
Calculus Calculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
July 11th, 2018, 12:15 PM  #1 
Newbie Joined: Jul 2018 From: United States Posts: 1 Thanks: 0  You know that P(10)=30, P'(10)=0.4, and P''(10)=0.0008.
There is a population of P(t) thousand bacteria in a culture at time t hours after the beginning of the experiment. So far, I have the point (10,30), and the slope of 10 at .4, from here I am lost any help will be appreciated. Last edited by skipjack; July 11th, 2018 at 08:14 PM. 
July 11th, 2018, 12:55 PM  #2 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894 
If all you know is the amount, the derivative, and the second derivative at a given t then the best you can say is that the quadratic $\displaystyle P= a(t 10)^2+ b(t 10)+ c$ has $\displaystyle P(10)= c= 30$. The derivative is $\displaystyle P'= 2a(t 10)+ b$ so $\displaystyle P'(10)= b= 0.4$. The second derivative is $\displaystyle P''= 2a$ so $\displaystyle P''(10)= 2a= 0.0008$. So $\displaystyle P(t)= 0.0004(t 10)^2+ 0.4(t 10)+ 30$. Now is there a question about this? (In the title you say "P'(10)= 0.4" but in the body you say "slope of 10 at 0.4". I have assumed you mean that the slope at 0.4 is 10, not the other way around.) Last edited by Country Boy; July 11th, 2018 at 12:59 PM. 
July 11th, 2018, 02:23 PM  #3 
Global Moderator Joined: May 2007 Posts: 6,628 Thanks: 622 
Typical experiments like this involve exponential increase. Do you have any information?

July 12th, 2018, 08:17 PM  #4  
Senior Member Joined: Sep 2016 From: USA Posts: 503 Thanks: 281 Math Focus: Dynamical systems, analytic function theory, numerics  Quote:
The question definitely has information missing. Presumably, it is supposed to mention that $P'$ is proportional to $P$. However, this IVP also would be overdetermined with a value for $P''$ so I'm not sure what is intended. What is definitely not intended is to presume it is a quadratic.  
July 13th, 2018, 04:16 AM  #5  
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 894  Quote:
Quote:
 