# Relative Change

1. Nov 1, 2005

### powp

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

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.

2. Nov 1, 2005

### EnumaElish

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.

3. Nov 1, 2005

### HallsofIvy

Staff Emeritus
"Relative change" is the change "relative to" the original value: i.e. the change divided by the orginal amount. If we use $$\Delta R$$ and $$\Delta F$$ to mean the changes in R and F respectively, then their "relative changes" are $$\frac{\Delta R}{R}$$ and $$\frac{\Delta F}{F}$$.
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 $$dR= \frac{\Delta R}{R}$$ so the actual change is $$\Delta R= Rdr$$ 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 $$\frac{dF}{dt}= 4kR^3\frac{dR}{dt}$$. dividing that by F= kR4, $$\frac{\frac{dF}{dt}}{F}= 4\frac{\frac{dR}{dt}}{R}$$ which says exactly that "the relative (rate of) change in F is equal to the relative (rate of) change in R".

Last edited: Nov 1, 2005
4. Nov 1, 2005

### powp

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?

5. Nov 2, 2005

### powp

Can sombody help with this part of the previous question

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

6. Nov 2, 2005

7. Nov 3, 2005

### powp

Thanks.

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

Thanks