ATAR Notes: Forum

Uni Stuff => Science => Faculties => Mathematics => Topic started by: QuantumJG on July 26, 2010, 09:59:00 pm

Title: Vector Calculus
Post by: QuantumJG on July 26, 2010, 09:59:00 pm: I'm starting a forum for my maths subject Vector Calculus and here is my first question.

Show that the following limit does not exist:

$\lim_{(x,y) \to (0,0)} \dfrac{x^2 y^2}{ (x^2 + y) ^{3}}$

I tried proving it by going along the path $y = kx^{2}$

So

Proof:

$\lim_{(x,y) \to (0,0)} \dfrac{x^2 y^2}{ (x^2 + y) ^{3}} = \lim_{x \to 0} \dfrac{k^{2} x^{6}}{ (x^2 + k x^2) ^{3}}$

$= \lim_{x \to 0} \dfrac{k^{2}}{ (1 + k) ^{3}}$

So the limit doesn't exist since the k dependency implies that the limit value will depend on what parabolic path you take.

But I'm not sure if this is right. :uglystupid2:
Title: Re: Vector Calculus
Post by: TrueTears on July 26, 2010, 10:18:55 pm: Hi there, this is my working:

Let's approach (0,0) from the x axis.

$\lim_{(x,0) \to (0,0)} \frac{0}{x^5} = 0$

Now let's approach (0,0) from y axis.

$\lim_{(0,y) \to (0,0)} \frac{0}{y^3} = 0$

Seems good so far...

Now let's approach (0,0) from the straight line y = mx

$\lim_{(x,mx) \to (0,0)} \frac{x^2(mx)^2}{(x^2+mx)^3} = 0$

So all 3 path lead to 0, now let's try parabolas.

The rest follows from your working.

It's correct.

However I'd change your final sentence to this: $\lim_{x \to 0} \dfrac{k^{2}}{ (1 + k) ^{3}} = \dfrac{k^{2}}{ (1 + k) ^{3}} = C$ where C is an arbitrary constant. So we have find a path which leads to a different limit. Thus the limit at (0,0) does not exist.

To be even more rigorous, you can try a $\epsilon - \delta$ proof if you want. That'd be an overkill, but fun :P
Title: Re: Vector Calculus
Post by: QuantumJG on July 26, 2010, 10:23:21 pm: Quote from: TrueTears on July 26, 2010, 10:18:55 pm
Hi there, this is my working:

Let's approach (0,0) from the x axis.

$\lim_{(x,0) \to (0,0)} \frac{0}{x^5} = 0$

Now let's approach (0,0) from y axis.

$\lim_{(0,y) \to (0,0)} \frac{0}{y^3} = 0$

Seems good so far...

Now let's approach (0,0) from the straight line y = mx

$\lim_{(x,mx) \to (0,0)} \frac{x^2(mx)^2}{(x^2+mx)^3} = 0$

So all 3 path lead to 0, now let's try parabolas.

The rest follows from your working.

It's correct.

Thanks TT!

In the proof I probably should start with linear equations and then move to higher order polynomials. Doing limits for functions of several variables is a bit more trickier because you aren't constrained to approaching the limit from the left or right, but you can approach it in more ways than you can imagine.
Title: Re: Vector Calculus
Post by: TrueTears on July 26, 2010, 10:44:52 pm: Yup, also if y=kx^2 doesnt work try x = ky^2 and if that also leads to a same limit, you should think that the limit might actually exist and use an e-d proof to show it does.
Title: Re: Vector Calculus
Post by: Ahmad on July 27, 2010, 09:33:43 pm: Just as an aside in the way of counter-examples, there are limits that don't exist but for which if you approach along any polynomial path you get the same value!
Title: Re: Vector Calculus
Post by: Cthulhu on July 27, 2010, 11:43:16 pm: Quote from: Ahmad on July 27, 2010, 09:33:43 pm
Just as an aside in the way of counter-examples, there are limits that don't exist but for which if you approach along any polynomial path you get the same value!
I ran into this last semester. Confused the hell out of me.
Title: Re: Vector Calculus
Post by: QuantumJG on July 28, 2010, 08:59:25 am: Quote from: Ahmad on July 27, 2010, 09:33:43 pm
Just as an aside in the way of counter-examples, there are limits that don't exist but for which if you approach along any polynomial path you get the same value!

Thanks I'll keep this in mind.
Title: Re: Vector Calculus
Post by: Ilovemathsmeth on July 28, 2010, 06:04:09 pm: WOW I'd love to do this subject. I don't think Actuarial requires it though :(
Title: Re: Vector Calculus
Post by: tram on July 28, 2010, 08:48:36 pm: ran into this question (or a very similar question) in umep....i don't get limits.... like, how do you know how you know which path you should approach the limit from to prove it dosen't exist? and if it's just a proof by contradiction, if the limit DOES exist, how do you prove it exists cos it's not like you can go though every single possibilty it and prove they ALL work.....

is there a thread explaining limits somewhere? soz to be all nooby >.<

Quote from: Ilovemathsmeth on July 28, 2010, 06:04:09 pm
WOW I'd love to do this subject. I don't think Actuarial requires it though :(

Vector caculus is the subject they recommend you do in first semester for acturial if you've done umep. Thus it would be highly likely that some acturial studes people have done this subject.
Title: Re: Vector Calculus
Post by: kamil9876 on July 28, 2010, 09:00:16 pm: there are some techniques which account for every possible path.

For example if I want the limit as (x,y) approaches 0 of:
$<br />\frac{x^4}{x^2+y^2}$

I bound it by:

$0\leq \frac{x^4}{x^2+y^2}\leq \frac{x^4}{x^2}=x^2$

and apply sandwhich theorem. (also to be rigorous you should mention that my inequality doesn't make sense when x=0, but when x=0 the function value is 0 anyway).

Read a precise definition of limits to remove any confusion.
Title: Re: Vector Calculus
Post by: TrueTears on July 28, 2010, 10:05:04 pm: Quote from: tram on July 28, 2010, 08:48:36 pm
ran into this question (or a very similar question) in umep....i don't get limits.... like, how do you know how you know which path you should approach the limit from to prove it dosen't exist? and if it's just a proof by contradiction, if the limit DOES exist, how do you prove it exists cos it's not like you can go though every single possibilty it and prove they ALL work.....

is there a thread explaining limits somewhere? soz to be all nooby >.<

Quote from: Ilovemathsmeth on July 28, 2010, 06:04:09 pm
WOW I'd love to do this subject. I don't think Actuarial requires it though :(

Vector caculus is the subject they recommend you do in first semester for acturial if you've done umep. Thus it would be highly likely that some acturial studes people have done this subject.

what i do is this, check your normal paths, x/y axis, linear lines, parabolas, if those all yield same limit then try an $\epsilon-\delta$ proof to see if the limit is actually what you conjecture. That is as rigorous as it gets.
Title: Re: Vector Calculus
Post by: QuantumJG on July 28, 2010, 11:19:57 pm: Quote from: tram on July 28, 2010, 08:48:36 pm
ran into this question (or a very similar question) in umep....i don't get limits.... like, how do you know how you know which path you should approach the limit from to prove it dosen't exist? and if it's just a proof by contradiction, if the limit DOES exist, how do you prove it exists cos it's not like you can go though every single possibilty it and prove they ALL work.....

is there a thread explaining limits somewhere? soz to be all nooby >.<

Quote from: Ilovemathsmeth on July 28, 2010, 06:04:09 pm
WOW I'd love to do this subject. I don't think Actuarial requires it though :(

Vector caculus is the subject they recommend you do in first semester for acturial if you've done umep. Thus it would be highly likely that some acturial studes people have done this subject.

In Vector Calculus we use the sandwhich theorem. Basically you find a function that is always less than your function and one that's greater than your function and you squeeze your function inbetween them and use it to prove the limit exists.

ε-δ proofs I just hate and apparently in vector calculus you aren't required to use them. Although I should probably refresh my memory on them.
Title: Re: Vector Calculus
Post by: tram on July 29, 2010, 09:16:16 am: *brain explodes*

ok i'm going and doing some reading on that e-d thingy proof and the sandwich theorem THEN comming back to annoy you guys :)
Title: Re: Vector Calculus
Post by: Mao on July 29, 2010, 01:19:30 pm: Simple answer to your 'what is a limit', it is a value you get as you approach a particular coordinate in a [possibly multivariable] function. Here, the keyword is 'approach'. You may want to know which way we approach it, and the answer to that is EVERY POSSIBLE DIRECTION. For a limit to exist, you must approach the same value from every direction.

For a univariable function (1D), you can only approach a coordinate from left and right. If you can show they go to the same value, then you are done.

For a multivariable function (2D or higher), you can approach a coordinate from an infinite number of paths, thus you can't use the above method. What we end up employing is the epsilon-delta proof, which incorporates the infinite number of paths. However, it is tedious, algebraically intensive and often very confusing, so we don't always want to do an epsilon-delta proof straight away. Instead, we test a few simple paths, which is much simpler than the e-d proof to try to find a counter-case (thus proving a limit does not exist), and if we cannot find a simple counter-case, we do a e-d proof.
Title: Re: Vector Calculus
Post by: QuantumJG on August 08, 2010, 09:36:44 pm: Using the matrix chain rule evaluate the derivative of:

f(g[f(x,y)]) at (0,1) where f(x,y) = (x², 2y, x - y) & g(x,y,z) = (x + z, y²)

I really need help since I'm finding this area hard.
Title: Re: Vector Calculus
Post by: Ilovemathsmeth on September 10, 2010, 12:23:08 am: tram - does that mean you can possibly choose it as a breadth subject? They said 'no-maths' though...
Title: Re: Vector Calculus
Post by: tram on September 10, 2010, 09:34:47 pm: Well yeah, i think the rule is bcs you're already 'using up' breath subjects in maths you can't choose any more maths as breath. Plus the point of breath is to do subjects DIFFERENT to what youa re doing in ur main cosue which is obvs very math heavy for act stud.
Title: Re: Vector Calculus
Post by: Ilovemathsmeth on September 19, 2010, 12:59:03 am: ahh no that sucks, got to find a breadth sub for next yr lol.
Title: Re: Vector Calculus
Post by: QuantumJG on September 30, 2010, 08:42:41 pm: I don't know where to start with solving this question.

BTW: The answer is 1/2.
Title: Re: Vector Calculus
Post by: /0 on September 30, 2010, 09:07:57 pm: The Jacobian is $\left|\begin{array}{cc} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v}\end{array}\right|=\left|\begin{array}{cc} 1 & 1 \\ 1 & -1 \end{array}\right|=-1-1=-2$

So we have:

$\int_{D'} 2u|-2|dudv$

In order to find the limits of integration for the region D',

The region $0 \leq y\leq x$ , $0 \leq x \leq 1$ is the triangle with sides $y=0$ , $y=x$ , $x=1$ .

The side $y=0$ transforms to $u-v=0\Rightarrow u=v$ in the u-v plane.
The side $y=x$ transforms to $u-v=u+v \Rightarrow v=0$ .
The side $x=1$ transforms to $u+v=1$ .
The region of integration is the enclosed triangle

If you graph this you will see you must integrate

$\int_0^\frac{1}{2} \int_0^u4u dvdu+ \int_\frac{1}{2}^1 \int_0^{1-u}4udvdu$
Title: Re: Vector Calculus
Post by: AzureBlue on September 30, 2010, 09:14:43 pm: Quote from: tram on September 10, 2010, 09:34:47 pm
Well yeah, i think the rule is bcs you're already 'using up' breath subjects in maths you can't choose any more maths as breath. Plus the point of breath is to do subjects DIFFERENT to what youa re doing in ur main cosue which is obvs very math heavy for act stud.
The breadth subjects for actuarial studies are not finalised yet - thus, they might end up being electives/breadth/or non-maths breadth. Better to wait until subject selection next year. But yes, vector calculus is used as breadth in maths. I love the economics plan actually - so many comm electives and breadth! ahhhh... :) If I end up doing Eco, I'll still probably accelerate vector calculus to breadth in first-year for the DipMSc.
Title: Re: Vector Calculus
Post by: QuantumJG on October 02, 2010, 01:54:26 pm: Ok I'm stuck on this:

Find the moment of inertia about the y-axis of a ball defined by x^² + y^² + z^² $\le$ R^² where the mass density ρ is a constant.

My problem is the x^² + z^² part in the integral. In spherical coordinates (what I want to use) it doesn't simplify.
Title: Re: Vector Calculus
Post by: kamil9876 on October 02, 2010, 02:18:01 pm: Post the rest of what you have done so far, and we will see what we can do. Cbf doing everything from scratch.
Title: Re: Vector Calculus
Post by: QuantumJG on October 02, 2010, 03:52:10 pm: Well I have the limits of integration ( 0 < r < R, 0 < θ < π, 0 < φ < 2π ).

But it's the ugly bit inside that's annoying me! Does r²sinθ*(x² + z²) simplify to something nice?
Title: Re: Vector Calculus
Post by: AzureBlue on October 02, 2010, 03:54:13 pm: Quote from: Ilovemathsmeth on September 19, 2010, 12:59:03 am
ahh no that sucks, got to find a breadth sub for next yr lol.
Considering you have Principles of Business Law this year, probably choose Corporate Law next year as breadth maybe because you will have met the pre-reqs.
Title: Re: Vector Calculus
Post by: kamil9876 on October 02, 2010, 06:38:49 pm: Quote from: QuantumJG on October 02, 2010, 03:52:10 pm
Well I have the limits of integration ( 0 < r < R, 0 < θ < π, 0 < φ < 2π ).

But it's the ugly bit inside that's annoying me! Does r²sinθ*(x² + z²) simplify to something nice?

Actually maybe try to use cylindrical co-ordinates, that way that $x^2+z^2$ bit is just merely $r$

Tho with what you have now, you have to find something to change $x^2+z^2$ to something nice using trigonometry. $(rcos \theta)^2$ or $(r cos \phi)^2$ . I forgot which letter corresponds to which angle.
Title: Re: Vector Calculus
Post by: TrueTears on October 02, 2010, 10:03:01 pm: phi for angle made with z axis.

theta for angle made with positive x axis.
Title: Re: Vector Calculus
Post by: Ilovemathsmeth on October 10, 2010, 01:53:06 am: thanks for the advice azureblue :)

i dunno about corporate law, pbl kinda scarred me for life... haha
Title: Re: Vector Calculus
Post by: /0 on October 10, 2010, 03:29:32 pm: Quote from: TrueTears on October 02, 2010, 10:03:01 pm
phi for angle made with z axis.

theta for angle made with positive x axis.

That's the way I always used to do spherical, until I read Griffiths electrodynamics. Now I use $\phi$ for angles with the x-axis in cylind and spher, unless it's only 2D in which case I use $\theta$ . Makes no sense, but whatever... when in Rome
Title: Re: Vector Calculus
Post by: TrueTears on October 10, 2010, 04:59:03 pm: Lol yeah hate it when conventional gets changed ><
Title: Re: Vector Calculus
Post by: QuantumJG on November 04, 2010, 11:25:19 am: Hey guys,

I keep getting this question wrong.

Let a surface be paramaterised by

Φ(u,v) = (u-v,u+v,uv)

Let D be a unit disc (I'm assuming u² + v² $\le$ 1)

Find the area of Φ(D).

I keep getting $\dfrac{2 \sqrt{2} \pi}{3}$ whereas the answer is $\dfrac{ \pi }{3} (6 \sqrt{6} - 8)$
Title: Re: Vector Calculus
Post by: /0 on November 04, 2010, 12:05:03 pm: $\mathbf{r}_u \times \mathbf{r}_v=(u-v)\mathbf{i}-(u+v)\mathbf{j}+2\mathbf{k}$

What we want is $\int_{disk} |\mathbf{r}_u \times \mathbf{r}_v| \,dA=\int_{disk} \sqrt{2u^2+2v^2+4}\,dA$

Now switching to polar works well:

$\int_0^{2\pi}\int_0^1 s\sqrt{2s^2+4}\,ds\,d\phi$
Title: Re: Vector Calculus
Post by: QuantumJG on November 04, 2010, 03:19:56 pm: I realized now where I stuffed up.

I did basically what you did, but after that to simplify things I let α = s² + 2 but didn't change my terminals!
Title: Re: Vector Calculus
Post by: QuantumJG on November 06, 2010, 06:39:55 pm: Another question:

F = (x-y,y-z,z-x)

Evaluate the surface integral of F if the surface S is the closed region:

x² + y² $\le$ z $\le$ 1 & x $\ge$ 0

Using the divergence theorem my answer was $\dfrac{3 \pi}{2}$ whereas the answer is $\dfrac{3 \pi}{4}$
Title: Re: Vector Calculus
Post by: /0 on November 06, 2010, 08:14:04 pm: In cylindrical coordinates, $s^2=x^2+y^2$ ,

$\int_0^1\int_{s^2}^1\int_0^\pi 3s\,d\phi\,dz\,ds$
Title: Re: Vector Calculus
Post by: QuantumJG on November 06, 2010, 09:22:58 pm: Ok now I get it. I stuffed up my terminals for z. Thanks /0.