Author Topic: TT's Maths Thread (Read 134745 times) Tweet Share

Mao · « **Reply #1080 on:** October 21, 2010, 11:49:52 pm »

Quote from: Cthulhu on October 17, 2010, 12:41:06 am

Quote from: Mao on October 17, 2010, 12:39:46 am
Quote from: TrueTears on October 16, 2010, 11:20:25 pm
Suppose a gambler with $n plans to bet $1 a time on the toss of a coin, until he reaches $100 or $0. What is the probability of going broke starting with $n? First make a recurrence relation to represent the situation and then solve it.

Hmmm how would I start this question?

Not helping, but I remember doing this when I was trying to get a strategy for winning roulette. Not sure about doing it analytically, but I ran a simulation 1,000,000 times and approximated the long-term probability
Link to arxiv or published paper or it didn't happen.

sif publish that. did you watch 21? people get beaten up for that shit

TrueTears · « **Reply #1081 on:** October 24, 2010, 08:37:25 pm »

Are the following true in general?

If a vector field F can be written into 2 parts, F_1 being conservative and F_2 is non conversative then:

$Curl \ \mathbb{F} = Curl \ \mathbb{F}_1 + Curl \ \mathbb{F}_2$

TrueTears · « **Reply #1082 on:** October 24, 2010, 08:43:15 pm »

Also with the same initial conditions on F, are these true?

$\int_C \mathbf{F}.d\mathbf{r} = \int_C \mathbf{F}_1.d\mathbf{r} + \int_C \mathbf{F}_2.d\mathbf{r}$

/0 · « **Reply #1083 on:** October 24, 2010, 08:44:54 pm »

You mean if $\mathbf{F}=\mathbf{F}_1+\mathbf{F}_2$ ? Then yep that's true (just from distributivity of the curl operator), and also $\nabla \times \mathbf{F_1}=0$ , so $\nabla \times \mathbf{F}=\nabla \times \mathbf{F_2}$

And also as far as I know ur next post should be true too (distributivity of the integral operator)

TrueTears · « **Reply #1084 on:** October 24, 2010, 08:46:57 pm »

Thanks man, lol yeah i forgot the latex for mathbf

TrueTears · « **Reply #1085 on:** October 24, 2010, 09:00:13 pm »

wth is up with this Q?

isn't the answer 0?

/0 · « **Reply #1086 on:** October 25, 2010, 02:04:42 am »

Hmm I get $\frac{\pi}{4}$

At $z=0$ , $\mathbf{F}=\frac{1}{3}y^3\mathbf{j}+y^2\mathbf{k}$

The surface normal pointing upwards is just $\mathbf{k}$ , so $\mathbf{F}\cdot \mathbf{k}=y^2$

So essentially you're integrating $y^2$ over the disc $x^2+y^2=1$ , probably a good idea to switch to polar coordinates.

TrueTears · « **Reply #1087 on:** October 27, 2010, 03:51:28 pm »

Quote from: /0 on October 25, 2010, 02:04:42 am

Hmm I get $\frac{\pi}{4}$

At $z=0$ , $\mathbf{F}=\frac{1}{3}y^3\mathbf{j}+y^2\mathbf{k}$

The surface normal pointing upwards is just $\mathbf{k}$ , so $\mathbf{F}\cdot \mathbf{k}=y^2$

So essentially you're integrating $y^2$ over the disc $x^2+y^2=1$ , probably a good idea to switch to polar coordinates.

Oh that's right, opps I thought it was the surface integral of a scalar function haha, silly me, thanks /0!!!

(Trust you to be good with this kinda maths

)

TrueTears · « **Reply #1088 on:** November 01, 2010, 05:17:49 pm »

Any help with this?

Thanks

Actually nevermind, I got it, but I was still wondering how do you approach these questions generally? Do you need to visualise what solid region you are integrating and then set the spherical coordinates?

Mao · « **Reply #1089 on:** November 01, 2010, 05:42:51 pm »

pretty much, yeah

Or you can treat it like a jacobian transformation, do a few simultaneous eqns to find the boundaries. This method is longer, but more general.

TrueTears · « **Reply #1090 on:** November 01, 2010, 08:08:15 pm »

Thanks Mao!

Another one...

[IMG]http://img408.imageshack.us/img408/4751/mth2010examhelp.jpg[/img]

I got part a) b) c) however I am just unsure on part d)

For a) I got $r(\theta, \phi) = \sin(\phi)\cos(\theta)i + \sin(\phi)\sin(\theta)j+\cos(\phi)k$

b) $r(x,y) = xi+yj+zk \ \text{but} \ z^2+y^2+x^2 = 1 \ \text{so} \ z = \sqrt{1-(x^2+y^2)}$

So $r(x,y) = xi+yj+\sqrt{1-(x^2+y^2)}k$

c) Not required for d)

d) So area of surface $= \int \int_D |r_x \times r_y| dA$

$|r_x \times r_y| = \frac{1}{\sqrt{1-x^2-y^2}}$ (I've double checked this, it's correct

So $\int \int_D \frac{1}{\sqrt{1-x^2-y^2}} dA$

But what are the bounds for the double integral?

Are we allowed to convert to polar coordinates? Cause I'm interpreting the question as set up the integral purely in cartesian axis, if it's in polar coordinates then it's quite easy and becomes trivial.

Thanks!

/0 · « **Reply #1091 on:** November 01, 2010, 08:54:50 pm »

$\frac{\pi}{6} \leq \phi \leq \frac{\pi}{3}$

$\Rightarrow \frac{1}{2} \leq z \leq \frac{\sqrt{3}}{2}$

$\Rightarrow \frac{1}{4} \leq x^2+y^2 \leq \frac{3}{4}$

I think... this is the annulus you might have to integrate over. If this is right it might require the sum of two integrals to get the whole area

e.g. $\int_{-1}^1 \int_{-\sqrt{(\frac{3}{4})^2-y^2}}^{\sqrt{(\frac{3}{4})^2-y^2}}....... dxdy-\int_{-1}^1 \int_{-\sqrt{(\frac{1}{4})^2-y^2}}^{\sqrt{(\frac{1}{4})^2-y^2}}....... dxdy$

But yeah polar coordinates would be nice

TrueTears · « **Reply #1092 on:** November 01, 2010, 08:59:10 pm »

yup I know that, it's between 2 circles right? So without converting to polar coordinates it's incredibly hard to write the integrals, you need like 4....

however with polar its so easy?

kamil9876 · « **Reply #1093 on:** November 01, 2010, 10:17:24 pm »

yeah the question was probably designed to test your parametrizing skills more than anything.

TrueTears · « **Reply #1094 on:** November 01, 2010, 10:27:39 pm »

thought so... thanks heaps guys!

Search the forums now!

We have moved!

Author Topic: TT's Maths Thread (Read 134745 times) Tweet Share

Mao

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

/0

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

/0

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Mao

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

/0

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

kamil9876

Re: TT's Maths Thread

TrueTears

Re: TT's Maths Thread

Recent Posts

User:
Password: