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/
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