Question

the flow of blood in a blood vessel is faster toward the center of the vessel and slower toward the outside. the speed of the blood v, in millimeters per second (mm/sec), is given by the following formula, where r is the radius of the blood vessel, r is the distance of the blood from the center of the vessel, and p, l, and v are physical constants related to pressure, length, and viscosity of the blood vessels, respectively. assume that r is a constant as well as p, l, and v. complete parts (a) and (b) below.

$v = \frac { p } { 4 l v } ( r ^ { 2 } - r ^ { 2 } )$

a) find the rate of change $\frac { d v } { d t }$ in terms of r (in mm) and $\frac { d r } { d t }$ (in mm/sec) when $l = 80 mm$, $p = 450 pa$, and $v = 0.003 pa \cdot sec$. select the correct answer below and fill in the answer box to complete your choice.

$\bigcirc$ a $\frac { d v } { d t } = \square \cdot \frac { d r } { d t }$

$\bigcirc$ b $\frac { d v } { d t } = \square \cdot \frac { d r } { d t }$

Explanation:

Step1: Substitute the given values into the formula

Given \(V=\frac{p}{4L
u}(R^{2}-r^{2})\), substitute \(L = 80\), \(p = 450\), \(
u=0.003\).

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Step2: Differentiate \(V\) with respect to \(t\) using the chain - rule

Since \(r\) is a constant, \(\frac{dV}{dt}=468.75\times2R\frac{dR}{dt}\) (using the chain - rule \(\frac{d}{dt}(u^{2}) = 2u\frac{du}{dt}\) where \(u = R\)).

Answer:

\(\frac{dV}{dt}=937.5R\frac{dR}{dt}\), so the answer is \(A\) and the box should be filled with \(937.5R\)