Author Topic: dcc + differential equations = atrocious (Read 10791 times) Tweet Share

Mao · « **Reply #30 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

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

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

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

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

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

dusty_girl1144 · « **Reply #31 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

NOTE: ummmm am i meant to get 4750 as the answer once i sub t=5 in the last line?

? coz i just did

Mao · « **Reply #32 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

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

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

dusty_girl1144 · « **Reply #33 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 $S= 4\pi r^2$

ell · « **Reply #34 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*} $

dcc · « **Reply #35 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)

/0 · « **Reply #36 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

dcc · « **Reply #37 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

another way of integrating that quadratic was making the substitution $t = u + \dfrac{1}{2}$ and arctanning, but I like what you did

/0 · « **Reply #38 on:** July 12, 2008, 12:12:09 pm »

lol thanks dcc, I guess I just did a roundabout way of arctanning

Mao · « **Reply #39 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

dusty_girl1144 · « **Reply #40 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.

phagist_ · « **Reply #41 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)

dusty_girl1144 · « **Reply #42 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

Mao · « **Reply #43 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)

dusty_girl1144 · « **Reply #44 on:** July 12, 2008, 09:51:52 pm »

LOL at phagist_

to smart Mao

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