Related Rates-blood vessel

[wind]

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 4 = 0.003 cm.

is the central axis the middle of the blood vessel

I don't get..whats the difference between R and r ? Can someone pleas explain this question? Thanks.

 

[tkhunny]

Read the definitions.

R is the radius of the vessel. It is constant.

r is the distance from the center of the vessel. It is not constant. AT ITS GREATEST VALUE, r = R. Otherwise 0 <= r < R

 

[wind]

Ok, thanks

so what do I do...I have all the informaion to plug into the equation so v= 1.375 right?
but then we want dv/dr...so do I differentiat the equation?

Find the velocity gradiaent ( the rate of change oh v with respect to r) where 4 = 0.003 cm.

thats supposed to be r=0.003cm, sorry

 

[tkhunny]

I'm a little puzzled by your equation. I suppose it is this: \(\displaystyle V = \frac{P}{4NL}*(R^{2}-r^{2})\).

You should be able to find the derivative of that.

Unfortunately, that is not what you have written.

 

[wind]

I tryed solving this but I got the wrong answer

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

f(x)= p / 4nl
f'(x)= -4p / (4nl)^2

g(x)= R^2 - r^2
g'(x)= 2R - 2r

v'= (-4p / (4nl)^2)(R^2 - r^2) + (2R - 2r)(p / 4nl)

would that be the expression for the derivitive?

then at the end do I sub in the values? and thats the answer?

Thanks

 

[tkhunny]

You seem to be a little confused by all the parameters. They are all constant. Their derivatives are zero.

Look at what you have written:

f(x) = p / 4nl <== Where is the 'x'?
f'(x) = 0

g(x) = R^2 - r^2 <== Where is the 'x'?
g'(x) = 0

The ONLY independent variable is 'r'. EVERYTHING else is constant.

V is a function of r

\(\displaystyle V(r) = \frac{p}{4nl}*(R^{2}-r^{2}) = \frac{p}{4nl}*R^{2} - \frac{p}{4nl}*r^{2}\)


\(\displaystyle \frac{dV}{dr} = -2 \frac{p}{4nl}*r\)

Now what?

 

[wind]

Thanks tkhunny