1. Sep 27, 2010

### johnq2k7

Model the arterial tree as a simple branching network in which each junction is composed of a parent vessel and daughter vessels, each having a diameter related to the parent's via the cube law (D_n-1)^3= (D_n*a)^3 + (D_n*b)^3 which is approx. equal to 2*(D_n)^3

a.) Show that D_n/D_0= 2^(-n/3)

b.) Show that 35 generations are required to model the arterial tree from aorta (D_0=2.6 cm) to the capillaries (D_34=10 um)

c.) Derive a formula for the mean transit time (ie vessel length divded by mean velociity) for an individual vessel. Based on this formula, how long does the model suggest it would take for blood to go from the aorta to the capillaries?

My work:

a.) since

(D_n-1)^3 is approximately equal to 2*(D_n)^3

therefore,

(D_n-1/D_n)^3= 2

therefore i Dn/Do= 2^(-n/3)

I'm not sure

i need help with the other questions, i'm confused how use this equation to solve the problem?

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Sep 27, 2010

### Staff: Mentor

I'd guess that you just plug in n=34, n=35, n=36 into the equation, and show that you need at least n=35 in order to get the large ratio in vessel sizes that they give in the question.

3. Sep 27, 2010

### johnq2k7

How do you prove part a.) properly though, and for part c.) how do u derive the mean transit time equation from the information?

4. Sep 28, 2010

### Staff: Mentor

What are a and b in the original equation? Are they somehow constrained to be close to 1 each, and that's why the simplified approximation works?

For the proof, I'd try doing something like this...

$${D_{n-1}}^3 = 2 {D_n}^3$$

So

$${D_0}^3 = 2 {D_1}^3$$

$${D_1}^3 = 2 {D_2}^3$$

$${D_2}^3 = 2 {D_3}^3$$

etc., So

$${D_0}^3 = 2 {D_1}^3 = 2^2 {D_2}^3 = 2^3 {D_3}^3 = ... = 2^n {D_n}^3$$

The rest of the proof should follow. Is that what you mean?

5. Sep 29, 2010

### johnq2k7

Do you simply

use, D_0= 2.6 cm

D(34)= 10 um

and substitute

(D_0)^3 / (2^34)= 10 um

How do you prove it?

I'm sort of confused?

6. Sep 29, 2010

### Staff: Mentor

I'd said:

So what is the ratio of the 2.6cm and 10um? What is the ratio of the following?

$$\frac{{D_0}^3}{{D_{33}}^3}$$

$$\frac{{D_0}^3}{{D_{34}}^3}$$

$$\frac{{D_0}^3}{{D_{35}}^3}$$