Author Topic: Bubble's specialist questions (Read 6615 times) Tweet Share

Mao · « **Reply #15 on:** August 22, 2008, 11:09:55 pm »

for definite integrals, it is often easier to use substitution (and then find the definite integral for the u value, rather than putting it back in to the original integral)

$\int_{t=2}^{t=3} \frac{t}{1+t^2}\; dt$

$u=1+t^2,\; \implies \frac{du}{dt}=2t,\; \implies \frac{dt}=\frac{du}{2t}$

$u(2)=5,\; u(3)=10$

$\implies \int_{u=5}^{u=10}\frac{t}{u}\cdot \frac{du}{2t}$

$\implies \frac{1}{2} \int_{u=5}^{u=10}\frac{1}{u}\; du$

$\implies \frac{1}{2} \cdot \left[ \mbox{ln}|u| \right]_5^{10}=\frac{1}{2}\mbox{ln}(2)$

as opposed to $\frac{1}{2}\cdot \left[ \mbox{ln}|1+t^2| \right]_2^3$

bubbles · « **Reply #16 on:** August 28, 2008, 08:34:26 pm »

This is related to a question I've previously posted:
Round to four decimal places
Important equations:
(1) $N={N}_{0}e^{kt}$
equations 1 and 2 are for growth of bacteria
(2) $\frac{dN}{dt}= KN ( 1 - \frac{N}{C})$

(3) $t=\frac{1}{k}log_e \left| \frac{N(N_0-C)}{N_0(N-C)}\right|$

Bacteria A:
$k\approx 3.2964$ (without optimal conditions)
$k_{A0}\approx 2k$ (optimal conditions)
N= 35000 when t=0 $\implies {N}_{0}=35000$
when t=3, $N= 6.9\times10^{8} cells$
Carrying capacity ${C}_{A}= 1.0\times10^{11}$
d) Using ${K}_{A},{C}_{A}$ , find the new equation for growth of bacteria. Assume colony starts with 35000 cells.
I get: $t=\frac{1}{3.2964}log_e \left| \frac{-10.0000\times10^{10}N}{35000(N-1.0\times10^{11})}\right|$ it doesn't look correct?
g) Exactly one day after the cells multiplied to $N=9.9\times10^{9}$ the number of cells was observed to have been reduced to $N=6.3\times10^{9}$ . Assume death occurs exponentially. Using (3) find k.

$\implies 1=\frac{1}{k}log_e \left|0.6119\right|$
$\implies k=-.04912$ ?

Bacteria B
A small colony of $6.9\times10^{7} cells$
$k\approx 2.0896$ (without optimal conditions)
$k_{B0}\approx 2.5k$ (optimal conditions)
N= 15000 when t=0 $\implies {N}_{0}=15000$
when $t\approx6.5502 days$ , $N= 1.1\times10^{9} cells$
Carrying capacity ${C}_{A}= 1.2\times10^{9}$
A colony of 15000 cells was allowed to reproduce at without optimal conditions for 3 days. The colony was then allowed to reproduce in optimal conditions until $N= 1.1\times10^{9}$ .
e) Using (3) estimate how long it would take bacteria B to reach $N= 1.1\times10^{9}$
I got: $t\approx4.14 days$ ??

3. Both strains of bacteria come into contact with each other. They continue to grow as normal. New combined carrying capacity is ${C}_{A}= 7.0\times10^{7} cells$ call this $C_{AB}$ . Once $N= 6.95\times10^{7} cells$ it takes exactly 2 days for the number to reproduce to a total of $3.0\times10^{5} cells$ , upon which times it resumes to normal growth pattern.
a) Convert your carrying capacity equation for both bacteria into N(t) form. Use these two equations to find $N_{A}+N_{B} = 6.95\times10^{7}$ , assuming both strains started with 25,000 cells (growth not optimal conditions). Hence determine % composition of each strain of bacteria when $N_{A}+N_{B} = 6.95\times10^{7}$
I AM LOST! I'm thinking you need to somehow get $t=\frac{1}{k}log_e \left| \frac{N(N_0-C)}{N_0(N-C)}\right|$ into the form of N(t) ??

Hopefully what i've wrote has made some sense to you guys. If it doesn't please ask me and i'll try to clarify.

xox.happy1.xox · « **Reply #17 on:** August 28, 2008, 09:03:06 pm »

Lol, this was part of our exact application task for specialist! I had NO CLUE what it was on about.

bubbles · « **Reply #18 on:** August 28, 2008, 09:40:38 pm »

Quote from: xox.happy1.xox on August 28, 2008, 09:03:06 pm

Lol, this was part of our exact application task for specialist! I had NO CLUE what it was on about.

LOL really? We got our marks back today. Nobody got question 3 correct and the majority of the class didn't even attempt it (it cost me 9 marks!!!). We weren't allowed to take it home (really strict teacher

) so i wrote bits and pieces down. This question is really annoying me, does anybody know how to do it?

Mao · « **Reply #19 on:** August 28, 2008, 10:08:14 pm »

on a glance, it seems largely senseless and extremely confusing...

~~I shall have a closer look maybe tomorrow =]~~

EDIT: no, i shall not explode my brain in this manner

bubbles · « **Reply #20 on:** August 29, 2008, 08:44:03 pm »

Quote from: Mao on August 28, 2008, 10:08:14 pm

it seems largely senseless and extremely confusing...

I COULDN'T AGREE WITH YOU MORE!

xox.happy1.xox · « **Reply #21 on:** August 29, 2008, 09:33:22 pm »

Maybe convert the equation into N= or something along those lines. Then, because you know that the CA and CB values are the same, let them =c,then you can see them when No=6.95*10^7, and t=2, and the different K values you found before into the different equations. So, into one equation you put KA=3.2964 and KB=...(I don't know from your working out

). But effectively what you are trying to do is get both equations to equal something, all with the same values, except for the K values. Then, let N=6.95*10^7, and find the time it takes to reach this (I think It's 2.9166... Don't trust me on that, I copies it from a girl who got an A+

). Then, percentage composition is just the equation, NA or NB divided by the total (in this case, 6.95*10^7), to see their percentage composition.

What did I just write

Oh well, I hope it still helped you, I ended up getting a bad mark (It's literally horrible, I don't even want to say it

)

Mao · « **Reply #22 on:** August 29, 2008, 09:44:45 pm »

oh god.. that last post made as much sense as the questions.... =P

jsimmo · « **Reply #23 on:** August 29, 2008, 09:47:02 pm »

Quote from: xox.happy1.xox on August 29, 2008, 09:33:22 pm

Maybe convert the equation into N= or something along those lines. Then, because you know that the CA and CB values are the same, let them =c,then you can see them when No=6.95*10^7, and t=2, and the different K values you found before into the different equations. So, into one equation you put KA=3.2964 and KB=...(I don't know from your working out ). But effectively what you are trying to do is get both equations to equal something, all with the same values, except for the K values. Then, let N=6.95*10^7, and find the time it takes to reach this (I think It's 2.9166... Don't trust me on that, I copies it from a girl who got an A+ ). Then, percentage composition is just the equation, NA or NB divided by the total (in this case, 6.95*10^7), to see their percentage composition.

What did I just write Oh well, I hope it still helped you, I ended up getting a bad mark (It's literally horrible, I don't even want to say it )

LOL! What the...!? That's really confusing. Lucky I only do further $:-\$

xox.happy1.xox · « **Reply #24 on:** August 29, 2008, 09:53:07 pm »

I only wish I had done further when I had the chance... Why oh why did I choose to do Methods and Specialist?

xox.happy1.xox · « **Reply #25 on:** August 29, 2008, 10:00:14 pm »

Oh, and it only gets worse, I think in a later question, you have to sketch a GRAPH of all of the information you had just found. Actually, I think the mark I got wasn't so bad after all, considering everything that was involved...

bubbles · « **Reply #26 on:** August 29, 2008, 10:57:11 pm »

Quote from: xox.happy1.xox on August 29, 2008, 10:00:14 pm

Oh, and it only gets worse, I think in a later question, you have to sketch a GRAPH of all of the information you had just found.

You are corret! Since I could NOT get my head around part a) there was no way in hell was i able to sketch any crappy graph -_-"

xox.happy1.xox · « **Reply #27 on:** August 30, 2008, 09:46:58 am »

Aww, hopefully you are still satisfied with what mark you eventually get...

bubbles · « **Reply #28 on:** September 02, 2008, 01:10:04 am »

Velocity-time graphs
Two tram stops, A and B, are 500 metres apart. A tram starts from A and travels with acceleration $a m/s^{2}$ to a certain point. It then decelerates at $4a m/s^{2}$ until it stops at B. The total time taken is two minutes. Sketch a velocity-time graph. Find the value of a and the maximum speed reached by the tram.
(the only thing im sure about is that x=500m)

Collin Li · « **Reply #29 on:** September 02, 2008, 07:36:53 am »

You will have two triangular components: one sloping up and one sloping down. You can get the areas in terms of $a$ and $t$ , where $t$ is the point at which the tram stops accelerating, and starts decelerating.

We need two pieces of information to solve for $a$ and $t$

1. The sum of the areas is 500.
2. The final velocity is zero. (assuming the tram starts from stop A at rest)

The first equation is just the sum of the areas of the triangles:

$500 = \frac{1}{2}t\cdot(at) + \frac{1}{2}(120-t)\times 4a(120-t)$ --- (1)

The second one is:

$v = at - 4a(120-t) = 0$

$\implies a(t - 480 + 4t) = 0$

$\implies a(5t - 480) = 0$

$\therefore\; t = 96\mbox{ s}$ (since $a>0$ otherwise it wouldn't move)

Substitute this back into (1) to get:

$500 = \frac{1}{2}\times96\times(96a) + \frac{1}{2}\times24\times 4a(24)$

$1000 = 9216a + 2304a$

$\therefore\; a = \frac{1000}{11520} = \frac{25}{288} \approx 0.0868\mbox{ ms}^{-2}$

Once you've found $a$ and $t$ : $v = at = \frac{25}{3} \approx 8.333\mbox{ ms}^{-1}$ should be your maximum speed.

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