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Mathematics 18 Online
OpenStudy (amenah8):

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?

OpenStudy (phi):

differential calculus ?

OpenStudy (amenah8):

yeah i'm in calc 1

OpenStudy (phi):

if we use y (instead of M) \[ \frac{dy}{dt} = .347 - 2.5 y \\ \frac{\frac{dy}{dt} }{ .347 - 2.5 y }=1 \\ \frac{dy}{.347 - 2.5 y }=dt\]

OpenStudy (phi):

we can integrate both sides \[ -\frac{2}{5} \int \frac{ - 2.5 \ dy}{0.347 - 2.5y}=\int dt \]

OpenStudy (phi):

can you integrate that ?

OpenStudy (amenah8):

where did 2/5 come from?

OpenStudy (phi):

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

OpenStudy (amenah8):

oh ok

OpenStudy (amenah8):

i'm forgetting how to integrate I haven't done it in a while XD

OpenStudy (phi):

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

OpenStudy (amenah8):

Yeah I get that

OpenStudy (amenah8):

but now i'm confused as to what I do next? I know I leave the -2/5 alone for now

OpenStudy (phi):

\[ \int \frac{dx}{x} = \ln x +C\]

OpenStudy (amenah8):

so is the x in lnx the -2.5dm / (.347-2.5m) ?

OpenStudy (phi):

the "x" would be the entire bottom dx is -2.5 dm

OpenStudy (amenah8):

so -2.5dm/(.347-2.5m) = ln(.347-2.5m) + C ??

OpenStudy (amenah8):

and what happens with the -2/5 ?

OpenStudy (phi):

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

OpenStudy (amenah8):

maybe it is. i had to come up with the equation but i wan't sure about it!

OpenStudy (amenah8):

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%

OpenStudy (phi):

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

OpenStudy (phi):

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

OpenStudy (amenah8):

oh i understand!

OpenStudy (amenah8):

could i use that equation to find the amount of morphine after 5 hours?

OpenStudy (phi):

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 \[ \frac{dM}{2.5 - 0.347 M } = dt \]

OpenStudy (phi):

multiply both sides by -0.347 (this will make the top the derivative of the bottom so that we have a dx/x form)

OpenStudy (amenah8):

dt(-.347)=dm/(2.5-m) ??

OpenStudy (phi):

your algebra is weak ? (calculus assumes you know algebra very well) \[ -0.347 \ dt = \frac{-0.347 \ dm}{2.5- 0.347 m} \]

OpenStudy (phi):

now integrate both sides \[ -0.347 \int \ dt = \int \frac{-0.347 \ dm}{2.5- 0.347 m} \\ -0.347 \ t+C = \ln \left(2.5- 0.347 m \right) \]

OpenStudy (amenah8):

and then i bring everything to one side right?

OpenStudy (phi):

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

OpenStudy (amenah8):

i can use either equation?

OpenStudy (phi):

\[ A e^{-0.347 t} - 2.5 = - 0.347 m \\ m = A e^{-0.347 t} + \frac{2.5}{0.347} \]

OpenStudy (phi):

notice when we divide A by -0.347 we just keep the answer A (A is arbitrary, and after dividing it's still arbitrary)

OpenStudy (amenah8):

but if i used the 1st equation would i divide the whole thing by -.347?

OpenStudy (phi):

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 \[ m= 92.8e^{-2.5 t} +7.205 \]

OpenStudy (phi):

** the last equation should read \[ m = 92.8 e^{-0.347 t} + 7.205 \]

OpenStudy (amenah8):

oh ok that's what i was confused about XD

OpenStudy (amenah8):

m=23.6 (rounded)

OpenStudy (phi):

yes, they don't say how accurate they want the answer, but that looks good.

OpenStudy (amenah8):

ok thank you so much! this makes much more sense

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