# Stupid Circle Problem

1. Mar 1, 2004

### Claire84

We're doing a bit about circles in Pure Maths at the mo and everything's fine about intersections etc until today the lecturer rammed up an easier eqt to tell if a line cuts a circle at 2 distint points or not. We had to show that the line 3y=x+5 cut the circle x^2 + y^2 -6x - 2y -15 in 2 distinct places. From before we found out that the centre of the circle was at (3,1) and the radius was five. It was then said tha the line would cut the circle in 2 distint places if its distance from the centre was less tha five. All that was fine until it was put up that this distance was the modulud of (-3 +3 -5)/(sqrt(3^2 + 1^2) whivh was clearly less than 5 so it cut it in 2 distinct places. How was this distance worked out? I asked some of the other people in the lecture about it and they were as confused as me. I mean it's obviously not rocket scientist stuff since the lecturer didn't even feel the need to tell us what was going on, so can someone please help? Thanks!

Also, what forum should I post in if I want to ask about solving a heat transfer problem using he Fourier series? It's just about the general formula. Anyone here know much about it?

2. Mar 1, 2004

### matt grime

The lecturer has just written down the formula for the shortest distance between a line and a point. It can be deduced easily from the vector equation for a line.

The vector equation for the line is

L= u+vt

u,v fixed vectors t a variable/parameter.

let v=(v_1,v_2) then a normal to the line is the vector w = (v_2,-v_1)

now, the vector from any point x to a point on the line call it l is some combination of v and w - you go along the line some units of v then perpendicular to the line along w some distance. here's where the vector stuff you've been doing is useful.

the displacement vector l to x we donote lx,

lx = av+bw

for some a,b in R

obviously the nearest point on the line to x has a=o, note we don't need to find the point l at all cos we can kill the av off easily - dot with w!

so pick some point on the line, work out lx the vector and dot with w the number we get is

b|w|^2

the quantity we need is b|w| so divide the number you've found by |w| - which is sqrt(w.w) remember

3. Mar 1, 2004

### Tom Mattson

Staff Emeritus
Differential Equations

Yep. Post away!

4. Mar 1, 2004

### himanshu121

this is the basic formula
perpendicualr Distance from a point (p,q) to the line ax+by+c=0 is
$$D= \frac{ap+bq+c}{\sqrt{a^2+b^2}}$$

vector method is best though u can do it with geometry

Let(h,k) be the point where perpendicular meets from the point (p,q)

then it would lie on the given line + the angle b/w the line and the line joining the 2 points are perpendicular

so from the above two equation u can find p and q and hence the distance This process is cumbersome I recommend u use vectors to prove it
&
$$D= \frac{ap+bq+c}{\sqrt{a^2+b^2}}$$
to remeber it

5. Mar 3, 2004

### Claire84

Crikey, I think I need a brain transplant.

Anyway, as for the Fourier series with heat transfer, I've got a tutorial tomorrow so I'll ask the girl about it and if I've any other probs with it I'll be here grovelling for help. It's like I've got the basic grasp of it but looking at the notes here at the examples he gave us, there look to be some mistakes, but then again, that is probably just me!

One final thing. For Pure Maths this semester we're doing Calculus and Analysis. At the moment we're on Calculus but we're nearly finished and the lecturer is about to have kittens about the Analysis since he says it is horrid. Honestly, everyday he harks on aout how awful it is and he said today 'enjoy your last days of freedom before delving into the icy depths of Analysis'- HELP! Is it really tht bad? Can someone please prepare me for the worst here or give me the names of any good books? I've been finding Calculus okay and I don't want to have heart-falure when Analysis starts. Is our lecturer bein a drama queen?

6. Mar 3, 2004

### matt grime

Why the distaste at the answers you got?

I must say that himanshu's post is the exact opposite of what every mathematician I know would say - there is no reason to memorize that formula at all - it is easily derivable from simple vector manipulation and knowing about dot products of orthoganal vectors.

Your lecturer is one of those people that I occasionally sympathize with, and occasionally want to thottle.

Analysis, as opposed to calculus, buries itself under a mound of notation, and trying to remember it is hard. Sometimes this overload gets in the way of the simple ideas that it is formalizing.

Here's an example

a sequence a(n) of real numbers tends to a if, for all $$\epsilon >0\ \exists n_0(\epsilon)\in \mathbb{N}$$ such that for all $$n>n_0(\epsilon)$$ $$|a(n)-a| < \epsilon$$

Now that's the formal definition (actually in one course I was supervising the standard was to use even more abstract notation than that with even fewer words).

Given that defintion one proves lots of things that make it easy to decide when things tend to a limit, and tries to minimize all these letters flying around.

1. What it means.

a(n) tends to a as n goes to infinity means. Imagine drawing the real line and marking a on it. We want to say a(n) tends to a if *eventually* all the terms of a(n) are as close to a as we like. eventually means after some term in the sequence, where that term comes, the cut off point depends on just how close to a we want to be. So on that real line, given any small distance, $$\epsilon$$ we draw the interval that extends epsilon either side. What we mean by all the n_o stuff is that if we plotted all the a(n) on the real line, there comes some finite time, n_0(epsilon), such that every term after that is plotted inside that little espilon interval.

Finding these quantities is quite difficult, so we get clever

lets show 1/n tends to zero by some obvious results that you'll prove later:

1/n is decreasing and bounded below. it must converge to something.

1/n tends to x say.

1/n^2 is a subsequence of 1/n, so it must converge, and converge to the same thing x

but it's obvious that 1/n^2 must also converge to x^2

so x=x^2 and x=0 or 1. It can't be 1 so it must be 0.

see, not an epsilon in sight. all the results i used we proved using epsilons, and will be in your lectures, though.

7. Mar 3, 2004

### Claire84

LOL, the answers I got were fine- it's just such a stupid thing of me to ask considering the fact we'd only covered that 2 weeks ago!

Anyway, Analysis looks a bit complicated with all the weird notation. I also tend to prefer geometry so it'll probably give me a few headaches. The lecturer did say that it wasn't quite so horrific once you got used to it, so at least we did get a little optimism out of him.

8. Mar 3, 2004

### matt grime

oh experience is great at analysis. the first time i did it i was appalling at it, second course it seemed as easy as falling off a log

9. Mar 3, 2004

### Claire84

Crikey, what hope is there for us since this is our first time and it'll be about half of our second semester paper? Argh! I guess I'll just have to work harder at it than anything else in ordet to do okay. I'm all worried now!

10. Mar 4, 2004

### Claire84

Okay, since help was offered with the Fourier stuff, I just want to know how to integrate sin(nx)sinx and cos(nx)sinx within the limits of 0 and pi

Is there an easy method for this? I can't think how you could do it via substituation or integration be parts. Is there some other methosd of doing it. I'm trying to work these put for the an and bn components of the fourier series for the function f(x)= 0 when pi<x<0 and f(x) is sinx when 0<x<pi

Can anyone give me a bit of help? I've been okay qith the others questions I've done but this one has left me a bit stumped. Thanks!

11. Mar 4, 2004

### matt grime

set I(n,m) = integral(cosnxsinmx)dx

and integrate by parts twice to get a relation that I(n,m) satisfies, you'll see that it is always zero

alternatively, change of variables, x to x-pi/2 and use that sin is odd, cos is even .

matt

12. Mar 4, 2004

### Claire84

So is it zero for both the ones I mentioned then? Like what about for the sin(nx)sinx one? I understand for the one with cosine in it but not that one.

13. Mar 4, 2004

### Claire84

Okay, I've got the bn term to be zero and the an term is what they want, as is the ao term, but there seems to be 0.5 sinx thrown in somehow that they'e after. Do you have any idea where it should have come from? I've checked my calculations again but to no avail.

14. Mar 4, 2004

### matt grime

I think you just need to check everything carefully.

My I(N,m) thing wasn't what you wanted - that was for the integral from -pi to pi, sorry.

the coefficient of sinx is the number a_1 or b_1 depending on how you define it.

you cannot have all the coeffs of sin zero because the function isn't odd, similarly it isnt' even so it can't ony consist of cosines either

15. Mar 5, 2004

### Claire84

Thanks for your help. After about 2 hours of writing down rubbish in he library at uni this afternoon the penny finally dropped about the whole thing and I managed to get the answer (possibly a miracle occurred, I don't know! ;)). If I've any more probs I'll post them here but fingers crossed the rest of the h/work will be fine.

I've just finished my pure h/work but the final question has me a bit baffled because in lectures we've only been dealing with parabols where the focus is like (a,0) and the directrix is just y=k where k is a real number. However, for this question we've to find the eqt of the parabola whose focus is at the point (1,1) and whose directrix has the eqt x + y = 0

I've drawn it out so that I've chosen an arbitrary point (x', y') and then worked out the distance between this point and the focus to be the square root of (x' - 1)^2 + (y' - 1)^2

For working out the distance between the point and the directrix I've got the modulus of (x' + y')/sqrt2

I've then set these two equal to each other by the definition of the parabola but it doesn't look quite right and the eqt looks much bigger and complicated than it should be. Any help would be much-appreciated as to how I should have approached this. Thanks! :)

16. Mar 5, 2004

### matt grime

in all honesty i have never done a question like this, but I would suggest that the thing will look a bit strange, cos what you've done is rotate the usual parabola by 45 degrees (or something). you can apply the transformation back the other way to see if you get an equation that looks more familiar.