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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: bubbles on August 16, 2008, 09:32:52 pm

Title: Bubble's specialist questions
Post by: bubbles on August 16, 2008, 09:32:52 pm: (http://i50.photobucket.com/albums/f313/mang0es/Specialistmathsassignment7.jpg)

What I have done so far:
(http://i50.photobucket.com/albums/f313/mang0es/Specialistmathsassignment7workingou.jpg)

How do you do 1. c) (i) ?
Title: Re: Bubble's specialist questions
Post by: cara.mel on August 16, 2008, 09:56:36 pm: In b) the volume is the integral from 0 to 30, not from -10 to 10, as you are rotating around y axis

For c i), I would guess do the integral again but from 0 to h, then differentiate it. There is probably an easier way and/or what I have said is wrong
Title: Re: Bubble's specialist questions
Post by: Mao on August 16, 2008, 10:21:10 pm: question 3:

$V=\pi \int_0^h \frac{(y+60)^2}{36}\; dy$

$\implies \frac{dV}{dh}=\frac{d}{dh}\left(\pi \int_0^h \frac{(y+60)^2}{36}\; dy\right)=\frac{\pi}{36}(h+60)^2$

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

$\frac{dh}{dt}=-\frac{3.6\sqrt{h}}{\pi (h+60)^2}$

$t=-\frac{\pi}{3.6}\int \frac{(h+60)^2}{\sqrt{h}}\; dh$

$\implies t= -\frac{\pi}{3.6}\int h^{3/2}+120h^{1/2}+3600h^{-1/2}\; dh$

$\implies t= -\frac{\pi}{3.6}\left(\frac{2}{5}h^{5/2}+80h^{3/2}+7200h^{1/2}\right)+C$

given (0,30) [very messy...]

$\implies C=\frac{\pi}{3.6}\left(\frac{2}{5}\cdot 30^{5/2}+80\cdot 30^{3/2}+7200\cdot 30^{1/2}\right)$

hence, when h=0, t...

$t=-(0)+C=\frac{\pi}{3.6}\left(\frac{2}{5}\cdot 30^{5/2}+80\cdot 30^{3/2}+7200\cdot 30^{1/2}\right)\approx 47606.6 s$ ?!

[i still have a feeling its wrong...]
Title: Re: Bubble's specialist questions
Post by: Mao on August 16, 2008, 10:47:00 pm: yes, of course. thank you

it still feels too whacky
haha
Title: Re: Bubble's specialist questions
Post by: shinny on August 16, 2008, 11:01:41 pm: It's a pretty ugly answer but it sorta sounds about right given that the rate of change of volume is in proportion to its height. So as it gets lower and lower, the rate of change gets infinitely small and the last few drops will take quite a while to clear. I'll verify the answer on CAS afterwards.

Now for the proper working out for Q2.
$\mbox{Let V be the volume of the bucket}$

$y=6x-60$

$y+60=6x$

$x=\frac{1}{6}y+10$

$x^{2}=\frac{1}{36}y^{2}+\frac{10}{3}y+100$

$V=\pi\int_{0}^{30}x^{2}dy$

$V=\pi\int_{0}^{30}(\frac{1}{36}y^{2}+\frac{10}{3}y+100)dy$

$V=\pi\int_{0}^{30}(\frac{1}{36}y^{2}+\frac{10}{3}y+100)dy$

$V=\pi\left[\frac{1}{108}y^{3}+\frac{5}{3}y^{2}+100y\right]_{0}^{30}$

$V=\pi\left[250+1500+3000\right]$

$\mbox{Therefore the volume of the bucket is 4500\pi cm^{3}}$

EDIT: yep, checked with CAS and theres no algebra mistakes in your working out. Pretty sure its the right answer then as there's no fault in your method I'm pretty sure =\
Title: Re: Bubble's specialist questions
Post by: bubbles on August 17, 2008, 12:37:37 am: yep i get it now, thank you guys!
Nearest minute therefore (47606.62s) / 3600 = 13 hrs and 13 minutes
I'll confirm the answer to c) iii. once i get a hold of it sometime this week
Title: Re: Bubble's specialist questions
Post by: bubbles on August 18, 2008, 06:49:54 pm: Quote from: Mao on August 16, 2008, 10:47:00 pm
$t=-(0)+C=\frac{\pi}{3.6}\left(\frac{2}{5}\cdot 30^{5/2}+80\cdot 30^{3/2}+7200\cdot 30^{1/2}\right)\approx 47606.6 s$ ?!

[i still have a feeling its wrong...]

yep that's the correct answer in seconds
Title: Re: Bubble's specialist questions
Post by: bubbles on August 18, 2008, 07:10:08 pm: (http://i50.photobucket.com/albums/f313/mang0es/Specialistmaths1c001.jpg)

1d) I don't have the question with me but from memory: Given $C_{A}$ = $1.0 \times 10^{11}$
$K_{A}$ = $1.0 \approx 3.964...$ found in part a) form a new rate of growth equation.
Any ideas?
Title: Re: Bubble's specialist questions
Post by: Mao on August 18, 2008, 07:45:49 pm: initially looking at it, I don't see anything especially wrong with your workings [or i might have overlooked something]. So heres my workings:

$\frac{dN}{dt}=kN\left(1-\frac{N}{C}\right)$

$\implies \frac{dt}{dN}=\frac{1}{kN\left(1-\frac{N}{C}\right)}=\frac{C}{kN(C-N)}=-\frac{C}{k}\cdot \frac{1}{N(N-C)}$

$\implies t=-\frac{C}{k}\int \frac{dN}{N(N-C)}$

partial fractions

$\frac{1}{N(N-C)}=\frac{A}{N}+\frac{B}{N-C}$

$\implies 1=A(N-C)+BN$

$\implies A=-\frac{1}{C},\; B=\frac{1}{C}$

$\implies t=-\frac{1}{k}\int -\frac{1}{N}+\frac{1}{N-C}\; dN$

$\implies t=-\frac{1}{k}log_e \left|\frac{N-C}{N}\right|+C$

given $(0,N_0)$

$\implies C=\frac{1}{k}log_e \left(\frac{N_0-C}{N_0}\right)$

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

1d) plug numbers into $\frac{dN}{dt}$
Title: Re: Bubble's specialist questions
Post by: Mao on August 18, 2008, 07:52:17 pm: okay, found your problem, it was when you tried to work out the partial fraction
what should have happened is:

$\begin{align*} \frac{C}{kN(C-N)} &= \frac{A}{kN}+\frac{B}{C-N} \\ &= \frac{A}{kN}-\frac{B}{N-C}\\ \implies C &= \frac{AkN(C-N)}{kN}-\frac{BkN(C-N)}{N-C}\\ \implies C &= A(C-N) - (-BkN)\\ \implies C &= -A(N-C) + B(kN)\\ \end{align*}$

which is the negative of your expression
Title: Re: Bubble's specialist questions
Post by: bubbles on August 22, 2008, 12:18:59 am: ~kinematics
(I dropped physics this year and now it has come back to haunt me -___- )

An object moves in a line with velocity v m/s given by $v = \frac{t}{1+t^{2}}$ . The object starts from the origin. Find:
b) the maximum velocity
max velocity occurs when v'(t) = 0
differentiate v via quotient rule, therefore t=1
sub t=1 into v(t), hence ${V}_{max}$ = 0.5m/s
c) the distance travelled in the third second
?
d) the position-time relationship
how do you anti-differentiate $v = \frac{t}{1+t^{2}}$ again ?
Title: Re: Bubble's specialist questions
Post by: enwiabe on August 22, 2008, 02:31:27 am: v = dx/dt = t/(1+t^2)

You want the time travelled in the third second, so integrate t/(1+t^2) from t = 2 to t = 3 (I.E. the third second).

I'll give you a hint for the integral of t/(1+t^2) - it looks eerily like f'(x)/f(x), eh? ;)

Also, you don't need Physics to do well in the Physics part of spec! Specialist Maths goes WAY above and beyond Physics in terms of kinematics, so don't sweat it. :P
Title: Re: Bubble's specialist questions
Post by: cara.mel on August 22, 2008, 09:08:42 am: Quote from: enwiabe on August 22, 2008, 02:31:27 am
You want the time travelled in the third second

I predict this is 1!
Title: Re: Bubble's specialist questions
Post by: bubbles on August 22, 2008, 10:54:02 pm: Quote from: enwiabe on August 22, 2008, 02:31:27 am
I'll give you a hint for the integral of t/(1+t^2) - it looks eerily like f'(x)/f(x), eh? ;)

Thanks enwiabe i understand a) but I haven't yet grasped how to integrate t/(1+t^2).
okay this is going to expose how inept i am with specialist maths :-[:

$\int \ \frac{f'(x)}{f(x)} .dx = log_{e} |f(x)|$

therefore $log_{e} (1+t^{2})$ ??

The answer is: $\frac{1}{2} log_{e} (1+t^{2})$ where does the half come from?

Using the correct answer at the back:
c) $\frac{1}{2} \left[ (log_{e} (10}) - (log_{e} (5}) \right]$

$\frac{1}{2} log_{e} (2}) metres$ (which is the correct answer)
Title: Re: Bubble's specialist questions
Post by: cara.mel on August 22, 2008, 11:02:15 pm: Make the substituion $u = 1 + t^2$

=> $\frac{du}{dt} = 2t$

$dt = \frac{du}{2t}$

Subbing this in to the equation

$\int \ \frac{t}{u} \times \frac{du}{2t}$

Then you can cancel the ts and finish it off and you see how the 1/2 ends up being there.
I would advise always using substitutions and writing out all the working vs trying to be clever and do it in your head, I stuff up too many times doing that :P
Title: Re: Bubble's specialist questions
Post by: Mao 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 $\frac{1}{2}\cdot \left[ \mbox{ln}|1+t^2| \right]_2^3$
Title: Re: Bubble's specialist questions
Post by: bubbles 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) (http://stuff.daniel15.com/cgi-bin/mathtex.cgi?t=\frac{1}{k}log_e%20\left|%20\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 (http://stuff.daniel15.com/cgi-bin/mathtex.cgi?t=\frac{1}{k}log_e%20\left|%20\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. :)
Title: Re: Bubble's specialist questions
Post by: 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. :P
Title: Re: Bubble's specialist questions
Post by: bubbles 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. :P
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?
Title: Re: Bubble's specialist questions
Post by: Mao 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 :P
Title: Re: Bubble's specialist questions
Post by: bubbles 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!
Title: Re: Bubble's specialist questions
Post by: 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 :()
Title: Re: Bubble's specialist questions
Post by: Mao on August 29, 2008, 09:44:45 pm: oh god.. that last post made as much sense as the questions.... =P
Title: Re: Bubble's specialist questions
Post by: jsimmo 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 :-\
Title: Re: Bubble's specialist questions
Post by: xox.happy1.xox 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? :P
Title: Re: Bubble's specialist questions
Post by: 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. Actually, I think the mark I got wasn't so bad after all, considering everything that was involved...
Title: Re: Bubble's specialist questions
Post by: bubbles 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 -_-"
Title: Re: Bubble's specialist questions
Post by: xox.happy1.xox on August 30, 2008, 09:46:58 am: Aww, hopefully you are still satisfied with what mark you eventually get...
Title: Re: Bubble's specialist questions
Post by: bubbles 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)
Title: Re: Bubble's specialist questions
Post by: Collin Li 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.
Title: Re: Bubble's specialist questions
Post by: Mao on September 02, 2008, 01:49:10 pm: an alternative method [that is slightly easier]:

the ratio of acceleration is 1:4, that implies both distance and time will be 4:1, or a fifth of the total [draw the v-t graph to convince yourself]

this means, the last 100m took 24s [a fifth of the total]

$d=\frac{v+u}{2}\times t$

$\implies u=\frac{2d}{t}-v$

since "u" in this case (initial velocity) is the maximum velocity, and v (final velocity) is 0:

$\implies v_{max}=\frac{2\times 100}{24}-0=\frac{25}{3}\mbox{ ms^{-1}}\approx 8.333 \mbox{ ms^{-1}}$

using ratios can save a lot of effort, and avoid large numbers such as those coblin encountered. however, it can get tricky as only in some cases can this method be used, and in most cases it cannot.
for example, where the v-t graph plateaus at a maximum velocity
Title: Re: Bubble's specialist questions
Post by: xox.happy1.xox on September 02, 2008, 09:53:07 pm: This is the maths I like! :)
Title: Re: Bubble's specialist questions
Post by: bubbles on September 02, 2008, 10:37:31 pm: Quote from: xox.happy1.xox on September 02, 2008, 09:53:07 pm
This is the maths I like! :)
LOL
Title: Re: Bubble's specialist questions
Post by: bubbles on September 02, 2008, 10:58:13 pm: I was heading in the right direction until i got stuck at step 2- finding the velocity equation.
Quote from: coblin on September 02, 2008, 07:36:53 am
The second one is:
$v = at - 4a(120-t) = 0$

All you did here was add the velocity from t= 0->96 and t= 96->120 to find the total velocity, is that correct?

Quote from: Mao on September 02, 2008, 01:49:10 pm
an alternative method [that is slightly easier]:

the ratio of acceleration is 1:4, that implies both distance and time will be 4:1, or a fifth of the total [draw the v-t graph to convince yourself]

this means, the last 100m took 24s [a fifth of the total]
I need to brush up on my ratios -"...Mao could you explain why both distance and time will be "4:1" instead of 1:4.
Title: Re: Bubble's specialist questions
Post by: Collin Li on September 02, 2008, 11:31:42 pm: Quote from: bubbles on September 02, 2008, 10:58:13 pm
All you did here was add the velocity from t= 0->96 and t= 96->120 to find the total velocity, is that correct?

Yes sort of. I added the "velocity gained," which is why I multiplied the time by acceleration (it's constant, so I can just go $at$ ).
Title: Re: Bubble's specialist questions
Post by: Mao on September 03, 2008, 01:15:59 pm: Quote from: bubbles on September 02, 2008, 10:58:13 pm
Quote from: Mao on September 02, 2008, 01:49:10 pm
an alternative method [that is slightly easier]:

the ratio of acceleration is 1:4, that implies both distance and time will be 4:1, or a fifth of the total [draw the v-t graph to convince yourself]

this means, the last 100m took 24s [a fifth of the total]
I need to brush up on my ratios -"...Mao could you explain why both distance and time will be "4:1" instead of 1:4.

the relationship between velocity, time and acceleration is $v=u+at$

rearranging: $a=\frac{v-u}{t}$ , assuming velocity(s) is constant, we will end up with $a\propto \frac{1}{t}$ , an inverse relationship.
hence, a 1:4 ratio of acceleration will become 4:1

now, we also know that $s=vt,\implies s\propto t$ , a direct variation
hence, a ratio of 4:1 of t will imply a ratio of 4:1 for distance.

you can also think about this in terms of what happens in real life: if you do someting 4 times as fast, you would take only a quarter of the time. 1:4 --> 1:0.25 == 4:1
Title: Re: Bubble's specialist questions
Post by: bubbles on September 03, 2008, 02:22:28 pm: Thanks I get it now ;D
Title: Re: Bubble's specialist questions
Post by: bubbles on September 03, 2008, 11:08:50 pm: Other expressions for acceleration
A particle moves in a horizontal line so that the velocity, v m/s, is given by $v=2\sqrt{1-x^{2}$ . Find:
a) the position, x m, in terms of time t seconds, given that, when t=0, x=1

$\frac{dx}{dt}= v$

$\therefore \frac{dx}{dt}=2\sqrt{1-x^{2}$

$\therefore \frac{dt}{dx}=\frac{1}{2\sqrt1-x^2}$

$\therefore t=\frac{1}{2}sin^{-1}(x) +c$
t=o, x=1
$\therefore c=-\frac{\pi}{4}$

$\therefore x= sin(2t + \frac{\pi}{2})$

... which is wrong --"
the correct answer is: $x = cos 2t$

(hmm on second thoughts.. are both answers correct. Is it one of those $cosx = sin(x+\frac{\pi}{2})$ ? If this is the case, which is the preferred form in the exam, does it matter?
Title: Re: Bubble's specialist questions
Post by: shinny on September 03, 2008, 11:33:44 pm: Yeh, both are correct. It's a pretty basic trig identity and you're expected to know it. In the case of an exam, I'd say simplify it to x=cos2t incase u get a pedantic examiner but all in all, i HEAVILY doubt you'll ever lose a mark for not simplifying it.
Title: Re: Bubble's specialist questions
Post by: bubbles on September 22, 2008, 09:09:10 pm: Resolution of forces and inclined planes
A particle of mass 5kg is being pulled up a slope inclined at 30° to the horizontal. The pulling force, F newtons, acts parallel to the slope, as does the resistance with a magnitude of one-fifth of the magnitude of the normal reaction.
a) Find the value of F, such that the acceleration is $1.5 m/s^2$ up the slope.
F - (Fr + 5gsin30) Fnet
R = 5gcos30
Fr = (1/5)(5gcos30)
Fr = gcos30
F = 5(1.5) + gcos30 + 5gsin30
F= 40.49 N

b) Also find the magnitude of the acceleration if this pulling force now acts at an angle of 30° to the slope (i.e. at 60° to the horizontal).

How do you do b) ?
Title: Re: Bubble's specialist questions
Post by: shinny on September 22, 2008, 10:41:09 pm: (http://img225.imageshack.us/img225/5712/img0792ty2.jpg)