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 July 11th, 2018, 11:15 AM #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 07:14 PM. July 11th, 2018, 11:55 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 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 11:59 AM. July 11th, 2018, 01:23 PM #3 Global Moderator   Joined: May 2007 Posts: 6,807 Thanks: 717 Typical experiments like this involve exponential increase. Do you have any information? Thanks from topsquark July 12th, 2018, 07:17 PM   #4
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 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, 03:16 AM   #5
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 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. Tags p10, p1004, p1030 Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode

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