Question

A \(\$ 200\) investment in a savings account grows according to \(A(t)=200 e^{0.0398 t}\), for \(t \geq 0,\) where \(t\) is measured in years. a. Find the balance of the account after 10 years. b. How fast is the account growing (in dollars/year) at \(t=10 ?\) c. Use your answers to parts (a) and (b) to write the equation of the line tangent to the curve \(A=200 e^{0.0398 t}\) at the point \((10, A(10))\)

Short answer

Answer: After 10 years, the balance is approximately $296.49. The equation of the tangent line representing the growth rate at that time is \(A-296.49 = 11.79(t-10)\).

Step-by-step solution

Calculate A(10)

To find the balance after 10 years, simply plug in \(t=10\) into the equation: \(A(10) = 200e^{0.0398(10)}\) Step 2: Evaluate the expression

Evaluate A(10)

Now, just calculate the value of the expression: \(A(10) = 200e^{0.398} \approx \$296.49\) Step 3: Find the growth rate (derivative)

Derivative of A(t)

To find the rate at which the account is growing, we need the derivative of \(A(t)\) concerning \(t\). Differentiate the equation with respect to \(t\): \(\frac{dA}{dt} = 200(0.0398)e^{0.0398t}\) Step 4: Calculate the growth rate at \(t=10\)

Find the growth rate at t=10

Plug in \(t=10\) into the derivative equation: \(\frac{dA}{dt}(10)=200(0.0398)e^{0.0398 \cdot 10} \approx \$11.79\) Step 5: Write the equation of the tangent line

Equation of the tangent line

Using the point-slope form of a line with the slope (\(\frac{dA}{dt}(10)\)) and the point \((10, A(10))\), write the equation of the tangent line at the point \((10, A(10))\): \(A-A(10) = \frac{dA}{dt}(10)(t-10)\) \(A-296.49 = 11.79(t-10)\)

Key concepts

Derivative Calculus

The concept of derivatives is fundamental in calculus, often thought of as the mathematical equivalent of finding the rate at which things change. When dealing with functions, the derivative at a certain point essentially provides us with the rate of change or the slope of the function at that point.

In our context, we are interested in the growth rate of an investment over time, which can be understood through the derivative of the exponential growth function. The derivative of an exponential function like \( A(t) = Pe^{rt} \) is \( \frac{dA}{dt} = Pr e^{rt} \), where \( P \) represents the principal amount, \( r \) is the growth rate, and \( t \) is time. In case of our exercise, \( P = 200 \) and \( r = 0.0398 \), making the derivative \( \frac{dA}{dt} = 200 \times 0.0398 \times e^{0.0398t} \), which shows how the account's balance changes over time.

Tangent Line Equation

A tangent line to a curve at a given point is a straight line that just 'touches' the curve at that point. This line represents the instantaneous rate of change of the function at that particular point, and its slope is given by the derivative of the function at that point.

To write the equation of this line, we use the point-slope form: \( y - y_1 = m(x - x_1) \) where \( m \) is the slope of the tangent, and \( (x_1, y_1) \) is the point of tangency. From our problem's Step 5, with \( A(10) \) being \( y_1 \) and the derivative at \( t=10 \) being \( m \), the equation of the tangent line is given by \( A - A(10) = \frac{dA}{dt}(10)(t - 10) \) which yields the linear equation \( A - 296.49 = 11.79(t - 10) \). This equation lets us predict the account balance at a moment close to \( t = 10 \) years.

Exponential Functions

Exponential functions are powerful tools for modeling situations where growth or decay is not constant, but rather changes at a rate proportional to the current amount. The general form of an exponential function is \( A(t) = Pe^{rt} \), where \( A(t) \) is the amount at time \( t \) and \( e \) is the base of the natural logarithm, approximately equal to 2.71828.

These functions are characterized by their 'J-shaped' curve, which represents rapid growth or decay. In our investment example, the function \( A(t) = 200e^{0.0398t} \) indicates that the investment grows exponentially over time. The base \( e \), raised to the power of the product of the growth rate \( r \) and time \( t \) dictates the shape of the growth. After 10 years, the balance \( A(10) \) reflects this exponential growth pattern.

Interest Rate Problems

Interest rate problems often involve calculating the future value of an investment or loan based on compound interest. The key to solving these problems is to either use the compound interest formula directly or to understand the exponential function that models the situation.

For instance, the exponential function in our example \( A(t) = 200e^{0.0398t} \) shows compound interest at play. The \( 0.0398 \) in the exponent represents the annual interest rate (3.98% in this case), and \( e \) indicates that we are dealing with continuously compounded interest. The compound interest growth is essential for accurately predicting how much money will be in the account after a certain number of years, which is a common type of interest rate problem encountered in finance and economics.