Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 38

Prove that the relative change in $F$ is about 4 times the relative change in $R$. State how will a 5% increase in the radius affect the flow of blood.
Taking the derivative of $F$ with respect to $R$, we have

$
\begin{equation}
\begin{aligned}
\frac{dF}{dR} &= K \frac{d}{dR} (R^4)\\
\\
\frac{dF}{dR} &= K (4R^3)\\
\\
\frac{dF}{dR} &= 4KR^2\\
\\
dF &= 4KR^3 \quad \text{or} \quad \Delta F = 4KR^3 \Delta R

\end{aligned}
\end{equation}
$


Hence, the relative error is,
$\displaystyle \frac{\Delta F}{F} = \frac{\Delta \cancel{K}R^3\Delta R}{\cancel{K}R^4} = 4 \frac{\Delta R}{R}$
When the increase in radius is 5%

$
\begin{equation}
\begin{aligned}
\frac{\Delta F}{F} &= 4 (5 \%)\\
\\
\frac{\Delta F}{F} &= 20 \%
\end{aligned}
\end{equation}
$


The flow of blood is increased yb 20%