Hi guys, am new here so go easy on me
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...
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...?
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
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...