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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Mao on May 23, 2008, 07:48:18 pm

Title: dcc + differential equations = atrocious
Post by: Mao on May 23, 2008, 07:48:18 pm: yep

Quote from: dcc on irc
[18:43] <dcc> ok
[18:44] there is 250 litres of liquid in a tank, and 30 grams of salt (initially)
[18:44] <dcc> A solution is being pumped in at 9 L/min with a concentration of x g/L
[18:44] The solution is kept uniform by mixing and is pumped out at a rate of 13L/min
[18:44] Find Q, the amount of salt in the tank at time T
[18:45] assume x consant

differential equation:

equation A: $\frac{dQ}{dt}=9x-\frac{13Q}{250-4t}$

$\frac{dQ}{dt} + \frac{13}{250-4t}Q = 9x$

let $I(t)=\mbox{Exp}\left( \int \frac{13}{250-4t}\; dt \right) = \mbox{Exp}\left( -\frac{13}{4}\log_e (250-4t)\right) = (250-4t)^{-13/4}$

$I(t)\cdot Q = \int I(t)\cdot 9x \; dt$

$\implies Q=\frac{9x\cdot \frac{1}{9}\cdot (250-4t)^{-9/4}+C}{(250-4t)^{-13/4}} = x\cdot (250-4t)+C(250-4t)^{13/4}=2x(125-2t)+C(250-4t)^{13/4}$

then solve for C using initial condition = VERY UGLY solution
Title: Re: dcc + differential equations = atrocious
Post by: dcc on May 23, 2008, 07:51:33 pm: (18:43:49) (dcc) there is 250 litres of liquid in a tank, and 30 grams of salt (initially)
(18:43:58) (Mao) dcc: is that the hardest?
(18:44:05) (dcc) A solution is being pumped in at 9 L/min with a concentration of x g/L
(18:44:19) (dcc) The solution is kept uniform by mixing and is pumped out at a rate of 13L/min
(18:44:31) (dcc) Find Q, the amount of salt in the tank at time T
(18:45:14) (dcc) assume x constant
Title: Re: dcc + differential equations = atrocious
Post by: dcc on May 23, 2008, 08:41:36 pm: $Q = \dfrac{-\left.\left(1 \cdot \log_{e}(k|250 - 4t|^{2.25 (250 -4t)x})}{\dfrac{W\left[-\left(1.68179 (125 -2t)^{3/4} \right.\right.\\ \left.\log_{e}\left[k |250 -4t|^{562\cdot5 x-9tx}\right]\right)}{ \left(3.90625\times 10^9 - 2.5\times 10^8 t+6.\times 10^6 t^2-64000t^3+256t^4\right)\right]}}$

lol :P
Title: Re: dcc + differential equations = atrocious
Post by: AppleXY on May 23, 2008, 08:44:05 pm: Quote from: dcc
(http://stuff.daniel15.com/cgi-bin/mathtex.cgi?Q%20=%20\dfrac{-\left.\left(1%20\cdot%20\log_{e}(k|250%20-%204t|^{2.25%20(250%20-4t)x})}{\dfrac{W\left[-\left(1.68179%20(125%20-2t)^{3/4}%20\right.\right.\\\left.\log_{e}\left[k%20|250%20-4t|^{562\cdot5%20x-9tx}\right]\right)}{\left(3.90625\times%2010^9%20-%202.5\times%2010^8%20t+6.\times%2010^6%20t^2-64000t^3+256t^4\right)\right]}}Q = \dfrac{-\left.\left(1 \cdot \log_{e}(k|250 - 4t|^{2.25 (250 -4t)x})}{\dfrac{W\left[-\left(1.68179 (125 -2t)^{3/4} \right.\right.\\\left.\log_{e}\left[k |250 -4t|^{562\cdot5 x-9tx}\right]\right)}{\left(3.90625\times 10^9 - 2.5\times 10^8 t+6.\times 10^6 t^2-64000t^3+256t^4\right)\right]}})
holy sweet mother.
Title: Re: dcc + differential equations = atrocious
Post by: Ahmad on May 23, 2008, 08:45:26 pm: http://en.wikipedia.org/wiki/Linear_differential_equation#First_order_equation
Title: Re: dcc + differential equations = atrocious
Post by: Mao on May 23, 2008, 08:56:53 pm: Quote from: dcc on May 23, 2008, 08:41:36 pm
$Q = \dfrac{-\left.\left(1 \cdot \log_{e}(k|250 - 4t|^{2.25 (250 -4t)x})}{\dfrac{W\left[-\left(1.68179 (125 -2t)^{3/4} \right.\right.\\ \left.\log_{e}\left[k |250 -4t|^{562\cdot5 x-9tx}\right]\right)}{ \left(3.90625\times 10^9 - 2.5\times 10^8 t+6.\times 10^6 t^2-64000t^3+256t^4\right)\right]}}$

lol :P
MATHEMATICA EXPOSED!!!!! :P
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 09, 2008, 11:44:46 pm: wow. took me 3 days to do differntial equations... for just one exercise. i feel so stupid :(
Title: Re: dcc + differential equations = atrocious
Post by: midas_touch on July 10, 2008, 12:19:47 am: I don't think youse will be required to solve these types of DEs in Spesh, you'll only have to be able to set them up afaik.
Title: Re: dcc + differential equations = atrocious
Post by: bigtick on July 10, 2008, 08:47:48 am: Note:
Is dt/dQ = f(Q) + constant the same as dQ/dt = 1/f(Q) + constant ?
Is $\frac{dt}{dQ}=f(Q)+constant$ the same as $\frac{dQ}{dt}=\frac{1}{f(Q)}+constant$ ?
Title: Re: dcc + differential equations = atrocious
Post by: beezy4eva on July 10, 2008, 12:39:30 pm: i told you dcc is evil
Title: Re: dcc + differential equations = atrocious
Post by: AppleXY on July 10, 2008, 12:58:18 pm: Quote from: beezy4eva on July 10, 2008, 12:39:30 pm
i told you dcc is evil

lol why is dcc evil. Is it because he's smarter than u :P
Title: Re: dcc + differential equations = atrocious
Post by: beezy4eva on July 10, 2008, 01:10:35 pm: Quote from: AppleXY on July 10, 2008, 12:58:18 pm
Quote from: beezy4eva on July 10, 2008, 12:39:30 pm
i told you dcc is evil

lol why is dcc evil. Is it because he's smarter than u :P
partially :P
also coz he keeps making poor mayo do long questions
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 10, 2008, 07:17:42 pm: Quote from: bigtick on July 10, 2008, 08:47:48 am
Note:
Is dt/dQ = f(Q) + constant the same as dQ/dt = 1/f(Q) + constant ?

did you mean $\frac{dQ}{dt}=\frac{1}{f(Q)+c}$ or $\frac{dQ}{dt}=\frac{1}{f(Q)}+c$ ?

the earlier would be true [even though some "real" mathematicians are against it], the latter will be false.

please learn how to use $\LaTeX$ :)

[and oh, i see what you mean now... yes, that's a big mistake =P ]
Title: Re: dcc + differential equations = atrocious
Post by: phagist_ on July 10, 2008, 07:20:06 pm: Quote from: bigtick on July 10, 2008, 08:47:48 am
Note:
Is dt/dQ = f(Q) + constant the same as dQ/dt = 1/f(Q) + constant ?
It would be the same if it wasn't an initial value problem (i.e you didnt' have to find the constant), since c represents ANY arbitrary constant.

I THINK.
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 10, 2008, 08:39:39 pm: so what IS an easy way to determine how to put together a differential equation. because this has been the worst topic for me... i dunno where to start!!! :(
Title: Re: dcc + differential equations = atrocious
Post by: phagist_ on July 10, 2008, 08:41:25 pm: well, do you have any specific problems in mind?
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 10, 2008, 11:53:10 pm: Quote from: phagist_ on July 10, 2008, 08:41:25 pm
well, do you have any specific problems in mind?

A colony of seals is declining in numbers so that there are "N" seals at the time "t" years. it is assumed that the rate of decline is proportional to "N". initially in the period of study there were 5000 seals in the colony but after 5 years, a 5% decline was noted. set up the differential equation and solve it, expressing "N" in terms of "t"

* wen it says proportional. am i right in saying:
dN/dt = -kN? where k is a constant?, and the -ve coz its a decline?

after this im stuck only because of the 5% thingy :D thanxs heaps!
Title: Re: dcc + differential equations = atrocious
Post by: phagist_ on July 11, 2008, 12:19:56 am: yep so you have

$\frac{dN}{dt}= -kN$
you are given (when $t=0~N=5000$ ) and using the 5% decline information you can get another set of data points (when $t=5~N=5000-(.05\times5000)$ ).

You need both sets since you need to find the constant of integration and the constant of proportionality $k$
So;

$\frac{dN}{dt}= -kN$

and $t(5000)=0~and~t(5)=4750$
Title: Re: dcc + differential equations = atrocious
Post by: bigtick on July 11, 2008, 09:28:14 am: Another question
Is $\int \frac{1}{Q}dQ\frac{dt}{dQ}=\frac{dt}{dQ}\int \frac{1}{Q}dQ$ ?
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 11, 2008, 09:41:04 am: Quote from: bigtick on July 11, 2008, 09:28:14 am
Another question
Is $\int \frac{1}{Q}dQ\frac{dt}{dQ}=\frac{dt}{dQ}\int \frac{1}{Q}dQ$ ?

at first glance it doesn't look like it's right,
but at the time this question popped up, someone did assure me it was basically just the chain rule, and I did somehow manage to convince myself with some reasoning or other

so to be quite frank, I don't have a clue right now :P
Title: Re: dcc + differential equations = atrocious
Post by: bigtick on July 11, 2008, 10:19:13 am: Try with $Q=\frac{1}{t}$ .
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 11, 2008, 10:23:26 am: Quote from: bigtick on July 11, 2008, 10:19:13 am
Try with $Q=\frac{1}{t}$ .

i see what you mean.
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 11, 2008, 08:56:52 pm: Quote from: phagist_ on July 11, 2008, 12:19:56 am
yep so you have

$\frac{dN}{dt}= -kN$
you are given (when $t=0~N=5000$ ) and using the 5% decline information you can get another set of data points (when $t=5~N=5000-(.05\times5000)$ ).

You need both sets since you need to find the constant of integration and the constant of proportionality $k$
So;

$\frac{dN}{dt}= -kN$

and $t(5000)=0~and~t(5)=4750$

i dont exactly get how u just pluged numbers in. lol.
like where does the intergration and the flip of dN/dt come in? :(

??? lol
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 11, 2008, 09:19:14 pm: Quote from: dusty_girl1144 on July 10, 2008, 11:53:10 pm
Quote from: phagist_ on July 10, 2008, 08:41:25 pm
well, do you have any specific problems in mind?

A colony of seals is declining in numbers so that there are "N" seals at the time "t" years. it is assumed that the rate of decline is proportional to "N". initially in the period of study there were 5000 seals in the colony but after 5 years, a 5% decline was noted. set up the differential equation and solve it, expressing "N" in terms of "t"

* wen it says proportional. am i right in saying:
dN/dt = -kN? where k is a constant?, and the -ve coz its a decline?

after this im stuck only because of the 5% thingy :D thanxs heaps!

firstly:

"it is assumed that the rate of decline is proportional to "N""
that is, rate of change of population $\frac{dN}{dt}$ is negative (decline) and is proportional to N

$\frac{dN}{dt}\propto -N\implies \frac{dN}{dt}=-kN$

$\implies \frac{dt}{dN}=-\frac{1}{kN}$

$\implies t=\int -\frac{1}{kN}\; dN$

$\implies t=-\frac{1}{k}\cdot log_e|N|+c$

at t=0, N=5000

$\implies 0=-\frac{1}{k}\cdot ln(5000) + c$

at t=5, N has dropped 5% $\implies N_{t=5}=5000*(100\%-5\%)=4750$

$\implies 5=-\frac{1}{k} \cdot ln(4750)+c$

solving these two simultaneously will give the values for k and c, which you can then just sub in and rearrange :)
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 11, 2008, 09:20:48 pm: Quote from: Mao on July 11, 2008, 09:19:14 pm
Quote from: dusty_girl1144 on July 10, 2008, 11:53:10 pm
Quote from: phagist_ on July 10, 2008, 08:41:25 pm
well, do you have any specific problems in mind?

A colony of seals is declining in numbers so that there are "N" seals at the time "t" years. it is assumed that the rate of decline is proportional to "N". initially in the period of study there were 5000 seals in the colony but after 5 years, a 5% decline was noted. set up the differential equation and solve it, expressing "N" in terms of "t"

* wen it says proportional. am i right in saying:
dN/dt = -kN? where k is a constant?, and the -ve coz its a decline?

after this im stuck only because of the 5% thingy :D thanxs heaps!

firstly:

"it is assumed that the rate of decline is proportional to "N""
that is, rate of change of population $\frac{dN}{dt}$ is negative (decline) and is proportional to N

$\frac{dN}{dt}\propto -N\implies \frac{dN}{dt}\propto -kN$

is that the same as dN/dt = -kN? (like without the fishy sign?)

if it is. yeah ive gotten that so far. :) whats next?
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 11, 2008, 09:31:24 pm: [i do a lot of edits, that was a mistake i made in copy-pasta] :)
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 11, 2008, 09:33:08 pm: Quote from: Mao on July 11, 2008, 09:19:14 pm
Quote from: dusty_girl1144 on July 10, 2008, 11:53:10 pm
Quote from: phagist_ on July 10, 2008, 08:41:25 pm
well, do you have any specific problems in mind?

A colony of seals is declining in numbers so that there are "N" seals at the time "t" years. it is assumed that the rate of decline is proportional to "N". initially in the period of study there were 5000 seals in the colony but after 5 years, a 5% decline was noted. set up the differential equation and solve it, expressing "N" in terms of "t"

* wen it says proportional. am i right in saying:
dN/dt = -kN? where k is a constant?, and the -ve coz its a decline?

after this im stuck only because of the 5% thingy :D thanxs heaps!

firstly:

"it is assumed that the rate of decline is proportional to "N""
that is, rate of change of population $\frac{dN}{dt}$ is negative (decline) and is proportional to N

$\frac{dN}{dt}\propto -N\implies \frac{dN}{dt}=-kN$

$\implies \frac{dt}{dN}=-\frac{1}{kN}$

$\implies t=\int -\frac{1}{kN}\; dN$

$\implies t=-k\cdot log_e|N|+c$

at t=0, N=5000

$\implies 0=-k\cdot ln(5000) + c$

at t=5, N has dropped 5% $\implies N_{t=5}=5000*(100%-5%)=4750$

$\implies 3=-k \cdot ln(4750)+c$

solving these two simultaneously will give the values for k and c, which you can then just sub in and rearrange :)

ok i get it up to t=0, N=5000.
but after that im kinda lost on how u get 3 on the last line
and like is something cut out after the 100? lol :s
Title: Re: dcc + differential equations = atrocious
Post by: *Roxxii* on July 11, 2008, 09:37:42 pm: Quote from: dusty_girl1144 on July 11, 2008, 09:20:48 pm

is that the same as dN/dt = -kN? (like without the fishy sign?)

if it is. yeah ive gotten that so far. :) whats next?

ROFL @ "fishy sign" !! hehee :D
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 11, 2008, 09:41:38 pm: Quote from: dusty_girl1144 on July 11, 2008, 09:33:08 pm
ok i get it up to t=0, N=5000.
but after that im kinda lost on how u get 3 on the last line
and like is something cut out after the 100? lol :s

yeah, it didn't like my % sign. fixed now.
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 11, 2008, 10:11:33 pm: [/tex][/tex]
Quote from: Mao on July 11, 2008, 09:19:14 pm
Quote from: dusty_girl1144 on July 10, 2008, 11:53:10 pm
Quote from: phagist_ on July 10, 2008, 08:41:25 pm
well, do you have any specific problems in mind?

A colony of seals is declining in numbers so that there are "N" seals at the time "t" years. it is assumed that the rate of decline is proportional to "N". initially in the period of study there were 5000 seals in the colony but after 5 years, a 5% decline was noted. set up the differential equation and solve it, expressing "N" in terms of "t"

* wen it says proportional. am i right in saying:
dN/dt = -kN? where k is a constant?, and the -ve coz its a decline?

after this im stuck only because of the 5% thingy :D thanxs heaps!

firstly:

"it is assumed that the rate of decline is proportional to "N""
that is, rate of change of population $\frac{dN}{dt}$ is negative (decline) and is proportional to N

$\frac{dN}{dt}\propto -N\implies \frac{dN}{dt}=-kN$

$\implies \frac{dt}{dN}=-\frac{1}{kN}$

$\implies t=\int -\frac{1}{kN}\; dN$

$\implies t=-k\cdot log_e|N|+c$

at t=0, N=5000

$\implies 0=-k\cdot ln(5000) + c$

at t=5, N has dropped 5% $\implies N_{t=5}=5000*(100\%-5\%)=4750$

$\implies 5=-k \cdot ln(4750)+c$

solving these two simultaneously will give the values for k and c, which you can then just sub in and rearrange :)

hummm ok. well so far my 2 simultaneous equations are

5=-1/k ln(4750)+ C
0=-1/k $log_e(5000)+ C$

it works out to be 5= -1/k ln(4750) + 1/k ln(5000)
therefore equals 5 = 1/k (ln(5000)/(4750)) ==> 5=1/k(ln(20/29))

k = 1/5 (ln(20/19))

subbed back in i get. 0= -5/ln(20/19) x ln(5000) + C

therefore C = 5(ln(5000))/ln(20/19)

..... t = -5/ln (20/19) X Ln (N) + 5(ln(5000))/ln(20/19)

am i CLOSE to correct? if so i sub t and find N? t=5?

btw how do u do the itergration sign and the devision things. coz my post looks like shit! lol :(
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 11, 2008, 10:15:48 pm: Quote from: dusty_girl1144 on July 11, 2008, 10:11:33 pm
hummm ok. well so far my 2 simultaneous equations are

5=-1/k ln(4750)+ C
0=-1/k ln(5000)+ C

it works out to be 5= -1/k ln(4750) + 1/k ln(5000)
therefore equals 5 = 1/k (ln(5000)/(4750)) ==> 5=1/k(ln(20/29))

k = 1/5 (ln(20/19))

subbed back in i get. 0= -5/ln(20/19) x ln(5000) + C

therefore C = 5(ln(5000))/ln(20/19)

..... t = -5/ln (20/19) X Ln (N) + 5(ln(5000))/ln(20/19)

am i CLOSE to correct? if so i sub t and find N? t=5?

btw how do u do the itergration sign and the devision things. coz my post looks like shit! lol :(

ahh, i did make a stupid mistake [again]

$\LaTeX$ typeset: http://vcenotes.com/forum/index.php/topic,3137.0.html

and yes, I think you are correct.

if we transpose $t=-\frac{1}{k}\cdot log_e|N|+c$

$\implies N=e^{-k(t-c)}$

substituting k and c

$\implies N=e^{-\frac{ln(20/19)}{5}\cdot \left(t-\frac{5\cdot ln(5000)}{ln(20/19)}\right)}$

$\implies N=e^{-\frac{ln(20/19)}{5}t+ln(5000)}$

$\implies N=5000e^{-\frac{ln(20/19)}{5}t}$

$\implies N=5000\left(e^{ln(20/19)}\right)^{ -(t/5)}$

$\implies N=5000\left(\frac{19}{20}\right)^{t/5}$
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 11, 2008, 10:24:56 pm: Quote from: Mao on July 11, 2008, 10:15:48 pm
Quote from: dusty_girl1144 on July 11, 2008, 10:11:33 pm
hummm ok. well so far my 2 simultaneous equations are

5=-1/k ln(4750)+ C
0=-1/k ln(5000)+ C

it works out to be 5= -1/k ln(4750) + 1/k ln(5000)
therefore equals 5 = 1/k (ln(5000)/(4750)) ==> 5=1/k(ln(20/29))

k = 1/5 (ln(20/19))

subbed back in i get. 0= -5/ln(20/19) x ln(5000) + C

therefore C = 5(ln(5000))/ln(20/19)

..... t = -5/ln (20/19) X Ln (N) + 5(ln(5000))/ln(20/19)

am i CLOSE to correct? if so i sub t and find N? t=5?

btw how do u do the itergration sign and the devision things. coz my post looks like shit! lol :(

ahh, i did make a stupid mistake [again]

$\LaTeX$ typeset: http://vcenotes.com/forum/index.php/topic,3137.0.html

and yes, I think you are correct.

if we transpose $t=-\frac{1}{k}\cdot log_e|N|+c$

$\implies N=e^{-k(t-c)}$

substituting k and c

$\implies N=e^{-\frac{ln(20/19)}{5}\cdot \left(t-\frac{5\cdot ln(5000)}{ln(20/19)}\right)}$

wow..... ur like so smart!!! lol.... jee wish i could do it as easy as u :( but glad to hear i did something right. im actually gettin the kinda hang of differential equations :)
thanxs heaps!

will keep u posted on harder ones :P ;D

NOTE: ummmm am i meant to get 4750 as the answer once i sub t=5 in the last line????? coz i just did :'(
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 11, 2008, 10:40:45 pm: Quote from: dusty_girl1144 on July 11, 2008, 10:24:56 pm
NOTE: ummmm am i meant to get 4750 as the answer once i sub t=5 in the last line????? coz i just did :'(

yes

$N=5000\left(\frac{19}{20}\right)^{t/5}$

$N_{t=5}=5000\cdot \left(\frac{19}{20}\right)^1 = 4750$
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 11, 2008, 11:07:02 pm: Quote from: Mao on July 11, 2008, 10:40:45 pm
Quote from: dusty_girl1144 on July 11, 2008, 10:24:56 pm
NOTE: ummmm am i meant to get 4750 as the answer once i sub t=5 in the last line????? coz i just did :'(

yes

$N=5000\left(\frac{19}{20}\right)^{t/5}$

$N_{t=5}=5000\cdot \left(\frac{19}{20}\right)^1 = 4750$

ok just checking :)

try this one out. i thought this was more related rates of change? how can it be done in differential equations form?

A sepherical ball of ice initially with a radius of 20cm , slowly melts so that the volume decreases at a constant rate of on $10 (cm^3)/min.$ find an expression for $\frac{dS}{dt}$ in terms of r where S $(cm^3)$ is the surface area and $S= 4\pi r^2$
Title: Re: dcc + differential equations = atrocious
Post by: ell on July 11, 2008, 11:54:02 pm: Quote from: dusty_girl1144 on July 11, 2008, 11:07:02 pm
ok just checking :)

try this one out. i thought this was more related rates of change? how can it be done in differential equations form?

A sepherical ball of ice initially with a radius of 20cm , slowly melts so that the volume decreases at a constant rate of on $10 (cm^3)/min.$ find an expression for $\frac{dS}{dt}$ in terms of r where S $(cm^3)$ is the surface area and $S= 4\pi r^2$

Yeah I think this one is just a related rates question, unless there's more parts to it.

$V=\frac{4}{3}\pi r^{3}$

$\Longrightarrow \frac{dV}{dr}=4\pi r^2$

We're told in the question that the volume is decreasing at a rate of $10\ cm^3/min$ , i.e. $\frac{dV}{dt}=-10\ cm^3/min$
Using this information we can now find $\frac{dr}{dt}$ :

$ \begin{align*} \frac{dr}{dt}&=\frac{dr}{dV}\cdot \frac{dV}{dt} \\ &= \frac{1}{4\pi r^2}\cdot -10 \\ &= -\frac{5}{2\pi r^2} \end{align*} $

Deriving surface area equation:
$S=4\pi r^2 \Longrightarrow \frac{dS}{dr}=8\pi r$

We can now find $\frac{dS}{dt}$ :

$ \begin{align*} \frac{dS}{dt}&=\frac{dS}{dr}\cdot \frac{dr}{dt} \\ &=8\pi r \cdot -\frac{5}{2\pi r^2} \\ &=-\frac{20}{r} \end{align*} $
Title: Re: dcc + differential equations = atrocious
Post by: dcc on July 12, 2008, 12:47:53 am: Find $\dfrac{du}{dx}$ given $u = x + y$ and hence find the general solution to the differential equation $\dfrac{dy}{dx} = \left(x + y\right)^2 + \left(x + y\right)$

(not too difficult but a slightly interesting question none the less p.s. ahmad + adib stay away stealers)
Title: Re: dcc + differential equations = atrocious
Post by: /0 on July 12, 2008, 01:38:23 am: Quote from: dcc on July 12, 2008, 12:47:53 am
Find $\dfrac{du}{dx}$ given $u = x + y$ and hence find the general solution to the differential equation $\dfrac{dy}{dx} = \left(x + y\right)^2 + \left(x + y\right)$

(not too difficult but a slightly interesting question none the less p.s. ahmad + adib stay away stealers)

$\frac{du}{dx} = 1 + \frac{dy}{dx}$

$\Rightarrow \frac{du}{dx} -1= u^2+u$

$du = (u^2+u+1)dx$

$\int \frac{du}{u^2+u+1}=\int dx$

$\int \frac{du}{(u+\frac{1}{2})^2+\frac{3}{4}}=\int dx$

Let $\frac{\sqrt{3}}{2}\tan \theta = u + \frac{1}{2}$ then $du = \frac{\sqrt{3}}{2}sec^2{\theta} d\theta$

$\Rightarrow \int \frac{\frac{\sqrt{3}}{2} \sec^2{\theta} d \theta}{\frac{3}{4} \tan^2 \theta + \frac{3}{4}}= \int dx$

$\frac{2\sqrt{3}}{3} \int \frac{\sec^2{\theta}d \theta}{\sec^2{\theta}}$

$\frac{2\sqrt{3}}{3} \theta + C = x + D$

$\frac{2\sqrt{3}}{3} \tan^{-1} \left(\frac{\sqrt{3}(2u+1)}{3}\right)+C = x+D$

$\frac{2\sqrt{3}}{3} \tan^{-1} \left(\frac{\sqrt{3}(2(x+y)+1)}{3}\right)+C = x+D$

$\frac{\sqrt{3}(2(x+y)+1)}{3} = \tan( \frac{3}{2\sqrt{3}}(x+c))$ where $c = D-C$

$y = \frac{1}{2} \left(\frac{3}{\sqrt{3}}\tan( \frac{3}{2\sqrt{3}}(x+c)) - 1 \right)-x$ waaaaaaaaah
Title: Re: dcc + differential equations = atrocious
Post by: dcc on July 12, 2008, 11:54:23 am: Quote from: DivideBy0 on July 12, 2008, 01:38:23 am

$\frac{du}{dx} = 1 + \frac{dy}{dx}$

$\Rightarrow \frac{du}{dx} -1= u^2+u$

$du = (u^2+u+1)dx$

$\int \frac{du}{u^2+u+1}=\int dx$

$\int \frac{du}{(u+\frac{1}{2})^2+\frac{3}{4}}=\int dx$

Let $\frac{\sqrt{3}}{2}\tan \theta = u + \frac{1}{2}$ then $du = \frac{\sqrt{3}}{2}sec^2{\theta} d\theta$

$\Rightarrow \int \frac{\frac{\sqrt{3}}{2} \sec^2{\theta} d \theta}{\frac{3}{4} \tan^2 \theta + \frac{3}{4}}= \int dx$

$\frac{2\sqrt{3}}{3} \int \frac{\sec^2{\theta}d \theta}{\sec^2{\theta}}$

$\frac{2\sqrt{3}}{3} \theta + C = x + D$

$\frac{2\sqrt{3}}{3} \tan^{-1} \left(\frac{\sqrt{3}(2u+1)}{3}\right)+C = x+D$

$\frac{2\sqrt{3}}{3} \tan^{-1} \left(\frac{\sqrt{3}(2(x+y)+1)}{3}\right)+C = x+D$

$\frac{\sqrt{3}(2(x+y)+1)}{3} = \tan( \frac{3}{2\sqrt{3}}(x+c))$ where $c = D-C$

$y = \frac{1}{2} \left(\frac{3}{\sqrt{3}}\tan( \frac{3}{2\sqrt{3}}(x+c)) - 1 \right)-x$ waaaaaaaaah

I love it! What an inspired substitution :P

another way of integrating that quadratic was making the substitution $t = u + \dfrac{1}{2}$ and arctanning, but I like what you did :P
Title: Re: dcc + differential equations = atrocious
Post by: /0 on July 12, 2008, 12:12:09 pm: lol thanks dcc, I guess I just did a roundabout way of arctanning
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 12, 2008, 05:15:19 pm: Quote from: DivideBy0 on July 12, 2008, 01:38:23 am

$\Rightarrow \frac{du}{dx} -1= u^2+u$

$du = (u^2+u+1)dx$

$\int \frac{du}{u^2+u+1}=\int dx$

naughty :P
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 12, 2008, 09:21:55 pm: need help!!....

A container with a conical shape has a height equal to the diameter of the top. Water is poured into the container at the rate of 20 L/min. if the depth is h cm after t mins, and the container was initially empty. express $\frac{dh}{dt}$ in terms of h

i guessin $\frac{dV}{dt}$ = 20 ?
but wat the? is next....
isant this related? how come i have it as a question relating to differential equations?

.xxx.
Title: Re: dcc + differential equations = atrocious
Post by: phagist_ on July 12, 2008, 09:32:48 pm: since $\frac{dV}{dt}=20 L/min$ converting to common units (cm $^3$ /min);
$\frac{dV}{dt}=20000 cm^3/min$

and we are looking for $\frac{dh}{dt}$

and $\frac{dh}{dt} = \frac{dV}{dt}\frac{dh}{dV}$ (chain rule)

using knowledge of the volume of a cone; $v=\frac{1}{3} \pi r^2 h$

since $h=2r$ this means $r=\frac{h}{2}$

=> $v=\frac{\pi h^3}{12}$

differentiating with respect to h; $\frac{dV}{dh} = \frac{\pi h^2}{4}$

so $\frac{dh}{dv} = \frac{4}{\pi h^2}$

subbing this into;

$\frac{dh}{dt} = \frac{dV}{dt}\frac{dh}{dV}$ yields;

$\frac{dh}{dt} = 20000 \times \frac{4}{\pi h^2} = \frac{80000}{\pi h^2}$
(20000 is used as h is in cm, and 1L = 1000 cm³)

Yeh, this is a related rates question.
(sorry if there are mistakes am hungover and tired)
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 12, 2008, 09:35:31 pm: Quote from: phagist_ on July 12, 2008, 09:32:48 pm
wait, what's s?

altered :) sorry my typo
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 12, 2008, 09:37:43 pm: volume of a cone: $V=\frac{1}{3}\pi r^2 h$

$h=2r \implies V=\frac{\pi}{3}\cdot \frac{h^2}{4}\cdot h = \frac{\pi h^3}{12}$

$\frac{dV}{dh}=\frac{\pi h^2}{4}$

$\frac{dh}{dV}=\frac{4}{\pi h^2}$

$\frac{dh}{dt}=\frac{dh}{dV}\cdot \frac{dV}{dt}=\frac{4}{\pi h^2}\cdot 20000 = \frac{80000}{\pi h^2}$

(20000 is used as h is in cm, and 1L = 1000 cm³)
Title: Re: dcc + differential equations = atrocious
Post by: dusty_girl1144 on July 12, 2008, 09:51:52 pm: LOL at phagist_

to smart Mao :P ;)
Title: Re: dcc + differential equations = atrocious
Post by: Mao on July 12, 2008, 11:27:08 pm: Quote from: phagist_ on July 12, 2008, 09:32:48 pm
(sorry if there are mistakes am hungover and tired)

hehe :)
Title: Re: dcc + differential equations = atrocious
Post by: dcc on July 14, 2008, 09:51:56 pm: A way of solving the OP:

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

$Q(0) = 30\; g$ .

$Q_{in} = 9\; L\; \mbox{min}^{-1}$

$V(t) = 250 - 4t$

$Q_{out} = \dfrac{13Q}{250 - 4t}$

Firstly: $\dfrac{dQ}{dt} = 9 - \dfrac{13Q}{250 - 4t}$

$\implies Q' + \dfrac{13}{250 - 4t}Q = 9$

Multiply through by:

$\left(250 - 4t\right)^{-\frac{13}{4}}$

and we get:

$\implies \left(250 - 4t\right)^{-\frac{13}{4}} Q' + 13\left(250 - 4t\right)^{-\frac{17}{4}}Q = 9 \left(250 - 4t\right)^{-\frac{13}{4}}$

$\implies \left[\left(250 - 4t\right)^{-\frac{13}{4}}Q \right]' = 9 \left(250 - 4t\right)^{-\frac{13}{4}}$

Therefore by the fundamental theorem of calculus:

$(250 - 4t)^{-\frac{13}{4}} \cdot Q = \int \dfrac{9}{\left(250 - 4t\right)^\frac{13}{4}}\; dt = \left(250 - 4t\right)^{-\frac{9}{4}} + C$

$Q = \left(250 - 4t\right)^{\frac{13}{4}} \left( \dfrac{1}{\left(250 - 4t\right)^{\frac{9}{4}}} + C \right)$

$Q = \left(250 - 4t\right) + C\left(250 - 4t\right)^\frac{13}{4}$

Using the initial condition Q(0) = 30g

$30 = 250 + C \cdot 250^\frac{13}{4} \implies C = \dfrac{30 - 250}{250^\frac{13}{4}}$

$\therefore \; Q = (250 - 4t) - \dfrac{220}{250^\frac{13}{4}} \cdot (250 - 4t)^\frac{13}{4}$