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Relative Change

  1. Nov 1, 2005 #1
    Hello All

    I have this problem that I have no idea how to do.

    F = flux or the volume of blood to flow past a point

    R = radius


    Show that the relative change in F is about four times the relative change in R. How will a 5% increase in radius affect the flow of Blood??

    How in the world do I do this?? What is Relative Change?? My text book does not have this term

    I think I need to have 2 equations F = kR^4 and R = (F/k)^(1/4). Not sure what to do.

    Please help.
  2. jcsd
  3. Nov 1, 2005 #2


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    I think by relative change they only mean percentage change, or absolute change divided by the level: rel. change (up to time t) = [x(t) - x(0)]/x(0) = x(t)/x(0) - 1.
  4. Nov 1, 2005 #3


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    "Relative change" is the change "relative to" the original value: i.e. the change divided by the orginal amount. If we use [tex]\Delta R[/tex] and [tex]\Delta F[/tex] to mean the changes in R and F respectively, then their "relative changes" are [tex]\frac{\Delta R}{R}[/tex] and [tex]\frac{\Delta F}{F}[/tex].
    I don't know how you should do this because I don't know what level you are at and what "mechanisms" you have available to you.

    Basic but harder way: Since you have F= kR4, if "dR" is the relative change in R, then [tex]dR= \frac{\Delta R}{R}[/tex] so the actual change is [tex]\Delta R= Rdr[/tex] and the new value for R (after the change) is R+ Rdr= R(1+dr). Then the new value for F is k(R(1+dr))4= kR4(1+ dr)4. Multiplying out (1+ dr)4= 1+ 4dr+ 6(dr)2+ 4(dr)3+ (dr)4 so the new value of F is kR4(1+ 4dr+ 6(dr)2+ 4(dr)3+ (dr)4). Subtracting of the old value, kR4 tells us that the actual change in F was kR4(4dr+ 6(dr)2+ 4(dr)3+ (dr)4) (we just removed that "1" inside the parentheses).
    The relative change then is kR4(4dr+ 6(dr)2+ 4(dr)3+ (dr)4) divided by kR4 which is 4dr+ 6(dr)2+ 4(dr)3+ (dr)4. If "dr" is relatively small, then those powers of dr will be even smaller- the largest term will be 4dr: that is, "about four times the relative change in R."

    More sophisticated and easier way. Differentiate F= kR4 with respect to time to get [tex]\frac{dF}{dt}= 4kR^3\frac{dR}{dt}[/tex]. dividing that by F= kR4, [tex]\frac{\frac{dF}{dt}}{F}= 4\frac{\frac{dR}{dt}}{R}[/tex] which says exactly that "the relative (rate of) change in F is equal to the relative (rate of) change in R".
    Last edited by a moderator: Nov 1, 2005
  5. Nov 1, 2005 #4
    I think this is the way we are expected to do it.

    This makes sense except for the one thing. This may be a silly question but can you just divide a function by another function without doing the same to both sides?

    It seems like you are do the following

    A = 2 + B is divided by C = D + 2 and you do the following

    A 2 + B
    - = ------
    C D + 2

    Don't you need to divide both sides by the same value? Or is it since C does equal D + 2 this is allowed?

    Thanks for your help
  6. Nov 2, 2005 #5
    Can sombody help with this part of the previous question

    How will a 5% increase in radius affect the flow of Blood??

    Please Please Pretty Please!!
  7. Nov 2, 2005 #6


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    Read HallsOfIvy's post C-A-R-E-F-U-L-L-Y.
  8. Nov 3, 2005 #7

    I have reread it and still have no clue. Can anybody give me a hint??

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