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Homework Help: Integration Question

  1. Nov 9, 2009 #1
    1. The problem statement, all variables and given/known data
    The rate at which a particular drug is absorbed by the body is proportional to the amount of the drug present (D) at any time t. If 50mL is initially administered to a patient and 50% remains after 4 hours, find

    A) D as a function of t.

    3. The attempt at a solution

    I'm confused as we have to make D=the amount present (not yet absorbed)

    This means that it would be:

    d(50-D)/dt [tex]\alpha[/tex] D

    Is this correct, in that the rate of absorption is Amount absorbed/time, so 50mL-D = amount absorbed?

    By doing this, I integrate and just get Ae^kt = D
  2. jcsd
  3. Nov 9, 2009 #2
    This is correct. I'd only write dD/dt though.

    Correct as well, now you just have two constants and two constraints on them: D(0) and D(4 hours).
  4. Nov 9, 2009 #3
    But the rate of absorption is the amount absorbed over time? D represents the amount left, so it would have to be d(50-D)/dt? Because 50-amount left = amount absorbed
  5. Nov 9, 2009 #4
    That's right, but you can certainly show that d(50-D)/dt = -dD/dt. What was your question again? :p
  6. Nov 9, 2009 #5
    How does d(50-D)/dt turn into -dD/dt???
  7. Nov 9, 2009 #6
    Differentiation is linear, so that d(50-D)/dt = d 50 /dt - dD/dt. Differentiation of a constant gives zero.
  8. Nov 9, 2009 #7
  9. Nov 9, 2009 #8


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    Homework Helper

    i don't think you need the 50 at all to formulate the differntial equation

    the statement
    is enough to give you the differential equation
    [tex]\frac{dD(t)}{dt} = -kD(t) [/tex]
    for some as yet undetermined constant k

    integrate, then use the given values to find the constants
    assume an initial value
    [tex]D(0) = d [/tex]
    value at time = 4hrs
    [tex]D(4) = d/2 [/tex]
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