Problem 40 - Calculus

Question

The function \(P(d)\) gives the total electricity, in \(\mathrm{kWh}\), that a solar array has generated between the start of the year and the end of the \(d^{\text {th }}\) day of the year. For each statement below, give a mathematical equation in terms of \(P,\) its inverse, or derivatives. (a) The array had generated 3500 kWh of electricity by the end of January 4. (b) At the end of January \(4,\) the array was generating electricity at a rate of \(1000 \mathrm{kWh}\) per day. (c) When the array had generated 5000 kWh of electricity, it took approximately half a day to generate an additional \(1000 \mathrm{kWh}\) of electricity. (d) At the end of January 30 , it took approximately one day to generate an additional 2500 kWh of electricity.

Step-by-Step Solution

Answer

(a) \( P(4) = 3500 \), (b) \( P'(4) = 1000 \), (c) \( P(d + 0.5) - P(d) = 1000 \), (d) \( P(31) - P(30) = 2500 \).

Step 1: Understand the Function P(d)

The function \( P(d) \) represents the total electricity generated by the solar array by the end of the \( d^{\text{th}} \) day. This means \( P(d) \) is a cumulative total, with \( d \) referring to a specific day of the year.

Step 2: Analyze Statement (a)

For statement (a), we need an equation representing the electricity generated by January 4. Since January 4 is the 4th day, the equation is \( P(4) = 3500 \).

Step 3: Analyze Statement (b)

Statement (b) refers to the rate of electricity generation at the end of January 4, which is a derivative of \( P(d) \) with respect to \( d \). Thus, the equation is \( P'(4) = 1000 \).

Step 4: Analyze Statement (c)

This statement involves the inverse of \( P(d) \), say \( P^{-1}(x) \), which indicates at which day \( x \) kWh was generated. The condition states that when \( P(d) = 5000 \), half a day later \( P(d+0.5) = 6000 \). Thus, \( P(d + 0.5) - P(d) = 1000 \) when \( P(d) = 5000 \).

Step 5: Analyze Statement (d)

Similar to step 4, statement (d) also deals with time taken to generate additional electricity. At day 30: \( P(30)+2500 = P(31) \), therefore \( P(31) - P(30) = 2500 \) when \( P(30) = x \).

Step 6: Presenting the Equations

Based on each analysis, the equations are: (a) \( P(4) = 3500 \); (b) \( P'(4) = 1000 \); (c) \( P(d + 0.5) - P(d) = 1000 \) when \( P(d) = 5000 \); (d) \( P(31) - P(30) = 2500 \).

Key Concepts

Derivative

Derivatives play an essential role in calculus, providing a way to measure how a function changes as its input changes. In this problem, we need to understand how fast the solar array generates electricity over time.

• The derivative of a function at a point measures the rate of change of the function at that specific point.
• If we consider the function \( P(d) \), which provides the total electricity generated by the solar array by the end of day \( d \), the derivative \( P'(d) \) tells us the rate at which electricity is being generated on the \( d^{\text{th}} \) day.

This is evident in statement (b) of the given exercise, where knowing the derivative \( P'(4) = 1000 \) helps us understand that on January 4, the solar array was producing electricity at a rate of 1000 kWh per day. Calculating derivatives can provide crucial insights into the behavior of functions, especially when interpreting real-world phenomena like the performance of solar panels.

Inverse Function

Inverse functions are a fascinating concept in mathematics where you essentially "reverse" a function’s operations. For the solar array problem, this would mean finding which day \( d \) corresponds to a specific amount of electricity \( x \) generated.

• An inverse function, denoted \( P^{-1}(x) \), helps find the input from the output.
• In this context, if \( P(d) \) gives total electricity by day \( d \), then \( P^{-1}(x) \) reveals what day the solar array reached \( x \) kWh.

Considering statement (c), when the solar array had created 5000 kWh by day \( d \), it indicated a subsequent rise of 1000 kWh over the next half-day. Hence, the rate of increase becomes vital to determining day and time specifics using the inverse function concept, allowing you to "reverse engineer" the day count from an electricity total.

Rate of Change

The concept of rate of change is a pivotal idea in understanding how different quantities vary with time or other factors. In calculus, the rate of change is usually represented by the derivative.

• It measures the speed at which one quantity changes relative to another.
• In our situation, \( P'(d) \) reflects the rate at which electricity is being generated at the end of the \( d^{\text{th}} \) day.

In example (b) of the exercise, the rate of change (1000 kWh per day) informs us about the instantaneous trend of electricity generation on January 4. This measurement provides insight into not only the efficiency but also the operation capacity of the solar array systems at any given point. Understanding the rate helps anticipate future outputs based on past performance observed through derivative calculations.

Cumulative Function

Cumulative functions provide a summary of how quantities add up over time or another process. For the solar array query, \( P(d) \) is a cumulative function indicating total electricity generation up to day \( d \).

• It represents the accumulation of a quantity, consolidating its progression.
• In this case, \( P(d) \) assures the sum of electricity generated day after day.

As demonstrated in statement (a) of the exercise, by calculating \( P(4)=3500 \), the cumulative function reveals the total electricity output by January 4. This function is helpful to analyze overall trends and set performance benchmarks in various fields involving accumulated progresses, such as energy production. Moreover, applying this function alongside its derivative can provide more comprehensive understanding, showing not just total amounts but also rates of change through time.