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July 11th, 2018, 12:15 PM   #1
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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.
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July 11th, 2018, 12:55 PM   #2
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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.)
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Last edited by Country Boy; July 11th, 2018 at 12:59 PM.
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July 11th, 2018, 02:23 PM   #3
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Typical experiments like this involve exponential increase. Do you have any information?
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July 12th, 2018, 08:17 PM   #4
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Quote:
Originally Posted by Country Boy View Post
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.
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July 13th, 2018, 04:16 AM   #5
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Quote:
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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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