Problem 31 - Applied Calculus

Question

Use the expressions for left and right sums on page 245 and Table 5.7 (a) If \(n=4,\) what is \(\Delta t ?\) What are \(t_{0}, t_{1}, t_{2}, t_{3}, t_{4} ?\) What are \(f\left(t_{0}\right), f\left(t_{1}\right), f\left(t_{2}\right), f\left(t_{3}\right), f\left(t_{4}\right) ?\) (b) Find the left and right sums using \(n=4\) (c) If \(n=2,\) what is \(\Delta t ?\) What are \(t_{0}, t_{1}, t_{2} ?\) What are \(f\left(t_{0}\right), f\left(t_{1}\right), f\left(t_{2}\right) ?\) (d) Find the left and right sums using \(n=2\) $$\begin{array}{c|r|r|r|r|r} \hline t & 0 & 4 & 8 & 12 & 16 \\ \hline f(t) & 25 & 23 & 22 & 20 & 17 \\ \hline \end{array}$$

Step-by-Step Solution

Answer

For \(n=4\), \( \Delta t = 4\), left sum = 360, right sum = 328. For \(n=2\), \( \Delta t = 8\), left sum = 376, right sum = 312.

Step 1: Calculate Δt for n=4

To find \( \Delta t \) when \( n = 4 \), use the formula \( \Delta t = \frac{b-a}{n} \), where \( a = 0 \) and \( b = 16 \) according to the given table. Thus, \( \Delta t = \frac{16-0}{4} = 4 \).

Step 2: Determine t values for n=4

The t-values are determined by starting at \( t_0 = 0 \) and adding \( \Delta t = 4 \) each time: \( t_0 = 0, t_1 = 4, t_2 = 8, t_3 = 12, t_4 = 16 \).

Step 3: Find f(t) for n=4

From the table, \( f(t_0) = 25, f(t_1) = 23, f(t_2) = 22, f(t_3) = 20, f(t_4) = 17 \).

Step 4: Compute Left Sum for n=4

The left sum is calculated as \( \Delta t(f(t_0) + f(t_1) + f(t_2) + f(t_3)) \). Substituting the values, we get:\[ 4(25 + 23 + 22 + 20) = 4 \times 90 = 360 \].

Step 5: Compute Right Sum for n=4

The right sum involves \( \Delta t(f(t_1) + f(t_2) + f(t_3) + f(t_4)) \). Substitute the values to get:\[ 4(23 + 22 + 20 + 17) = 4 \times 82 = 328 \].

Step 6: Calculate Δt for n=2

For \( n = 2 \), \( \Delta t = \frac{16-0}{2} = 8 \).

Step 7: Determine t values for n=2

With \( \Delta t = 8 \), the t-values are \( t_0 = 0, t_1 = 8, t_2 = 16 \).

Step 8: Find f(t) for n=2

Using the table for the corresponding t-values, \( f(t_0) = 25, f(t_1) = 22, f(t_2) = 17 \).

Step 9: Compute Left Sum for n=2

The left sum is \( \Delta t(f(t_0) + f(t_1)) \). \[ 8(25 + 22) = 8 \times 47 = 376 \].

Step 10: Compute Right Sum for n=2

For the right sum, use \( \Delta t(f(t_1) + f(t_2)) \). \[ 8(22 + 17) = 8 \times 39 = 312 \].

Key Concepts

Left Sum

The concept of the Left Sum is tied to the method of approximating the area under a curve using rectangles. In a left Riemann Sum, the height of each rectangle is determined by the function value at the left endpoint of each subinterval. This approach is particularly helpful when calculating the area under a decreasing function, as it tends to give an overestimation of the area. If the function is increasing, it might undercut the true value.

To compute the Left Sum, you will need to:

• Identify the width of each interval, which is referred to as \( \Delta t \).
• Determine the left endpoint values of the function, like \( f(t_0), f(t_1), \ldots, \) up to the second to last \( f(t_{n-1}) \).
• Multiply these values by \( \Delta t \) and sum them up to approximate the total area.

Consider the example where \( n=4 \), and the function values are drawn from the table. The Left Sum is computed as: \[ 4(25 + 23 + 22 + 20) = 360. \]
This sum represents the approximate area based on the function values at the left endpoints.

Right Sum

The Right Sum, much like the Left Sum, is a technique used in Riemann Sums to estimate the area under a curve. However, for a Right Sum, the rectangle heights are determined by the function values at the right endpoints of each interval. Because of this, the Right Sum tends to provide an underestimate for a decreasing function and an overestimate for an increasing one.

Here's how you can calculate the Right Sum:

• First, find \( \Delta t \), the width of each interval.
• Use the function values at each right endpoint, which is from \( f(t_1) \) to \( f(t_n) \).
• Multiply these values by \( \Delta t \) and sum them to get the total.

For example, with \( n = 4 \), the Right Sum evaluates to: \[ 4(23 + 22 + 20 + 17) = 328. \]
This value tells us the approximate cumulative area based on the function values at the right edge of each subinterval.

Interval Partitioning

Interval Partitioning is a fundamental concept in the Riemann Sum methodology. It involves splitting the area under a curve into smaller segments or intervals. The idea is to create multiple, manageable pieces that make approximating the total area easier. The number of partitions, indicated by \( n \), and their width, \( \Delta t \), play a crucial role in determining the precision of approximation.

To partition an interval:

• Determine the total length of the interval: this can be calculated by subtracting the starting value \( a \) from the ending value \( b \).
• Decide on the number of partitions \( n \), which defines \( \Delta t \) as \( \Delta t = \frac{b-a}{n} \).
• Divide the interval into equal partitions, yielding endpoints like \( t_0, t_1, \ldots, t_n \).

Consider the original problem with \( n=4 \). The interval \( [0, 16] \) is divided with a \( \Delta t \) of 4, resulting in \{0, 4, 8, 12, 16\}. Similarly, if \( n = 2 \), \( \Delta t \) is 8, resulting in \{0, 8, 16\}.
This subdivision allows for detailed calculation of Riemann Sums, either left or right,
and enhances our understanding of how integrals represent areas under curves.