Chapter 3 - Section 3.10 - Linear Approximations and Differentials. - 3.9 Exercises - Page 257: 40

$\frac{dF}{F} = 4~\frac{dR}{R}$ The relative change in $F$ is 4 times the relative change in $R$ A 5% increase in the radius would result in a 20% increase in the flow of blood.

$F = kR^4$ $dF = 4kR^3~dR$ We can find an expression for the relative change in $F$: $\frac{dF}{F} = \frac{4kR^3~dR}{kR^4}$ $\frac{dF}{F} = 4~\frac{dR}{R}$ Therefore, the relative change in $F$ is 4 times the relative change in $R$ A 5% increase in the radius would result in a 20% increase in the flow of blood.