When blood flows through an artery (which can be thought of as cylindrical tube)

Question

When blood flows through an artery (which can be thought of as cylindrical tube) its velocity is greast at the center of the artery. Because of friction along the walls of the tube, the bloods velocity decrease at the istance r from the center of the artery increas, finally becoming 0 at the wall of the artery. The velocity (in centimeters per second) is given by the function v=18,500(.000065-r^2), where r is measured in centimeters. Find the average rate of ange of the velocity as the distance from the center changes from

r= .001 to r=.002

r=.002 to r=.003

r=0 to r=.005

Explanation / Answer

Let r1 = 0.001, r2 = 0.002, r3 = .003, r0 = 0 and r5 = .005.

We are given that v = 18500(0.000065 - r2) . Therefore, v1 = 18500(0.000065 - r12) = 18500(0.000065 - 0.000001) =

18500(0.000064 ) = 1.184 Similarly, v2 = 18500(0.000065 - r22) = 18500(0.000065 -  0.000004) = 16500(0.000061) = 1.1285 and v3 = 18500(0.000065 - r32) = 18500(0.000065 -  0.000009) = 18500(0.000056) = 1.036 and

v5= 8500(0.000065 - r52) = 18500(0.000065 -  0.000025) = 18500(0.000040) = 0.74 Finally, v0=

18500(0.000065 - r02) = 18500(0.000065 - 0 ) = 18500(0.000065) = 1.2025

The desired average rates of changes of velocity are as under:

a1 = (v1 - v2) / ( r1 - r2) = (1.184 - 1.1285) / ( .001 - .002) = - 0.0555/0.001 = - 55.5 Similarly,

a2 =  (v2 - v3) / ( r2 - r3) = (1.1285 - 1.036) / ( 0.002 - 0.003) = - 0.0925 / 0.001 = - 92.5 and

a0 = (v0 - v5) / ( r0 - r5) = (1.2025 - 0.74)/ ( 0 - 0.005) = - 0.4625 / 0.005 = - 92.5

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