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Heinsbergrelatz

Posted: Tue Mar 02, 2010 9:59 am Post subject: Integration-separable

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Joined: 01 Aug 2009
Posts: 633
Location: Singapore

can anyone try and solve this question, i can solve it, but i cant get the correct answer, which tells me something is wrong;

the number of bacteria present in a culture increases at a rate proportional to the number present. If the number increases by 10% in one hour, what percentage increase occurs after a further 5 hours??

thank you in advance~
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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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MagiMaster

Posted: Tue Mar 02, 2010 12:26 pm Post subject:

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A 10% increase means that the total number is multiplied by 1.1. This happens each hour, and the question asks for the answer after six total hours (although it's badly worded). How far did you get on this?

Arcane_Mathematician

Posted: Tue Mar 02, 2010 1:20 pm Post subject:

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it should be modeled as 1.1^xb

yes?
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DrRocket

Posted: Tue Mar 02, 2010 2:46 pm Post subject: Re: Integration-separable

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Posts: 3213

Heinsbergrelatz wrote:

I know that this is not homework.

Let x(t)

be the number of bacteria after t hours

$\frac{dx}{dt} = C x(t)$ for some

So $x(t) = Ce^{At}$ for some A.

$\frac {x(1)}{x(0)} = \frac {11}{10} = e^A$

Hence $A = ln(\frac{11}{10})$ and

is the number of bacteria at time 0, which is arbitrary.

So after 5 hours $\frac {x(5)}{x(0)} = e^{5A}=1.1^5$ .

Heinsbergrelatz

Posted: Tue Mar 02, 2010 11:24 pm Post subject:

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Joined: 01 Aug 2009
Posts: 633
Location: Singapore

thank you for all the replies, but last night i really concentrated and eventually solved this question, this is how i got the answer, but im not sure if its correct. Please correct me if im wrong.

Quote:

How far did you get on this?

well first of all $\frac{dN}{dt} = kN$ as i know that it is direct proportional.

$\displaystyle \int\frac{1}{N}\frac{dN}{dt}$ = $\displaystyle \int k\,dx$

this gives:

ln|N| = kt + c

$N= e^{kt+c}$

$N=Ae^{kt}$ $A=e^{c}$ (this is how i symbolized the "c")

well i think i have to firstly find the value "c".

we can find one information in the text, when the time is 0, then the number of bacteria present will also be unaffected remaining 100%

t=0

so the equation will now stand like this:

ln|N|=kt+ln|N_0|

- now we have to find the "K"

t=1

$k=ln|\frac{1.1N_0}{N_0}|$ = ln|1.1|

so another 5 hours now mean, 6 hours in total.

$N=N_0\times(1.1.)^{t}$ N_0=100%

$N=N_0(1.1)^{6}$

$1.7716...\times N_0$

177.16..%

has increased by 77.16%

does it look right? please correct me if necessary. Thank you in advance
_________________
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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

-------------------

Heinsbergrelatz

Posted: Fri Mar 05, 2010 5:06 am Post subject:

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Dr Rocket wrote:

So after 5 hours

Dr Rocket, i think it should have been $1.1^{6}$ not $1.1.^{5}$ because, i have approached your way, but i dont seem to get a reasonable conclusion.
_________________
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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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DrRocket

Posted: Fri Mar 05, 2010 9:22 am Post subject:

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Heinsbergrelatz wrote:

Dr Rocket wrote:

So after 5 hours

Dr Rocket, i think it should have been $1.1^{6}$ not $1.1.^{5}$ because, i have approached your way, but i dont seem to get a reasonable conclusion.

After 6 hours the factor is $1.1^{6}$ . I agree that the problem by asking for the population after "an additional 5 hours" is asking for the propoulation at 6 hours. I did misread the question in that regard (didn't notice "additional").

However, the derivation that I provided for you gives the population as a function of time, for any time, and that derivation is correct. It also agrees with your work, which would do the same.

Heinsbergrelatz

Posted: Fri Mar 05, 2010 11:03 am Post subject:

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Joined: 01 Aug 2009
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Location: Singapore

DrRocket wrote:

After 6 hours the factor is . I agree that the problem by asking for the population after "an additional 5 hours" is asking for the propoulation at 6 hours. I did misread the question in that regard (didn't notice "additional").

However, the derivation that I provided for you gives the population as a function of time, for any time, and that derivation is correct. It also agrees with your work, which would do the same.

thank you for the clarification.
_________________
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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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DrRocket

Posted: Fri Mar 05, 2010 11:17 am Post subject:

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Posts: 3213

Heinsbergrelatz wrote:

DrRocket wrote:

thank you for the clarification.

Since you are studying for exams, you should go back and compare what you did with what I did and understand why they are essentially the same thing.

Heinsbergrelatz

Posted: Fri Mar 05, 2010 11:29 am Post subject:

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Joined: 01 Aug 2009
Posts: 633
Location: Singapore

Quote:

Since you are studying for exams, you should go back and compare what you did with what I did and understand why they are essentially the same thing.

yes i understand your working

which was $e^{t ln|1.1|}$ where t=5+1

but i realized your working does not convert it in to a % form. apart from that if you just multiply from your solution $e^{ln|1.1| \times 6} \times 100$ it should come to an agreeable answer.
_________________
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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

-------------------

Arcane_Mathematician

Posted: Fri Mar 05, 2010 12:27 pm Post subject:

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Joined: 21 Feb 2009
Posts: 2575
Location: San Jose, California

Heinsbergrelatz wrote:

Quote:

Since you are studying for exams, you should go back and compare what you did with what I did and understand why they are essentially the same thing.

yes i understand your working

which was $e^{t ln|1.1|}$ where t=5+1

but i realized your working does not convert it in to a % form. apart from that if you just multiply from your solution $e^{ln|1.1| \times 6} \times 100$ it should come to an agreeable answer.

Almost. You have to subtract you're initial amount (in this case it's 1) from your final amount, then divide that new number by your initial amount (again, it's 1 in this case) to get the percentage rise.

1.1^6

=177.156% which means you have a 77.1561% increase. you have to subtract the initial amount.
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Plato wrote:

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Heinsbergrelatz

Posted: Fri Mar 05, 2010 11:04 pm Post subject:

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Joined: 01 Aug 2009
Posts: 633
Location: Singapore

Oo yes, to get the difference `
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$G_\mu_\nu + \Lambda g_\mu_\nu=\frac{8\pi G}{c^{4}} T_\mu_\nu$

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