the flow of blood in a blood vessel is faster toward the center of the vessel and slower toward the outside…

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.\n\n$v = \\frac { p } { 4 l v } ( r ^ { 2 } - r ^ { 2 } )$\n\na) 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.\n\n$\\bigcirc$ a $\\frac { d v } { d t } = \\square \\cdot \\frac { d r } { d t }$\n\n$\\bigcirc$ b $\\frac { d v } { d t } = \\square \\cdot \\frac { d r } { d t }$\n

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.\n\n$v = \\frac { p } { 4 l v } ( r ^ { 2 } - r ^ { 2 } )$\n\na) 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.\n\n$\\bigcirc$ a $\\frac { d v } { d t } = \\square \\cdot \\frac { d r } { d t }$\n\n$\\bigcirc$ b $\\frac { d v } { d t } = \\square \\cdot \\frac { d r } { d t }$\n

Answer

Explanation:

Step1: Substitute the given values into the formula

Given (V=\frac{p}{4L\nu}(R^{2}-r^{2})), substitute (L = 80), (p = 450), (\nu=0.003). [ \begin{align*} V&=\frac{450}{4\times80\times0.003}(R^{2}-r^{2})\ &=\frac{450}{0.96}(R^{2}-r^{2})\ & = 468.75(R^{2}-r^{2}) \end{align*} ]

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)