# Single Variable Calculus, Chapter 3, 3.7, Section 3.7, Problem 25

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The equation $\displaystyle \nu = \frac{P}{4\eta \ell}(R^2 - r^2)$ represents the law of Laminar flow.

Where $\eta$ is the viscosity of the blood and $P$ is the pressure difference between the ends of the tube with length $\ell$ and radius $R$. While $r$ is the distance from the axis. Consider a blood vessel with radius 0.01cm, length 3cm, pressure difference $\displaystyle 3000 \frac{\text{dynes}}{\text{cm}^2}$ and viscosity $\eta = 0.027$

a.) Determine the velocity of blood along the centerline $r = 0$, at radius $r = 0.005$cm, and at the wall $r = R = 0.01$cm.

b.) Find the velocity gradient at $r=0$, $r=0.005$, and $r=0.01$

c.) Where is the velocity the greatest? Where is the velocity changing most?

a.) @ $r = 0$

\begin{equation} \begin{aligned} \nu(r) &= \frac{P}{4\eta \ell} (R^2 - r^2)\\ \\ \nu(0) &= \frac{3000}{4(0.027)(3)}\left( (0.01)^2 - (0)^2\right)\\ \\ \nu(0) &= 0.9259 \frac{cm}{s} \end{aligned} \end{equation}

@ $r = 0.005$cm

\begin{equation} \begin{aligned} \nu(0.005) &= \frac{3000}{4(0.027)(3)} \left[ (0.01)^2 - (0.005)^2 \right]\\ \\ \nu(0.005) &= 0.6944 \frac{\text{cm}}{s} \end{aligned} \end{equation}

@ $r=0.01$cm

\begin{equation} \begin{aligned} \nu(0.01) &= \frac{3000}{4(0.027)(3)} \left[ (0.01)^2 - (0.01)^2 \right]\\ \\ \nu(0.01) &= 0 \frac{\text{cm}}{s} \end{aligned} \end{equation}

b.) The velocity gradient $\displaystyle = \frac{d\nu}{dr}$

\begin{equation} \begin{aligned} \nu &= \frac{P}{4\eta \ell} \left( R^2 - r^2 \right)\\ \\ \nu &= \frac{PR^2}{4 \eta \ell} - \frac{Pr^2}{4 \eta \ell}\\ \\ \frac{d \nu}{dr} &= \frac{d}{dr} \left( \frac{PR^2}{4 \eta \ell}\right) - \frac{P}{4 \eta \ell} \frac{d}{dr} (r^2)\\ \\ \frac{d \nu}{dr} &= 0 - \frac{P}{4\eta \ell} (2r)\\ \\ \frac{d \nu}{dr} &= \frac{-Pr}{ 2 \eta \ell} \end{aligned} \end{equation}

Velocity gradient at $r=0$

\begin{equation} \begin{aligned} \frac{d \nu}{dr} &= \frac{-3000 (0)}{2(0.027)(3)}\\ \\ \frac{d \nu}{dr} &= 0 \frac{\frac{\text{cm}}{s}}{\text{cm}} \end{aligned} \end{equation}

Velocity gradient at $r = 0.005$

\begin{equation} \begin{aligned} \frac{d \nu}{dr} &= \frac{-3000(0)}{2(0.027)(3)}\\ \\ \frac{d \nu}{dr} &= -92.5926 \frac{\text{cm}/s}{\text{cm}} \end{aligned} \end{equation}

The velocity gradient at $r = 0.01$

\begin{equation} \begin{aligned} \frac{d \nu}{dr} &= \frac{-3000(0.01)}{(0.0027)(3)}\\ \\ \frac{d \nu}{dr} &= -185.1852 \frac{\text{cm}/s}{\text{cm}} \end{aligned} \end{equation}

c.) The velocity is greatest at $r = 0$. While the velocity is changing most (decreasing) at when $r = R = 0.01$cm.