1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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


    User Avatar
    Science Advisor
    Homework Helper

    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


    User Avatar
    Science Advisor

    "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


    User Avatar
    Science Advisor
    Homework Helper

    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??

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook