per year when \( t = 10 \).
4Step 4: Calculate Future Value at t = 10
To calculate the percentage rate of change, first find \( F(10) \) by substituting \( t = 10 \) into the original function: \( F(10) = 1500 \times 1.0407^{10} \). Compute this as \( F(10) \approx 1500 \times 1.4863 \approx 2229.45 \). Thus, the future value at \( t = 10 \) is approximately \$2229.45.
5Step 5: Calculate Percentage Rate of Change
The percentage rate of change is given by \( \text{percentage rate} = \frac{F'(10)}{F(10)} \times 100\% \). Substitute the known values: \( \text{percentage rate} = \frac{88.71}{2229.45} \times 100\% \approx 3.98\% \). Therefore, the percentage rate of change at \( t = 10 \) is approximately 3.98\%.
Key Concepts
Rate of ChangePercentage Rate of ChangeExponential Growth
Rate of Change
The concept of "rate of change" is essentially how fast a quantity increases or decreases over a certain period. In mathematical terms, it is often expressed as the derivative of a function. The derivative measures the instantaneous rate of change at any point of a function. Here, we are examining the rate of change of a bank account's future value over time, modeled by an exponential function.
In this case, the function is given as \( F(t) = 1500 \times 1.0407^t \). To find the rate of change of the savings account at year 10, we differentiate this function with respect to \( t \). The derivative \( F'(t) \) shows us how much \( F(t) \) changes as \( t \) changes. Differentiating an exponential function like this involves using the rule \( F'(t) = a \ln(b) b^t \), where \( a \) is the initial amount, and \( b \) is the base of the exponent.
After deriving, we substitute \( t = 10 \) into the derivative to find the numerical rate of change. This reflects the yearly increase in the account's balance at that specific point in time, which is approximately $88.71 per year. This means that after 10 years, the account's balance is growing by that much each year.
In this case, the function is given as \( F(t) = 1500 \times 1.0407^t \). To find the rate of change of the savings account at year 10, we differentiate this function with respect to \( t \). The derivative \( F'(t) \) shows us how much \( F(t) \) changes as \( t \) changes. Differentiating an exponential function like this involves using the rule \( F'(t) = a \ln(b) b^t \), where \( a \) is the initial amount, and \( b \) is the base of the exponent.
After deriving, we substitute \( t = 10 \) into the derivative to find the numerical rate of change. This reflects the yearly increase in the account's balance at that specific point in time, which is approximately $88.71 per year. This means that after 10 years, the account's balance is growing by that much each year.
Percentage Rate of Change
The percentage rate of change provides an understanding of a change relative to the size of the value at that point in time. It is calculated by dividing the rate of change by the original value, multiplied by 100 to convert it to a percentage.
First, we determine the rate of change at \( t = 10 \), which was found to be \( 88.71 \) dollars per year. Then, we find the future value at \( t = 10 \), \( F(10) \), by substituting into the original function to get approximately \( 2229.45 \) dollars.
To find the percentage, we use the formula:
This means that in the tenth year, the account grows by 3.98% relative to its size at that point. This rate provides a clearer picture of the growth's significance, illustrating the growth of the investment relative to its total value.
First, we determine the rate of change at \( t = 10 \), which was found to be \( 88.71 \) dollars per year. Then, we find the future value at \( t = 10 \), \( F(10) \), by substituting into the original function to get approximately \( 2229.45 \) dollars.
To find the percentage, we use the formula:
- \( \text{percentage rate} = \frac{F'(10)}{F(10)} \times 100\% \)
This means that in the tenth year, the account grows by 3.98% relative to its size at that point. This rate provides a clearer picture of the growth's significance, illustrating the growth of the investment relative to its total value.
Exponential Growth
Exponential growth describes a process where quantities increase by a consistent relative rate. This is often seen in scenarios where growth is compounded, like interest in bank accounts. In our example, the future value function \( F(t) = 1500 \times 1.0407^t \) models how an initial amount of \( \\(1500 \) grows over time with compound interest at 4.07% per annum.
Unlike linear growth, exponential growth doubles at regular intervals, becoming increasingly significant over longer periods. This is characterized by the base of the exponent \( 1.0407 \), which reveals that the account grows by 4.07% each year.
By ten years, the principal has grown to approximately \( \\)2229.45 \), significantly larger than the linear growth would provide. This modeling shows why starting to invest or save early is advantageous, as exponential growth can drastically increase the value over many years, even with a modest initial amount.
Unlike linear growth, exponential growth doubles at regular intervals, becoming increasingly significant over longer periods. This is characterized by the base of the exponent \( 1.0407 \), which reveals that the account grows by 4.07% each year.
By ten years, the principal has grown to approximately \( \\)2229.45 \), significantly larger than the linear growth would provide. This modeling shows why starting to invest or save early is advantageous, as exponential growth can drastically increase the value over many years, even with a modest initial amount.
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