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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: d0minicz on May 15, 2009, 09:01:35 pm

Title: Differential Equations
Post by: d0minicz on May 15, 2009, 09:01:35 pm: A city with the population P, at time t years after a certain date, has a population which increases at a rate proportional to the population at that time.
a)i) ~~Set up a differential equation to describe the situation.~~ $\frac {dP}{dt} = kp$
ii) ~~Solve to obtain a general solution.~~ $t= \frac {1}{k} log_e (P) + C , P>0$
b) If the initial population was 1000 and after two years the population had risen to 1100:
i) find the population after five years
ii) sketch a graph of P against t

need workings for part b)
thanks =]
Title: Re: Differential Equations
Post by: /0 on May 15, 2009, 09:43:56 pm: $t = \frac{1}{k}\log_e(P)+C$

$\Rightarrow P= e^{k(t-C)}$ . Let this be $f(t)$

$f(0) = 1000 \Rightarrow 1000 = e^{k(-C)} \Rightarrow 1000 = e^{-Ck}$ --------------------(1)

$f(2) = 1100 \Rightarrow 1100 = e^{k(2-C)}$ -----------------------(2)

Dividing equation 2 by equation 1,

$1.1 = e^{k(2-C) - (-kC)} = e^{-Ck+Ck+2k} = e^{2k}$

$\therefore k = \frac{1}{2}\log_e {1.1}$

$\Rightarrow 1000 = e^{-C\left(\frac{1}{2}\log_e(1.1)\right)}$

$1000 = (e^{\log_e(1.1)})^{\frac{-C}{2}}$

$1000 = (1.1)^{\frac{-C}{2}}$

$1000000=(1.1)^{-C}$

$\therefore C = -\log_{1.1}1000000=-\log_{1.1} 10^6=-6\log_{1.1} 10$

$\therefore P = e^{(\frac{1}{2}\log_e1.1)(t-(-6\log_{1.1}10))}=(e^{\log_e1.1})^{\frac{1}{2}(t+6\log_{1.1}10)}=(1.1)^{\frac{1}{2}(t+6\log_{1.1}10)}$

So b) i): After five years, $t = 5$ , and $P =(1.1)^{\frac{1}{2}(149.95)} = 1269$ , when rounded down.

For b) ii)

The graph is:

$P = (1.1)^{\frac{1}{2}(t+6\log_{1.1}10)}$

It has asymptote at y = 0.
You can solve to find intercepts. It is the same basic shape as $(1.1)^t$ just dilated by factor 2 from y-axis then translated $6\log_{1.1}10$ in the negative x-direction.
Title: Re: Differential Equations
Post by: d0minicz on May 15, 2009, 09:53:18 pm: yo, for b)i) answr says 1269
Title: Re: Differential Equations
Post by: /0 on May 15, 2009, 10:05:25 pm: oops in your initial question i read that the pop after two years would be 10000 when it's actually 1000

but then how does the population increase?
Title: Re: Differential Equations
Post by: d0minicz on May 15, 2009, 10:06:40 pm: fucken hell sincere apologies =_______________="
edited
you dont have to answer it if u dont want lol alrdy wasted your time =]
Title: Re: Differential Equations
Post by: /0 on May 15, 2009, 10:24:52 pm: nah man, I need to read the quetsion more carefully next time, anyway i edited my post :P
Title: Re: Differential Equations
Post by: d0minicz on May 21, 2009, 05:53:18 pm: An island has a population of rabbits of size P, t years after 1 Jan 2000. Due to a virus the population is decreasing at a rate proportional to the square root of the population at that time.
a)i) Set up a differential equation to describe this situation
ii) solve to obtain a general solution
b) If the initial population was 15000 and the population decreased to 13 500 after five years:
i) find the population after 10 years
ii) sketch the graph of P against t

need to see workkings for part b)
thanks =]
Title: Re: Differential Equations
Post by: kamil9876 on May 21, 2009, 06:38:55 pm: $\frac{dP}{dt}=kP^{0.5}$ k is negative because the population is decreasing.
$\frac{dt}{dP}=K P^{-0.5}$ I made up a new constant for conveneince, K, the recirpical of k.

$t=2K P^{0.5} + c$ (*)

$P=(\frac{t-c}{2K})^2$

b.)

sub in t=0, P=15000:

$15000=\frac{c^2}{4K^2}$ (1)

Now the other time co-ordinate, except this time to * because it's more convenient:
$ 5=2K\sqrt{13500}+c$
$c=5-2K\sqrt{13500}$

Now sub into (1):

$15000=\frac{(5-2K\sqrt{13500})^2}{4K^2}$
$60000K^2=25-20K\sqrt{13500}-13500K^2$

Let's hope I havn't made a mistake so far... :P
When you solve that quadratic equation, remember to take the negative solution since K is negative as stated in first line of post.
Title: Re: Differential Equations
Post by: d0minicz on May 23, 2009, 12:38:21 pm: hey i need help with solving this differential equation
$\frac {dm}{dt} = 0.25 - \frac {m}{100} kg/min$

Information: A tank holds 100L of pure water. A sugar solution containing 0.25kg per litre is being run into the tank at the rate of one litre/minute. The liquid in the tank is continuously stirred, and at the same time, liquid from the tank is being pumped out at the rate of one litre per minute. After t minutes, there are m kg of sugar dissolved in the solution.

thanks
Title: Re: Differential Equations
Post by: kamil9876 on May 23, 2009, 01:04:39 pm: $\frac{dt}{dm}=\frac{1}{0.25-0.01m}$

$\frac{dt}{dm}=\frac{100}{25-m}$

$\frac{dt}{dm}=-100\frac{1}{m-25}$

Which is a simple logarithm in the end. (exponential when expressing m in terms of t)
Title: Re: Differential Equations
Post by: TrueTears on May 23, 2009, 01:39:15 pm: As requested:

16. a) Rate in : 0.25 kg/min

b) Rate out $\frac{m}{100}$ kg/min

c) $\frac{dm}{dt} = 0.25 - \frac{m}{100} = \frac{25-m}{100}$

$\frac{dt}{dm} = \frac{100}{25-m}$

d) $t = 100 \int (25 - m)^{-1} dm = -100log_e|25-m| + c$

when $t = 0 m = 0$

$c = 100log_e(25)$

$t = 100log_e\frac{25}{25-m}$

$m = 25-25e^{\frac{-t}{100}}$
Title: Re: Differential Equations
Post by: d0minicz on June 28, 2009, 12:35:28 pm: Construct but do not solve a differential equation for:
a) An inverted cone with depth 50cm and radius 25cm is initially full. Water drains out at 0.5 litres per minute. The depth of water in the cone is h cm at t minutes. (Find $\frac {dh}{dt}$ )

b) A cylindrical tank 4m high with base radius 1.5m is initially full of water. Water starts flowing out through a hole at the bottom of the tank at the rate of $\sqrt {h} m^3/hour$ , where h m is the depth of water remaining in the tank after t hours. (Find $\frac {dh}{dt}$ ).

thanks =]
Title: Re: Differential Equations
Post by: Damo17 on June 28, 2009, 01:09:57 pm: Quote from: d0minicz on June 28, 2009, 12:35:28 pm
Construct but do not solve a differential equation for:
a) An inverted cone with depth 50cm and radius 25cm is initially full. Water drains out at 0.5 litres per minute. The depth of water in the cone is h cm at t minutes. (Find $\frac {dh}{dt}$ )

thanks =]

a) $\frac{dV}{dt}=-0.5 L/min = -500cm/min$ -----EDIT: $\frac{dV}{dt}= -500cm^3/min$

$V= \frac{\pi r^2 h}{3}$

as $r= \frac{h}{2}$
$ V=\frac{1}{3} \pi \frac{h^3}{4} = \frac{\pi h^3}{12}$

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

$ \frac{dh}{dt} = -500 (\frac{4}{\pi h^2}) = \frac{-2000}{\pi h^2}$
Title: Re: Differential Equations
Post by: Damo17 on June 28, 2009, 01:23:14 pm: Quote from: d0minicz on June 28, 2009, 12:35:28 pm
Construct but do not solve a differential equation for:
b) A cylindrical tank 4m high with base radius 1.5m is initially full of water. Water starts flowing out through a hole at the bottom of the tank at the rate of $\sqrt {h} m^3/hour$ , where h m is the depth of water remaining in the tank after t hours. (Find $\frac {dh}{dt}$ ).

thanks =]

b) $\frac{dV}{dt} = -\sqrt{h} m^3/hour$

$V= \pi r^2 h$

as $r=1.5$

$V= \frac{9 \pi h}{4}$

$ \frac{dV}{dh}= \frac{9\pi}{4}$ $\therefore$ $\frac{dh}{dV}= \frac{4}{9\pi}$

$ \frac{dh}{dt}= -\sqrt{h} \cdot \frac{4}{9\pi} = \frac{-4\sqrt{h}}{9\pi}$
Title: Re: Differential Equations
Post by: d0minicz on June 29, 2009, 12:07:01 pm: Find the general solution for:
$\frac {dy}{dx} = cos^2 y$
ty
Title: Re: Differential Equations
Post by: Damo17 on June 29, 2009, 12:17:56 pm: Quote from: d0minicz on June 29, 2009, 12:07:01 pm
Find the general solution for:
$\frac {dy}{dx} = cos^2 y$
ty

$\frac{dy}{dx}= cos^2 y$

$\frac{dx}{dy}=\frac{1}{cos^2 y} = sec^2 y$

$x= \int sec^2 y \ dy = tan y +c$

$\therefore$ $y= tan^{-1} (x-c)$
Title: Re: Differential Equations
Post by: dcc on June 29, 2009, 01:12:47 pm: Quote from: d0minicz on June 29, 2009, 12:07:01 pm
Find the general solution for:
$\frac {dy}{dx} = cos^2 y$

Or one could use seperation of variables to tackle this ( $\int \dfrac{dy}{\cos^2 y} = \int dx$ noting that we require $\cos y \neq 0$ ).
Title: Re: Differential Equations
Post by: d0minicz on June 29, 2009, 04:58:23 pm: The rate of decay of a radioactive substance is proportional to the amoutn of Q of matter present at any time, t. The differntial equation for this situation is $\frac {dQ}{dt} = -kQ$ where k is a constant. If Q=50 when t=0 and Q=25 when t=10 , find the time taken for Q to reach 10.
thanks :)
Title: Re: Differential Equations
Post by: Damo17 on June 29, 2009, 05:18:52 pm: Quote from: d0minicz on June 29, 2009, 04:58:23 pm
The rate of decay of a radioactive substance is proportional to the amoutn of Q of matter present at any time, t. The differntial equation for this situation is $\frac {dQ}{dt} = -kQ$ where k is a constant. If Q=50 when t=0 and Q=25 when t=10 , find the time taken for Q to reach 10.
thanks :)

$\frac{dt}{dQ}= -\frac{1}{kQ}$ $\therefore$ $t= -\frac{1}{k} log_e Q +c$

when $t=0$ , $Q=50$ $\therefore$ $c= \frac{1}{k} log_e 50$

$t= \frac{1}{k} log_e \frac{50}{Q}$

When $t=10$ , $Q=25$ $\therefore$ $10=\frac{1}{k} log_e 2$ $\implies$ $k= \frac{log_e 2}{10}$

$t= \frac{10log_e \frac{50}{Q}}{log_e 2}$

when $Q=10$

$t= \frac{10log_e 5}{log_e 2}=23.22s$
Title: Re: Differential Equations
Post by: d0minicz on June 29, 2009, 05:30:04 pm: cheers damo
The number, n, of bacteria in a colony grows according to the law $\frac {dn}{dt} = kn$ , where k is a positive constant. If the number increases from 4000 to 8000 in four days, find, to the nearest hundred, the number of bacteria after three days more.
i need help interpreting this question.
thanks
Title: Re: Differential Equations
Post by: Damo17 on June 29, 2009, 05:43:54 pm: Quote from: d0minicz on June 29, 2009, 05:30:04 pm
cheers damo
The number, n, of bacteria in a colony grows according to the law $\frac {dn}{dt} = kn$ , where k is a positive constant. If the number increases from 4000 to 8000 in four days, find, to the nearest hundred, the number of bacteria after three days more.
i need help interpreting this question.
thanks

np :)

$\frac{dt}{dn}= \frac{1}{kn}$ $\implies$ $t= \frac{1}{k}log_e n +c$

When $t=0$ , $n=4000$

put these into t and solve for c to get: $c= -\frac{1}{k} log_e 4000$

$t= \frac{1}{k} log_e \frac{n}{4000}$

When $t=4$ , $n=8000$

sub these into t and solve for k: $k= \frac{log_e 2}{4}$

$t= \frac{4}{log_e 2} log_e \frac{n}{4000}$

when $t= 7$ , find n: sub $t=7$ and solve for n:

$n= 4000e^\frac{7log_e 2}{4} = 13,500$
Title: Re: Differential Equations
Post by: d0minicz on June 29, 2009, 06:01:22 pm: thanks again. sorry more q's, im just really bad at differentials.
A town had a population of 10 000 in 1990 and 12 000 in 2000. If the population is N at a time t years after 1990, find the predicted population in the year 2010 assuming;
a) $\frac {dN}{dt} = \frac {k}{N}$

thanks in advance !!!

edit: solved b)
Title: Re: Differential Equations
Post by: d0minicz on June 29, 2009, 06:23:19 pm: wait do you nmean 13 711 ? thanks for the workings =]

A tank intially contains 200L of pure water. A salt solution containing 5kg of salt per litre is added at the rate of 10 L/min, and the mixed solution is drained simultaenously at the rate of 12 L/min. There is m kg of salt in the tank after t mins. Find $\frac {dm}{dt}$
thanks again
Title: Re: Differential Equations
Post by: Damo17 on June 29, 2009, 06:28:27 pm: Quote from: d0minicz on June 29, 2009, 06:01:22 pm
thanks again. sorry more q's, im just really bad at differentials.
A town had a population of 10 000 in 1990 and 12 000 in 2000. If the population is N at a time t years after 1990, find the predicted population in the year 2010 assuming;
a) $\frac {dN}{dt} = \frac {k}{N}$

thanks in advance !!!

edit: solved b)

$ t= \frac{N^2}{2k} +c$

t=0, N=10000 : solve for c

$t= \frac{N^2}{2k} - \frac{(10000)^2}{2k}$

t=10, N=12000 : solve for k
$ t= \frac{N^2 - 10^8}{4.4 \times 10^6}$

sub in t=20 and you get N= 13711
Title: Re: Differential Equations
Post by: pHysiX on June 29, 2009, 06:29:35 pm: oops made a silly mistake sorry
Title: Re: Differential Equations
Post by: Damo17 on June 29, 2009, 06:37:59 pm: Quote from: d0minicz on June 29, 2009, 06:23:19 pm
wait do you nmean 13 711 ? thanks for the workings =]

A tank intially contains 200L of pure water. A salt solution containing 5kg of salt per litre is added at the rate of 10 L/min, and the mixed solution is drained simultaenously at the rate of 12 L/min. There is m kg of salt in the tank after t mins. Find $\frac {dm}{dt}$
thanks again

$V(t)= 200+10t-12t= 200-2t$

$IR= 10L/min \times 5kg/L= 50kg/min$

$OR= \frac{m}{V(t)} \times 12 = \frac{6m}{100-t}$

$\frac{dm}{dt} = IR-OR = 50 - \frac{6m}{100-t}$

EDIT: Restriction on t: $0 \le t < 100$
Title: Re: Differential Equations
Post by: d0minicz on June 30, 2009, 11:11:31 am: A partially filled tank contains 200L of water in which 1500g of salt have been dissolved. Water is poured into the tank at a rate of 6L/min. The mixture, which is kept uniform by stirring, leaves the tank through a hole at a rate of 5L/min. There are x grams of salt in the tank after t mins. Find $\frac {dx}{dt}$
thank you

edit: solved ; but thanmks for any attemps :) :)
Title: Re: Differential Equations
Post by: d0minicz on June 30, 2009, 12:48:13 pm: A country's population N at time t years after 1 Jan 2000 changes according to the differential equation $\frac {dN}{dt} = 0.1N - 5000$ .

(Five thousand people leave the country every year and there is a 10% growth rate).

a) Given that the population was 5 000 000 at the start of 2000, find N in terms of t.

b)In which year will the country have a population of 10 million?

Basically need help with a) mostly. Need the equation before i can do b).

thanks alot
Title: Re: Differential Equations
Post by: kamil9876 on June 30, 2009, 12:53:38 pm: Use the same process as here:

http://vcenotes.com/forum/index.php/topic,13737.msg151711.html#msg151711

or even better, TT's post just below it.
Title: Re: Differential Equations
Post by: d0minicz on June 30, 2009, 12:55:19 pm: thanks man but can soemone please show the working :(
Title: Re: Differential Equations
Post by: pHysiX on June 30, 2009, 01:10:27 pm: ok here's another attempt...god i hope i don't fail =]

a)

$\frac{dt}{dN} = \frac{10}{N-50000}$

implying that t=10 ln(N-50000) + c: c is an element of R

implying
N = $Ae^{\frac{t}{10}}$ + 50000 : A is an element of R

At t=0, N = 5 x $10^6$
A = 4950000

Therefore, N = 4950000 $e^{\frac{t}{10}}$ + 50000

edit: silly mistake again >.>
Title: Re: Differential Equations
Post by: Damo17 on June 30, 2009, 01:14:47 pm: Quote from: d0minicz on June 30, 2009, 12:48:13 pm
A country's population N at time t years after 1 Jan 2000 changes according to the differential equation $\frac {dN}{dt} = 0.1N - 5000$ .

(Five thousand people leave the country every year and there is a 10% growth rate).

a) Given that the population was 5 000 000 at the start of 2000, find N in terms of t.

Basically need help with a) mostly. Need the equation before i can do b).

thanks alot

a) $\frac{dt}{dN}= \frac{1}{\frac{N}{10} - 5000} = \frac{10}{N-50000}$

$t= 10log_e (N-50000) + c$

When $t=0$ , $N=5,000,000$

$\therefore$ $c= -10log_e (4.95 \times 10^6)$

$t= 10log_e \frac{N-50000}{4.95 \times 10^6}$

$N= (4.95 \times 10^6)e^ {\frac{t}{10}} + 50000$

$= 50000(99e^{ \frac{t}{10}} +1) , t \ge 0$
Title: Re: Differential Equations
Post by: d0minicz on June 30, 2009, 01:40:03 pm: can you please show me the working for b)
thanks alot :)
Title: Re: Differential Equations
Post by: Damo17 on June 30, 2009, 01:43:51 pm: Quote from: d0minicz on June 30, 2009, 01:40:03 pm
can you please show me the working for b)
thanks alot :)

$10,000,000= 50,000(99e^{ \frac{t}{10}}+1)$

$\frac{199}{99} = e^ {\frac{t}{10}}$

$t= 7 \ years$

so answer is $2007$ .

Note: the book answer is $1996$ but is obviously wrong as $t$ is years after $2000$ .
Title: Re: Differential Equations
Post by: d0minicz on June 30, 2009, 02:00:33 pm: no wonder haha
thanks mate
Title: Re: Differential Equations
Post by: d0minicz on July 03, 2009, 02:00:51 pm: Quote from: Damo17 on June 28, 2009, 01:09:57 pm
Quote from: d0minicz on June 28, 2009, 12:35:28 pm
Construct but do not solve a differential equation for:
a) An inverted cone with depth 50cm and radius 25cm is initially full. Water drains out at 0.5 litres per minute. The depth of water in the cone is h cm at t minutes. (Find $\frac {dh}{dt}$ )

thanks =]

a) $\frac{dV}{dt}=-0.5 L/min = -500cm/min$

$V= \frac{\pi r^2 h}{3}$

as $r= \frac{h}{2}$
$ V=\frac{1}{3} \pi \frac{h^3}{4} = \frac{\pi h^3}{12}$

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

$ \frac{dh}{dt} = -500 (\frac{4}{\pi h^2}) = \frac{-2000}{\pi h^2}$

Hey where did the = -500cm /min part come from? im confused :( haha
thanks =]
Title: Re: Differential Equations
Post by: Damo17 on July 03, 2009, 02:11:28 pm: Quote from: d0minicz on July 03, 2009, 02:00:51 pm
Hey where did the = -500cm /min part come from? im confused :( haha
thanks =]

Sorry it should be $cm^3$ .
$1L=1000cm^3$
$1L/min = 1000cm^3/min$ so $-0.5L/min = -500cm^3/min$
Title: Re: Differential Equations
Post by: d0minicz on July 03, 2009, 02:43:36 pm: oh okay thanks
why would it be necessary to convert it ?
also need help on this: sorry not too good at these.

A tank with a flat bottom and vertical sides has a constant horizontal cross-section of A square metres. The tank has a tap in the bottom through which water is leaving at a rate of $c\sqrt {h}$ cubic metres per minute, where h metres is the height of the water in the tank, and c is a constant. Water is being poured in at a rate of Q cubic metres per minute. Find an expression for $\frac {dh}{dt}$
thanks alot.
Title: Re: Differential Equations
Post by: Damo17 on July 03, 2009, 02:53:13 pm: Quote from: d0minicz on July 03, 2009, 02:43:36 pm
oh okay thanks
why would it be necessary to convert it ?
also need help on this: sorry not too good at these.

A tank with a flat bottom and vertical sides has a constant horizontal cross-section of A square metres. The tank has a tap in the bottom through which water is leaving at a rate of $c\sqrt {h}$ cubic metres per minute, where h metres is the height of the water in the tank, and c is a constant. Water is being poured in at a rate of Q cubic metres per minute. Find an expression for $\frac {dh}{dt}$
thanks alot.

you convert it because the other values are in cm.

for your question:

$\frac{dV}{dt}= (Q-c\sqrt{h})m^3/min$

$V= Ah$

$\frac{dh}{dV}= \frac{1}{A}$

$\frac{dh}{dt}= \frac{Q-c\sqrt{h}}{A} , h>0$
Title: Re: Differential Equations
Post by: d0minicz on July 03, 2009, 03:40:20 pm: cheers damo once again
Hey for the Essentials book exercise 9E question 2; i need help getting started with it.
thanks heaps

The question is:
A conical tank has a radius length at the top equal to its height. Water, initially with a depth of 25cm, leaks out through a hole in the bottom of the tank at the rate of $5\sqrt {h} cm^3/min$ where the depth is h cm at time t minutes.
a) Construct a differential equation expressing $\frac {dh}{dt}$ as a function of h, and solve it.
b) Hence find how long it will take for the tank to empty.

=]
Title: Re: Differential Equations
Post by: Damo17 on July 03, 2009, 03:53:53 pm: Quote from: d0minicz on July 03, 2009, 03:40:20 pm
cheers damo once again
Hey for the Essentials book exercise 9E question 2; i need help getting started with it.
thanks heaps

The question is:
A conical tank has a radius length at the top equal to its height. Water, initially with a depth of 25cm, leaks out through a hole in the bottom of the tank at the rate of $5\sqrt {h} cm^3/min$ where the depth is h cm at time t minutes.
a) Construct a differential equation expressing $\frac {dh}{dt}$ as a function of h, and solve it.
b) Hence find how long it will take for the tank to empty.

=]

My pleasure to help.

for a)

$V= \frac{1}{3} \pi r^2 h$ , sub in $r=h$ to get $V$ in terms of just $h$ and find $\frac{dV}{dh}$

$\frac{dV}{dt}= -5\sqrt{h}$

$\frac{dh}{dt} = \frac{dh}{dV} \cdot \frac{dV}{dt}$

then flip this to get $\frac{dt}{dh}$ and integrate to get $t=??? +c$

then when $t=0$ , $h=25$ so find $c$ .

for b)

let $h=0$ and solve for $t$ , convert value to hours and minutes.
Title: Re: Differential Equations
Post by: d0minicz on July 04, 2009, 01:29:26 pm: A water tank of uniform cross-sectional area $A cm^2$ is being filled by a pipe which supplies Q litres of water every minute. The tank has a small hole in its base through which water leaks at a rate of kh litres every minute where h cm is the depth of water in the tank at time t minutes. Initially the depth of the water is $h_0$ .
a) Construct the differential equation expressing $\frac {dh}{dt}$ as a function of h.
b) Solve the differential equation if $Q > kh_0$
c) Find the time taken for the depth to reach $\frac {Q+kh_0}{2k}$

thank you
Title: Re: Differential Equations
Post by: kamil9876 on July 04, 2009, 02:54:31 pm: $V=Ah$
$ \frac{dV}{dh}=A $
$\frac{dV}{dt}=Q-kh$

$\implies \frac{dh}{dt}=\frac{Q-kh}{A}$

b.)
$\frac{dt}{dh}=\frac{-A}{k}\frac{1}{h-\frac{Q}{k}}$

$t=\frac{-A}{k} ln|h-\frac{Q}{k}| + c$

$However, Q>kh_0 \implies 0>h_0-\frac{Q}{k}$ and because h_0 is a value of h and t is continous(because differentiable) then the thing inside the modulus is negative.

$\therefore t=\frac{-A}{k} ln(\frac{Q}{k}-h) + c$
$0=\frac{-A}{k}ln(\frac{Q}{k}-h_o) + c$
$c=\frac{A}{k}ln(\frac{Q}{k}-h_o)$

$t=\frac{A}{k}ln(\frac{\frac{Q}{k}-h_o}{\frac{Q}{k}-h})$
$t=\frac{A}{k}ln(\frac{Q-kh_0}{Q-kh})$ **

You should rearange ** to have it as a function of h.

c.)

You shoud sub in that value into **:

The argument of the ln is:

$\frac{Q-kh_o}{Q-k(\frac{Q-kh_o}{2k})}$
$=\frac{(Q-kh_o)}{Q-0.5Q-0.5kh_o}$
$=\frac{(Q-kh_o)}{0.5Q-0.5kh_o}$
$=\frac{Q-kh_o}{0.5(Q-kh_o)}$
$=\frac{1}{0.5}$
$=2$

and so $t=\frac{A}{k}ln(2)$
Title: Re: Differential Equations
Post by: d0minicz on July 13, 2009, 05:17:25 pm: Use Euler's method with steps of 0.01 to find an approximate value of y at x=0.5 if $\frac {dy}{dx} = cos^{-1}x$ and y= 0 when x = 0.

thanks...
Title: Re: Differential Equations
Post by: TrueTears on July 13, 2009, 05:20:17 pm: Use the program Mao put up. Pointless to just redo the thing 50 times over.

http://vcenotes.com/forum/index.php/topic,3629.msg42179.html#msg42179

Get it on your calc.
Title: Re: Differential Equations
Post by: d0minicz on July 13, 2009, 05:30:37 pm: lols i have TI-84
Title: Re: Differential Equations
Post by: d0minicz on July 13, 2009, 07:15:20 pm: How would you do it by hand?
Title: Re: Differential Equations
Post by: TrueTears on July 13, 2009, 07:17:11 pm: You're gonna have to do linear approximation 50 times.

GL mate.
Title: Re: Differential Equations
Post by: d0minicz on July 13, 2009, 07:22:52 pm: wow this is gonna be fun
Title: Re: Differential Equations
Post by: /0 on July 13, 2009, 08:18:22 pm: You can input:

$y_n = y_0 + \sum_{k=0}^{n-1} hy(x_0+kh)$

Where $n = \frac{\Delta x}{h}$

(i think)
Title: Re: Differential Equations
Post by: dcc on July 13, 2009, 08:41:24 pm: Quote
Use Euler's method with $h= 0.01$ to find the approximate value of $y$ at $x = \frac{1}{2}$ if $\frac{dy}{dx} = \arccos(x)$ given that $y_0 = 0$ .

Euler's method:
$\begin{align*} y_{n+1} &= y_n + h \cdot \arccos\left(hn\right)\\ \implies \left(y_{n+1} - y_n\right) &= h \cdot \arccos\left(hn\right)\\ \implies \sum_{n=0}^{49} \left(y_{n+1} - y_{n}\right) &= h \sum_{n=0}^{49} \arccos\left(hn\right)\\ \left(y_1 - y_0\right) + \left(y_2 - y_1\right) + \cdots + \left(y_{50} - y_{49}\right) &= h \sum_{n=0}^{49} \arccos\left(hn\right)\\ \implies y_{50} = y_0 + h \sum_{n=0}^{49} \arccos\left(hn\right)\\ \end{align*} $

In this case, that means $y_{50} = 0.01 \sum_{n=0}^{49} \arccos\left(0.01n\right)\\$ , which is the answer you SEEK.

note: Perhaps I have used a poor choice for the index of summation, but the final sum is correct, if you ignore the multiple uses of n throughout.
Title: Re: Differential Equations
Post by: QuantumJG on July 17, 2009, 02:34:15 pm: Quote from: d0minicz on May 15, 2009, 09:01:35 pm
A city with the population P, at time t years after a certain date, has a population which increases at a rate proportional to the population at that time.
a)i) ~~Set up a differential equation to describe the situation.~~ $\frac {dP}{dt} = kp$
ii) ~~Solve to obtain a general solution.~~ $t= \frac {1}{k} log_e (P) + C , P>0$
b) If the initial population was 1000 and after two years the population had risen to 1100:
i) find the population after five years
ii) sketch a graph of P against t

need workings for part b)
thanks =]

a) part ii) is actually P = Ae^kt, A is an constant

b) if P(0) = 1000 and P(2) = 1100

.: A = 1000

1100 = 1000e^2k

k = 0.5ln(1.1)

i) P(5) = 1000e^2.5ln(1.1)

= 1000*1.269

= 1269

ii) this is pretty much self done! You have an exponential graph going through (0,1000), (2,1100) and (5,1269)
Title: Re: Differential Equations
Post by: kamil9876 on July 17, 2009, 05:21:43 pm: Quote from: QuantumJG on July 17, 2009, 02:34:15 pm
Quote from: d0minicz on May 15, 2009, 09:01:35 pm
A city with the population P, at time t years after a certain date, has a population which increases at a rate proportional to the population at that time.
a)i) ~~Set up a differential equation to describe the situation.~~ $\frac {dP}{dt} = kp$
ii) ~~Solve to obtain a general solution.~~ $t= \frac {1}{k} log_e (P) + C , P>0$
b) If the initial population was 1000 and after two years the population had risen to 1100:
i) find the population after five years
ii) sketch a graph of P against t

need workings for part b)
thanks =]

a) part ii) is actually P = Ae^kt, A is an constant

b) if P(0) = 1000 and P(2) = 1100

.: A = 1000

1100 = 1000e^2k

k = 0.5ln(1.1)

i) P(5) = 1000e^2.5ln(1.1)

= 1000*1.269

= 1269

ii) this is pretty much self done! You have an exponential graph going through (0,1000), (2,1100) and (5,1269)

If you havn't already noticed, latest posts are found at the bottom of the last page, not the top of the first. Furthermore, notice that this topic hasn't had a reply in 4 days and ussually all hw problems are solved at most 30min after the question is being posted.
Title: Re: Differential Equations
Post by: TrueTears on July 17, 2009, 05:23:09 pm: kamil let him have his fun, he think he's a pr0
Title: Re: Differential Equations
Post by: TonyHem on July 17, 2009, 05:42:43 pm: He did only join 3 days ago. Probably just didn't realise that they were old Q's.
Title: Re: Differential Equations
Post by: zzdfa on July 17, 2009, 05:52:27 pm: Quote from: TrueTears on July 17, 2009, 05:23:09 pm
kamil let him have his fun, he think he's a pr0
he's just new&enthusiastic :)