# 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

### Tide

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 $F = k R^4$. 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 $\delta F$ and $\delta R$. Using the relation between F and R, those changes result in

$$F + \delta F = k \left(R + \delta R\right)^4$$

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

$$F + \delta F = k \left(R^4 + 4 R^3 \delta R + \cdot \cdot \cdot\right)$$

Snce $$F = k R^4[/itex] those two terms cancel and to first order in the small variations we find [tex]\delta F = 4 k R^3 \delta R$$

Finally, divide both sides by $F = k R^4$ so that

$$\frac {\delta F}{F} = 4 \frac {\delta R}{R}$$

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.

3. Sep 29, 2009

### HighTek

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