Question

Can someone please help me with this problem...
Coal gas is being produced at a gasworks. Pollutants in the gas are removed by scrubbers which become less efficient as time goes by. The pollutant rate is measured at the start of every other month for a year and is given in the table. Since the scrubbers are becoming less efficient we can assume that the pollutant rate is always increasing.
Estimate the highest total amount of pollutants that could have escaped over the year. Write the answer in a complete sentence that explains the meaning of the number you find.

Answer 1

@electrokid or @phi can either of you please help me with this problem

Answer 2

do you understand the problem

Answer 3

yes, but I don't now what you have been studying, and how they expect you to do this problem

Answer 4

hold on let me get the exact title out of my book

Answer 5

We have been working on definite integral

Answer 6

does that help

Answer 7

Here is how I would do it (which may or not make sense)

the numbers are going up by a factor of about 1.3. That means it is exponential, and we could try fitting an exponential curve through the data points.

\[ y = a \ e^{m \cdot b} \]
where m is the month, b is some unknown growth, and a is the initial value.
based on the table we have
\[ 3.7 = a \ e^{0 \cdot b} \\ a= 3.7 \]

Answer 8

using the last entry in the table
\[ 20.2 = 3.7 e^{12 b} \]
solve for b:
divide both sides by 3.7
and take the natural log of both sides
\[ \ln(20.2/3.7)= 12 b\]
\[ b = 0.1414 \]

Answer 9

our model is
\[ y = 3.7 e^{0.1414\ m} \]
now integrate from 0 to 12 to get the total emissions over the course of 12 months

Answer 10

so then for the next one put 4.9 where m is. Is that right?

Answer 11

no, m is the month. if we put m=2, we should get the number in the table, 4.9 if this model is accurate.

Answer 12

oh ok hold on let me do the math

Answer 13

I got 8.5

Answer 14

Is that what you got

Answer 15

Oh wait I just redid it and I did get 4.9

Answer 16

if you mean what do you get when you evaluate
3.7*exp(2*0.1414)=
you can cut and paste that into google and you get 4.9

Answer 17

So know what do I do

Answer 18

that curve represents how much is being emitted over time.
If it were a constant number like 5 per month, we could just multiply by the number of months.
But it is constantly changing so we integrate

if we use x for the number of months the problem is
\[total\ pollutants = 3.7 \int\limits_{0}^{12}e^{0.1414\ x} dx\]

Answer 19

Have you learned how to integrate ?

Answer 20

yes ok hold on whie I integrate it

Answer 21

I got 8813.342

Answer 22

as a sanity check. at month 12 the rate is 20 tons/month
for 12 months that give 240 tons total.
and the rate is less than 20 tons/month for most of the year, so the answer is less than 240

Answer 23

would the answer be 130.83

Answer 24

I get something slightly different.

the integral of 3.7 exp(0.1414 x) is
\[ \frac{3.7}{0.1414} e^{0.1414 \ x} \]
evaluate between 0 and 12:

\[ \frac{3.7}{0.1414} \left(e^{0.1414 \cdot 12}- e^{0.1414 \cdot 0}\right)\]

Answer 25

116.72

Answer 26

the other way to estimate the max amount is sum up rectangular areas