Problem 54 - Applied Calculus

Question

A biotechnology investment, originally worth \(\$ 20,000\), grows continuously at the rate of \(1000 e^{0.10 t}\) dollars per year, where \(t\) is the number of years since the investment was made. a. Find a formula for the value of the investment after years. b. Use your formula to find the value of the investment after 7 years.

Step-by-Step Solution

Answer

a) The formula is \( V(t) = 20000 + 10000(e^{0.10t} - 1) \). b) After 7 years, it's \$30,137.5.

Step 1: Understanding Continuous Growth

The formula for continuously growing investment is given by \( A(t) = A_0 imes e^{rt} \), where \( A_0 \) is the initial amount, \( r \) is the rate of growth, and \( t \) is the time in years. Here, the growth contribution per year is directly given by \( 1000 e^{0.10 t} \).

Step 2: Set Initial Investment Value

The initial investment \( A_0 \) is \$20,000. Since the growth per year is already given by a separate expression \( 1000 e^{0.10 t} \), we will incorporate it into the continuous growth formula. The total value after \( t \) years becomes the sum of the initial investment and the integral of the annual growth.

Step 3: Derive the Formula for Investment Value

To find the total value after \( t \) years, we integrate the growth rate: \( \int_0^t 1000 e^{0.10 x} \, dx \). This integration gives the additional growth on the investment over time. Evaluating this integral, we find \( 1000 \times [10 e^{0.10 x}]_0^t = 1000 \times (10e^{0.10t} - 10) = 10000(e^{0.10t} - 1) \). Adding this to the original investment, the formula becomes \( V(t) = 20000 + 10000(e^{0.10t} - 1) \).

Step 4: Evaluate the Investment Value After 7 Years

Plug \( t = 7 \) into the function \( V(t) = 20000 + 10000(e^{0.10t} - 1) \). Calculating \( e^{0.10 \times 7} \approx 2.01375 \), we substitute it to find \( V(7) = 20000 + 10000(2.01375 - 1) \). The investment value after 7 years thus is \( 20000 + 10000 \times 1.01375 = 30137.5 \).

Key Concepts

Exponential Growth

Exponential growth is a fascinating and powerful concept that describes a process where the quantity increases continuously at a constant rate relative to the current amount. Unlike linear growth, where things increase steadily over time, exponential growth rapidly accelerates.

This is best understood through the formula:

• \( A(t) = A_0 \times e^{rt} \)

Here, \( A_0 \) is the initial amount, \( e \) is the base of natural logarithms, \( r \) is the growth rate, and \( t \) is the time elapsed.

In our example, the growth in the investment is modeled by the expression \( 1000 e^{0.10 t} \). This indicates that every year, the amount by which the investment grows multiplies exponentially, showcasing how small changes in time can yield significant increases in value.

Investment Calculation

Calculating investment growth over time involves understanding how contributions and compound growth accumulate. In our example, the investment starts at \( \$20,000 \) and experiences continuous growth based on a function \( 1000 e^{0.10 t} \).

To determine the future value of this investment, we don't just look at where it started, but also at the additional value generated each year. This involves integrating the growth function over the period of investment.

Here, we compute the integral of \( 1000 e^{0.10 x} \) to find out how much the investment's value increases over time. Finally, this integrated growth is added to the initial investment to yield the total future value. This approach effectively captures both the original value and the subsequent growth, reflecting the real-world dynamics of investments.

Integration in Calculus

Integration is a fundamental concept in calculus that helps us find the area under a curve, which in turn can represent accumulated quantities like growth over time. In this exercise, we used integration to calculate the total accrued growth of an investment.

Why Integration?
When faced with a variable rate of growth that changes continuously, integration becomes crucial. It allows us to sum up tiny contributions of growth over an interval, generating a comprehensive picture of the overall increase.

In the problem, we integrated the function \( 1000 e^{0.10 x} \) over the time \( t \) to derive the formula for additional growth on top of the initial investment. The result of this integration forms a component of the investment calculation:

• \( \int_0^t 1000 e^{0.10 x} \, dx = 10000(e^{0.10t} - 1) \)

This represents the total growth over the time interval considered, highlighting the power of integration in solving real-life problems where quantities change dynamically over time.