Questions: According to Poiseuille's laws, the velocity V of blood is given by V=k(R^2-r^2) where R is the radius of the blood vessel, r is the distance of a layer of blood flow from the center of the vessel, and k is a constant. A patient takes blood vessel dilation medication so that her blood velocity is modeled by Poiseuille's law with k=597.4. Find the rate of change of the blood velocity when R=0.03 mm and R is changing at 0.006 mm per minute. Assume r is constant. The blood velocity is increasing at mm per minute each minute. (Round to the nearest thousandth as needed.)

Solution Steps

According to Poiseuille's law, the velocity \( V \) of blood flow is given by the equation: \[ V = k(R^2 - r^2) \] where \( k = 597.4 \), \( R \) is the radius of the blood vessel, and \( r \) is the distance from the center of the vessel.

To find the rate of change of the blood velocity with respect to time \( t \), we differentiate \( V \) with respect to \( t \): \[ \frac{dV}{dt} = \frac{dV}{dR} \cdot \frac{dR}{dt} \] Since \( r \) is constant, its derivative is zero, and we only consider the change in \( R \).

We are given:

• \( R = 0.03 \, \text{mm} \)
• \( \frac{dR}{dt} = 0.006 \, \text{mm/min} \)

Substituting these values into the differentiated equation, we find: \[ \frac{dV}{dt} = 597.4 \cdot 2R \cdot \frac{dR}{dt} \] Calculating this at \( R = 0.03 \, \text{mm} \): \[ \frac{dV}{dt} = 597.4 \cdot 2 \cdot 0.03 \cdot 0.006 \]

Evaluating the expression gives: \[ \frac{dV}{dt} = 0.006 \]

Final Answer