Problem 76 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

The total number of acres of genetically modified crops grown worldwide from 1997 through 2003 was changing at the rate of $$ R(t)=2.718 t^{2}-19.86 t+50.18 \quad(0 \leq t \leq 6) $$ million acres/year. The total number of acres of such crops grown in 1997 ( \(t=0\) ) was \(27.2\) million acres. How many acres of genetically modified crops were grown worldwide in \(2003 ?\)

Step-by-Step Solution

Answer

In 2003, there were approximately 103.08 million acres of genetically modified crops grown worldwide.

Step 1: Integrate the rate function

In this step, we will find the anti-derivative of the given rate function, \(R(t)=2.718 t^{2}-19.86 t+50.18\) to get the total number of acres at any time t. To do this, integrate \(R(t)\) with respect to t: $$ A(t) = \int R(t) dt = \int (2.718 t^{2} - 19.86 t + 50.18) dt $$

Step 2: Evaluate the integral

Now, we will integrate each term in the function: $$ A(t) = \frac{2.718}{3} t^3 - \frac{19.86}{2} t^2 + 50.18t + C $$ Here, C is the constant of integration.

Step 3: Use the initial condition to find the constant of integration

We know that in 1997 (\(t=0\)), the number of acres of genetically modified crops was 27.2 million acres. Plugging this information into our function, we get: $$ 27.2 = \frac{2.718}{3} (0)^3 - \frac{19.86}{2} (0)^2 + 50.18(0) + C \Rightarrow C = 27.2 $$ Now our function for total number of acres looks like this: $$ A(t) = \frac{2.718}{3} t^3 - \frac{19.86}{2} t^2 + 50.18t + 27.2 $$

Step 4: Find the number of acres in 2003

To find the number of acres of genetically modified crops in 2003, we need to evaluate the function A(t) at \(t=6\): $$ A(6) = \frac{2.718}{3} (6)^3 - \frac{19.86}{2} (6)^2 + 50.18(6) + 27.2 $$ Calculate the value: $$ A(6) = 103.08 $$ So there were approximately 103.08 million acres of genetically modified crops grown worldwide in 2003.

Key Concepts

Integration

Integration is a fundamental concept in calculus. It allows us to find the area under the curve of a function or, in this case, to determine the cumulative quantity over time, such as acres of crops grown.

When we integrate a function, we reverse the process of differentiation. This means we find a function whose derivative will give us the function we started with. In the exercise, the function given is a rate of change, specifically how the number of acres of crops grows over time. Integration helps us find the total number of acres by determining the antiderivative of the rate function.

• Integration is often symbolized with the integral sign \(\int\).
• After integrating a function, we generally add a constant, known as the constant of integration, denoted as \(C\).

Essentially, integration accumulates quantities and helps solve real-world problems involving totals.

Rate of Change

The rate of change refers to how one quantity changes in relation to another. In this exercise, it describes how the number of acres of genetically modified crops changes per year. This rate function is given by \( R(t) = 2.718 t^{2} - 19.86 t + 50.18 \).

Understanding the rate of change is crucial because it informs how quickly or slowly the area of crops is expanding or contracting over time. In calculus, rates of change are analyzed using derivatives, while their cumulative effects are analyzed using integrals.

• The rate function can be seen as the slope or steepness of the function at any given point.
• Physically, it tells us how the function values (acres in this case) advance as the time value, \( t \), progresses.

By integrating this rate function, we obtain the total acreage at any particular point in time.

Initial Conditions

Initial conditions are values that define the starting point of a function. These values are vital in solving problems that involve integration because they help determine the constant of integration (\( C \)). Without initial conditions, we can't get the specific solution we need.

In this problem, the initial condition is given as the number of acres in 1997, which is 27.2 million. When integrating a function, this initial value allows us to solve for \( C \), fine-tuning our integrated function to match real-world conditions at \( t = 0 \).

• Initial conditions ground the mathematical model in reality, ensuring it matches known data points.
• They are often presented at \( t = 0 \), making calculations simpler.

Using initial conditions makes sure that our solutions to calculus problems are not just general but also practical and accurate.

Antiderivative

An antiderivative is a function that, when derived, gives the original function. In other words, if \( F(t) \) is the antiderivative of \( f(t) \), then the derivative of \( F(t) \) is \( f(t) \). Calculating the antiderivative effectively reverses the derivative process—it's how integration works.

For this problem, finding the antiderivative of the rate function \( R(t) \) results in the function \( A(t) \), which represents the total number of acres over time.

• The process involves integrating each term of the rate function separately, then adding the constant of integration.
• The general form of an antiderivative includes "+ C", representing all possible constants added to one particular solution.

Antiderivatives are essential for solving calculus problems that require determining total quantities from rate functions, such as the total acreage of a crop field over time.