Dave

The problem is one of the vastness of both time and space. The chances of aliens being intelligent enough to transmit radio waves or try to listen in on them from space in our timeframe and from somewhere close enough that we can hear them are fairly minimal.

I love that hat

twas me. I read the post through and clearly it was a load of inane drivel so it was removed.

eh? Money does have an influence on their decisions to perform some disciplinary actions, but quite frankly a university's reputation is never going to be something you can buy. They're not stupid enough to start randomly accepting people in order of the size of their wallet. As MrL pointed out, if they start accepting people because their parents are completely loaded, their reputation is going to go down the toilet rather quickly. If a university has a bad reputation, then people aren't going to pay their fees in the first place and hence it'll just degrade from there.

Also, moved this to geometry because it's more suitable there.

For all we know it might've been God messing around with some fireworks.

I can't quite believe just how quickly this thread deteriorated I'd close it but someone might actually have something sensible to say.

From what I gather, these are very similar to polar co-ordinates. They have an angle between the x and y axes (theta), a distance to the point (rho, I use r myself) and another angle between the xy plane and the elevation of the point in question (phi). You can specify any 3d point in space like this.

Absolutely. They rock

Hmm, I don't know. I may be wrong, I've never touched this subject before

He writes strange books.

If you really don't want to spend any money, you can get a free domain from eu.org, free DNS hosting from Granite Canyon or ZoneEdit, free hosting from somewhere else (there's loads of places, but you might need to pay for somewhere with php capabilities) and a free bulletin board system such as phpBB or Invision Board. Failing that you could try somewhere like ezboard - which sucks really.

I would say the answer in the book is probably wrong, yes. If you differentiate the answer that they give, I don't think you get the equation quoted in the question because of that c that's knocking about.

Hmm, it appears that you may be wrong on this one. http://mathworld.wolfram.com/ExactFirst-OrderOrdinaryDifferentialEquation.html I have to say I've not heard of an exact first order ordinary differential equation, but apparently it is p(x,y)dx + q(x,y)dy = 0 => p(x,y) = -2xy, q(x,y) = y^2 - x^2 The partial derivatives of each of these are -2x, and hence it's exact. Apparently the general solution is this, and the answer comes out as: -x^2 * y + y^3/3 = c where c is a constant. Well done Aleph-Null, this one had been bothering me for some time

The rate of change of area isn't constant - it depends on the width of the rectangle. The answer quoted is only for a particular instant in time.

I think I can see where you're coming from. In the question it tells you that "A diagonal of the rectangle is the diameter of the circle." and also that the rectangle is inscribed in the circle (i.e. always in it). So basically, when the width of the rectangle decreases, it's going to mean that the height will increase to ensure that it is still inscribed. As for the other question about the rate of change of height - it doesn't matter. We basically have one important formula, which is the area of the rectangle, A = w*h. We don't want the h in there, so we're going to have to find a formula to get it. We know that the rectangle is always going to be touching the circle so we have one length of 5 inches from the circle's circumference to its centre, O. From O to the upper line of the rectangle the height is going to be (h/2). From the upper line to the circle's circumference we have a width (w/2). So by using Pythagoras's Theorem, we can find a formula of h in terms of w and substitute it into the formula for A, differentiate and do all the other stuff. If you need more help, just ask.

Hmm. I did this question and got their answer. I drew the rectangle inscribed in the circle and called the width w. If you do a bit of trig etc (which I can help you with if you want - it's not all that hard) then you get the area of the rectangle (which I called A) to be: A = 2w*sqrt(25-(w^2/4)) = w * sqrt(100 - w^2) Now by differentiating, dA/dw = sqrt(100 - w^2) - w^2/(sqrt(100 - w^2)) But dA/dw = dA/dt * dt/dw (by the chain rule), and we know at this particular instant in time that dw/dt = -2 and w = 6, so by substituting in, we have: dA/dt = -2 * (8 - 9/2) = -7 sq. inches per sec I'm not entirely sure what you've done tbh, but it may have something to do with assuming starting values and whatnot. You definately need to differentiate - perhaps you're getting confused with the unit length and the rate of change of the unit length. I don't really know

Just separate the variables out: int((1+2y^2)/y dy) = int(cos(x) dx) and solve it that way.

It was split into two papers, the main one was about 135 pages from what I remember, and the smaller one wasn't very long, but the major one depended upon it. As for the rest of the thread; this is probably why I don't read pseudoscience apart from when I'm bored :\

Sorry for the wait (I'm sure you all missed it really), but I decided to do another question for this week. It's quite good fun, but you'll definately need to think about it. Enjoy! mathsprobs3.pdf

someone definately needs to get out more

I've got one logging tapes down at a company in Oxford. Boring as hell but it pays well.

what's that, sorry?

My name's dave. Occasionally (like on IRC) i'll use davem because my surname begins with 'm'. Just shows how creative I am

Too much haggis for my liking - and then there's the deep fried pizzas and mars bars and all that stuff