Expert: Paul Klarreich - 10/17/2008

Question
1) The concentration (in milligrams per cubic centimeter) of a certain drug in a patient's body t hours after injections is given by C(t)=t^2/2t^3+1.  0 is less than or equal to t and t is less than or equal to 4.  When is the concentration of the drug increasing, and when is it decreasing? Use the quotient rule.

Answer
Questioner:   Loan
Category:  Calculus
Private:  No
 
Subject:  optimaization problem
Question:  1) The concentration (in milligrams per cubic centimeter) of a certain drug in a patient's body t hours after injections is given by C(t)=t^2/2t^3+1.  0 is less than or equal to t and t is less than or equal to 4.  When is the concentration of the drug increasing, and when is it decreasing? Use the quotient rule.
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Hi, Loan,
            t^2
If C(t) = ---------, with 0 <= t <= 4
         2t^3 + 1

..........................................
{BTW, that is how to write these things.}

You wrote C(t)=t^2/2t^3+1, which means
       t^2
C(t) = ----- + 1
      2t^3

and I don't think you meant that.  Always parenthesize carefully, like

C(t)=t^2/(2t^3+1)
.........................................

Increasing means  C' > 0, decreasing means C'< 0.

So find C', then set it EQUAL to zero, to find the boundary point, between positive and negative, then solve your inequality.  [Look up and practice solving inequalities in your old algebra book.]

To find C'(t), use the quotient rule, as it is said.
        ()() - ()()
C'(t) = -------------  << write that first.
            ()^2

        (2t^3 + 1)(2t) - (t^2)((6t^2)
C'(t) = -----------------------------------  << fill in the stuff.
            (2t^3 + 1)^2

        4t^4 + 2t - 6t^4
C'(t) = ------------------ << simplify
          (2t^3 + 1)^2


        + 2t - 2t^4
C'(t) = ---------------
         (2t^3 + 1)^2

Set 2t - 2t^4 = 0 and solve,

t = 0 and  t = 1.


Now the function is increasing or decreasing on (0,1) and increasing or decreasing on (1,4).

I'll leave that to you.