Poiseuille's Law: flux of blood flow is proportional to R^4

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Jakotheshadows

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"When blood flows along a blood vessel, the flux F (the volume of blood per unit time that flows past a given point) is proportional to the fourth power of the radius R of the blood vessel: F = kR^4. A partially clogged artery can be expanded by an operation called angioplasty, in which a balloon-tipped catheter is inflated inside the artery in order to widen it and restore the normal blood flow. Show that the relative change in F is about four times the relative change in R. How will a 5% increase in the radius affect the flow of blood?"

The chapter doesn't illustrate what it might mean by "relative change", so I'm just going to go with the change that I know.

Starting with F = kR^4.. dF = (4kR^3)*(dR).. And that is about as far as I can seem to take it without mindlessly manipulating the expression to no end.
 


I think that we may read "relative change" as "percent change". It's the ratio of change to starting amount.

If R2 is some ending radius, and R1 is some starting radius, then function notation gives the respective fluxes as F(R2) and F(R1).

The relative change in this senario is given by [F(R2) - F(R1)]/F(R1).

REFERENCE

 


I would start by looking at the changes in F and R with respect to time.

Differentiate F = k * R^4 with respect to time to get \(\displaystyle \frac{dF}{dt} \; = \; 4k R^3 \; \frac{dR}{dt}\).

dF/dt is the change in F with respect to time, and dividing this change by F gives the relative change in F: (dF/dt)/F.

"Divide" the equation in black (above) by the equation F= k * R^4 (i.e, leftside divided by leftside and rightside divided by rightside).

That result makes a statement about the relative change in F compared to the corresponding relative change in R.

 
Re: Poiseuille's Law

so dF/F(dt) is the relative change in F
and dR/R(dt) is the relative change in R?
 
Re: Poiseuille's Law

If I am correct in the last post, how should I apply a 5% increase in the radius?
if dF/F(dt) = 4(dR)/R(dt)
and dR/R(dt) = 1.05R
then dF/F(dt) = 4.2R? Am I on the right track with this?
So if there is a 5% increase in the radius, then the relative change in blood flow is 4.2 times the radius?
(The book gives that if there is a 5% increase in radius, then the blood flow increases by 20%)
I'm stuck here.
 
Re: Poiseuille's Law

\(\displaystyle F = kR^{4}, \frac{dF}{dR} = 4kR^{3}\)

\(\displaystyle F = k(1.05R)^{4}, \frac{dF}{dR} = 4.20k(1.05R)^{3}\)

\(\displaystyle Hence, \ a \ change \ in \ radius \ of \ .05 \ results \ in \ an \ increase \ of \ blood \ flow \ of \ .20.\)
 
Jakotheshadows said:
so dF/F(dt) is the relative change in F

and dR/R(dt) is the relative change in R?
Click to expand...


No. I already told you that the relative change in F is (dF/dt)/F.

This is not the same as (dF/F)(dt). (I'm not sure how you got that.)

Follow my instructions; divide dF/dt by F, and divide (4k R^3 dR/dt) by k R^4.

The resulting equation will show that the relative change in F is 4 times the relative change in R. (You need to know how to simplify a compound fraction.)

 
Re: Poiseuille's Law

You misread my post.

I didn't write (dF/F)(dt).. I wrote dF/F(dt).. perhaps to be clear I should write dF/[F(dt)].

I did follow your instructions already, and came to dF/[F(dt)] = 4dR/[R(dt)]..

I don't understand how to apply the 5% increase to get the increase in blood flow, despite the attempt in a previous post to illustrate the process.
 
Re: Poiseuille's Law

BigGlenntheHeavy said:
\(\displaystyle F = kR^{4}, \frac{dF}{dR} = 4kR^{3}\)

\(\displaystyle F = k(1.05R)^{4}, \frac{dF}{dR} = 4.20k(1.05R)^{3}\)

\(\displaystyle Hence, \ a \ change \ in \ radius \ of \ .05 \ results \ in \ an \ increase \ of \ blood \ flow \ of \ .20.\)
Click to expand...

I see how you did that, but I don't see how an increase in one constant when you differentiate F with respect to R gets a total increase in F by .20. Could you elaborate any more than that?
 
Jakotheshadows said:
You misread my post. How do you know what I do, without asking first? :?

I read what you typed.

Without grouping symbols, dF/F(dt) typed means (dF/F)(dt).



… to be clear I should [have written] dF/[F(dt)] … Exactly.

Check out "Karl's Notes - Typing Math" under Forum Help, for more information about how to type mathematical expressions. 8-)

Click to expand...


I was trying to lead you to the following equation's form because it shows both relative changes (as I previously explained: the ratio of change to original amount).

\(\displaystyle \frac{\frac{dF}{dt}}{F} \; = \; 4 \cdot \frac{\frac{dR}{dt}}{R}\)

\(\displaystyle \frac{dF}{dt}\) is some change in F, per some change in time.

\(\displaystyle \frac{\frac{dF}{dt}}{F}\) is the resulting relative change in F.

\(\displaystyle \frac{dR}{dt}\) is some change in R, per the same change in time.

\(\displaystyle \frac{\frac{dR}{dt}}{R}\) is the resulting relative change in R.

Do you "see" how the equation above states that the relative change in F is four times the relative change in R? Your exercise asks you to show the given relationship between relative changes in F and R; I think this equation does that.

I could be wrong, of course. What do you think?

 
If I were to do the problem - I would do it the following way:

F = k * R^4

?F = k * 4 * R^3 * ?R

(?F)/F = (k * 4 * R^3 * ?R)/( k * R^4)

(?F)/F = 4 * (?R)/(R) ...Show that the relative change in F is about four times the relative change in R

(?R)/(R) = 0.05

(?F)/F = 4 * (?R)/(R) = 4 * 0.05 = 0.2 = 20%
 
Thanks Subhotosh Khan.

I had already understood how the relative change in F was 4 times the relative change in R once mmm4444bot explained how to show that the first time. I needed to understand how to apply the 5% increase in radius to get the 20% increase in blood flow. Yes, it should be obvious once the above is shown, but somehow I wasn't making the connection. Thanks again for your help and patience guys.
 
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