Related rates-blood vessel

[Aka]

Hi, I am having trouble with this question

 

Assume that a blood vessel, such as a vein or an artery, has the shape of a cylindrical tube with radious R and lenght L. because of the friction at the walls of the tube, the velocity, v, of blood flowing throught the blood vessel is greatest along the central axis of the tube and decreses as the distance r from the axis increses unitl v becomes 0 at the wall. the relationship between v and r is givin by the law of laminar flow, discovered by the French physician Poiseuille in 1840. the law

 

V= p / 4nl (R^2 - r^2)

 

where n is the velosity of the blood and P is the pressure difference between the ends of the tube. If P and L are constant, then V is a function of r. in a typical human artery, the values are n= 0.002 R= 0.008cm, L = 2cm and P= 400 Pa.s. Find the velocity gradiaent ( the rate of change oh v with respect to r) where r= 0.003 cm.

 

can someone tell me the steps to solving thsi problem? Thanks

[the tree]

[math]V=\frac{P}{4nL}(R^2 - r^2)[/math]

 

Well, I was about to say factoring in the constants is always a good start, but actually no, it isn't. Instead let's multiply out the brackets.

 

[math]V=\frac{P}{4nL}R^2 - \frac{P}{4nL}r^2[/math]

 

Now then, we know that [math]\frac{P}{4nL}R^2[/math] is all constants, so they don't affect the rate of change, so we can forget about them.

 

[math]\frac{dV}{dr}=\frac{d}{dr}\frac{P}{4nL}r^2[/math]

 

Now we can factor in the constants.

 

[math]\frac{P}{4nL}r^{2}=\frac{400}{4\times 0.002 \times 2}r^{2}=2500r^{2}[/math]

 

And so we now know that:

 

[math]\frac{dV}{dr}=\frac{d}{dr}2500r^2=5000r[/math]

[Aka]

Thanks