Answered

Suppose that a volcano is erupting and readings of the rate [tex]\( r(t) \)[/tex] at which solid materials are spewed into the atmosphere are given in the table. The time [tex]\( t \)[/tex] is measured in seconds and the units for [tex]\( r(t) \)[/tex] are measured in tonnes (metric tons) per second. Note that [tex]\( r(t) \)[/tex] is increasing over the interval [tex]\([0,6]\)[/tex].

[tex]\[
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
$t$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 \\
\hline
$r(t)$ & 2 & 12 & 18 & 34 & 48 & 52 & 60 \\
\hline
\end{tabular}
\][/tex]

(a) Give upper and lower estimates (in tons) for the total quantity [tex]\( Q(6) \)[/tex] of erupted materials after six seconds. (Use six equal subintervals.)

- Lower estimate [tex]\( Q(6) = \, \square \, \)[/tex] tons
- Upper estimate [tex]\( Q(6) = \, \square \, \)[/tex] tons

(b) Use the midpoint rule to estimate [tex]\( Q(6) \)[/tex] (in tons). (Use three equal subintervals.)

- [tex]\( Q(6) = \, \square \, \)[/tex] tons



Answer :

GinnyAnswer
Sure! Let's break down the problem step by step:

### (a) Lower and Upper Estimates

Lower Estimate:

The lower estimate is calculated by summing up the rates [tex]\( r(t) \)[/tex] for each interval except the last one and multiplying by the width of each interval (which is 1 second in this case).

Given data:
- [tex]\( t = [0, 1, 2, 3, 4, 5, 6] \)[/tex]
- [tex]\( r(t) = [2, 12, 18, 34, 48, 52, 60] \)[/tex]
- The interval width, [tex]\( \Delta t = 1 \)[/tex] second.

To find the lower estimate:
[tex]\[ Q_{\text{lower}} = \sum_{i=0}^{5} r(t_i) \times \Delta t = (2 + 12 + 18 + 34 + 48 + 52) \times 1 \][/tex]

Summing those values:
[tex]\[ 2 + 12 + 18 + 34 + 48 + 52 = 166 \][/tex]

Therefore:
[tex]\[ Q_{\text{lower}} = 166 \text{ tons} \][/tex]

Upper Estimate:

The upper estimate is calculated by summing up the rates [tex]\( r(t) \)[/tex] for each interval except the first one and multiplying by the width of each interval.

To find the upper estimate:
[tex]\[ Q_{\text{upper}} = \sum_{i=1}^{6} r(t_i) \times \Delta t = (12 + 18 + 34 + 48 + 52 + 60) \times 1 \][/tex]

Summing those values:
[tex]\[ 12 + 18 + 34 + 48 + 52 + 60 = 224 \][/tex]

Therefore:
[tex]\[ Q_{\text{upper}} = 224 \text{ tons} \][/tex]

### (b) Midpoint Rule Estimate

For the midpoint rule with three equal subintervals, we divide the interval [tex]\([0, 6]\)[/tex] into three subintervals:
- [tex]\([0, 2]\)[/tex]
- [tex]\([2, 4]\)[/tex]
- [tex]\([4, 6]\)[/tex]

The midpoints of these subintervals are:
- Midpoint of [tex]\([0, 2]\)[/tex] is [tex]\( t = 1 \)[/tex]
- Midpoint of [tex]\([2, 4]\)[/tex] is [tex]\( t = 3 \)[/tex]
- Midpoint of [tex]\([4, 6]\)[/tex] is [tex]\( t = 5 \)[/tex]

We use the corresponding values of [tex]\( r(t) \)[/tex] at these midpoints:
- [tex]\( r(1) = 12 \)[/tex]
- [tex]\( r(3) = 34 \)[/tex]
- [tex]\( r(5) = 52 \)[/tex]

Each subinterval has a width of 2 seconds. We multiply the rate by the width of each subinterval and sum up the results:
[tex]\[ Q_{\text{midpoint}} = (r(1) + r(3) + r(5)) \times \Delta t = (12 + 34 + 52) \times 2 \][/tex]

Summing those values:
[tex]\[ 12 + 34 + 52 = 98 \][/tex]

Thus:
[tex]\[ Q_{\text{midpoint}} = 98 \times 2 = 196 \text{ tons} \][/tex]

So, the answers are:
- Lower Estimate: [tex]\( 166 \text{ tons} \)[/tex]
- Upper Estimate: [tex]\( 224 \text{ tons} \)[/tex]
- Midpoint Estimate: [tex]\( 194.0 \text{ tons} \)[/tex]

Other Questions