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Math Help - Avg. Value of a function: Word Problem

  1. #1
    Member RedBarchetta's Avatar
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    Avg. Value of a function: Word Problem

    1. The problem statement, all variables and given/known data

    The velocity v of blood that flows in a blood vessel with radius R and length l at a distance r from the central axis is v(r), where P is the pressure difference between the ends of the vessel and η is the viscosity of the blood.

    (a)Find the average velocity v_ave (with respect to r) over the interval 0 ≤ r ≤ R.

    (b)Compare the average velocity v_ave with the maximum velocity v_max.(v_ave/v_max)

    So I need to set the problem up, into an integral form, to find the avg velocity for part A. The integration limits for the integral should be from zero to Big R, multiplied by (1 over big R) to get our average value. Now the problem is, every thing in the equation except the (1/4) is a variable. How would I even begin to integrate this?

    Thanks.
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  2. #2
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    Quote Originally Posted by RedBarchetta View Post
    1. The problem statement, all variables and given/known data

    The velocity v of blood that flows in a blood vessel with radius R and length l at a distance r from the central axis is v(r), where P is the pressure difference between the ends of the vessel and η is the viscosity of the blood.

    (a)Find the average velocity v_ave (with respect to r) over the interval 0 ≤ r ≤ R.

    (b)Compare the average velocity v_ave with the maximum velocity v_max.(v_ave/v_max)

    So I need to set the problem up, into an integral form, to find the avg velocity for part A. The integration limits for the integral should be from zero to Big R, multiplied by (1 over big R) to get our average value. Now the problem is, every thing in the equation except the (1/4) is a variable. How would I even begin to integrate this?

    Thanks.
    \int_{0}^{R}\frac{P}{4\eta l}(R^2-r^2)dr

    The only variable in the integrand is little r the rest are parameters(measured constants) so we are integrating with repsect to little r.

    \frac{P}{4\eta l} \int_{0}^{R} (R^2-r^2)dr=\frac{P}{4\eta l}\left( rR^2-\frac{r^3}{3}\right)_{0}^{R}=\frac{P}{4\eta l}\left( R^3-\frac{R^3}{3}\right)=\frac{2R^3P}{12\eta l}=\frac{R^3P}{6\eta l}
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  3. #3
    Member RedBarchetta's Avatar
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    Ahhh, so the other letters are just constants. Well that certainly makes the intergrating _a lot_ easier.

    For part B, where do you get the v_max from?
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  4. #4
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    Quote Originally Posted by RedBarchetta View Post
    Ahhh, so the other letters are just constants. Well that certainly makes the intergrating _a lot_ easier.

    For part B, where do you get the v_max from?
    You need to maximize your velocity function

    v(r)=\frac{P}{4\eta l}(R^2-r^2)

    so taking the derivative we get

    \frac{dv}{dr}=\frac{P}{4\eta l}(-2r)

    setting equal to zero and solving for r, we get r=0. The maximum velocity occurs when r=0.

    v_{max}=v(0)=\frac{P}{4\eta l}(R^2-0^2)=\frac{PR^2}{4\eta l}
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