Morphine is administered to a patient intravenously at a rate of 2.5 mg per hour. About 34.7% of the morphine is metabolized and leaves the body each hour. M'(t)=34.7% - (2.5*M(t)) If there is initially 100mg of morphine in the body, how much morphine will there be after 5 hours?
differential calculus ?
yeah i'm in calc 1
if we use y (instead of M) \[ \frac{dy}{dt} = .347 - 2.5 y \\ \frac{\frac{dy}{dt} }{ .347 - 2.5 y }=1 \\ \frac{dy}{.347 - 2.5 y }=dt\]
we can integrate both sides \[ -\frac{2}{5} \int \frac{ - 2.5 \ dy}{0.347 - 2.5y}=\int dt \]
can you integrate that ?
where did 2/5 come from?
if we make the bottom u u = 0.347 - 2.5y then du = -2.5 dy and we would have du/u ...which integrates to ln( u ) so I put in a -2.5 i.e. -5/2 up top and a -2/5 out front to compensate
oh ok
i'm forgetting how to integrate I haven't done it in a while XD
or you could do a "change of variables" (which is really the same thing) let u = 0.347 - 2.5 y du = -2.5 dy= - 5/2 dy dy = -2/5 du and replace dy/(0.347 -2.5y) with -2/5 du / u
Yeah I get that
but now i'm confused as to what I do next? I know I leave the -2/5 alone for now
\[ \int \frac{dx}{x} = \ln x +C\]
so is the x in lnx the -2.5dm / (.347-2.5m) ?
the "x" would be the entire bottom dx is -2.5 dm
so -2.5dm/(.347-2.5m) = ln(.347-2.5m) + C ??
and what happens with the -2/5 ?
before going on, are you sure about that equation, because it's not making sense to me I'm thinking it would be M' = -0.347M + 2.5
maybe it is. i had to come up with the equation but i wan't sure about it!
yours makes more sense because i though we would have to subtract a number from the equation, but it makes more sense to add the 34.7%
M' = 2.5 says: the change in morphine is +2.5 mg (per hour) assuming it was not metabolized. i.e. it goes up at that rate but some of it "disappears" how much ? 34.7% of how much there is i.e. 34.7% of M
so the total change is M' = 2.5 - 0.347M that says we add 2.5 mg, but subtract off 34.7% of what's in the body
oh i understand!
could i use that equation to find the amount of morphine after 5 hours?
now we solve it the same as before. write it as \[ \frac{ \frac{dM}{dt}}{2.5- 0.347M} = 1\] multiply both sides by "dt" (this is a dubious step mathematically , but we are "separating the variables") we get \[ \frac{dM}{2.5 - 0.347 M } = dt \]
multiply both sides by -0.347 (this will make the top the derivative of the bottom so that we have a dx/x form)
dt(-.347)=dm/(2.5-m) ??
your algebra is weak ? (calculus assumes you know algebra very well) \[ -0.347 \ dt = \frac{-0.347 \ dm}{2.5- 0.347 m} \]
now integrate both sides \[ -0.347 \int \ dt = \int \frac{-0.347 \ dm}{2.5- 0.347 m} \\ -0.347 \ t+C = \ln \left(2.5- 0.347 m \right) \]
and then i bring everything to one side right?
we could put the unknown constant of integration on either side. make each side the exponent of "e" (to get rid of the natural logarithm) we get \[ e^{-0.347 \ t + C} = 2.5 - 0.347 m \] or, writing e^C as an (unknown) constant A \[ A e^{-0.347 \ t}= 2.5 - 0.347 m \] now solve for m
i can use either equation?
\[ A e^{-0.347 t} - 2.5 = - 0.347 m \\ m = A e^{-0.347 t} + \frac{2.5}{0.347} \]
notice when we divide A by -0.347 we just keep the answer A (A is arbitrary, and after dividing it's still arbitrary)
but if i used the 1st equation would i divide the whole thing by -.347?
we get roughly \[ m = A e^{-0.347 t} + 7.205\] at t=0 we have 100 mg putting in t=0 and m=100 we get 100= A+7.205 A= 100-7.205 = 92.8 (roughly) and finally \[ m= 92.8e^{-2.5 t} +7.205 \]
** the last equation should read \[ m = 92.8 e^{-0.347 t} + 7.205 \]
oh ok that's what i was confused about XD
m=23.6 (rounded)
yes, they don't say how accurate they want the answer, but that looks good.
ok thank you so much! this makes much more sense
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