#### To determine

**To find:**

(i) The average velocity over the interval 0≤r≤R

(ii) Compare the average velocity with the maximum velocity.

#### Answer

(i)

PR26ηl

(ii) The average velocity is 23 of the maximum velocity.

#### Explanation

**1) Concept:**

Use the mean value theorem for integral to find the average velocity over the interval 0≤r≤R

**2) Mean Value Theorem for Integrals**

If f is continuous on a,b then there exists a number c in a,b such that

fc=fave=1b-a∫abfx dx

that is,

∫abfx dx=fcb-a

**3) Given:**

The velocity v of blood that flows in a blood vessel with radius R and length l at a distance r from the central axis is

vr=P4ηlR2-r2

where P is the pressure difference between the ends of the vessel and η is the viscosity of the blood.

**4) Calculation:**

(i)

The average velocity on 0,R is

Vave=1R-0∫0Rvrdr

=1R∫0RP4ηlR2-r2 dr

=P4Rηl∫0RR2-r2 dr

=P4RηlR2r-r330R

=P4RηlR3-R330R

=P4Rηl2R33-0

=PR26ηl

Therefore,

Vave=PR26ηl

(ii)

vr=P4ηlR2-r2 is a polynomial of degree 2 in r

vr=PR24ηl-Pr24ηl

Differentiate vr with respect to r.

v'r=0-Pr2ηl

By using the first derivative test,

The maximum or minimum velocity is reached when v'r=0.

-Pr2ηl=0

As P,η,l ≠0

⇒r=0

Therefore, at r=0 average velocity becomes maximum velocity, since vr is decreasing on (0, R].

Substitute r=0 in vr

The maximum velocity is

Vmax=v0=PR24ηl-P04ηl=PR24ηl

The average velocity is

Vave=PR26ηl

Multiply and divide by 2.

=2·PR22·6ηl

=2PR212ηl

By rearranging,

=2PR23·4ηl

=23PR24ηl

=23Vmax

Therefore,

Vave=23Vmax

The average velocity is 23 of the maximum velocity.

**Conclusion:**

(i) The average velocity over the interval 0≤r≤R is

PR26ηl .

(ii) The average velocity is 23 of the maximum velocity.