Author Topic: Dekoyl's Questions (Read 25847 times) Tweet Share

kamil9876 · « **Reply #45 on:** June 14, 2009, 01:35:05 pm »

let $I_o$ be the original intensity(hence the intensity when it hasn't penetrated at all, so at x=0)

rate of loss(with respect to x) is proportional to I:
$ \frac{dI}{dx}=kI where k<0$

If the whole length is L. then at $x=0.1L, I=0.95I_o$

It's just another ordinary exponential one wehre you have two pieces of info $( x=0, I=I_o and x=0.1L,I=0.95I_o)$

Mao · « **Reply #46 on:** June 15, 2009, 10:10:07 am »

let I(x) be the intensity of the light at the depth x.

$\frac{dI}{dx} \propto -I$ , given that $I(10) = 95$ , find $I(100)$

kamil9876 · « **Reply #47 on:** June 15, 2009, 02:13:32 pm »

Quote from: Mao on June 15, 2009, 10:10:07 am

let I(x) be the intensity of the light at the depth x.

$\frac{dI}{dx} = -I$ , given that $I(10) = 95$ , find $I(100)$

That's a good way of avoiding mess. Just considering the initial Intensity to be 100 and total length to be 100. However one thing you cannot avoid is the constant in the differential equation. You have assumed that it is -1 without justification and in fact it isn't so.

Mao · « **Reply #48 on:** June 15, 2009, 03:00:14 pm »

Quote from: kamil9876 on June 15, 2009, 02:13:32 pm

Quote from: Mao on June 15, 2009, 10:10:07 am
let I(x) be the intensity of the light at the depth x.

$\frac{dI}{dx} = -I$ , given that $I(10) = 95$ , find $I(100)$

That's a good way of avoiding mess. Just considering the initial Intensity to be 100 and total length to be 100. However one thing you cannot avoid is the constant in the differential equation. You have assumed that it is -1 without justification and in fact it isn't so.

ahhh, oooooops. I meant $\propto$

dekoyl · « **Reply #49 on:** June 15, 2009, 08:57:40 pm »

Q: A body initially at room temperature of 20 C, is heated so that its temperature would rise by 5 C/min if no cooling took place. Cooling does occur in accordance with Newton' Law of Cooling and the maximum temperature the body could attain is 120 C. How long would it take to reach a temperature of 100 C?

Thanks

Mao · « **Reply #50 on:** June 15, 2009, 09:53:38 pm »

$\frac{dT}{dt} & = 5 + k(20 - T)$

since the maximum temperature is 120 degrees, this means when T = 120, the rate of heating = rate of cooling, or $\frac{dT}{dt} = 0$

$\implies 0 = 5 + k (20 - 120) \implies k = 0.05$

$\begin{align*} \frac{dT}{dt} & = 5 + \frac{1}{20}(20 - T)\\ \frac{dT}{dt} & = -\frac{1}{20}T + 6 \\ \implies \frac{dt}{dT} & = \frac{20}{-T + 120} \\ t & = -20\log_e |120-T| + C \\ e^{-(t + C)/20} & = |120-T|,\; \mbox{ since }T<120 \\ e^{-t/20 - C/20} & = 120-T \\ T & = 120 - Ae^{-t/20},\; \mbox{ where } A = e^{-C/20}\\ \end{align*}$

given when t=0, T=20, $20 = 120 - A e^0 \implies A = 100$

$T = 120 - 100e^{-t/20}$

hence when T = 100

$\implies 20 = 100e^{-t/20} \implies -\frac{t}{20} = -\log_e 5 \implies t = 20 \log_e 5$

dekoyl · « **Reply #51 on:** June 16, 2009, 10:03:46 pm »

Okay a question to get me started with these types of questions..
Q:A cylindrical vessel is vertically divided into two equal compartments by a porous membrane. One compartment is filled with 12cm and the other is empty. Liquid flows from one compartment to the other at a rate which is proportional to the difference in level of the liquid in the two compartments.
If there is 1 cm of liquid in the second compartment after 1 hour, find how long it will take one-quarter of the liquid to penetrate the membrane.

/0 · « **Reply #52 on:** June 16, 2009, 10:40:36 pm »

Haha soz I haven't really done many of these problems so this whole method might be wrong but i'm just going with what feels right lol

Let the level in the second compartment be $x$ .

The volume in the second compartment will be $V = \frac{1}{2}\pi R^2x\Rightarrow \frac{dV}{dx} = \frac{1}{2}\pi R^2$

$\frac{dV}{dt} = k[(12-x)- x ]=k( 12 - 2x)$

$\Rightarrow \frac{dV}{dx} \times \frac{dx}{dt}=k(12-2x)$

$\frac{1}{2} \pi R^2 \times \frac{dx}{dt} = k(12 - 2x)$

$\frac{dx}{dt} = \frac{2k(12-2x)}{\pi R^2}$

$\frac{dt}{dx} =\frac{\pi R^2}{2k(12-2x)}$

$t = \int \frac{\pi R^2}{2k(12-2x)}dx= \frac{\pi R^2}{4k}\int \frac{1}{6-x}dx=-\frac{\pi R^2}{4k}\log_e|x-6|+C$

When $t = 0$ , $x = 0$

$\Rightarrow 0 = -\frac{\pi R^2}{4k} \log_e(6)+C$

When $t = 1$ , $x = 1$

$\Rightarrow 1 = -\frac{\pi R^2}{4k}\log_e(5)+C$

$\therefore C = \frac{\log_e6}{\log_e \frac{5}{6}}$ and $k = \frac{-\log_e\frac{5}{6} \pi r^2}{4}$

$\therefore t =\frac{1}{\log_e\frac{5}{6}}\log_e|x-6|-\frac{\log_e6}{\log_e\frac{5}{6}}$

$t=\frac{1}{\log_e\frac{5}{6}}\left(\log_e\frac{|x-6|}{6}\right)$

When one-quarter of the liquid penetrates the membrance, $V = \frac{1}{4}\left(\frac{1}{2} \pi R^2 (12)\right)$

$\Rightarrow x = 3$

$t =\frac{-\log_e2}{\log_e\frac{5}{6}} = 3.8\ hours$

kamil9876 · « **Reply #53 on:** June 16, 2009, 10:49:32 pm »

SO the 12cm one will have W cm liquid, whereas the other one has V cm liquid :

$\frac{dV}{dt}=k(W-V)$

However: $W+V=12$ (by conservation of matter

)

$\frac{dV}{dt}=k(12-2V)$

And i think you can handle the rest.

Note: Because a cylinder is one of those... umm... i think they are called "prisms" the height is a direct measure of Volume hence the reason why V+W=12 works. If the two cylinders had different radius's this equation would be invalid, not to mention if we were dealing with cones or other objects with non-constant cross sectional area.

Mao · « **Reply #54 on:** June 16, 2009, 11:03:11 pm »

since the two compartments are equally divided, you don't need to associate volume with radius and things of that nature, as $V \propto h$ where h is the liquid level.

flow out of the first compartment: $\frac{dh}{dt} = -k ( h - (12-h) ) = -2k (h - 6)$ where k is positive. [liquid in the second compartment is 12 - h, and so long as this is less than h, flow will be outwards from first compartment]

$\implies t = -\frac{1}{2k} \log_e |h-6| + C$

$\implies |h-6| = Ae^{-2kt}$ , since h > 6 (second compartment can never have greater liquid level than first), the exponential term is positive. Also since when t=0, h=12, this implies A=6.

$\implies h = 6e^{-2kt} + 6$

When t=1, h=11, this implies $e^{-2k} = \frac{5}{6}$

$\implies h = 6\cdot \left(\frac{5}{6}\right) ^t + 6$ , solve for t when h=9

$\implies \frac{1}{2} = \left(\frac{5}{6}\right) ^t \implies t = \frac{\log_e 2}{\log_e 6 - \log_e 5} = 3.80$ hours

dekoyl · « **Reply #55 on:** July 24, 2009, 09:47:23 pm »

An old question: A particle is projected vertically upwards against air resistance and its acceleration at any time, t seconds, after projection is given by:
$a = -(g+\frac{1}{10}v^2)$ where $v m/s$ is the velocity. If the initial velocity is 20 m/s, find the greatest height reached.

Thanks!

kamil9876 · « **Reply #56 on:** July 24, 2009, 10:10:29 pm »

$\frac{dv}{dt}=-(g+0.1v^2)$

$\frac{dt}{dv}=\frac{-1}{g+0.1v^2}$

$t=-10 \int \frac{1}{10g+v^2} dv$

Perhaps arctan may help.

TrueTears · « **Reply #57 on:** July 24, 2009, 10:17:05 pm »

Would it be better to sub $a = v\frac{dv}{dx}$ because question requires to work out something that relates height and velocity. So...

kamil9876 · « **Reply #58 on:** July 24, 2009, 10:19:25 pm »

yep that works too. Probably even quicker because it doesn't require you to find the time at which the thing reached max height.

TrueTears · « **Reply #59 on:** July 24, 2009, 10:20:26 pm »

Quote from: kamil9876 on July 24, 2009, 10:19:25 pm

yep that works too. Probably even quicker because it doesn't require you to find the time at which the thing reached max height.

Yeah exactly what I was thinking, whatever floats your boat, both ways work just one I think would be a bit more time-saving.

Search the forums now!

We have moved!

Author Topic: Dekoyl's Questions (Read 25847 times) Tweet Share

kamil9876

Re: Dekoyl's Questions

Mao

Re: Dekoyl's Questions

kamil9876

Re: Dekoyl's Questions

Mao

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

Mao

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

/0

Re: Dekoyl's Questions

kamil9876

Re: Dekoyl's Questions

Mao

Re: Dekoyl's Questions

dekoyl

Re: Dekoyl's Questions

kamil9876

Re: Dekoyl's Questions

TrueTears

Re: Dekoyl's Questions

kamil9876

Re: Dekoyl's Questions

TrueTears

Re: Dekoyl's Questions

Recent Posts

User:
Password: