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Math Help - Average speed over an interval.

  1. #1
    Member sinewave85's Avatar
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    Average speed over an interval.

    Blood flows through an artery of radius R. At a distance r from the central axis of the artery, the speed of the blood flow is given by S(r)=k(Rē - rē). Show that the average speed of the blood flow is one-half the maximum speed.

    My problem is that I can't prove that -- I keep proving that it is two-thirds the maximum speed. Can anyone tell me where I am going wrong?

    The maxiumum speed is kRē at r = 0 and the function reaches zero when r = R.

    AverageSpeed = \frac{1}{b-a}\int_{a}^{b}f(x)dx

    AS = \frac{1}{R-0}\int_{0}^{R}k(R^{2}-r^{2})dr

    AS = \frac{k}{R}\left(R^{2}r-\frac{1}{3}r^{3}\right)|_{0}^{R}

    AS=\frac{k}{R}\left(\frac{2}{3}R^{3}\right)

    AS= (2/3)kR^{2}

    Where did I go wrong? I just can't see it. Thanks for the help!
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  2. #2
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    Hi

    Because you are working in polar coordinates

    AS = \frac{1}{\pi \:R^2}\:\int_{0}^{2\pi} \int_{0}^{R} k(R^{2}-r^{2})\:rdr \: d\theta
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by running-gag View Post
    Hi

    Because you are working in polar coordinates

    AS = \frac{1}{\pi \:R^2}\:\int_{0}^{2\pi} \int_{0}^{R} k(R^{2}-r^{2})\:rdr \: d\theta
    Ok, bear with me. I am not that comfortable working with polar coordinates, so I have spent the last few hours with one of my old textbooks brushing up, but that has not clarified all of my questions.

    1. First of all, how do you know that solving this problem requires the use of polar coordinates? (I warned you that this would be involved ) The example problems in the book that use this equation don't involve polar coordinates, at least not by name, so if you could explain the your mental process in addressing the problem, that would be wonderful.

    2. Second, where does that extra r here: AS = \frac{1}{\pi \:R^2}\:\int_{0}^{2\pi} \int_{0}^{R} k(R^{2}-r^{2})r dr\: d\theta
    come from?

    3. Third, how do you integrate an expression twice? Do you integrate the expression and then combine the limits of integration? a = 0, b = 2piR?

    I know this is a lot to ask, but you guys give more thorough explanations that what my teacher can provide and I want to understand what I am doing.
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  4. #4
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    I used polar coordinates because S is given as an expression of radius r

    If you do not know about polar coordinates maybe this link could help you
    Polar coordinate system - Wikipedia, the free encyclopedia

    I have also performed a sketch (see below)
    A point M is given by 2 coordinates r and \theta


    Starting from M, when doing a little displacement dr along \vec{e_r} point M is sliding along dr \vec{e_r}

    When doing a little dispalcement d \theta point M is sliding along r d \theta \vec{e_\theta}



    The elementary surface dS (which is dx dy in cartesian coordinates) is therefore
    dS = dr \times r d \theta = r dr d \theta
    This is why there is an "extra r" in the expression of the integral

    For instance if you want to compute the area of the disk

    A = \int_{0}^{2\pi} \int_{0}^{R} r dr\: d\theta

    A = \int_{0}^{2\pi} d\theta\:\int_{0}^{R} r dr

    A = 2\pi \frac{R^2}{2} = \pi R^2


    Integrating S(r) over the disk gives

    \int_{0}^{2\pi} \int_{0}^{R} S(r) r dr\: d\theta

    S being only an expression with r, the double integral can be split into 2 parts

    \int_{0}^{2\pi} d\theta \: \int_{0}^{R} S(r) r dr

    I don't know if my explanations are useful ... and understandable
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  5. #5
    Member sinewave85's Avatar
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    Quote Originally Posted by running-gag View Post
    I don't know if my explanations are useful ... and understandable
    Very -- but unfortunately you are way out of my league mathematically. I thought to check and double integrals and integrals of polar coordinates are a course ahead of where I am, introduced late in Calculus III at my school. I am just starting Calculus II and would probably have some explaining to do if I turned that solution in -- even to a practice problem.

    You have been more than helpful and very patient , and your solution certainly works beautifully (I went ahead and read the chapter on integrals of polar coordinates and they certainly seem to offer the most straightforward solution to this problem). However, if you will bear with me a little longer, is there a way to work this problem that does not involve these more advanced topics?
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    Since S is only r-dependent, I do not see anything else more simple than

    AS = \frac{\int_{0}^{R} S(r) r dr}{\int_{0}^{R} r dr}
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  7. #7
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    Quote Originally Posted by sinewave85 View Post
    Blood flows through an artery of radius R. At a distance r from the central axis of the artery, the speed of the blood flow is given by S(r)=k(Rē - rē). Show that the average speed of the blood flow is one-half the maximum speed.

    My problem is that I can't prove that -- I keep proving that it is two-thirds the maximum speed. Can anyone tell me where I am going wrong?

    The maxiumum speed is kRē at r = 0 and the function reaches zero when r = R.

    AverageSpeed = \frac{1}{b-a}\int_{a}^{b}f(x)dx

    AS = \frac{1}{R-0}\int_{0}^{R}k(R^{2}-r^{2})dr

    AS = \frac{k}{R}\left(R^{2}r-\frac{1}{3}r^{3}\right)|_{0}^{R}

    AS=\frac{k}{R}\left(\frac{2}{3}R^{3}\right)

    AS= (2/3)kR^{2}

    Where did I go wrong? I just can't see it. Thanks for the help!
    there is no need for polar coordinates in doing this problem. in fact, you did it correctly.

    the result of the average speed being 1/2 the max would be true if the speed were S(r) = k(R-r) instead of k(R^2 - r^2)

    I suggest there may be a mistake in the model equation for S(r).
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  8. #8
    Member sinewave85's Avatar
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    Running-gag, thanks so much for your dedication to helping me with this problem. You really gave me a lot to think about and I broadened my mathmatical horizons.

    Skeeter, thanks for your input as well. I am always hesitant to think that there is something wrong with the origianl problem, but in this case it looks like there may be, at least in the context of the lesson it is given in.

    I think that it looks like I need to bring this problem to the teacher's attention, since it does not seem to work as a Calc II level problem.

    Once again, thank you for all of your help and time!
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  9. #9
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    This is a wise decision
    Please keep us informed !
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