Arterial Branching Calculus

Date: 05/28/2003 at 08:47:46
From: Daniel
Subject: Arterial branching calculus

In the surgical connection of a small artery to an existing larger 
artery, attention must be paid to minimising the viscous resistance 
to the blood flow. 

A surgeon must attach a small blood vessel to an existing artery AB 
in order to get a blood supply from point A to point C.

Blood flow suffers resistance proportional to the length of tube 
travelled and inversely proportional to the fourth power of the radius 
of the tube, i.e. R=(kd)/r^4 where R is the resistance, k is a 
constant, d is the distance travelled along the tube, and r is the 
radius of the tube.

The surgeon's problem is where to make the attachment. If it is close 
to A the path travelled by the blood is shortened, but the length 
travelled in the narrower tube is increased. If the join is closer to 
B, the path the blood travels is longer, but less of it is in the 
narrow tube.

Investigate the distance from B of the join and hence the angle of 
attachment that minimises the resistance to flow.

I've used the following variable names
x distance from B to the join
D the distance from A to B
d the distance form B to C
r1 radius of the artery (larger blood vessel)
r2 radius of the blood vessel to be attached

I formed an expression for the total resistance
Rtotal = R1 + R2
where R1 = k(D-x)/(r1)^4
R2 = k(sqrt(d^2 + x^2))/(r2)^4
I tried to find a value of x that minimises the total resistance but 
it got very messy.

I found a website: 

   The Mathematician's Quest for Superlatives - Joseph MacDonnell
   http://www.faculty.fairfield.edu/jmac/ther/superlatives.htm 

which says the optimum angle is cos q = (r2/r1)
I'm not sure how they derive it...

but on another site

   The Ant and the Blade of Grass - Tewodros Amdeberhan
   http://www.nj.devry.edu/~amdberha/Maple217.html 

Maple V Project 4
Hint: The optimal angle is: a=arccos(r^4/R^4). 

Now I'm not sure what arccos is. Is it like 1/cosx = secx so arccos = 
sec? Or is it the inverse of cosx ie cos^-1(x)?

Date: 05/28/2003 at 12:57:57
From: Doctor Peterson
Subject: Re: Arterial branching calculus

Hi, Daniel.

It sounds as if you are taking B to be the nearest point on the vessel 
to C; that isn't stated, but is reasonable, since it doesn't matter 
to the problem where B is. We get this picture:

                 C
                /|
               / |
              /  |d
             /   |
            / x  |
   A-------X-----B
    <-----D----->

You have correctly written the expression to be minimized,

  Rtotal = k(D-x)/(r1)^4 + k(sqrt(d^2 + x^2))/(r2)^4

We might simplify this by temporarily defining some constants:

  A = k/(r1)^4
  B = k/(r2)^4

so we want to minimize

  A(D-x) + Bsqrt(d^2 + x^2)

Now you just have to differentiate this and solve for x when the 
derivative is zero.

Alternatively, you might take a clue from the sites you found, despite 
their apparent disagreement about the actual answer: use the angle of 
the attachment, rather than the distance back from the nearest point, 
as your variable. That would make your picture like this:

                 C
                /|
               / |
              /  |d
             /   |
            /T   |
   A-------X-----B
    <-----D----->

Then XB is d cot(T), XC is d sec(T), and AX is D - d cot(T), and you 
can write your total resistance from this.

As for what arccos is, it is the same as cos^-1, the inverse cosine:

   Trigonometry Terminology
   http://mathforum.org/library/drmath/view/54185.html 

If you need more help, please write back and show me how far you got.

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/