Problem 58 - Calculus and its applications

Question

Nadia's fashions discovers that the number of items sold \(t\) days after launching a new sales promotion is given by $$ N(t)=2 t^{3}-3 t^{2}+2 t $$ a) Find \(N^{\prime}(1), N^{\prime}(2),\) and \(N^{\prime}(4)\). b) Find \(N^{\prime \prime}(1), N^{\prime \prime}(2),\) and \(N^{\prime \prime}(4)\). c) Interpret the meaning of your answers to parts (a) and (b).

Step-by-Step Solution

Answer

N'(t) shows sales increase rate per day; N''(t) shows acceleration of sales rate. Sales rate is increasing faster over time.

Step 1: Differentiate to Find N'(t)

First, find the first derivative of the function \(N(t) = 2t^3 - 3t^2 + 2t\). To do this, apply the power rule of differentiation. The power rule states that \(\frac{d}{dt}[t^n] = nt^{n-1}\).- Differentiate \(2t^3\), which results in \(6t^2\).- Differentiate \(-3t^2\), which results in \(-6t\).- Differentiate \(2t\), which results in \(2\).Thus, the first derivative is \(N'(t) = 6t^2 - 6t + 2\).

Step 2: Evaluate N'(t) at Specific Points

Now evaluate \(N'(t)\) at specific time values as required:- For \(t = 1\): Substitute \(t = 1\) into \(N'(t)\). \[ N'(1) = 6(1)^2 - 6(1) + 2 = 6 - 6 + 2 = 2 \]- For \(t = 2\): Substitute \(t = 2\) into \(N'(t)\). \[ N'(2) = 6(2)^2 - 6(2) + 2 = 24 - 12 + 2 = 14 \]- For \(t = 4\): Substitute \(t = 4\) into \(N'(t)\). \[ N'(4) = 6(4)^2 - 6(4) + 2 = 96 - 24 + 2 = 74 \]

Step 3: Differentiate to Find N''(t)

Find the second derivative \(N''(t)\) by differentiating \(N'(t) = 6t^2 - 6t + 2\) again.- Differentiate \(6t^2\), which results in \(12t\).- Differentiate \(-6t\), which results in \(-6\).- The derivative of a constant \(2\) is \(0\).Thus, the second derivative is \(N''(t) = 12t - 6\).

Step 4: Evaluate N''(t) at Specific Points

Evaluate \(N''(t)\) at the specific time values:- For \(t = 1\): Substitute \(t = 1\) into \(N''(t)\). \[ N''(1) = 12(1) - 6 = 12 - 6 = 6 \]- For \(t = 2\): Substitute \(t = 2\) into \(N''(t)\). \[ N''(2) = 12(2) - 6 = 24 - 6 = 18 \]- For \(t = 4\): Substitute \(t = 4\) into \(N''(t)\). \[ N''(4) = 12(4) - 6 = 48 - 6 = 42 \]

Step 5: Interpret Results

Interpret the results from parts (a) and (b):- **Part (a):** \(N'(t)\) represents the rate at which items are sold per day. So, \(N'(1) = 2, N'(2) = 14, N'(4) = 74\) indicate increasing rates of sales on days 1, 2, and 4.- **Part (b):** \(N''(t)\) represents the acceleration of sales (how quickly the rate changes). Values \(N''(1) = 6, N''(2) = 18, N''(4) = 42\) show that the sales rate is increasing faster each day.

Key Concepts

First Derivative

The concept of the first derivative is a cornerstone in calculus. A derivative essentially represents the rate of change of a function. When we're talking about the first derivative, symbolized as \(N'(t)\), we are observing how the number of items sold changes over time, specifically from one day to the next. In mathematical terms, the first derivative of a function is found using differentiation rules, such as the power rule. The power rule helps us differentiate expressions by lowering the power of each term and multiplying by the original power. This operation tells us how steep the graph of the function is at any given point.

In this exercise, the function describing sales, \(N(t) = 2t^3 - 3t^2 + 2t\), is differentiated to yield \(N'(t) = 6t^2 - 6t + 2\). This new function, \(N'(t)\), gives the rate at which items are being sold per day. Evaluating it at specific times, like day 1, 2, and 4, reveals how sales are changing as time progresses, indicating an increase in sales rate over the chosen days.

Second Derivative

The second derivative is often used to understand the curvature or concavity of a function. In real-world terms, it can indicate the acceleration of a phenomenon. By taking the derivative of the first derivative, we uncover how the rate of change itself is changing over time. In this context, it reveals not just the increase in sales but how rapidly the increase in sales is occurring.

From the initial sales function, the second derivative, denoted as \(N''(t)\), is calculated as \(N''(t) = 12t - 6\). This derivative is crucial because it informs us about the acceleration of sales. Evaluating it at days 1, 2, and 4 shows how much faster the sales rate grows with each passing day. Positive values of \(N''(t)\) indicate that the sales rate is improving at an increasing pace.

Rate of Change

The rate of change is essentially what the first derivative measures. It's about determining how quickly one quantity changes in relation to another. For Nadia's Fashions, the rate of change shows how quickly items are sold as each day passes. This is important for businesses to understand trends and manage supply versus demand efficiently.

The rate of change can be seen directly in \(N'(t)\), the first derivative of the sales function. When \(N'(t)\) is calculated for specific days, it provides exact numbers that represent the sales rate for those days. For example, \(N'(1) = 2\) tells us that on day one, two items were sold, whereas \(N'(4) = 74\) indicates a significant surge in sales by day four.

Acceleration of Sales

The term "acceleration of sales" refers to how quickly the rate of sales changes over time. This can be vital for strategizing in marketing and management because it highlights whether promotional tactics are gaining momentum or not. The acceleration is quantified by the second derivative, \(N''(t)\), derived from the sales function.

When analyzing the acceleration of sales, higher positive values of \(N''(t)\) imply that each subsequent day sees a greater increase in sales rate than the previous one. For instance, \(N''(4) = 42\) signifies that sales are growing at a rapid pace by day four. Thus, tracking the acceleration gives insights into how dynamic and responsive the market is to promotional efforts.