Problem 54
Question
Suppose that a volcano is erupting and readings of the rate \(r(t)\) at which solid materials are spewed into the atmosphere are given in the table. The time \(t\) is measured in seconds and the units for \(r(t)\) are tonnes (metric tons) per second. $$\begin{array}{|c|c|c|c|c|c|c|}\hline t & {0} & {1} & {2} & {3} & {4} & {5} & {6} \\ \hline r(t) & {2} & {10} & {24} & {36} & {46} & {54} & {60} \\\ \hline\end{array}$$ (a) Give upper and lower estimates for the total quantity \(Q(6)\) of erupted materials after 6 seconds. (b) Use the Midpoint Rule to estimate \(Q(6)\)
Step-by-Step Solution
Verified Answer
Lower: 172 tonnes, Upper: 230 tonnes, Midpoint: 201 tonnes.
1Step 1: Understand the Problem
We need to estimate the total erupted material over 6 seconds using provided rate data. First, let's analyze the data to understand how it might be used to get these estimates.
2Step 2: Lower Estimate Using Left Endpoint Approximation
We will approximate the integral using left endpoint of each subinterval. To do this, we multiply each rate value (excluding the last one) by the change in time between consecutive readings, which is 1 second.\[Q_{lower} = (r(0) \times 1) + (r(1) \times 1) + (r(2) \times 1) + (r(3) \times 1) + (r(4) \times 1) + (r(5) \times 1)\]\[ = (2 \times 1) + (10 \times 1) + (24 \times 1) + (36 \times 1) + (46 \times 1) + (54 \times 1) = 172 \]Therefore, the lower estimate of the total quantity is 172 tonnes.
3Step 3: Upper Estimate Using Right Endpoint Approximation
In this step, we use the right endpoint of each interval, which means starting with the second rate value and continuing until the last.\[Q_{upper} = (r(1) \times 1) + (r(2) \times 1) + (r(3) \times 1) + (r(4) \times 1) + (r(5) \times 1) + (r(6) \times 1)\]\[ = (10 \times 1) + (24 \times 1) + (36 \times 1) + (46 \times 1) + (54 \times 1) + (60 \times 1) = 230 \]Therefore, the upper estimate of the total quantity is 230 tonnes.
4Step 4: Estimate Using the Midpoint Rule
For this estimate, we find the average rate in each interval and multiply by the interval length:\[Q_{midpoint} = (\frac{r(0) + r(1)}{2} \times 1) + (\frac{r(1) + r(2)}{2} \times 1) + (\frac{r(2) + r(3)}{2} \times 1) + (\frac{r(3) + r(4)}{2} \times 1) + (\frac{r(4) + r(5)}{2} \times 1) + (\frac{r(5) + r(6)}{2} \times 1) \]\[ = (\frac{2 + 10}{2} \times 1) + (\frac{10 + 24}{2} \times 1) + (\frac{24 + 36}{2} \times 1) + (\frac{36 + 46}{2} \times 1) + (\frac{46 + 54}{2} \times 1) + (\frac{54 + 60}{2} \times 1)\]\[ = (6) + (17) + (30) + (41) + (50) + (57) = 201 \]The estimated total quantity using the Midpoint Rule is 201 tonnes.
Key Concepts
Midpoint RuleUpper and Lower EstimatesEstimation TechniquesIntegral Approximation
Midpoint Rule
The Midpoint Rule is a numerical technique used to estimate the value of an integral, especially when calculating the total accumulation of a continuous rate like the one observed during a volcanic eruption. In simpler terms, it helps us estimate the total quantity of material erupted over a period.
The process involves dividing the time into intervals and using the midpoint of each interval for calculation. For example, with readings at every second from 0 to 6, the midpoints are calculated between each pair of consecutive time values.
The process involves dividing the time into intervals and using the midpoint of each interval for calculation. For example, with readings at every second from 0 to 6, the midpoints are calculated between each pair of consecutive time values.
- Calculate each midpoint: These are the values between consecutive readings, for instance, the midpoint between time 0 and 1 seconds is 0.5 seconds.
- Use these midpoints to estimate the average rate: This involves taking the function values at these midpoints.
- Multiply the average by the width of each interval: In our case, each interval is 1 second long.
Upper and Lower Estimates
Upper and Lower Estimates create bounds when approximating an integral, which can be visualized as the total area under a rate curve. These bounds are essential because they provide a range within which the actual integral value lies.
The Lower Estimate uses the Left Endpoint Approximation:
The Upper Estimate uses the Right Endpoint Approximation:
The Lower Estimate uses the Left Endpoint Approximation:
- Choose the left value of each interval. For example, from 0 to 1 second, use the rate at 0 seconds.
- Multiply this rate by the interval width (1 second in this case).
The Upper Estimate uses the Right Endpoint Approximation:
- Select the right value of each interval, meaning from 0 to 1 second, use the rate at 1 second, and so on.
- Again, multiply by the width of each interval.
Estimation Techniques
Various techniques are used for numerical integration, and selecting the appropriate one can depend on the data and situation. Let's highlight the most common approaches, like the ones involved in this exercise:
- **Midpoint Rule:** Utilizes the midpoint of intervals for a more balanced estimate, ideal when reaction time and fluctuations occur evenly across intervals.
- **Left and Right Endpoint Approximations:** These use the start or end of intervals which might cause either underestimation or overestimation as they assume the rate remains constant based on endpoint readings.
- Diverse methodologies exist like Trapezoidal and Simpson’s Rule that further balance the estimation by simulating more realistic curves over sections.
Integral Approximation
Integral Approximation is a fundamental mathematical concept that estimates total values from rate data. When precise functions cannot be used or are unavailable, we depend on approximation.
These approximations are essential in scenarios involving real-world data, like volcanic eruptions where actual function data isn't precisely calculable. Estimation methods help fill this gap.
These approximations are essential in scenarios involving real-world data, like volcanic eruptions where actual function data isn't precisely calculable. Estimation methods help fill this gap.
- Approximation forms include numerical integration techniques like those discussed, converting discrete data points into continuous estimates.
- Monte Carlo and Riemann Sum techniques are further robust strategies used across fields like physics, engineering, and anywhere interim data points exist.
Other exercises in this chapter
Problem 53
If oil leaks from a tank at a rate of \(r(t)\) gallons per minute at time \(t,\) what does \(\int_{0}^{120} r(t) d t\) represent?
View solution Problem 53
Evaluate \(\int_{-2}^{2}(x+3) \sqrt{4-x^{2}} d x\) by writing it as a sum of two integrals and interpreting one of those integrals in terms of an area.
View solution Problem 54
Evaluate \(\int_{0}^{1} x \sqrt{1-x^{4}} d x\) by making a substitution and interpreting the resulting integral in terms of an area.
View solution Problem 55
Water flows from the bottom of a storage tank at a rate of \(r(t)=200-4 t\) liters per minute, where 0\(\leqslant t \leqslant 50\) . Find the amount of water th
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