Riemann Sum to find the time to fill a container

[songoku]

Homework Statement

The function p = f(t); 2 ≤ t ≤ 36 shows the rate (liter/hour) at which water flows into a container at time t hours.
(a) Write an equation you would solve to find when the container holds 22 fewer liters than it does at time t = 9. Do not use indefinite integral

(b)
(i) Use the graph of p = 12 - 3√t to approximate the time when the container holds 22 fewer liters than it does at time t = 9.

(ii) Find the exact times. Do not use integral

Relevant Equations

Riemann Sum

(a) I imagine there are several rectangles to represent the area under graph of p vs t then I try to make equation for the total area. Since the question asks about time when the container holds 22 fewer liters than it does at time t = 9, I think the total area of rectangles starting from t = b until t = 9 will be 22 liters, where b is the time I want to find. So:
$$\Sigma_{t=b}^{9} ~ f(t).\Delta t=22$$

Is that correct?

(b) I tried to calculate the area of rectangles using the right end-point and add the area starting from t = 9 and going backwards until I get the total area = 22. I did it twice, one by using Δt = 1 but I felt this is not accurate enough so I did it again using Δt = 0.5

t	f(t)	Cumulative Area
9	3	1.5
8.5	3.2535	3.12675
8	3.5147	4.8841
7.5	3.7841	6.77615
7	4.0627	8.8075
6.5	4.3514	10.9832
6	4.6515	13.30895
5.5	4.9643	15.7911
5	5.2917	18.43695
4.5	5.636	21.25495
4	6	24.25495

So the approximate time is t = 4 hours

Is that correct?

(ii) To find the exact time, I take the interval for t to be [b, 9] and $\Delta t = \frac{9-b}{n}$ where $n$ is the number of rectangles

$$22=\lim_{n \to \infty}~\Sigma f(t).\Delta t$$
$$22=\lim_{n \to \infty}~\Sigma (12-3\sqrt {t}).\left(\frac{9-b}{n} \right)$$
$$22=\lim_{n \to \infty}~\left(\frac{9-b}{n} \right) (\Sigma_{t=b}^{9} 12 - 3\Sigma_{t=b}^{9} \sqrt{t})$$
$$22=\lim_{n \to \infty}~\left(\frac{9-b}{n} \right) (12 (10-b) - 3\left(\sqrt{b} + \sqrt{b+\left(\frac{9-b}{n}\right)}+\sqrt{b+2\left(\frac{9-b}{n} \right)}+\sqrt{b+3\left(\frac{9-b}{n} \right)}+...+\sqrt{9}\right)$$

Is this correct? If yes, how to simplify the sum of root?

Thanks

 

[BvU]

Hello again,

I understand the function is given as $\ p(t) = 3-\sqrt t\$ in the exercise, not only for part b), but also for part a) ?

I wonder why your table is ascending order in time (somewhat confusing , at least for me ).

From column 3 I gather you approach the integral as a lower Riemann sum (red line). There is also an upper Riemann sum, and the integral is more closely approached if you take the average of the two ...

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[Office_Shredder]

When it says do not use an indefinite integral in the first part, is that because they want you to write down a definite integral instead?

 

[songoku]

Hello again,

I understand the function is given as $\ p(t) = 3-\sqrt t\$ in the exercise, not only for part b), but also for part a) ?

Hello BvU

I am not sure but I think the equation $p=12-3\sqrt{t}$ (not $p=3-\sqrt{t}$) is for question (b) only, not for question (a)

I wonder why your table is ascending order in time (somewhat confusing , at least for me ).

You mean descending in time? I tried to find the total area to be equal to 22 starting from t = b to t = 9 so I think it will be easier for me if I do it backwards, starting from t = 9

From column 3 I gather you approach the integral as a lower Riemann sum (red line). There is also an upper Riemann sum, and the integral is more closely approached if you take the average of the two ...

View attachment 282106​

You mean the approximation will be better if I use the midpoint of the rectangle (rather than the left end-point or the right end-point) ?

When it says do not use an indefinite integral in the first part, is that because they want you to write down a definite integral instead?

I am not so sure about this one, the instruction from the question is only that. I tried to avoid integration so I tried to find the total area of rectangles. Is it possible to answer without involving integral or must be related to integration?

Thanks

 

[BvU]

I am not sure but I think the equation p=12−3t (not p=3−t) is for question (b) only, not for question (a)

Yes, of course (not $p=3−\sqrt t$) -- typo.

And I think you are correct. So the answer for part a) is most likely something in the form of $$22\; \text {L} \ = \int_{t_1}^{t_2}f(t)\, dt$$ ...

So in part b i) you are asked to approximate and that means you can do as you did. Indeed, average of upper and lower sum is more accurate. (Even more acurate will be using Simpson's rule)

I then do not understand how an exact answer can be given in part b ii) without actually using the integral.

And again I think I found a serious flaw in the exercise. Inspired by the $p=12-3\sqrt{t}\$ in combination with $2 ≤ t ≤ 36$ !

 

[songoku]

I then do not understand how an exact answer can be given in part b ii) without actually using the integral.

If we only consider the working, is my approach correct?

Thanks

 

[Office_Shredder]

The right approach to compute the exact time is to compute the integral of p(t) and use that. Writing it as the limit of an infinite sum and trying to compute that from first principles is crazy.

Your approximation method seems fine, you could probably be a bit more careful about whether you like t=4 or 4.5 or maybe just pick some point between them but the principle of what you did looks ok.

 

[BvU]

The right approach to compute the exact time is to compute the integral of p(t) and use that. Writing it as the limit of an infinite sum and trying to compute that from first principles is crazy.

I completely agree, but what can be meant with

(ii) Find the exact times. Do not use integral

Must the poor student really execute the Riemann sum limit (which we know IS doing the integration) ?(In view of the last line in my #5 I do find the plural 'times' ironic )

So the approximate time is t = 4 hours

Is that correct?

Doing the integral shows it is the exact time, so: yes.

Let's dive into this Riemann sum with the knowledge that $b$ is equal to 4

(ii) To find the exact time, I take the interval for t to be [b, 9] and $\Delta t = \frac{9-b}{n}$ where $n$ is the number of rectangles

$$22=\lim_{n \to \infty}~\Sigma f(t).\Delta t$$

Perhaps more clearly written as $$\lim_{n \to \infty}\sum_{i = 0}^{n-1} \, f(t_i) \,\Delta t $$
Where you also want to tell us that $\ t_i = b + (i+{1\over 2}) \Delta t\$ -- (or use lower or upper Riemann sum, whatever: be precise !)

$$22=\lim_{n \to \infty}~\Sigma (12-3\sqrt {t}).\left(\frac{9-b}{n} \right)$$
$$22=\lim_{n \to \infty}~\left(\frac{9-b}{n} \right) (\Sigma_{t=b}^{9} 12 - 3\Sigma_{t=b}^{9} \sqrt{t})$$
$$22=\lim_{n \to \infty}~\left(\frac{9-b}{n} \right) (12 (10-b) - 3\left(\sqrt{b} + \sqrt{b+\left(\frac{9-b}{n}\right)}+\sqrt{b+2\left(\frac{9-b}{n} \right)}+\sqrt{b+3\left(\frac{9-b}{n} \right)}+...+\sqrt{9}\right)$$

Is this correct? If yes, how to simplify the sum of root?

No. The $\ {9-b\over n} \$ should not precede the $\ 12(9-b) \$ (not 10 but 9!) . The constant $12$ simply yields $n$ times $\ 12\Delta t\$.

Then the 'sum of root' as you designate it: from the integral we know $$\int _b^9 \, 3\sqrt t \,dt = 2\, \left. t^{3\over 2} \right \rvert_b^9 \ = 2(27-8) $$ and there may (must be) a way to end up with that using limits and Taylor, but I leave that to the interested reader ...

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[songoku]

No. The $\ {9-b\over n} \$ should not precede the $\ 12(9-b) \$ (not 10 but 9!) . The constant $12$ simply yields $n$ times $\ 12\Delta t\$.

I understand why it should be 9, not 10 but sorry I don't understand what you mean by "The $\ {9-b\over n} \$ should not precede the $\ 12(9-b) \$ "

 

[songoku]

And again I think I found a serious flaw in the exercise. Inspired by the $p=12-3\sqrt{t}\$ in combination with $2 ≤ t ≤ 36$ !

By this, do you mean the rate will be negative after t = 16 hours?

Thanks

 

[BvU]

Correct. And since the claim is that the expression is valid for a much longer period (perhaps by mistake, 36 instead of 16?), there is a second point where the holdup is 22 L

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[songoku]

Thank very much for the help and explanation BvU and Office_Shredder

 

[BvU]

I understand why it should be 9, not 10 but sorry I don't understand what you mean by "The $\ {9-b\over n} \$ should not precede the $\ 12(9-b) \$ "

$$22=\lim_{n \to \infty}~\left(\frac{9-b}{n} \right) \Biggl ( \Sigma_{t=b}^{9} \; 12 \quad ... $$
The summation of $n$ identical terms yields $12n$. With the factor in front you get $12(9-b)$.

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