Problem 17 - Applied Calculus

Question

The table gives the number of passenger cars, \(C=f(t)\) in millions, \(^{24}\) in the US in the year \(t\) (a) Do \(f^{\prime}(t)\) and \(f^{\prime \prime}(t)\) appear to be positive or negative during the period \(1975-1990 ?\) (b) Do \(f^{\prime}(t)\) and \(f^{\prime \prime}(t)\) appear to be positive or negative during the period \(1990-2000 ?\) (c) Estimate \(f^{\prime}(2005) .\) Using units, interpret your answer in terms of passenger cars. $$\begin{array}{c|c|c|c|c|c|c|c}\hline t & 1975 & 1980 & 1985 & 1990 & 1995 & 2000 & 2005 \\ \hline C & 106.7 & 121.6 & 127.9 & 133.7 & 128.4 & 133.6 & 136.6 \\\\\hline\end{array}$$

Step-by-Step Solution

Answer

(a) \(f'(t)\) positive, \(f''(t)\) negative; (b) \(f'(t)\) slightly positive, \(f''(t)\) negative; (c) \(f'(2005) \approx 0.6\) million cars per year.

Step 1: Understanding f'(t) for 1975-1990

To analyze if \(f'(t)\) is positive or negative during 1975-1990, observe the change in the number of passenger cars \(C\). From 1975 (106.7 million) to 1990 (133.7 million), the number of cars increases consistently. Therefore, \(f'(t)\) appears to be positive because the rate of change is positive.

Step 2: Analyzing f''(t) for 1975-1990

\(f''(t)\) refers to the acceleration or the change in the rate of \(f'(t)\). From 1975 to 1990, the increment from one period to the next decreases over time: 106.7 to 121.6 (14.9 increase), 121.6 to 127.9 (6.3 increase), 127.9 to 133.7 (5.8 increase). Since these increments are decreasing, \(f''(t)\) appears to be negative, indicating a slowing growth rate.

Step 3: Understanding f'(t) for 1990-2000

From 1990 to 2000, compare the number of passenger cars: 133.7 million in 1990 to 133.6 million in 2000 indicates a decrease and then a slight recovery, suggesting \(f'(t)\) first was negative but turns slightly positive near the end. Overall the net change suggests a nearly zero or slightly positive \(f'(t)\).

Step 4: Analyzing f''(t) for 1990-2000

\(f''(t)\) over 1990-2000 primarily depends on how \(f'(t)\) changes. Since \(f'(t)\), indicated by the change in \(C\), fluctuates with a decrease followed by a slight increase but overall stagnates, \(f''(t)\) appears to be negative, highlighting a non-uniform rate leading to stagnation.

Step 5: Estimating f'(2005)

To estimate \(f'(2005)\), look at changes from 2000 to 2005: 133.6 to 136.6 million cars. The increase is 3 million cars over 5 years. Thus, \(f'(2005)\) can be estimated as \(\frac{136.6 - 133.6}{2005 - 2000} = \frac{3}{5} = 0.6\) million cars per year.

Step 6: Interpretation of f'(2005)

The result \(0.6\) million cars per year indicates that during the year 2005, the number of passenger cars in the US was increasing at an approximate rate of 0.6 million cars per year.

Key Concepts

Rate of Change

When we talk about the rate of change in calculus, we are discussing how one quantity changes in relation to another. In the context of the exercise provided, this involves analyzing the number of passenger cars over different years and understanding how this number has changed over time. The rate of change is expressed through the derivative, noted as \( f'(t) \), which helps us understand this change.

- A positive rate of change means the quantity is increasing over time. For example, between 1975 and 1990, the number of cars increased from 106.7 million to 133.7 million, suggesting a positive rate of change, \( f'(t) > 0 \).- A negative rate of change implies the quantity is decreasing. From 1990 to 2000, however, we see a slight decrease in the number before stabilizing, making the rate of change less positive, or almost zero.

In real-world terms, understanding the rate of change helps analysts and policymakers make predictions about trends, plan for resources, or adjust strategies to accommodate changing conditions.

Derivatives

Derivatives are a fundamental concept in calculus that help in measuring how a function changes as its input changes. In simpler terms, it tells us how fast something is happening at a specific point. For our exercise, the derivative \( f'(t) \) represents the rate at which the number of passenger cars in the US is changing at any given year \( t \).

There are two main aspects covered:
- **First Derivative (\( f'(t) \)):** It gives the instantaneous rate of change of a function, which in our case means how quickly the number of cars is increasing or decreasing each year.
- **Second Derivative (\( f''(t) \)):** This provides information on the acceleration or deceleration concerning the rate of change. From 1975 to 1990, though \( f'(t) \) was positive, \( f''(t) \) was negative due to decreasing increments, indicating a slowdown in growth rate.

Understanding derivatives helps in multiple fields such as physics, engineering, and economics to determine trends, optimize processes, and enhance performance.

Data Analysis

Data analysis involves examining datasets to draw out meaningful information, patterns, or conclusions. In our passenger cars scenario, the process includes looking at given data across several decades and deciphering trends in the growth rate of car ownership.

Key steps include:- **Data Interpretation:** Assess numbers from different years to see how they evolve. For instance, examining car count changes from 1975 through 2005 using tools like derivatives helps to visualize growth patterns.
- **Trend Identification:** By calculating \( f'(t) \) and \( f''(t) \), analysts can identify whether there's consistent growth, decline, or stagnation.
- **Forecasting:** By estimating \( f'(2005) \), we predict future behavior, which helps in planning or decision-making for future infrastructure needs or market behavior.

Analyzing such data empowers businesses, governments, and researchers to make informed decisions by understanding historical trends and predicting future changes.