Problem 39
Question
Let \( P \) represent the percentage of a city's electrical power that is produced by solar panels \( t \) years after January 1, 2000. (a) What does \( dP/dt \) represent in this context? (b) Interpret the statement $$ \frac{dP}{dt} \bigg|_{t = 2} = 3.5 $$
Step-by-Step Solution
Verified Answer
(a) Rate of change of solar power percentage over time.
(b) Solar power use increases by 3.5 percentage points per year at \( t = 2 \).
1Step 1: Understanding the Derivative in Context
The expression \( \frac{dP}{dt} \) represents the derivative of the percentage \( P \) of the city's power that comes from solar panels with respect to time \( t \). In this context, it describes how quickly the percentage \( P \) is changing over time, specifically, the rate of change of solar power percentage as time progresses.
2Step 2: Specific Interpretation of the Derivative at a Given Time
For the statement \( \frac{dP}{dt} \bigg|_{t = 2} = 3.5 \), this means that 2 years after January 1, 2000, the percentage of the city's electric power generated by solar panels is increasing at a rate of 3.5 percentage points per year. This indicates a positive growth in the proportion of solar power usage by the city at that specific time.
Key Concepts
DerivativeRate of ChangeInterpretation of Derivatives
Derivative
The concept of a derivative in calculus is central to understanding how quantities change. In our exercise, the derivative is represented by \( \frac{dP}{dt} \). This notation tells us that we are looking at how one quantity, the percentage of electrical power from solar panels \( P \), is changing over another variable, time \( t \).
Simply put, the derivative offers a way to measure change. It answers the question: How does \( P \) change as \( t \) progresses? This allows us to see not only the amount but also the direction of change, whether \( P \) increases or decreases as time moves forward.
Derivatives have many practical uses beyond this scenario, such as determining the speed of a car (change in position over time) or finding out how much an economy grows over a quarter of the year. In conclusion, any time you hear the term "rate of change," you're dealing with a derivative.
Simply put, the derivative offers a way to measure change. It answers the question: How does \( P \) change as \( t \) progresses? This allows us to see not only the amount but also the direction of change, whether \( P \) increases or decreases as time moves forward.
Derivatives have many practical uses beyond this scenario, such as determining the speed of a car (change in position over time) or finding out how much an economy grows over a quarter of the year. In conclusion, any time you hear the term "rate of change," you're dealing with a derivative.
Rate of Change
The rate of change, as given by our derivative \( \frac{dP}{dt} \), is a crucial metric in analyzing dynamic situations. This rate indicates how much a certain variable, such as the percentage of solar power, changes over time.
In our specific context, when we refer to \( \frac{dP}{dt} \), we're asking: How quickly is the city's reliance on solar power increasing or decreasing each year? A positive rate means an increase; a negative rate points to a decrease.
Understanding the rate of change helps cities and planners to project future trends, make informed decisions about sustainable energy investments, and adjust to the growing or shrinking reliance on solar power.
In our specific context, when we refer to \( \frac{dP}{dt} \), we're asking: How quickly is the city's reliance on solar power increasing or decreasing each year? A positive rate means an increase; a negative rate points to a decrease.
Understanding the rate of change helps cities and planners to project future trends, make informed decisions about sustainable energy investments, and adjust to the growing or shrinking reliance on solar power.
- Positive rate: indicates growth or increase
- Negative rate: indicates reduction or decrease
- Zero rate: stable, no change
Interpretation of Derivatives
Interpreting derivatives, especially in specific contexts like ours, helps to draw meaningful conclusions about the situation under study. In the given exercise, the derivative \( \frac{dP}{dt} \bigg|_{t = 2} = 3.5 \) informs us that, two years after January 1, 2000, the percentage of electrical power from solar panels was increasing at a steady rate of 3.5% per year.
This is indicative of positive growth in solar power reliance, showing successful adoption and integration of solar technology by the city up to that time.
Such an interpretation is essential for policymakers and engineers as it reflects the effectiveness of current strategies and the need to either maintain or adjust efforts. It's all about understanding the implications of these changes over time to drive good decision-making.
This is indicative of positive growth in solar power reliance, showing successful adoption and integration of solar technology by the city up to that time.
Such an interpretation is essential for policymakers and engineers as it reflects the effectiveness of current strategies and the need to either maintain or adjust efforts. It's all about understanding the implications of these changes over time to drive good decision-making.
- Growth rate provides insight into technological adoption
- Helps in assessing effectiveness of policies
- Guides future planning and investments
Other exercises in this chapter
Problem 38
If \( 2x \le g(x) \le x^4 - x^2 +2 \) for all \( x \), evaluate \( \displaystyle \lim_{x \to 1}g(x) \).
View solution Problem 38
Determine the infinite limit. \( \displaystyle \lim_{x \to \pi^-}\cot x \)
View solution Problem 39
Each limit represents the derivative of some function \( f \) at some number \( a \). State such an \( f \) and \( a \) in each case. \( \displaystyle \lim_{x \
View solution Problem 39
Find the limit or show that it does not exist. \( \displaystyle \lim_{x \to \infty} (e^{-2x}\cos x) \)
View solution