Problem 28 - Applied Calculus

Question

World meat \(^{12}\) production, \(M=f(t)\), in millions of metric tons, is a function of \(t\), years since 2000 . (a) Interpret \(f(5)=249\) and \(f^{\prime}(5)=6.5\) in terms of meat production. (b) Estimate \(f(10)\) and interpret it in terms of meat production.

Step-by-Step Solution

Answer

(a) In 2005, production was 249 million tons, increasing at 6.5 million tons/year. (b) Estimated production in 2010 is 281.5 million tons.

Step 1: Understand f(t)

Consider the function \(M=f(t)\), where \(f(t)\) represents the meat production in millions of metric tons \(t\) years after the year 2000. So when interpreting \(f(t)\), we need to consider \(t\) as years after 2000.

Step 2: Interpret f(5)=249

Given \(f(5)=249\), this means that in the year 2005, the meat production was 249 million metric tons. This provides us with a specific data point for the meat production 5 years after the year 2000.

Step 3: Interpret f'(5)=6.5

The derivative \(f'(5)=6.5\) signifies that in 2005, the rate of change of meat production was 6.5 million metric tons per year. In other words, in the year 2005, the production of meat was increasing by 6.5 million metric tons each year.

Step 4: Estimate f(10) Using Linear Approximation

To estimate \(f(10)\), we use the linear approximation method which involves using the known value of \(f(5)\) and the rate of change at \(t=5\) (i.e., \(f'(5)\)). The formula for linear approximation is \(f(a) + f'(a)\times(t-a)\). Here, we use \(f(5)\) and estimate \(f(10) = f(5) + f'(5)\times(10-5)\).

Step 5: Calculating f(10)

Using the linear approximation formula: \[f(10) = f(5) + f'(5) \times (10-5) = 249 + 6.5 \times 5 = 249 + 32.5 = 281.5\]. Thus, the estimated meat production in 2010 is 281.5 million metric tons.

Step 6: Interpret f(10) in Terms of Meat Production

The approximation \(f(10) = 281.5\) means that, according to the linear projection, the world meat production is estimated to be 281.5 million metric tons in the year 2010. This is based on the rate of change observed in 2005.

Key Concepts

Interpretation of Derivatives

In calculus, understanding the derivative of a function is a key skill. It gives insight into how a function behaves at a particular point.
In the context of the world meat production problem, the derivative is noted as \(f'(t)\). This represents the rate at which meat production is changing at a specific time \(t\).

• For \(f'(5) = 6.5\), it indicates that in the year 2005, the production of meat was increasing at a rate of 6.5 million metric tons per year. This tells us not just about the production amount but also about the speed of its growth.
• The positive value of \(6.5\) reflects an upward trend in meat production, meaning there was a consistent increase in production.

Understanding these nuances helps in predicting and strategizing future production changes.

Linear Approximation

Linear approximation is a method used in calculus to estimate the value of a function near a certain point using its derivative.
The basic idea is to approximate a function with a linear function (a straight line) that closely follows the curve of the original function at a point.

• In the meat production scenario, we use linear approximation to estimate \(f(10)\), representing production in 2010. We start with the known value \(f(5)=249\) and use the derivative \(f'(5)=6.5\) as our slope.
• The formula \(f(a) + f'(a)\times(t-a)\) was used, meaning we take the initial value of the function at \( t = 5 \), and add the change in production over 5 years, \(6.5 \times 5 = 32.5\), to get \(f(10) = 281.5\).

This approach might not be accurate over long periods, but it's excellent for short-term estimates.

Function Interpretation

Interpreting a function involves understanding both its numerical values and what those values actually represent in real terms.
For the exercise provided, \(f(t)\) is a real-life representation of world meat production, making it vital to connect these numerical values to the actual scenario.

• When we say \(f(5) = 249\), it translates to 249 million metric tons of meat were produced in the year 2005.
• This function gives us concrete data points to analyze trends and inform decisions regarding meat production policies, forecasts, or investments.

Thus, interpreting functions in this manner aids in making informed decisions based on mathematical insights.