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

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**[QUES19LGND]**

*Italicized =* *What is the Question asking*

**Bold = Important notes and steps that need to be noted**

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**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]**

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

**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:

**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.

Need Further Assistance?

**[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.

**[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.

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.

**ques10pd**

**P****ART 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.

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.**

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.

**[QUES10ANS]**

**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.**