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Question 10

Page history last edited by Rence 11 years, 5 months ago

A Calculator MAY be used to help answer this question.

Be certain to limit yourself to the four permissable calculator functions.

 

A factory has been releasing pollutants into Lake Winnipeg at a rate, tons per week. After 1 week, a clean up crew arrives and begins removing pollutants from the lake at a rate  tons per week.

 

(a) How much pollutant is in the lake after 1 week?

 

(b) Write a function that calculates the amount of pollutant in the lake at any time t ≥ 1.

 

(c) For what time t, in weeks, is the amount of pollutant in the lake at a maximum?

 

(d) 12 weeks after the factory opened, t = 12, the factory is shut down. How many more weeks will it take to remove all the pollutants left by the factory?


 

Quick Find Code For This Problem.

 

Legend: [QUES10LGND]

Solution For Part A.): [QUES10PA]

Solution For Part B.): [QUES10PB]

Solution For Part C.): [QUES10PC]

Solution For Part D.): [QUES10PD]

Solutions For All Parts: [QUES10ANS

 

How To Use Quick Find.

 

-To Use Quick Find, copy (Ctrl+C/Command+C) the code (All the stuff INSIDE the square brackets in the code region above) for which ever section you'd like to go to, then, hit in this order;

 

Ctrl+F

Ctrl+V

Enter/Return

 

or if your on a Mac You'd hit

 

Command+F

Command+V

Enter/Return.

 

Then, your webbrowser should bring you directly to the beginning of the solution indicated by the code, making navigation through longish solutions much easier. Enjoy :]

 

 

Solution


[QUES19LGND]

 

Legend:

 

Italicized =  What is the Question asking

Bold = Important notes and steps that need to be noted

Normal = Fillers, Normal transition text to get from one point of problem to the next.

Colour Coded = Coloured to follow each Equation without getting lost in Fillers.

Green - Denotes that A question has been Solved.

 


 

FUNCTIONS:

Equation 1:               [Pollutants entering the Lake at any time t > 0]

Equation 2:                 [Pollutants being removed from the Lake at any time t > 1]

 

Equation 3:              [Pollutants in the lake at any time > 1}


[QUES10PA]

 

PART A) How much pollutant is in the lake after 1 week?

     To solve this, integration by parts will be used because we want to know HOW MUCH Pollutant is going to be in the Lake at any time. So to solve Part A, we need to use integration by parts for Equation 1.

 

Note: We will ignore Equation 2, because the second equation doesn't take place until t is greater than 1.

 

Integration by Parts Explained:

Let:

u = t               

 

du = 1dt          

Substituting Values:

By Substituting the values back in from what we had, we get:

 

 

Continuing with the Integration, the constant of negative ten over 3 is moved outside the integration symbol. However, moving the fraction outside turns it into a positive fraction.

 

 

Further Integrating, another fraction of -10/3 is brought down from the exponent of "V"

(DON'T FORGET THE PLUS C!!)

 

 

Multiplying the two fractions together, we now get

 

So now to solve A, we evaluate from 0 to 1 and we get a value of 0.4104 tons.

 

 

PART  A  SOLVED

 

Need Further Assistance?

YouTube plugin error


[QUES10PB]

 

PART  B) Write a function that calculates the amount of pollutant in the lake at any time t ≥ 1.

 So we are asked to create a function that calculates the amount of pollutant in the lake at anytime after one week.  The equation WILL look like this:

 

 

where T is any time > 1

 

The reason the equation is P - C is because C is an amount of pollutant being removed from the Lake so that Value will be removed from P to get the Total amount of Pollution remaining in the Lake.

 

 

Blown up, it looks like this:

 

 

NOTE  THAT  I  USED (X-1)^1/2 IN PLACE OF A SQUARE ROOT AS IT IS EASIER TO  WORK  WITH

 

 

The intermediate step would be to integrate them separately, making it easier and less likely to make a mistake.

 

 

Now, the equation will look like this after integration and being evaluated from 1 to any time T is greater than 1

This is the equation that will find the amount of Pollutants in the Lake for any time >1.

 

PART  B  SOLVED


[QUES10PC]

 

PART  C) For what time t, in weeks, is the amount of pollutant in the lake at a maximum?

Now, for this part, to find t manually by hand is not impossible, but it would take a long time, say, more than approximately ten minutes (for me anyway) and that isn't being very efficient, so we'll solve this by calculator!

 

First, we have to realize that the Rate of Pollutants Added to the Lake is P-C,so we'll find t by setting the equation to zero like so.

P-C = 0

 

 

So unfortunately, it would be too long and tedious to find t by hand.

 

So, we turn to the calculator!

 

First, input P-C into Y1 of your calculator.

Second, Set Y2 to 0 on your calculator.

Graph it, and you get this:

 

It's important to note that t = the point where y = 0

 

So to find that accurately, Press 2nd, then Press Calc, and Press 5(Option 5 is Intersect).

 

Now, move the tracer approximately to zero (It won't get to zero on the graph) and press Enter

 

Next, the cursor will already be at 0 because that's what we set Y2 to if you recall.

and out pops a number which would be 11.1004 weeks.

          Therefore, we can conclude that the pollution in the Lake is at a maximum at approximately 11 weeks.

 

PART  C SOLVED


ques10pd

 

PART  D) 12 weeks after the factory opened, t = 12, the factory is shut down. How many more weeks will it take to remove all the pollutants left by the factory?

 

Now we're getting into over-time. This question is by far the longest, but in no circumstance hard, so pay attention,.

Remember, there is a difference between long and tedious compared to hard and difficult.

 

So what we're trying to find out here is the Total Pollution for 0 < t < 12

Knowing that, we'll use the knowledge we gained from Part A and Part B

 

STEP  A

 

 

So this basically means that for the time from 1 to 12, we get this equation to find how much Pollution is in the Lake.

 

= 0.4104 + 6.239

= 6.6494 Tons

 

After 12 Weeks:

After 12 weeks, the factor stops putting pollution into the Lake.

The Clean-Up crew from then takes care of the rest.

So to figure out how many more week's it'll take to clean up,

we let the Total Pollution from Step  A = C and solve for t.

 

STEP  B

Therefore,

 

We let 6.6494 equal C and solve t that is within C.

(NOTE: I skipped a step and set it up as an integral already and also replaced the square root symbol again)

 

Continuing on, we Evaluate the integral from 12 to t.

 

 

Here, we the function at t minus the function at 12 (it's 11 because of t-1).

From here, it's just simple Algebra.

 

After calculating, we get these values. Next, we'll be isolating t.

 

Isolating t, 79.7928 was multiplied by 12, and then 36.4829 was added to it to get 116.2757

 

To get rid of the fraction from the left side, both sides were squared.

116.2757 squared equals 13520.0384

 

Next, we cube root both sides, and we get t-1 = 23.8228.

 

Therefore, by adding one to both sides, t = 24.8228 weeks.

 

Finally, we find that it will take approximately 25 weeks to remove all of the pollution from the Lake.

 

PART  D  SOLVED


[QUES10ANS]

In Closing:

 

PART  A)

 

PART  B)

 

 

PART  C)

 

11.1004 WEEKS

 

 

PART  D)

 

t = 24.8228 weeks.

 


 

 

CONCLUSION.

 

After One week, 0.4104 Tons of Pollutants were dumped into Lake Winnipeg. Following a week, the amount of Pollutants was tracked using the Integral of P-C. The Lake would reach a Maximum amount of Pollutants approximately a little after 11 weeks. The following week, it took approximately 25 additional weeks to remove the remaining Pollutants from the Lake.

 

 

 

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