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Expert: Steve Holleran - 8/4/2008


Question

question on rates
Hi Steve,
we were given analysis booklet and the last question has stumped me can you please help!!! the question is attached if you could just help with question a,b c that would be a real big help. i think i know for question a you have to anti-derive it to get p(t) which is just a proof i think but i don't know how to get question b i think A = e^-kc or something but i don't know where to go from there from there. thanks. Jaime

Answer
Hi Jaime,

Wow, this one is a killer!  I haven't come across one like this in a long time, because its a form of logistic function which I don't deal with in intro calc I or II, but I think I found the way through it.  I just hope I can write it clearly enough:

We have              dP/dt = .02P(1 - P/1200)

so            dP / P(1 - P/1200) = .02 dt

now, on the left, the integrand is really

           1 / P(1 - P/1200) * dP

so we need to break the fraction down using partial fractions:

 1 / P(1 - P/1200) = a / P + b / (1200-P)

Multiplying through by P(1200-P), we have :

  1200 = a(1200-P) + bP

  1200 = 1200a -aP + bP  = 1200a + P(b - a)

since the constants must be =, 1200 = 1200a so a = 1.

Then since there are no P terms on the left, the coefficient of P on the right, b-a = 0   ----> b - 1 = 0  so b = 1 and we have the integrand

  INT[1/P + 1/(1200-P)] * dP = INT[.02 * dt]  and that gives

  ln|P| - ln|1200 - P| = .02t + C

or,   ln|1200 - P| - ln|P| = -.02t - C     (mult each side by -1)

Now change to exponential form:

   |(1200-P)/P| = e^(-.02t-C) = e^-.02t * e^-C

and let A = e^-C.

That gives us:     |(1200-P)/P| = Ae^-.02t

Okay, now at t = 0, P = 400, so

                  |800/400| = A e^-.02*0 = A * 1 = A

so                     2 = A.  (This is the answer to part b)

Now to finish part a, We have (dropping the absolute values, since the  values will always be positive,

                  1200-P / P = 2e^-.02t

                 1200 - P = 2Pe^-.02t

                 1200 = P + 2Pe^-.02t

                 1200 = P(1 + 2Pe^-.02t)

so                    P = 1200 / (1 + 2e^-.02t)    Whew!!!

For part c, we have

                800 = 1200 / (1 + 2e^-.02t)

so                 1+2e^-.02t = 1200/800 = 1.5

                    2e^-.02t = 0.5

                     e^-.02t = 0.25   so -.02t = ln.25

and                         t = ln.25 / -.02 = 6.93 yrs

For part d, set up

                    P = 1200 / (1 + Ae^-.02t)

and at t=0, P=3000 so     3000 = 1200 / (1 + Ae^0) = 1200 / (1+A)

so                        1+A = 1200 / 3000 = 0.4

                           A = -0.6


I sure hope you can follow all this--its why it took me so long to get back to you.  This is a pretty advanced problem!

Good luck!

Steve
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    Commentthanks so much Steve, you have know idea how long i have sat down and just looked at that question, fingers crossed its right, only a few had even attempted it. hopefully i don't meet to many more of those questions, thanks again.


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Steve Holleran

Expertise

I can help with all math questions from basic math to Calculus. Whether it`s consumer questions, or questions from high school or college students, I have probably dealt with it at some time in my career.

Experience

33 years teaching experience in NJ public schools

Education/Credentials
B.S. Mathematics : Wake Forest University 1972 M.S. Mathematics : Monmouth University 1981

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