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Troublesome Integral Part two

  1. Mar 30, 2014 #1

    I was not certain as to whether I should begin a new thread or not, but I am still working on the integral found in the post https://www.physicsforums.com/newreply.php?do=postreply&t=743661 [Broken]

    If a moderator deems this should be merged with the linked thread, please do so. I felt as though I should begin a new thread, because the other starting becoming littered; and also, I have a lot more information I need to provide.

    Okay, so I have just revisited this integral, and I still can't seem to get the correct answer. So, here is my work, starting form the beginning.

    Here is the problem:

    An outdoor decorative pond in the shape of a hemispherical tank is to be filled with water pumped into the tank through an inlet in its bottom. Suppose that the radius of the tank is R10 ft, that water is pumped in at a rate of pi ft^3/min, and that the tank is initially empty. See Figure 3.2.6. As the tank fills, it loses water through evaporation. Assume that the rate of evaporation is proportional to the area A of the surface of the water and that the constant of proportionality is k = 0.01

    (a) The rate of change dV/dt of the volume of the water at time tis a net rate. Use this net rate to determine a differential equation for the height h of the water at time t.The volume of the water shown in the figure is [itex]V = \pi R [h(t)]^2 - \frac{1}{3} \pi [h(t)]^3[/itex], where R=10. Express the area of the surface of the water A=pi*r^2 terms of h.

    (b) Solve the differential equation in part (a). Graph the

    (c) If there were no evaporation, how long would it take
    the tank to fill?

    (d) With evaporation, what is the depth of the water at
    the time found in part (c)? Will the tank ever be
    filled? Prove your assertion.


    We are given the volume as a function of the height: [itex]V = \pi R [h(t)]^2 - \frac{1}{3} \pi [h(t)]^3[/itex], where R=10, the radius of the tank. I shall denote h(t) as h, for convenience. Differentiating this with respect to time, I get

    [itex]\frac{dV}{dt} = 2 \pi R h \frac{dh}{dt} - \pi h^2 \frac{dh}{dt}[/itex]

    We are given that [itex]\frac{dV}{dt} = \pi - kA[/itex], where A=pi*r^2. So, the differential equation becomes,

    [itex]\pi - k \pi r^2 = 2 \pi R h \frac{dh}{dt} - \pi h^2 \frac{dh}{dt}[/itex]

    Dividing by pi, and solving for [itex]\frac{dh}{dt}[/itex],

    [itex]1 - kr^2 = (2 R h - h^2 )\frac{dh}{dt}[/itex]

    Dividing both sides by [itex]2 R h - h^2 [/itex],

    [itex]\frac{1 - kr^2}{2 R h - h^2} = \frac{dh}{dt} \implies[/itex]

    [itex]\frac{dh}{dt} = \frac{1 - kr^2}{2 R h - h^2} [/itex]

    Using right-triangle trigonometry, I can relate the height of the water to the radius of the water at any given time, which gives me

    [itex]R^2 = \left( R^2 - h \right)^2 + r^2 \implies[/itex]

    [itex]r^2 = 2Rh - h^2 [/itex]

    Therefore, the differential equation becomes

    [itex]\frac{dh}{dt} = \frac{1 - k(2Rh - h^2)}{2 R h - h^2}[/itex]

    Substituting in k=1/100 and R=10,

    [itex]\frac{dh}{dt} = \frac{1 - \frac{1}{100}(20h - h^2)}{20 h - h^2}[/itex].

    Multiplying the numerator and denominator by 100,

    [itex]\frac{dh}{dt} = \frac{100 - 20h + h^2}{100(20 h - h^2)}[/itex].

    Factoring the numerator,

    [itex]\frac{dh}{dt} = \frac{(h-10)^2}{100(20 h - h^2)}[/itex]

    Multiplying both sides of the equation by the reciprocal of the right-hand side of the equation, we get

    [itex]\frac{100(20 h - h^2)}{(h-10)^2} \frac{dh}{dt} = 1[/itex] (1)


    Now, according the answer key, the numerator should actually be [itex]100h(h-20)[/itex]. I have tried, and tried--and tried!--but I can not reproduce what the answer has. Somewhere I must be missing a negative sign, and I can not find it.

    I am pretty certain that I am wrong, because when I went to go answer parts (c) and (d), after having used separation of variables on (1), which gave me a solution, I got wrong answers.

    For part (c), I was able to calculate the time 666.7, which is correct according to the answer key; however, when I went to calculate the height at t=666.7, using the incorrect equation I derived from equation (1), I got a height of around 25. The reason why this is wrong is because you are supposed to show in part (d) that the height of the water in the tank can never exceed 10.

    Could someone possibly help me?

    Attached Files:

    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Mar 30, 2014 #2


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    With a numerator 100h(h-20), what would the sign of dh/dt be when h < 10? Does that make sense?

    Please post your working that led to wrong answers for (c) and (d).

    For part (d), what does your equation say about dh/dt when h = 10?
  4. Mar 30, 2014 #3
    If h < 10, then the sign would be negative, but what does that have to do with anything? So, are you agreeing that equation (1) was incorrectly derived, and that the numerator should actually be 100h(h-20)?
  5. Mar 30, 2014 #4


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    So is the water level going up or down?
  6. Mar 31, 2014 #5
    It would be decreasing. So, because we know that it is decreasing, we can arbitrarily insert a negative sign to change the numerator?
  7. Mar 31, 2014 #6

    D H

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    It has a whole lot to do with it. Rhetorical question: Does a negative ##\dot h## make any sense here? The answer is no. Developing a qualitative feel for physical problems is important. There won't be any answer keys once you graduate from college.

    Your equation (1) looks quite fine to me. (Aside: I would have used (10-h)2 rather than (h-10)2 for the denominator, but that's personal preference.) What haruspex was hinting at was that the answer key was incorrect, or perhaps you read it incorrectly.

    Your problem lies somewhere else, in the work that you haven't shown us.
  8. Mar 31, 2014 #7
    And why can't [itex]\dot{h}[/itex] be negative? Wouldn't the physical interpretation be that the height of the water is decreasing with time?

    Oh, wait. Does the actual problem lie with the fact that the initial height is 0, and that it wouldn't make sense for the height to be decreasing for all time, when there isn't any height initially?

    Okay, he is he rest of my work for part (b), (c), and (d):


    Part (b): If I integrate (1) with respect to time, I get

    [itex]\int \frac{100(20h-h^2)}{(h-10)^2} \frac{dh}{dt}~dt = \int dt \implies[/itex]

    [itex]100 \int \frac{20h-h^2}{(h-10)^2} ~dh = \int dt[/itex]

    Let [itex]u = h-10 \implies \frac{du}{dh} = 1 \implies du = dh[/itex]. Solving for [itex]h[/itex], we get [itex]h = u+10[/itex]. Substituting this into the integral, we get

    [itex]100 \int \frac{20(u+10)-(u+10)^2}{(u)^2} du = \int dt \implies[/itex]

    [itex]100 \int \frac{20u +200-u^2-20u-100}{u^2} ~dh = t + c \implies[/itex]

    [itex]100 \int \left(\frac{100}{u^2}-\frac{u^2}{u^2} \right)~du = t + c \implies[/itex]

    [itex]100 \int \left(\frac{100}{u^2}-1 \right)~du = t + c \implies[/itex]

    [itex]100\left(-\frac{100}{u} - u \right) = t+ c [/itex] I think I

    Putting back in u = h - 10,

    [itex]100 \left(-\frac{100}{h-10}-(h-10) \right) = t+c \implies[/itex]

    [itex]100 \left(\frac{100}{10-h} + 10 -h \right) = t + c \implies[/itex]

    [itex]100 \left( \frac{100}{10-h} + \frac{(10-h)^2}{10-h} \right) = t + c \implies[/itex]

    [itex]100 \left( \frac{100+(10-h)^2}{10-h} \right) = t + c[/itex].

    Using the initial condition, that h(0)=0, I can determine the constant of integration:

    [itex]100 \left( \frac{100+(10-0)^2}{10-0} \right) = 0 + c \implies[/itex]

    [itex]100 \left( \frac{200}{10} \right) = c \implies[/itex]


    Again, according to the textbook, c should actually be 1000, but you mentioned that the answer key may be incorrect. With this in mind, the equation becomes

    [itex]100 \left( \frac{100+(10-h)^2}{10-h} \right) = t + 2000[/itex].

    I had my computer solve this equation for h; although I realized afterwards that the equation was quadratic, and I could have used the quadratic formula to find h. At any rate, my computer gave me

    [itex]h(t) = \frac{1}{2} \sqrt{\frac{t^2 +4000t}{1000}} - \frac{t}{200}[/itex].

    This is the positive root. I neglected the negative root, seeing as we generally measure heights with positive values.


    Part (c): If there is no evaporation, then there is no water leaving the system at any time; so, the net rate is due wholly to the water flowing in, at a rate of pi cubic feet per minute. As such, the differential equation in this case is

    [itex]\frac{dh}{dt} = \frac{1}{20 h - h^2}[/itex]

    This is a separable differential equation. Using this idea, we can solve it:

    [itex](20h-h^2)\frac{dh}{dt}=1 \implies[/itex]

    [itex]\int (20h-h^2)\frac{dh}{dt}~dt = \int dt \implies[/itex]

    [itex]10h^2 - \frac{1}{3} h^3 = t + c [/itex]

    Applying the same initial condition as before,

    [itex]10(0)^2-\frac{1}{3}(0)^3=0+c \implies[/itex]

    [itex]c=0 \implies[/itex]

    [itex]10h^2 - \frac{1}{3} h^3 = t[/itex]

    I tried using my computer to solve this, but it said that there was no explicit form.

    To find the time at which the water occupies the entire, we have to solve the equation for h=10:

    [itex]10(10)^2 - \frac{1}{3}(10)^3 = t \implies[/itex]

    [itex]t = 667[/itex], which is the correct answer, according to the textbook.


    Part (d): When we consider the effects of evaporation, the height of the water at the time calculated above is

    [itex]h(667) = \frac{1}{2} \sqrt{\frac{(667)^2 +4000(667)}{1000}} - \frac{667}{200} = 25[/itex]

    And this is where I stopped, because according to the answer key, the height of the water cannot exceed 10, when there is evaporation .
    Last edited: Mar 31, 2014
  9. Mar 31, 2014 #8

    D H

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    Your result ##100\left(\frac {100}{10-h}+10-h\right) = t+2000## is correct. Your answer key may have integrated the left hand side differently. If the answer key has integrated the differential equation as ##100\left(\frac {100}{10-h}-h\right) = t+c## then their c should be 1000. Your constant of integration and their constant of integration have to differ if that's what they did.

    Let's see what your result ##100\left(\frac {100}{10-h}+10-h\right) = t+2000## says t should be when h=5. Evaluating at h=5 yields ##100\left(\frac {100}{10-5}+10-5\right) = t+2000##, or ##t=500##.

    That's obviously wrong. Plug in t=500. The result won't be 5. You did something wrong in inverting your correct expression for time as function of height.
  10. Mar 31, 2014 #9
    So, are you saying that I had my computer solve the wrong equation?
  11. Mar 31, 2014 #10

    D H

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    There's no telling because you didn't tell us how you went from ##100\left(\frac {100+(10-h)^2}{10-h}\right)=t+2000## to ##h=\frac 1 2 \sqrt{\frac{t^2+4000t}{1000}} - \frac t {200}##.
  12. Mar 31, 2014 #11
    Oh, okay. Here is the matlab code I used

    Code (Text):
    syms h t c h1

    eq = 100*((100)/(10-h)+10-h)-t-2000;

    s = solve(eq,h);
    where s is a matrix containing the positive and negative root, and, as I mentioned above, I neglected the negative root.
  13. Mar 31, 2014 #12
  14. Mar 31, 2014 #13

    D H

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  15. Mar 31, 2014 #14
    Really? Oh, so I was typing it into matlab incorrectly, but I typed it in wolfram correctly?
  16. Mar 31, 2014 #15

    D H

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    You missed a possible explanation: You typed it into matlab correctly but you incorrectly copied the result from matlab.
  17. Mar 31, 2014 #16
    Okay, everything is as it should be. Thank you very much for the help.
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