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Hathlan
I usually have my own protractor if I suspect I'm going to need one. (Heehee).


I had a feeling this was coming.
Here's something interesting I discovered a few days ago. (But it's probably in the lecture notes I haven't revised...)

Suppose you have a quadric surface S described by a quadratic form: Let S={xR3:xTAx=k}S = \{ \mathbf{x} \in \mathbb{R}^3 : \mathbf{x}^T \mathbf{A} \mathbf{x} = k \}, where A\mathbf{A} is a constant symmetric matrix and k is a constant. Show that Ax\mathbf{A}\mathbf{x} is normal to S at every point xS\mathbf{x} \in S.

Spoiler

Zhen, surely you just take the grad of it and it drops out in 1 line?
Yes, but it was interesting nonetheless. I didn't expect
1. that it was so simple, and
2. that it would give an interpretation of A as something other than a matrix of coefficients.
Simplicity
How would you actually solve the equation using a matrix?

Also, how you solve a equation like
x2y2=dx^2-y^2=d
and
x+y=ex+y=e
lets say e=1 and d=2

As I got a similar question in FP4 and couldn't anwser it as it was not in the books or the other pasts papers, I really don't want to read the uni book I got on matrices to find out.

( - of 2^2)
Which of these is correct and why:

ddx(dydx)=ddθ(dθdxdydx) \frac{d}{dx}(\frac{dy}{dx})=\frac{d}{d\theta}( \frac{d\theta}{dx}\frac{dy}{dx})

ddx(dydx)=dθdxddθ(dydx) \frac{d}{dx}(\frac{dy}{dx})=\frac{d\theta}{dx} \frac{d}{d\theta}(\frac{dy}{dx})

? I think it's the latter but I was just doing a question and did the top one, then realised I wasn't actually sure which is correct.
Latter. Easy to check: e.g. take y = x^2, t = x^2.

d/dx (dy/dx) = 2.

d/dt(dt/dx dy/dx) = d/dt (2x.2x) = d/dt(4t) = 4.

dt/dx d/dt (dy/dx) = 2x d/dt (2x) = 2x d/dt 2 \sqrt{t} = 2x / sqrt(t) = 2x / x = 2.

I sometimes resort to checks like this when I'm rusty about how bits of A-level calculus work...
The bottom one is just a clear example of the chain rule... isnt it ?

Sort of like this...
[ddydxdx=dθdxddydxdθ \displaystyle [\frac{d\frac{dy}{dx}}{dx} = \frac{d\theta}{dx} \frac{d\frac{dy}{dx}}{d\theta}
8 Horizontal
The bottom one is just a clear example of the chain rule... isnt it ?Yes. But when you've got stuff with a few partial derivatives mixed in it can get a bit more tricky. (Or integration, where I am more rusty).

My point wasn't "this is difficult enough that you really need to check", but more "when you have a problem like this and you don't know which is right, then there's nothing stopping you cooking up an example to test the methods".

After all, in an exam, you can't go back to a textbook to check which method to use.
DFranklin

I sometimes resort to checks like this when I'm rusty about how bits of A-level calculus work...


This is in fact, what I did, it was more the why that I was interested in.
8 Horizontal
The bottom one is just a clear example of the chain rule... isnt it ?

Sort of like this...
ddx(dydx)=dθdx×(dydx)dθ \displaystyle \frac{d}{dx}(\frac{dy}{dx})=\frac{d\theta}{dx} \times \frac{(\frac{dy}{dx})}{d\theta}


You're missing a d, but thanks, that helped. Basically you just consider dydx \frac{dy}{dx} as one symbol and it's obvious.
toasted-lion
You're missing a d, but thanks, that helped. Basically you just consider dydx \frac{dy}{dx} as one symbol and it's obvious.


Oops :redface: ... I'll edit it, thanks
8 Horizontal
Oops :redface: ... I'll edit it, thanks


Personally, I'd write: ddydxdx=dθdxddydxdθ \frac{d\frac{dy}{dx}}{dx} = \frac{d\theta}{dx} \frac{d\frac{dy}{dx}}{d\theta}
Reply 1773
how do i join this society :biggrin:?
Reply 1774
dluineo
how do i join this society :biggrin:?


Post an interesting problem in this thread and then request to join

http://www.thestudentroom.co.uk/socs.php?do=join&socid=136
Reply 1775
Hi, I was working at a Step Paper at home, and was attempting to solve the following inequality and came across some problems:

[br]sinθ+1cosθ1[br][br]0θ2π[br][br]cosθ0[br][br][br]\frac{\sin\theta +1}{\cos\theta}\leq1[br][br]0\leq \theta \leq2\pi[br][br]\cos \theta\neq0[br][br]

I can get this far

[br]sinθ+1cosθ1[br][br]tanθ+secθ1[br][br]\frac{sin\theta+1}{cos\theta}\leq 1[br][br]tan\theta+sec\theta\leq 1[br]

I can then draw the graph of this ok in the given region, the only thing is it seems to be somewhat undefined at 3pi/2 rads, where tan (3pi/2) = +infinity and sec (3pi/2)=-infinity - do these cancel out, I just feel a bit dubious about it and want to check with someone - my teacher at school isn't too sure either


P.S. I'm new on here and would like to join the maths society
Reply 1776
jbeacom600
Hi, I was working at a Step Paper at home, and was attempting to solve the following inequality and came across some problems:

[br]sinθ+1cosθ1[br][br]0θ2π[br][br]cosθ0[br][br][br]\frac{\sin\theta +1}{\cos\theta}\leq1[br][br]0\leq \theta \leq2\pi[br][br]\cos \theta\neq0[br][br]

I can get this far

[br]sinθ+1cosθ1[br][br]tanθ+secθ1[br][br]\frac{sin\theta+1}{cos\theta}\leq 1[br][br]tan\theta+sec\theta\leq 1[br]

I can then draw the graph of this ok in the given region, the only thing is it seems to be somewhat undefined at 3pi/2 rads, where tan (3pi/2) = +infinity and sec (3pi/2)=-infinity - do these cancel out, I just feel a bit dubious about it and want to check with someone - my teacher at school isn't too sure either


P.S. I'm new on here and would like to join the maths society


There are several ways to approach this problem:

1. Half-tangent angles
2. Considering cases on the sign of cos x (to multiply it up)
Reply 1777
Hopefully this hasn't been posted before. It's stolen off a supervisor, but I think it's reasonably well known.

By considering an integral or otherwise (there exists at least one nice otherwise), prove that a rectangle tiled with rectangles of at least one integer side each must also have at least one integer side.

Integral hint

Otherwise hint

Simplicity
Also, how you solve a equation like

and

lets say e=1 and d=2
x2-y2=(x+y)(x-y)
Or substitute y=e-x
Here is another identity I came across doing some probability work: Show that r=kn[(nk)!n!(r1)!(rk)!]=1k\displaystyle \sum_{r=k}^{n} \left[ \frac{(n-k)!}{n!} \frac{(r-1)!}{(r-k)!} \right] = \frac{1}{k}.

(Again, I have not figured out a way of doing the sum directly.)
Zhen Lin
Here is another identity I came across doing some probability work: Show that r=kn[(nk)!n!(r1)!(rk)!]=1k\displaystyle \sum_{r=k}^{n} \left[ \frac{(n-k)!}{n!} \frac{(r-1)!}{(r-k)!} \right] = \frac{1}{k}.

(Again, I have not figured out a way of doing the sum directly.)

Spoiler

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