# High blood pressure results from constriction of the arteries. To maintain a normal flow rate (flux), the heart has to pump harder, thus increasing the blood pressure. Use Poiseuille's Law to show that if R0 and P0 are normal values of the radius and pressure in an artery and the constricted values are R and P, then for the flux to remain constant, P and R are related by the equation: (P/P0) = (R0/R)^4. If the radius of an artery is reduced to three-fourths of its former value, what is the pressure in terms of P0?

According to the Medical definition of Poiseuille's Law, the flux or the velocity of the steady flow of a liquid (or blood) through a narrow tube (in this case, arteries) is directly proportional to the pressure and fourth power of the radius of the tube and inversely proportional to the length of the tube and coefficient of viscosity of the liquid.

Mathematically, it can be represented as:

F prop \frac{PR^{4}}{ eta l}

or F=\frac{\pi PR^{4}}{8\eta l}

where F is the Flux, P is the pressure , R is the radius ,  eta  is the coefficient of viscosity of blood, and l is the length of the artery.

Now we have to show that if R0 and P0 are normal values of the radius and pressure in an artery and the constricted values are R and P, then for the flux to remain constant, P and R are related by the equation: \frac{P}{P_0}=(\frac{R_0}{R})^{4}

For the flux F to remain constant, the equation using P and R must equal the equation using P0 and R0:

\frac{\pi PR^4}{8\eta l}=\frac{\pi P_0 R_0^4}{8\eta l}

PR^4=P_0R_0^4

or, \frac{P}{P_0}=(\frac{R_0}{R})^4 ------>(1)

Next we have to find the pressure in terms of P0 if the radius of the artery is reduced to three-fourths of its former value.

By substituting R=\frac{3}{4}R_0 in equation (1),

\frac{P}{P_0}=(\frac{R_0}{\frac{3}{4}R_0})^4

\frac{P}{P_0}=(\frac{4}{3})^4

\frac{P}{P_0}=\frac{256}{81}

i.e. P=\frac{256}{81}P_0

P\approx 3.2P_0

Hence if the radius is reduced to 3/4ths of its former value, then the pressure is more than tripled.

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