Problem 63 - Calculus and its applications

Question

Raggs, Ltd., estimates that its sales are growing continuously at a rate given by $$S^{\prime}(t)=10 e^{t}$$ where \(S^{\prime}(t)\) is in dollars per day, on day \(t\). a) Find the accumulated sales for the first 5 days. b) Find the sales from the 2 nd day through the 5 th day. (This is the integral from 1 to \(5 .)\)

Step-by-Step Solution

Answer

Accumulated sales for the first 5 days: $1474.13; sales from 2nd to 5th day: $1456.95.

Step 1: Understanding the Problem

Raggs Ltd. gives a continuous sales growth function: \(S^{\prime}(t)=10 e^{t}\), where \(S^{\prime}(t)\) represents the rate of sales in dollars per day. We need to find the total sales over specific intervals of time by integrating this rate.

Step 1: Setting Up the Integral for Part (a)

To find the accumulated sales for the first 5 days, calculate the integral of \(S^{\prime}(t) = 10e^t\) from \(t = 0\) to \(t = 5\). This will give the total sales from day 0 to day 5: \[\int_{0}^{5} 10 e^{t} \, dt\]

Step 2: Solving the Integral for Part (a)

Evaluate the integral step by step: The antiderivative of \(10e^t\) is \(10e^t\). Thus, the integral becomes: \[10e^{t}\bigg\rvert_{0}^{5} = 10e^{5} - 10e^{0}\].Calculate the values: \[10e^{5} - 10e^{0} = 10e^{5} - 10\].

Step 3: Calculating Final Value for Part (a)

Evaluate the expression \(10(e^{5} - 1)\) using a calculator or approximation for \(e\). Assume \(e^{5}\) is approximately \(148.413\): \[10 imes (148.413 - 1) \approx 10 imes 147.413 = 1474.13\] dollars.

Step 4: Setting Up the Integral for Part (b)

To find the sales from the 2nd day through the 5th day, set up the integral from \(t = 1\) to \(t = 5\):\[\int_{1}^{5} 10 e^{t} \, dt\]

Step 5: Solving the Integral for Part (b)

Evaluate the integral to find the accumulated sales:The antiderivative remains \(10e^t\), so the integral becomes: \[10e^{t}\bigg\rvert_{1}^{5} = 10e^{5} - 10e^{1}\].Substitute the values: \[10e^{5} - 10e^{1} = 10(e^{5} - e^{1})\].

Step 6: Calculating Final Value for Part (b)

Calculate \(10(e^{5} - e^{1})\) using an approximation for \(e\) where \(e \approx 2.718\) and \(e^{5} \approx 148.413\):\[10(148.413 - 2.718) = 10 \times 145.695 = 1456.95\] dollars.

Key Concepts

Accumulated Sales

Accumulated sales refer to the total sales collected over a certain period. In this context, Raggs Ltd. wants to calculate how much revenue has been generated from their continuous sales over specific days. They provide a continuous growth rate for this calculation, given by the function \(S'(t) = 10e^{t}\). Understanding accumulated sales requires determining the total amount of sales from day 0 to day 5 for the first part of the exercise and from day 2 to day 5 for the second part.To find the accumulated sales, we need to integrate the sales rate function. This process transforms the rate of sales into total sales over a period, effectively summing up the continuous flow of sales day by day. By integrating the continuous sales growth function over the specified time intervals, we calculate how sales accumulate over time, converting a dynamic rate into a comprehensive sales figure.

Integration

Integration is a fundamental concept in calculus, commonly used to find accumulated quantities or areas under curves. In this exercise, integration is used to calculate the accumulated sales from a continuous growth rate. When you have a function representing a rate of change, integrating this function over an interval will give you the total change over that interval.For Raggs Ltd., the integral of the rate function \(S'(t) = 10e^{t}\) over a specified interval provides the number of sales over that period.

• To find accumulated sales from 0 to 5 days, we set up the integral: \[\int_{0}^{5} 10e^{t} \, dt\]
• To find accumulated sales from 1 to 5 days, we adjust the limits as follows:\[\int_{1}^{5} 10e^{t} \, dt\]

Solving these integrals helps bridge the gap between a varying rate of sales and the total sales accumulated over those days.

Exponential Function

Exponential functions are mathematical functions of the form \(f(t) = ae^{kt}\), where \(e\) is the base of natural logarithms, approximately equal to 2.718. These functions are pivotal when analyzing scenarios involving continuous growth or decay, as they perfectly model situations where change happens proportionally to the current value.In this problem, the exponential function appears as part of the sales rate \(S'(t) = 10e^{t}\). This indicates that the rate of sales grows continuously and exponentially over time. The constant 10 scales the rate, ensuring that at \(t = 0\), the initial rate is 10 dollars per day.Understanding exponential functions in economics and business contexts, like sales growth, is crucial. The exponential growth represented provides a realistic model for how sales can increase rapidly when growth is continually compounded over time. Recognizing this allows businesses to forecast future sales and strategize effectively.