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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: beezy4eva on July 19, 2008, 04:35:07 pm

Title: differential equation help
Post by: beezy4eva on July 19, 2008, 04:35:07 pm: In order to prevent temperature falling too low, a heating system is installed in a greenhouse. While switched on it operates continuously and causes the temperature to change according to the model:
$\frac{dT}{dt} = -k \left( T - T_{0} \right) + H$
where T is the temperature inside the greenhouse at time t, $T_{0}$ is the temperature outside the greenhouse and H is a positive constant relating to the available power of the heating system.
Take k= 0.3662 and $T_{0}$ =5 (assume that temperature outside remains constant)

Using H=4, solve the differential equation and hence find the temperature inside the greenhouse at t= 3

Help would be much appreciated. The correct answer is supposed to be 9.something, but I keep getting way off.
Title: Re: differential equation help
Post by: dcc on July 19, 2008, 05:38:17 pm: What are the initial conditions for the internal temperature of the Greenhouse?
Title: Re: differential equation help
Post by: beezy4eva on July 19, 2008, 05:42:57 pm: Ooops thought i forgot to include something
at t=0 temperature is 20 degrees
Title: Re: differential equation help
Post by: ganges on July 19, 2008, 05:52:45 pm: heres another one.......

A tank contains 2000L of salt solution with concentration of 0.3kg per L.Pure water runs into tanks at 50L per min and the well mixed solution runs out at the same rate. find amount of salt in tank after 5 minutes.
Title: Re: differential equation help
Post by: dcc on July 19, 2008, 06:11:58 pm: $\dfrac{dT}{dt} = -0.3662(T-5) + 4$

$\dfrac{dt}{dT} = \dfrac{1}{-0.3662(T - 5) + 4} = \dfrac{1}{-0.3662T + 5.831} = -\dfrac{1}{0.3662} \cdot \dfrac{-0.3662}{-0.3662T + 5.831}$

$\implies t = -\dfrac{1}{0.3662} \cdot \log_{e} \left| -0.3662T + 5.831 \right| + C$

$t = -2.73075 \log_{e} \left| -0.3662T + 5.831 \right| + 1.09445$

$T = 15.923 + 2.73075 \; \exp \left(0.400788 - 0.3662t \right)$

$T(3) = 17.282 ^\circ$
Title: Re: differential equation help
Post by: beezy4eva on July 19, 2008, 06:16:26 pm: dcc thats what i got but my book says its wrong :(
Title: Re: differential equation help
Post by: ganges on July 19, 2008, 06:17:23 pm: It says 529.5
Title: Re: differential equation help
Post by: dcc on July 19, 2008, 06:17:57 pm: Quote from: ganges on July 19, 2008, 05:52:45 pm
heres another one.......

A tank contains 2000L of salt solution with concentration of 0.3kg per L.Pure water runs into tanks at 50L per min and the well mixed solution runs out at the same rate. find amount of salt in tank after 5 minutes.

Let Q be the amount of salt in the tank at time t minutes.

$\dfrac{dQ}{dt} = -50 \cdot \dfrac{Q}{2000} = -\dfrac{Q}{40}$

$\dfrac{dt}{dQ} = -\dfrac{40}{Q} \implies t = -40 \log_{e} Q + C$

At $t = 0$ , $Q = 2000 \cdot 0.3\; kg = 600 \; kg$

$0 = -40 \log_{e} 600 + C \implies C = 40 \log_{e} 600$

$t = 40 \log_{e} {600} - 40 \log_{e} Q = 40 \log_{e} \dfrac{600}{Q} \implies \dfrac{600}{Q} = \exp \left(\dfrac{t}{40}\right)$

$Q = \dfrac{600}{\exp \left(\dfrac{t}{40}\right)} \implies Q(5) = \dfrac{600}{\exp \left(\dfrac{1}{8}\right)} = 529.50\; kg \;$
Title: Re: differential equation help
Post by: phagist_ on July 19, 2008, 06:18:33 pm: @beezy
$\frac{dx}{dt}$ should be $\frac{-50 x}{2000} = \frac{-x}{40}$
Title: Re: differential equation help
Post by: dcc on July 19, 2008, 06:22:41 pm: Quote from: beezy4eva on July 19, 2008, 06:16:26 pm
dcc thats what i got but my book says its wrong :(

I didn't like that question, I do not think they would put that on a VCE exam.
Title: Re: differential equation help
Post by: ganges on July 20, 2008, 05:30:51 pm: A theremometer is taken from a house at 21 degrees to the outside. One minute later it reads 27 degrees, another minute later it reads 30 degrees. find temperature outside house.

answers like 33 degress, how?
Title: Re: differential equation help
Post by: Mao on July 20, 2008, 06:03:19 pm: Newton's law of cooling - the rate at which a body cools is proportional to the difference between its temperature and that of its immediate surroundings

$\frac{dT}{dt}\propto T_o-T$ where $T<T_o$

$\frac{dT}{dt}=k(T_o-T)$

$t=\int \frac{dT}{k(T_o-T)}$

$t=\frac{1}{k}log_e|T_o-T|+C$

when t=0, T=21

$0=\frac{1}{k}log_e|T_o-21|+C$

$C=-\frac{1}{k}log_e|T_o-21|$

$t=\frac{1}{k}log_e \left| \frac{T_o-T}{T_o-21}\right|$

when t=1, T=27
when t=2, T=30

$1=\frac{1}{k}log_e \left( \frac{T_o-27}{T_o-21}\right)$ -- [1]

$2=\frac{1}{k}log_e \left( \frac{T_o-30}{T_o-21}\right)$ -- [2]

[2]-2*[1]

$0=\frac{1}{k}log_e \left(\frac{T_o-30}{T_o-21} \cdot \frac{(T_o-21)^2}{(T_o-27)^2}\right)$

$\frac{(T_o-30)(T_o-21)}{(T_o-27)^2}=1$

$(T_o-30)(T_o-21)=(T_o-27)^2$

$T_o^2-51T_o+630=T_o^2 - 54T_o+729$

$3T_o=99$

$\boxed{T_o=33}$

not sure if there's an easier way, this does seem a bit too much...
Title: Re: differential equation help
Post by: perfectscore on August 05, 2008, 10:06:21 pm: what is a differential equation? can somebody give me a short wrap on what it is? how is it different to just diffing or antidiffing a function?