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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

    F=kR^4

    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

    Tide

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    Science Advisor
    Homework Helper

    Relative change is the ratio of how much a quantity changes to its starting value.

    In your problem, F varies as the 4th power of R which you wrote as [itex]F = k R^4[/itex]. Here's one way to get the needed "formula."

    Clearly, if F changes, there will be a corresponding change in the value of R and vice versa. Suppose these changes are [itex]\delta F[/itex] and [itex]\delta R[/itex]. Using the relation between F and R, those changes result in

    [tex]F + \delta F = k \left(R + \delta R\right)^4[/tex]

    Now expand the expression on the right (we don't have to do the whole thing because we're going to assume that the changes are small!) so we have, approximately

    [tex]F + \delta F = k \left(R^4 + 4 R^3 \delta R + \cdot \cdot \cdot\right)[/tex]

    Snce [tex]F = k R^4[/itex] those two terms cancel and to first order in the small variations we find

    [tex]\delta F = 4 k R^3 \delta R[/tex]

    Finally, divide both sides by [itex]F = k R^4[/itex] so that

    [tex]\frac {\delta F}{F} = 4 \frac {\delta R}{R}[/tex]

    which compares the relative error in F to the relative error in R. We could have accomplished the same thing using calculus but I wasn't sure if that is in your toolkit.
     
  4. Sep 29, 2009 #3
    This post is pretty old but I came across the same problem. Using calculus, the relative change would be

    F'(R)/F(R)

    if F(R) = kR^4, then F'(R) = 4KR^3 dR

    So the relative change is:

    4kR^3 dR
    -----------
    kR^4

    What you are left with is

    4 * (dR/R)

    This means that the relative change in F is 4 times the relative change in R.

    Now, if there is 5% increase, then the change in radius is dR/R = 5% = 0.05. So you would plug that back into your relative change equation:

    4 * (0.05) = .2 = 20%

    If there is a 5% increase in the radius, then the flow increases by 20%. Which follows that the relative change in F (20%) is 4 times the change in radius ( 4 * 5%).


    Is this forum LaTeX supported?
     
    Last edited: Sep 29, 2009
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