Author Topic: Differential Equations (Read 10130 times) Tweet Share

pHysiX · « **Reply #30 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 >.>

Damo17 · « **Reply #31 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 $t=0$ , $N=5,000,000$

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

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

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

$= 50000(99e^{ \frac{t}{10}} +1) , t \ge 0$

d0minicz · « **Reply #32 on:** June 30, 2009, 01:40:03 pm »

can you please show me the working for b)
thanks alot

Damo17 · « **Reply #33 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 $2000$ .

d0minicz · « **Reply #34 on:** June 30, 2009, 02:00:33 pm »

no wonder haha
thanks mate

d0minicz · « **Reply #35 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 =]

Damo17 · « **Reply #36 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 $-0.5L/min = -500cm^3/min$

d0minicz · « **Reply #37 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.

Damo17 · « **Reply #38 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:

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

$V= Ah$

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

$\frac{dh}{dt}= \frac{Q-c\sqrt{h}}{A} , h>0$

d0minicz · « **Reply #39 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.

=]

Damo17 · « **Reply #40 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.

d0minicz · « **Reply #41 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

kamil9876 · « **Reply #42 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 $t=\frac{A}{k}ln(2)$

d0minicz · « **Reply #43 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...

TrueTears · « **Reply #44 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.

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Author Topic: Differential Equations (Read 10130 times) Tweet Share

pHysiX

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