Problem 29 - Applied Calculus

Question

The rate of change of a quantity is given by \(f(t)=t^{2}+1\) Make an underestimate and an overestimate of the total change in the quantity between \(t=0\) and \(t=8\) using (a) \(\quad \Delta t=4\) (b) \(\Delta t=2\) (c) \(\Delta t=1\) What is \(n\) in each case? Graph \(f(t)\) and shade rectangles to represent each of your six answers.

Step-by-Step Solution

Answer

Underestimates: 72, 120, 148. Overestimates: 328, 248, 212.

Step 1: Understanding the Problem

We need to approximate the total change in a quantity given its rate of change, \(f(t) = t^2 + 1\), over the interval from \(t = 0\) to \(t = 8\). The task is to find both underestimates and overestimates by using Riemann sums with different \(\Delta t\) values: 4, 2, and 1. For each \(\Delta t\), calculate the number of rectangles \(n\) as \(n = \frac{8 - 0}{\Delta t}\).

Step 2: Calculate for \(\Delta t = 4\)

\(n = \frac{8}{4} = 2\). Subdivide the interval [0,8] into 2 subintervals. Lower Riemann sum (underestimate): evaluate \(f(t)\) at the left end of each subinterval: \(f(0) = 1\), \(f(4) = 17\). Total is \(4(1) + 4(17) = 72\). Upper Riemann sum (overestimate): at right end: \(f(4) = 17\), \(f(8) = 65\). Total is \(4(17) + 4(65) = 328\).

Step 3: Calculate for \(\Delta t = 2\)

\(n = \frac{8}{2} = 4\). Subdivide into 4 subintervals. Underestimate: \( f(0) = 1\), \(f(2) = 5\), \(f(4) = 17\), \(f(6) = 37\). Total is \(2(1 + 5 + 17 + 37) = 120\). Overestimate: \(f(2) = 5\), \(f(4) = 17\), \(f(6) = 37\), \(f(8) = 65\). Total is \(2(5 + 17 + 37 + 65) = 248\).

Step 4: Calculate for \(\Delta t = 1\)

\(n = \frac{8}{1} = 8\). Subdivide into 8 subintervals. Underestimate: evaluate \(f(t)\) at each left endpoint (0 to 7): \(f(t) = 1, 2, 5, 10, 17, 26, 37, 50\). Total is \(1 \times (1 + 2 + 5 + 10 + 17 + 26 + 37 + 50) = 148\). Overestimate evaluate at right endpoint (1 to 8): \(f(t) = 2, 5, 10, 17, 26, 37, 50, 65\). Total is \(1 \times (2 + 5 + 10 + 17 + 26 + 37 + 50 + 65) = 212\).

Step 5: Summary of Results

The calculated underestimates are 72, 120, and 148 for \(\Delta t = 4, 2, 1\) respectively, while the overestimates are 328, 248, and 212. As \(\Delta t\) decreases, both estimates converge towards a more accurate value of the total change. Graphically, draw \(f(t) = t^2 + 1\), and shade the rectangles representing the Riemann sums computed with each \(\Delta t\).

Key Concepts

Rate of Change

The rate of change measures how a quantity increases or decreases over time. In this exercise, we're given a specific rate of change function, \( f(t) = t^2 + 1 \). This represents how quickly the quantity changes with respect to time \( t \). When we calculate the total change from \( t = 0 \) to \( t = 8 \), we're essentially summing up small changes over very short intervals.
To approximate this sum, we use Riemann sums which involve breaking down the time interval into smaller pieces of a constant length. The choice of \( \Delta t \), the width of these small intervals, affects the accuracy of our sum. Smaller \( \Delta t \) values result in more intervals, corresponding to finer calculations and thus more accurate approximations of the total change.

Underestimate and Overestimate

Underestimates and overestimates are methods for approximating integrals, using Riemann sums. In the context of our problem, the underestimate is calculated by assuming the rate of change \( f(t) \) at the beginning of each time subinterval (left endpoints). This gives a lower value as it doesn't account for the increase in \( f(t) \) over the duration of each subinterval. Conversely, overestimates are calculated using the rate of change at the end of each subinterval (right endpoints). This assumes the highest rate present in the subinterval, thus providing an overstatement of the true change.
For example, with a large \( \Delta t \) like 4, fewer but wider intervals provide rough estimations, resulting in a high discrepancy between under- and overestimates (72 and 328 respectively). As \( \Delta t \) decreases (1 or 2), these estimates come closer together (148 and 212 for \( \Delta t = 1 \)), better reflecting the true total change.

Graphical Representation

Visualizing Riemann sums through graphs provides a clearer comprehension of underestimates and overestimates. On a graph of \( f(t) = t^2 + 1 \), Riemann sums are represented as rectangles under the curve on the interval \( t = 0 \) to \( t = 8 \).
For an underestimate, rectangles are drawn such that their height corresponds to \( f(t) \) at the start of every subinterval. They sit under the curve, leaving "gaps" between the top of the rectangle and the curve of \( f(t) \). An overestimate's rectangles are taller, as their heights reach up to \( f(t) \) at the end of each subinterval, often extending past the actual curve.

• Shaded areas of these rectangles are the sum of all these individual estimates, either under or over the curve.
• Smaller intervals (lower \( \Delta t \)) mean the rectangles' heights are closer to the curve of \( f(t) \), providing richer shaded detail and precision.

Graphs make these discrepancies visual, offering a tangible method for comparing different estimation techniques.