My Math Forum You know that P(10)=30, P'(10)=0.4, and P''(10)=0.0008.

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 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: 895 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.) Thanks from topsquark 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: 6,660 Thanks: 647 Typical experiments like this involve exponential increase. Do you have any information? Thanks from topsquark
July 12th, 2018, 08:17 PM   #4
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Quote:
 Originally Posted by Country Boy 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.)
This is extremely misleading. How are you solving a differential equation given only the values for $P,P',P''$ at some moment in time? You seem to be assuming that $P''$ is constant for no reason I can determine.

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
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Quote:
 Originally Posted by SDK This is extremely misleading. How are you solving a differential equation given only the values for $P,P',P''$ at some moment in time? You seem to be assuming that $P''$ is constant for no reason I can determine.
What differential equation are you talking about? There was no differential equation given in the problem. What I gave was the simplest function that satisfies the given data.

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.
Yes, there was information missing. So you decide that it is correct to assume one thing but not another. I did not read the problem that way.

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