Author Topic: Differential Equations (Read 11489 times) Tweet Share

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 =]

/0 · « **Reply #1 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.

d0minicz · « **Reply #2 on:** May 15, 2009, 09:53:18 pm »

yo, for b)i) answr says 1269

/0 · « **Reply #3 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?

d0minicz · « **Reply #4 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 =]

/0 · « **Reply #5 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

d0minicz · « **Reply #6 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 =]

kamil9876 · « **Reply #7 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...

When you solve that quadratic equation, remember to take the negative solution since K is negative as stated in first line of post.

d0minicz · « **Reply #8 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

kamil9876 · « **Reply #9 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)

TrueTears · « **Reply #10 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 $t = 0 m = 0$

$c = 100log_e(25)$

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

$m = 25-25e^{\frac{-t}{100}}$

d0minicz · « **Reply #11 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 =]

Damo17 · « **Reply #12 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}$

Damo17 · « **Reply #13 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}$

d0minicz · « **Reply #14 on:** June 29, 2009, 12:07:01 pm »

Find the general solution for:
$\frac {dy}{dx} = cos^2 y$
ty

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