Problem 44 - College Algebra and Calculus: An Applied Approach

Question

The balance \(A\) (in dollars) in a savings account is given by \(A=5000 e^{0.08 t}\), where \(t\) is measured in years. Find the rates at which the balance is changing when (a) \(t=1\) year, (b) \(t=10\) years, and (c) \(t=50\) years.

Step-by-Step Solution

Answer

The rates at which the balance is changing are \(400 e^{0.08}\) dollars per year, \(400 e^{0.8}\) dollars per year, and \(400 e^{4}\) dollars per year when \(t=1\), \(t=10\), and \(t=50\) years respectively.

Step 1: Identify the function and the variable

The function given is \(A=5000 e^{0.08 t}\) and the variable is \(t\), which represents time.

Step 2: Compute the derivative

The derivative of the function with respect to \(t\) gives us the rate of change of the balance \(A\) over time. By applying the chain rule of differentiation to \(e^{rt}\), where \(r\) is a constant, we get its derivative to be \(r e^{rt}\). Therefore, the derivative of \(A\) with respect to \(t\) is \(\frac{dA}{dt}=5000 \cdot 0.08 e^{0.08 t}\).

Step 3: Evaluate the derivative at \(t=1\) year

By substituting \(t=1\) into the derivative, we get \(\frac{dA}{dt}=5000 \cdot 0.08 \cdot e^{0.08 \cdot 1}=400 e^{0.08}\) dollars per year.

Step 4: Evaluate the derivative at \(t=10\) years

By substituting \(t=10\) into the derivative, we get \(\frac{dA}{dt}=5000 \cdot 0.08 \cdot e^{0.08 \cdot 10}=400 e^{0.8}\) dollars per year.

Step 5: Evaluate the derivative at \(t=50\) years

By substituting \(t=50\) into the derivative, we get \(\frac{dA}{dt}=5000 \cdot 0.08 \cdot e^{0.08 \cdot 50}=400 e^{4}\) dollars per year.

Key Concepts

Derivative of Exponential Functions

Understanding the derivative of exponential functions is crucial when analyzing the growth or decay processes in various disciplines. Exponential functions are in the form of \( f(x) = a e^{kx} \), where \( e \) is Euler's number (approximately 2.71828), \( a \) is a constant that represents the initial value and \( k \) is the growth or decay rate.

When differentiating exponential functions, we apply the rule which states that the derivative of \( a e^{kx} \) with respect to \( x \) is \( ak e^{kx} \). This property allows us to predict how rapidly quantities grow or decrease over time. For example, in our exercise, we calculated the rate at which the balance of a savings account changed, which involved the differentiation of an exponential function.

Rate of Change

The rate of change is a measure that describes how one quantity changes in relation to another. When we talk about the rate of change in the context of calculus, we are often referring to the derivative of a function at a particular point, which represents the instantaneous rate of change at that point.

In the context of our savings account example, the rate at which the balance increases is found by differentiating the balance formula with respect to time. The resulting derivative provides us with the speed of the account's growth at any moment in time. Computing the derivative for different values of \( t \) reveals the rate of change at specific times, such as 1 year, 10 years, or 50 years.

Chain Rule of Differentiation

The chain rule is a fundamental technique in calculus used for finding the derivative of composite functions. It states that the derivative of a composite function is the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

In our exercise, the savings account balance is a composite function where an exponential function is multiplied by a constant. Using the chain rule, we find the derivative by keeping the constant as it is and differentiating the exponential part, then multiplying the two. Understanding the chain rule is essential for differentiation, especially when functions are more complex than simple polynomials or basic exponentials.

Applied Mathematical Modeling

Applied mathematical modeling involves creating mathematical representations of real-world situations to make predictions or gain insights into complex phenomena. In our savings account scenario, we used a mathematical model to represent the balance over time. This type of modeling often relies on understanding the underlying physical, financial, or biological processes to construct an accurate model.

Once a model is established, tools like calculus allow us to analyze the model and make predictions. In this case, by differentiating the balance equation, we predicted the rate at which money will accumulate in the account over time. Applied mathematical modeling thus plays a critical role in decision-making, forecasting, and optimizing outcomes in various fields.