#### To determine

**To show:** The relative change in *F* is the four times the relative change in *R* and to find how a 5% increase in the radius affect the flow of blood.

#### Explanation

Let *F* denote the flux and *R* denote the radius.

It is given that the flux *F* is proportional to the fourth power of the radius *R* of the blood vessel that is F=kR4.

The differential is defined by the equation, dF=f′(R)dR.

Obtain the derivative of the function f(R)=kR4 as follows,

f′(R)=ddR(kR4)=kddR(R4)=k(4R3)=4kR3

Substitute f′(R)=4kR3 in the equation for the differential dF=f′(R)dR as,

dF=4kR3dRdFdR=4kR3

When ΔR is small the above equation becomes,

ΔFΔR≈4kR3ΔF≈4kR3ΔR

Divide on both sides by kR4 as,

ΔFkR4≈4kR3ΔRkR4ΔFF≈4⋅ΔRkR4 [∵F=kR4]

Therefore, the relative error in *F* is the four times of the relative error in *R.*

The percentage error in *F* is the product of relative error and 100 that is,

Percentage error in F≈4×Percentage error in R

If the percentage error in *R* increase 5% then the percentage error in *F* is 4×5%=20%.

Therefore, the 5% increase in the radius corresponds to 20% increase in the blood flow.