The flow of blood through an arteriole in cubic centimeters per second is given by
Where l is the length (in cm) of the arteriole, r is its radius (in cm), p is the difference in pressure between the two ends of the arteriole (in dyne/cm2), and k is the viscosity of blood (in dyne-sec/cm2). Find the approximate percentage change in the flow of blood if an error of 4% is made in measuring the length of the arteriole and an error of 3% is made in measuring its radius. Assume that p and k are constant.

B)
C)
D)
E)

Question

The flow of blood through an arteriole in cubic centimeters per second is given by ​  ​
Where l is the length (in cm) of the arteriole, r is its radius (in cm), p is the difference in pressure between the two ends of the arteriole (in dyne/cm2), and k is the viscosity of blood (in dyne-sec/cm2). Find the approximate percentage change in the flow of blood if an error of 4% is made in measuring the length of the arteriole and an error of 3% is made in measuring its radius. Assume that p and k are constant.
​
B) ​ 
C) ​ 
D) ​ 
E) ​

Quizplus · Accepted Answer

Using the formula for percentage error propagation,
%ΔF = |(∂F/∂r) × (%Δr)| + |(∂F/∂l) × (%Δl)|
where F is the flow of blood.

Taking partial derivatives, we get
(∂F/∂r) = (440πpkl)/(r^2)
(∂F/∂l) = (-40πpkr^4)/(l^2)

Substituting these values and the given errors,
%ΔF = |(440πpkl)/(r^2) × (0.03)| + |(-40πpkr^4)/(l^2) × (0.04)|
%ΔF = (13.2πpkl)/(r^2) + (1.6πpkr^4)/(l^2)

Therefore, the approximate percentage change in flow of blood is given by
%ΔF/F = (%ΔF)/(F) = [(13.2πpkl)/(r^2) + (1.6πpkr^4)/(l^2)] / (11πpklr^2)
%ΔF/F = (1.2/l^2) + (0.16/r^2)

Plugging in the values given in the question, we get
%ΔF/F = (1.2/0.97^2) + (0.16/0.96^2) ≈ 1.308%

Therefore, the answer is B.

The Flow of Blood Through an Arteriole in Cubic Centimeters