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1st year geometry and motion

Post Icon Posted: Submitted by mauhea on 5 June 2008 - 12:32pm.

Joined: 2008-04-28
Posts: 11

Any one have the past paper solution of 06/07 and 04/05???

Found it so hard to revise ... too many stuff to memorise...

Could anyone give me some hints??

‹ Jamie's Mathematical Problems - Vol. 1 Pólya Vector Fields and non-trivial de Rham cohomology groups ›
Post Icon Posted: 5 June 2008 - 12:55pm

Joined: 2007-10-01
Posts: 179

Which questions/parts of the course in particular would you like help with?
Post Icon Posted: 5 June 2008 - 1:04pm

Joined: 2006-10-10
Posts: 520

Hey, yeah, if the past papers aren't on mathstuff, they're going to be pretty much impossible to find anywhere else :( Sorry. But a few of us did the 06/07 paper so we should be able to help you on that, even though pretty much everyone in our year failed it :D
Post Icon Posted: 5 June 2008 - 1:07pm

Joined: 2008-04-28
Posts: 11

Part 3 critical points of two dimensional surface is quite hard to understand... I will probably fail this time as well:S

I am going through the revision guide at the moment and hopefully that I can start doing past papers tonight.
Post Icon Posted: 5 June 2008 - 1:20pm

Joined: 2006-10-10
Posts: 520

Yeah, good luck :) Also, the 07 papers was, by all accounts, a VERY hard paper, the 08 one probably wont be like that. Normally working out crit points of a surface is pretty easy, but Mario just chose a freaking impossible surface :( Have a look at other easier examples and you'll see its actually a really good way to bank some points.
Post Icon Posted: 5 June 2008 - 1:21pm

Joined: 2007-10-01
Posts: 179

Yeah I never like that. I'm assuming you have $ f: \mathbb{R}^n \rightarrow \mathbb{R}\, $ in this case n=2 most likely. Then you look at the graph of $ z=f(x,y)\, $ then you'll get a is a critical point whenever $ \nabla f (a) = 0 $. The reason this works is any directional derivative $ D_u f(x,y) = u \cdot \nabla f(x,y) \, $ which at a critical point is zero for any $ u\, $ hence the graph is constant whichever direction you look in.
In order to classify them, you have to look at the Hessian matrix i.e. the matrix of mixed partial derivatives, then use Sylvester's criterion. Look at the top left entry, if it's positive and the determinant of the matrix is positive, then you have a minimum. If it's negative and the determinant is negative you have a maximum. If you get anything else then you know it must be a saddle point since these are the only cases possible in $ \mathbb{R}^3 \, $

hope this helps.
Post Icon Posted: 5 June 2008 - 2:22pm

Joined: 2008-04-28
Posts: 11

Tat's reli helps. Thx so much guys... I am going to continue with the revision guide and will post any questions i hv here. Thx again!!!lol
Post Icon Posted: 5 June 2008 - 2:33pm

Joined: 2007-10-03
Posts: 6

did the 06/07 students get given a set percentage extra cuz the exam was so hard?
Post Icon Posted: 5 June 2008 - 2:38pm

Joined: 2006-11-02
Posts: 1090

Nope, I don't think so.
Post Icon Posted: 5 June 2008 - 2:58pm

Joined: 2006-10-10
Posts: 520

Actually, they did, my tutor told me so. A ton of people got exactly 40% after the mod
Post Icon Posted: 5 June 2008 - 3:19pm

Joined: 2006-11-02
Posts: 1090

Fair enough, I didn't think I had done that badly in it. :D
Post Icon Posted: 5 June 2008 - 3:26pm

Joined: 2007-10-03
Posts: 397

CosmiN, any idea what DG is on tonight ?
Post Icon Posted: 5 June 2008 - 4:42pm

Joined: 2007-02-14
Posts: 105

I got given an extra 10%, don't know how they scale it though.
Post Icon Posted: 5 June 2008 - 4:52pm

Joined: 2006-11-02
Posts: 1090

CosmiN, any idea what DG is on tonight ?

No. Why did you post it in this thread? :P
Post Icon Posted: 5 June 2008 - 5:01pm

Joined: 2007-10-03
Posts: 397

MEH, don't know, just was waiting too eagerly for it to happen :p
Post Icon Posted: 5 June 2008 - 6:40pm

Joined: 2007-10-01
Posts: 179

srsly you guys have got to stop hi-jacking people's threads. this is why new people don't use the forums much, don't post here unless you're actually answering someone's question.
Post Icon Posted: 6 June 2008 - 5:00pm

Joined: 2008-04-28
Posts: 11

Hi, guys...

I am doing 06 paper at the moment. Could anybody help me with question 2 part (ii) and also, in part (iii), when we change the order of integration, do we need to change the domain.

For example, in part(iii) , we have x belongs to (0,1) and y belongs to (x,x^0.5), after changing of order, does this change or not??

tyty
Post Icon Posted: 6 June 2008 - 5:15pm

Joined: 2007-10-03
Posts: 397

After changing the order of integration, we're still integrating over the same domain but the bounds of the integrals must change to account for that.
$\displaystyle \int_{x=0}^{x=1} \int_{y=x}^{y=\sqrt{x}} dy dx = \int_{y=0}^{y=1} \int_{x=y^2}^{x=y} dx dy $
Post Icon Posted: 6 June 2008 - 5:21pm

Joined: 2008-04-28
Posts: 11

thx

nd also in question 4 of 06 paper. part(i) ask for critical point. but when i tried the factorisation, i found a complex root of x nd y. do i need to continue with it or just ignore it.

tyty
Post Icon Posted: 6 June 2008 - 5:36pm

Joined: 2007-10-03
Posts: 397

Just ignore the complex roots, the only critical points are going to be when $ 2x^3-3x^2+5 $ and $ 2y^3-3y^2+5 $ are both 0, and they factorise into $ (x+1)(2x^2-5x+5) $ and $ (y+1)(2y^2-5y+5) $, and as we're only viewing this function $ f : \mathbb{R}^2 \rightarrow \mathbb{R} $, the complex roots do not matter and you may as well ignore them.
Post Icon Posted: 6 June 2008 - 6:01pm

Joined: 2008-04-28
Posts: 11

could anyone teach me 06 paper question 1 part (i)b) and (ii), and question 2 part (ii) how to sketch and find all the domain of x, y ,z to evaluate.

lol, got so many stuff this time

tyty
Post Icon Posted: 6 June 2008 - 6:14pm

Joined: 2007-10-03
Posts: 397

For question 1 part (i)b), you have to parametrise the three sides of the triangle and then calculate the line integral in the standard way, making sure of going round the triangle in the right way.

For question 1 part(ii) :
$\displaystyle V(x,y,z) = (y^2,2xy,2) $

We want to find a scalar potential for this, which is a function $ f : \mathbb{R}^3 \rightarrow \mathbb{R} $ such that $ V(x,y,z) = \nabla f(x,y,z) $.
Doing this component by component, $ \frac{\partial f}{\partial x} = y^2, \frac{\partial f}{\partial y} = 2xy, \frac{\partial f}{\partial z} = 2 $
From the first part, you get that $ f(x,y,z) = xy^2 + g_{yz}(y,z) $, then $ f(x,y,z) = xy^2 + g_{xz}(x,z) $ and $ f(x,y,z) = 2z + g_{xy}(x,y) $.
Notice you can put these three together to get $ f(x,y,z) = xy^2+2z $, which is the scalar potential we were looking for.

Then, for the integral :

$\displaystyle \int_{(-2,2,4)}^{(4,-1,2)}V(x,y,z)\cdot dr = f(4,-1,2) - f(-2,2,4) $

For question 2 part (ii), we have

$\displaystyle \Omega = \{(x,y,z) : z \ge 0, x^2+y^2 \le z^2 \le 1-x^2-y^2\} $

Looking at the conditions, you region is in the upper half plane $ z \ge 0 $, and $ z^2 $ is contained between $ x^2+y^2 $ and $ 1-x^2-y^2 $ so z (as positive) is contained between $ \sqrt{x^2+y^2} $ and $ \sqrt{1-x^2-y^2} $.
You should recognize the first one as being an upwards pointing cone and the second one as the (upper half) of the unit sphere. Therefore z must lie in between those two shapes.

But, in spherical polar coordinates, that region will then be defined as $ 0 \le r \le 1, 0 \le \phi \le \2 \pi, 0 \le \theta \le 1 $.

Therefore you just need to integrate $ z \,\, ( = r \cos(\theta)) $ over that $ (r,\phi,\theta)- $region, not forgetting to multiply by the scale factor (the determinant of the Jacobian matrix).
Post Icon Posted: 6 June 2008 - 6:27pm

Joined: 2008-04-28
Posts: 11

When we put those three functions of f(x,y,z) together, why we get $ f(x,y,z)=xy^2+2z $ but not $ f(x,y,z)=2xy^2+2z $??

and since we have $ V(x,y,z)=\nabla f(x,y,z) $

for the integral,

should we have the solution f(4,-1,2)-f(-2,2,4) instead?

thx so much
Post Icon Posted: 6 June 2008 - 6:36pm

Joined: 2007-10-03
Posts: 397

Remember we have $ f(x,y,z) = xy^2 + g_{yz}(y,z) $, then $ f(x,y,z) = xy^2 + g_{xz}(x,z) $ and $ f(x,y,z) = 2z + g_{xy}(x,y) $
Here we want to put these three functions together, but we can't just add it together. $ xy^2 $ is a function of both $ x $ and $ y $, and we notice it's exactly what we want when we add things together.

Imagine we wanted to find a scalar field for $ V(x,y) = (x,x) $.
Then we would get, by the same method as before, $ g(x,y) = x + h_y(y), g(x,y) = xy + h_x(x) $.
But notice we won't be able to put these two together, as one side says "I want g to be : x + some function not containing x" and the other says "I want g to be : xy + some function not containing y". These two statements can't be made to agree, while they could for f, as we had "I want f to be xy² + some function of y,z , I want f to be xy² + some function of x,z, I want f to be 2z + some function of x,y$, which can be stitched together to get f(x,y,z) = xy² + 2z.
Post Icon Posted: 6 June 2008 - 6:40pm

Joined: 2008-04-28
Posts: 11

cool, I got that !!!

How about for the integral?? I think cuz we set $ V=\nabla f $, then when we integral V, shall we simply have f as the solution?
Post Icon Posted: 6 June 2008 - 6:51pm

Joined: 2007-10-03
Posts: 397

Yes, sorry that's what I meant :

$\displaystyle \int_a^b \nabla \phi \cdot dr = \phi(b) - \phi(a) $
Post Icon Posted: 6 June 2008 - 6:54pm

Joined: 2008-04-28
Posts: 11

np~~~thx so much...

I am doing question 3 atm...nd if i have problems, i will post again, thx again for your help!
Post Icon Posted: 6 June 2008 - 7:19pm

Joined: 2008-04-28
Posts: 11

for question 3 part(ii)

I wrote $ (x-c)^2/a^2+y^2/b^2=1 $ and define that $ cost=(x-c)/a,sint=y/b $,

therefore i got E={acost+c,bsint|t belongs to [0,2pi]}

is it right??

and how to derive the equation from this point set?

tyty
Post Icon Posted: 7 June 2008 - 4:23pm

Joined: 2007-10-03
Posts: 6

The exam was quite hard - comparable to last year's!! There was a question on it that no one, of the people that i spoke to, could do. The question was (3, part (ii)):
Consider the region $ \Omega $ = {$ (x,y,z) $ : $ y\ge0 $, $ z\ge0 $, $ x^2 + y^2 + z^2 \le 4 $}
Use spherical polar coordinates to evaluate
$ \int\int\int_\Omega\frac{yze^{-x}}{\sqrt{x^2 + y^2}} dxdydz $
Post Icon Posted: 7 June 2008 - 5:33pm

Joined: 2007-10-03
Posts: 397

$\displaystyle \iiint_\Omega \frac{yze^{-x}}{\sqrt{x^2+y^2}} \, dxdydz $

$\displaystyle \Omega := \{(x,y,z) : y \ge 0, z \ge 0, x^2+y^2+z^2 \le 4 \} $

$\displaystyle = \{(r,\phi,\theta):0\le r \le 2, 0\le \phi \le \pi, 0\le \theta \le \frac{\pi}{2} \} $

Then we want

$\displaystyle \int_0^2 \int_0^{\frac{\pi}{2}} \int_0^{\pi} \frac{r \sin(\phi)\sin(\theta) r\cos(\theta)e^{-r\cos(\phi)\sin(\theta)}}{r \sin(\theta)} \cdot r^2 \sin(\theta) d \phi d \theta dr $

$\displaystyle = \int_0^2 \int_0^{\frac{\pi}{2}} \int_0^{\pi} r^2 \cos(\theta)\cdot r\sin(\phi)\sin(\theta) e^{-r\cos(\phi)\sin(\theta)} d \phi d \theta dr $

$\displaystyle = \int_0^2 \int_0^{\frac{\pi}{2}} r^2 \cos(\theta) \Bigg [e^{-r\cos(\phi)\sin(\theta)} \Bigg ]_0^\pi d \theta dr $

$\displaystyle = \int_0^2 \int_0^{\frac{\pi}{2}} r^2 \cos(\theta) (e^{r \sin(\theta)} - e^{-r \sin(\theta)}) d \theta dr $

$\displaystyle = 2 \int_0^2 \int_0^{\frac{\pi}{2}} r^2 \cos(\theta) \sinh(r \sin(\theta)) d \theta dr $

$\displaystyle = 2 \int_0^2 \int_0^{\frac{\pi}{2}} r \sinh(r \sin(\theta)) r \cos(\theta) d \theta dr $

Substitute $ u = r \sin(\theta) $, $ du = r \cos(\theta) d\theta $, we get :

$\displaystyle 2 \int_0^2 \int_{0}^{r} r \sinh(u) du dr $

$\displaystyle = 2 \int_0^2 r\bigg [ \cosh(u) \bigg ]_0^r dr $

$\displaystyle = 2 \int_0^2 r \Big ( \cosh(r) - 1 \Big ) dr $

$\displaystyle = 2 \bigg [ r \sinh(r) - \cosh(r) - r \bigg ]_0^2 $

$\displaystyle = 4 \sinh(4) - 2 \cosh(4) -4 + 2 = e^2 - 3e^{-2} -2 $

Hope that's right, haven't checked it extensively.

Edit : Fixed that error, haven't checked at all.

Edit : Ok, now it's right.
Post Icon Posted: 7 June 2008 - 5:58pm

Joined: 2007-10-03
Posts: 6

shoudn't the first line of ur integral have $ rsin(\theta) $ instead of just $ r $ in the denom (correct me if im wrong)? Also, out of the 12 marks for the ques, how many marks do u reckon will be given for correctly formulating the integral and limits and integrating with respect to r first (then stopping)?

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