1. ## Interesting problem about blood...Help please :)

Hi guys, am new here so go easy on me

Help plzzz

2. I may have messed up on the way so here is the full problem...

The drug is removed by the patient's kidneys a rate which is proportional to the concentration of the drug in the patients blood...

From this i thought...Rate Out = k * conc. of drug

and conc. of drug is r/v1 where r (as above) is the mg of drug in the blood and v1 is the constant ammount of blood in the body

so Rate Out = k * r/v1 ...is this the right thinking?

and dr/dt = Rate in - Rate out...using the rate out k*r/v1 and the rate in given in the question below

Rest of question...

Derive a differential equation describing the dependence on time of the drug concentration in the patient's blood. You can assume the volume of the patient's blood is a constant and that the rate of drug addition per unit volume of blood is a1 + a2*sin(k1*t) where a1, a2 and k1 are constants and t is time.

The rest of the questions gives values and asks to plug them in etc which is straight forward...just stuck on integrating the diff. equation and whether or not I even have the correct diff. equation in the first place...

3. Originally Posted by shabz
I may have messed up on the way so here is the full problem...

The drug is removed by the patient's kidneys a rate which is proportional to the concentration of the drug in the patients blood...

From this i thought...Rate Out = k * conc. of drug

and conc. of drug is r/v1 where r (as above) is the mg of drug in the blood and v1 is the constant ammount of blood in the body

so Rate Out = k * r/v1 ...is this the right thinking?

and dr/dt = Rate in - Rate out...using the rate out k*r/v1 and the rate in given in the question below

Rest of question...

Derive a differential equation describing the dependence on time of the drug concentration in the patient's blood. You can assume the volume of the patient's blood is a constant and that the rate of drug addition per unit volume of blood is a1 + a2*sin(k1*t) where a1, a2 and k1 are constants and t is time.

The rest of the questions gives values and asks to plug them in etc which is straight forward...just stuck on integrating the diff. equation and whether or not I even have the correct diff. equation in the first place...
The ODE is:

$\frac{d}{dt}r=-k\frac{r}{v}+v(a_1+a_2\sin(k_1 t))$

rearrange:

$\frac{d}{dt}r+k\frac{r}{v}=v(a_1+a_2\sin(k_1 t))$

and this is a first order linear constant coefficient ODE and you should be able to solve this.

CB

4. thank you so much! been a long time since i've done ODEs

will go and finish off the question now

5. ok now its getting real hairy...the next step is i need to integrate the following

integrate: exp(k2*t/v).sin(k1*t) dt, where k1 and k2 are constants

how can i integrate it by parts when exp will keep giving exp and sin will keep giving cos/sin/cos/sin etc...?

6. ok i got it...you chose the exp as your u twice and it comes out with the original I which you substitute back in and divide! lol...

this is gunna be interesting to check over

7. Originally Posted by shabz
ok now its getting real hairy...the next step is i need to integrate the following

integrate: exp(k2*t/v).sin(k1*t) dt, where k1 and k2 are constants

how can i integrate it by parts when exp will keep giving exp and sin will keep giving cos/sin/cos/sin etc...?
exp(k2*t/v).sin(k1*t) is the imaginary part of:

exp(k2*t/v)*exp(k1*t*i) = exp((k2/v+k1*i)*t)

CB

8. not sure i'm with you there? why do we need imaginary numbers?

i integrated by parts twice chosing u as the exp part and v' as the sin part...after that i got back to the original integral and sub'd that back in, factorising and divided through...i think that worked? i got something that looked feasible as my final solution at least lol

9. Originally Posted by shabz
not sure i'm with you there? why do we need imaginary numbers?

i integrated by parts twice chosing u as the exp part and v' as the sin part...after that i got back to the original integral and sub'd that back in, factorising and divided through...i think that worked? i got something that looked feasible as my final solution at least lol
The point is the integral is trivial if regarded as the imaginary part of the integral of a complex exponential.

CB

10. I woke up this morning and a thought occured...

for the rate in...why did you multiply by v?

if dr/dt = rate in - rate out

...the rates are both measured in mg per litre blood per hour...so do we need to multiply the rate in by v? if we multiply the rate in by v our new units will be mg per hour and the rate out will still be mg per litre per hour...since in the question it tells us that k's units are measured in per hour...

11. have added the question to the first post to make life easier

12. Originally Posted by shabz
I woke up this morning and a thought occured...

for the rate in...why did you multiply by v?

if dr/dt = rate in - rate out

...the rates are both measured in mg per litre blood per hour...so do we need to multiply the rate in by v? if we multiply the rate in by v our new units will be mg per hour and the rate out will still be mg per litre per hour...since in the question it tells us that k's units are measured in per hour...
Because you are given a dose rate in whatever (milligm?) per unit volume of blood. To get the actual rate in milligms or whatever you need to multiply by the blood volume.

CB

13. but the rate in and rate out units don't match up? i've uploaded the question to post 1...here's my working aswell..