Expert: Paul Klarreich - 2/25/2011

Question
Thank you for answering my question yesterday.

Researchers used a version of the Gompertz curve to model the rate that children learn with the equation y(t)=A^(C^t), where y(t) is the portion of children of age t years passing a certain mental test, A=0.3982x10^-291, and C=0.4252. Find the inflexion point and describe what it signifies.

This is my working:
Let y(t)=y

ln y = ln A^(C^t)
ln y = C^t ln A
(1/y)y' = ln A . C^t ln C
y' = ln A . ln C . C^t . y

y" = ln A . ln C. (y(C^t.ln C)+ C^t(y"))
         = ln A . ln C . (A^(C^t) . C^t .ln C
           + C^t . ln A . ln C . C^t . A^(C^t)

In order to find the inflexion point, I know I must take y" and make it equal 0 and find the value of t, but I'm totally stuck at this point because ln A = undefined.

Answer
Questioner: Jasmine
Category: Calculus
Private: No
Subject: Calculus
Question: Thank you for answering my question yesterday.

>>>>> Sorry -- I don't recall doing that.  Did you send it under a different name?  That is a definite no-no.

Researchers used a version of the Gompertz curve to model the rate that children learn with the equation y(t)=A^(C^t), where y(t) is the portion of children of age t years passing a certain mental test, A=0.3982x10^-291, and C=0.4252. Find the inflexion point and describe what it signifies.

This is my working:
Let y(t)=y

ln y = ln A^(C^t)
ln y = C^t ln A
(1/y)y' = ln A . C^t ln C
y' = ln A . ln C . C^t . y

y" = ln A . ln C. (y(C^t.ln C)+ C^t(y"))
        = ln A . ln C . (A^(C^t) . C^t .ln C
          + C^t . ln A . ln C . C^t . A^(C^t)

In order to find the inflexion point, I know I must take y" and make it equal 0 and find the value of t, but I'm totally stuck at this point because ln A = undefined.

......................................

1. ln A is not undefined -- you just haven't calculated it yet.
2. A=0.3982x10^-291 ?? Really?  You have to be kidding!

Nonetheless, you can simplify the work a bit:

Write y = a^u, with u = c^t   << chain rule, right?

We need:

u' = ln c c^t   and    u'' = ln c ln c c^t
.......................

y = a^u

y' = ln a a^u u'  <<< standard differentiation formula

y'' = ln a D(a^u u')

y'' = ln a(ln a a^u u' u' + a^u u'')  << product rule

y'' = ln a(ln a a^u ln c c^t ln c c^t + a^u ln c ln c c^t)

WRITE A = ln a, C = ln c  

(Yes, this is backwards from your symbols. I have 'a' where you have 'A'; you will hve to cope.)

y'' = A(A a^u C^2 c^t c^t + a^u C^2 c^t)

y'' = A C^2a^u c^t (A c^t + 1)

Now a^u and c^t are never zero. So all you need is:

A c^t + 1 = 0

c^t = -1/A, which you can solve for t.
[Note: A = ln a, and a is a small number, so A < 0 and we are OK.]

c^t = -1/A

t ln c = ln(-1/A)

t  = ln(-1/A)/ ln c