DFranklin
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Insparato:This may be an artifact of your LaTeXing more than anything else, but I'm really struggling to follow what you're doing with each of the substitutions. Mainly because you don't actually link the integral before the substitution with the one after. It's pretty obvious what the "before and after" must be for the first substitution, but in the 2nd one I really can't tell.

If it's just the LaTeX, don't worry about it, but in the actual exam, try to make it clearer. If nothing else, start off by writing I = \int \text{stuff} \,dx and then later you can write I = \int \text{different stuff}\, dy. Makes the examiner's life a lot easier. And a happy examiner is a generous examiner...

As far as the end (guessed) bit goes, an important observation is \sin(\pi-x) = \sin(x).
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Rabite
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I'm a bit behind, but I did that last bit differently still! (In that ancient STEP I question...)
1+\sin x = \sin^2 ½x + 2\sin ½x \cos ½x + \cos^2 ½x = (\sin ½x+\cos ½x)^2
and of course 1+\cos x = 2\cos^2 ½x
So by log rules, the integral becomes
\int 2 \ln(\frac{sin ½x+\cos ½x}{\sqrt{2} \cos ½x})
And dividing through by cos and taking the sqrt2 outside (as a -ln2) via the log(a/b) rule gives your 1+tan½x, and so it's been rewritten as the problem a few inches above it.
---

As for the current challenge...Perhaps sin2t = 2sintcost?
That would certainly evaluate the integral (ln2 + lnsint + lncost which means I=-pi ln2/2 ?) but only given that result...hm.
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insparato
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#83
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(Original post by DFranklin)
Insparato:This may be an artifact of your LaTeXing more than anything else, but I'm really struggling to follow what you're doing with each of the substitutions. Mainly because you don't actually link the integral before the substitution with the one after. It's pretty obvious what the "before and after" must be for the first substitution, but in the 2nd one I really can't tell.

If it's just the LaTeX, don't worry about it, but in the actual exam, try to make it clearer. If nothing else, start off by writing I = \int \text{stuff} \,dx and then later you can write I = \int \text{different stuff}\, dy. Makes the examiner's life a lot easier. And a happy examiner is a generous examiner...

As far as the end (guessed) bit goes, an important observation is \sin(\pi-x) = \sin(x).
Sorry, hopefully ive made it more clearer now. I dont know what ive done is correct at the end. I was thinking because Sin2x is shortened so that it intersects the x axis at pi/2. That twice the area between 0 and pi/4 is the same as the area between the limits of 0 and pi/2. Even if theres a ln infront.

I do try to make it as clear as possible in exams. Although i think i can see that slipping after June considering i probably wont be taking any formal exams ever again.
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Rabite
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Oh wait I think I did it.
I had some integral from 0 to pie, but you can split that up to an integral over (0,½pi) and (½pi,pi) and use u=pi-x on the second bit.
Let me just make sure I'm not missing anything though...

--
Let
\displaystyle I = \int _0 ^{½ \pi} \sin 2x dx
If u=pi-2x then the limits are (top) 0 and (bottom) pi
du = -2dx so...
\displaystyle  I = -½ \int _\pi ^0 {\sin{(\pi - u)}} du
Sorta gave the game away, but I think I'd have gotten it - sin(pi-u)=sinu.
\displaystyle  = ½ \int _0 ^{\pi} {\sin{u}} du
\displaystyle  = ½ \int _0 ^{½ \pi} \sin{u} du + ½ \int _{½\pi} ^{\pi} \sin{u} du
And for the second bit there, let's try
t = pi-u
So the limits are (top) 0 and (bottom) ½pi
dt = -du so we can just flip the limits upside down

\displaystyle  I = ½ \int _0 ^{½ \pi} \sin {u} du + ½ \int _{0} ^{½ \pi} \sin {t}dt

Which are both the same thing, so the half goes and we have what we want.
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insparato
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Oh well it was worth a try .
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ad absurdum
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STEP I question 9

Let (x,y) be a point on the cannon-ball's trajectory, then:
x=ut\cos\alpha
y=ut\sin\alpha - \frac{1}{2}gt^2

Now if the distance from the origin is \sqrt{x^2 + h^2}, then we have the point (x,h), so writing the x and y coordinates with the trig function the subject:
\cos\alpha = \frac{x}{ut}
\sin\alpha = \frac{h + \frac{1}{2}gt^2}{ut}
But \cos^2\alpha + \sin^2\alpha = 1, so:
x^2 + h^2 + hgt^2 + \frac{1}{4}g^2t^4 = u^2t^2
\frac{1}{4}g^2t^4 - (u^2-gh)t^2 + h^2 + x^2 = 0, as required.

Since this equation arose from considering a point we assumed to be on the trajectory, then there has to exist a real value of t at which the cannon-ball is at this point. For a real value of t, we need a real value of t^2, so the solutions to the quadratic must be real, so using the discriminant:
u^4 - 2ghu^2 + g^2h^2 - g^2(h^2+x^2) \geq 0
u^2(u^2-2gh) \geq g^2x^2
x \leq \frac{u\sqrt{u^2 -2gh}}{g}, as required.

Okay, now we need to show that there is an angle of firing such that we have  x= \frac{u\sqrt{u^2 - 2gh}}{g} when y=h, so going back to the equations for x and y we get:
x=ut\cos\alpha \Rightarrow t=\frac{\sqrt{u^2 - 2gh}}{g\cos\alpha}
and h=ut\sin\alpha - \frac{1}{2}gt^2
substitute in the time from the x into the y:
\frac{g(u^2-2gh)}{2g^2} \sec^2\alpha - \frac{u\sqrt{u^2-2gh}}{g}\tan\alpha + h=0
(u^2-2gh)\tan^2\alpha - 2u\sqrt{u^2-2gh}\tan\alpha + u^2 = 0
For there to exist an angle of firing as described, we again need real roots to this quadratic, 4u^2(u^2-2gh) - 4u^2(u^2-2gh)\geq 0, which is of course true (and it really is an angle of firing).
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generalebriety
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Are these going on the wiki? If not, does anyone think it might be a nice idea to put them there?
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eponymous
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Step I Q1


Since all planes are at the same angle \theta to the horizontal, the roof ridge is central.

The height (h) of the roof (see image 1) is equal to p\tan\theta. Therefore x (see diagram) = \frac{h}{\tan\theta} = \frac{p\tan\theta}{\tan\theta} = p. Since the ridge is central, this will be the same at the other end of the roof.

So R (ridge length) = 2q - 2p, since \theta is not present in this expression, the ridge length is independent of \theta.

Volume of the roof:

Slicing up the roof, we can join the two ends to make a pyramid and leave a prism. (see image 2)

The volume of a pyramid = \frac{1}{3} \times area of base \times height(h)

Area of base =  p^{2}

So its volume = \frac{1}{3} \times p^{2} \times p\tan\theta = \frac{p^{3}}{3} \tan\theta.

Volume of a prism = area of face \times length

Area of face = \frac{1}{2} \times 2p \times h = p^{2} \tan\theta.

So the volume of the prism = p^{2} \tan\theta \times (2q - 2p) = 2p^{2}(q - p)\tan\theta

So the total volume of the roof = volume of prism + volume of pyramid = \frac{p^{3}}{3} \tan\theta + 2p^{2} (q - p)\tan\theta = \frac{p^{2}}{3} (6q - 5p) \tan\theta.



there probably is a much simpler way to do this

EDIT: actually now i'm not sure, should theta be in the volume?

*surface area is on its way *
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nota bene
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STEP III Question 8


\begin{pmatrix}z_1&z_2\\-z_2^*&z_1^*\end{pmatrix}\begin{pmatrix}z_3&z_4\\-z_4^*&z_3^*\end{pmatrix}=\begin{pmatrix}z_1z_3-z_2z_4^*&z_1z_4+z_2z_3^*\\-z_2^*z_3-z_1^*z_4^*&-z_2^*z_4+z_1^*z_3^*\end{pmatrix}
The determinant of this matrix is
(z_1z_3-z_2z_4^*)(z_1^*z_3^*-z_2^*z_4)-(z_1z_4+z_2z_3^*)(-z_2^*z_3-z_1^*z_4^*)= \newline z_1z_1^*z_3z_3^*-z_1z_3z_2^*z_4-z_2z_4^*z_1^*z_3^*+z_2z_2^*z_4z_4^*+z_1z_4z_2^*z_3+z_1z_1^*z_4z_4^*+z_2z_2^*z_3z_3^*+z_1^*z_2z_3^*z_4^* = \newline z_1z_1^*(z_3z_3^*+z_4z_4^*)+z_2z_2^*(z_3z_3^*+z_4z_4^*)= \newline (z_1z_1^*+z_2z_2^*)(z_3z_3^*+z_4z_4^*)=(|z_1|^2+|z_2|^2)(|z_3|^2+|z_4|^2)

(In hindsight it would have been less algebra to use |AB|=|A||B|)

Now, chose e.g.


z_1=p+qi\newline z_2=r+si \newline z_3=a+bi \newline\text{and} \newline z_4=c+di
and the determinant of the matrix product will be the desired LHS, i.e.
(a^2+b^2+c^2+d^2)(p^2+q^2+r^2+s^2).

(Provided z_1\not=-z_2 \text{and} z_3\not=-z_4, because then there is a zero-determinant, which we don't want)

Now, to get the RHS of the equality we are to prove, take a look at the matrix \begin{pmatrix}z_1z_3-z_2z_4^*&z_1z_4+z_2z_3^*\\-z_2^*z_3-z_1^*z_4^*&-z_2^*z_4+z_1^*z_3^*\end{pmatrix}again.
You can see that if we let w_1=z_1z_3-z_2z_4^* and w_2=z_1z_4+z_2z_3^* we have:
'
\begin{pmatrix}w_1& w_2\\-w_2^*&w_1^*\end{pmatrix}
This matrix also has a real determinant (i.e. |w_1|^2+|w_2|^2).
Now, if we set w_1=L_1+iL_2 and w_2=L_3+iL_4 this means the matrix has the determinant L_1^2+L_2^2+L_3^2+L_4^2, which is the RHS in the equality.


Now to the messy algebra of determining L_1, L_2, L_3 and L_4.

We have L_1+iL_2=z_1z_3-z_2z_4^* and L_1-iL_2=z_1^*z_3^*-z_2^*z_4

Therefore follows that 2L_1=z_1z_3-z_2z_4^*+z_1^*z_3^*-z_2^*z_4
So  L_1=\frac{1}{2}((p+qi)(a+bi)-(r+si)(c-di)+(p-qi)(a-bi)-(r-si)(c+di))=\frac{1}{2}(2ap-2rc-2sd-2bq)=ap-bq-cr-ds


L_2=-i(z_1z_3-z_2z_4^*-L_1)
So L_2=-i((p+qi)(a+bi)-(r+si)(c-di)-ap+bq+cr+ds)=-i(aqi+bip+rdi-sic)=aq+bp-cs+dr


We also have L_3+iL_4=z_1z_4+z_2z_3^* and L_3-iL_4=z_1^*z_4^*+z_2^*z_3
Therefore follows that
L_3=\frac{1}{2}((p+qi)(c+di)+(r+si)(a-bi)+(p-qi)(c-di)+(r-si)(a+bi)=\frac{1}{2}(2pc+2ar+2bs-2qd)=pc+ar+bs-qd


L_4=-i(z_1z_4-z_2z_3^*-L_3)
So
L_4=-i((p+qi)(c+di)+(r+si)(a-bi)-pc-ar-bs+qd)=-i(pid+qic-rbi+asi)=pd+qc-rb+as



I'd be grateful if someone checked my interpretation of "linear in a, b, c and d and also linear in p, q, r and s" (David?)


Overall a surprisingly nice question I think (latexing it was pain and there is likely a few silly things:/).
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DFranklin
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(Original post by nota bene)
I'd be grateful if someone checked my interpretation of "linear in a, b, c and d and also linear in p, q, r and s" (David?)
Yes that's fine.

Overall a surprisingly nice question I think (latexing it was pain and there is likely a few silly things:/).
It's nothing but an algebra grind really - there's nothing deep there at all.
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brianeverit
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Paper II number 12.
I am not too sure about the last part but I think most of it is correct.

Solution by Geriatric
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brianeverit
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1994 Paper II numbers 7 and 8
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brianeverit
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1994 Paper II numbers 12,13 and 14
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DFranklin
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I've only been glancing at your answers, and up 'til now I've not seen any mistakes I could spot on cursory inspection.

But with 1994 STEP II, Q14, I think you need to be a little more careful at the end: just because m = 2 gives better results than m=1 and m=3 doesn't prove m=2 is optimum.
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brianeverit
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(Original post by khaixiang)
STEP III, Question 2

2(i)
\displaystyle \\x^4+10x^3+26x^2+10x+1=0\\x^2+10x+26+\frac{10}{x}+\frac{1}{x^2}=0\\ (x+\frac{1}{x})^2-2+10y+26=0\\y^2+10y+24=0\\(y+6)(y+4)=0\\x+\frac{1}{x}=-4\\x=-2\pm\sqrt{3}\\x+\frac{1}{x}=-6\\x=-3\pm\sqrt{10}




2(ii)
\displaystyle \\x^4+x^3-10x^2-4x+16=0\\x^2+x-10-\frac{4}{x}+\frac{16}{x^2}=0\\ \text{ set } y=x-\frac{4}{x}\\(x-\frac{4}{x})^2+8+(x-\frac{4}{x})-10=0\\y^2+y-2=0\\(y-1)(y+2)=0\\x-\frac{4}{x}=1\\x=\frac{1\pm\sqrt{17}}{2}\\x-\frac{4}{x}=-2\\x=-1\pm\sqrt{5}

Shouldn't the last line of (i) be -3\pm\2sqrt(2)?
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brianeverit
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(Original post by nota bene)
STEP III Question 8


\begin{pmatrix}z_1&z_2\\-z_2^*&z_1^*\end{pmatrix}\begin{pmatrix}z_3&z_4\\-z_4^*&z_3^*\end{pmatrix}=\begin{pmatrix}z_1z_3-z_2z_4^*&z_1z_4+z_2z_3^*\\-z_2^*z_3-z_1^*z_4^*&-z_2^*z_4+z_1^*z_3^*\end{pmatrix}
The determinant of this matrix is
(z_1z_3-z_2z_4^*)(z_1^*z_3^*-z_2^*z_4)-(z_1z_4+z_2z_3^*)(-z_2^*z_3-z_1^*z_4^*)= \newline z_1z_1^*z_3z_3^*-z_1z_3z_2^*z_4-z_2z_4^*z_1^*z_3^*+z_2z_2^*z_4z_4^*+z_1z_4z_2^*z_3+z_1z_1^*z_4z_4^*+z_2z_2^*z_3z_3^*+z_1^*z_2z_3^*z_4^* = \newline z_1z_1^*(z_3z_3^*+z_4z_4^*)+z_2z_2^*(z_3z_3^*+z_4z_4^*)= \newline (z_1z_1^*+z_2z_2^*)(z_3z_3^*+z_4z_4^*)=(|z_1|^2+|z_2|^2)(|z_3|^2+|z_4|^2)

(In hindsight it would have been less algebra to use |AB|=|A||B|)

Now, chose e.g.


z_1=p+qi\newline z_2=r+si \newline z_3=a+bi \newline\text{and} \newline z_4=c+di
and the determinant of the matrix product will be the desired LHS, i.e.
(a^2+b^2+c^2+d^2)(p^2+q^2+r^2+s^2).

(Provided z_1\not=-z_2 \text{and} z_3\not=-z_4, because then there is a zero-determinant, which we don't want)

Now, to get the RHS of the equality we are to prove, take a look at the matrix \begin{pmatrix}z_1z_3-z_2z_4^*&z_1z_4+z_2z_3^*\\-z_2^*z_3-z_1^*z_4^*&-z_2^*z_4+z_1^*z_3^*\end{pmatrix}again.
You can see that if we let w_1=z_1z_3-z_2z_4^* and w_2=z_1z_4+z_2z_3^* we have:
'
\begin{pmatrix}w_1& w_2\\-w_2^*&w_1^*\end{pmatrix}
This matrix also has a real determinant (i.e. |w_1|^2+|w_2|^2).
Now, if we set w_1=L_1+iL_2 and w_2=L_3+iL_4 this means the matrix has the determinant L_1^2+L_2^2+L_3^2+L_4^2, which is the RHS in the equality.


Now to the messy algebra of determining L_1, L_2, L_3 and L_4.

We have L_1+iL_2=z_1z_3-z_2z_4^* and L_1-iL_2=z_1^*z_3^*-z_2^*z_4

Therefore follows that 2L_1=z_1z_3-z_2z_4^*+z_1^*z_3^*-z_2^*z_4
So  L_1=\frac{1}{2}((p+qi)(a+bi)-(r+si)(c-di)+(p-qi)(a-bi)-(r-si)(c+di))=\frac{1}{2}(2ap-2rc-2sd-2bq)=ap-bq-cr-ds


L_2=-i(z_1z_3-z_2z_4^*-L_1)
So L_2=-i((p+qi)(a+bi)-(r+si)(c-di)-ap+bq+cr+ds)=-i(aqi+bip+rdi-sic)=aq+bp-cs+dr


We also have L_3+iL_4=z_1z_4+z_2z_3^* and L_3-iL_4=z_1^*z_4^*+z_2^*z_3
Therefore follows that
L_3=\frac{1}{2}((p+qi)(c+di)+(r+si)(a-bi)+(p-qi)(c-di)+(r-si)(a+bi)=\frac{1}{2}(2pc+2ar+2bs-2qd)=pc+ar+bs-qd


L_4=-i(z_1z_4-z_2z_3^*-L_3)
So
L_4=-i((p+qi)(c+di)+(r+si)(a-bi)-pc-ar-bs+qd)=-i(pid+qic-rbi+asi)=pd+qc-rb+as



I'd be grateful if someone checked my interpretation of "linear in a, b, c and d and also linear in p, q, r and s" (David?)


Overall a surprisingly nice question I think (latexing it was pain and there is likely a few silly things:/).
To find the L's why not simply equate real and imaginary parts instead of solving simultaneous equations.
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brianeverit
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(Original post by *bobo*)
if the particles are perfectly elastic then no energy should be lost in the collision, hence the total energy of the particles must be equal to 2ghk(1 + alpha^2). This was deduced by forming the energy equations of each particle and the constants.
This is hardly a description of the subsequent motion of the particles.
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squeezebox
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#98
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(Original post by brianeverit)
This is hardly a description of the subsequent motion of the particles.
It is good that you are pointing out mistakes in some of the solutions, but if you are going to you could atleast suggest improvement. This comment is not the slightest bit helpful. :rolleyes:
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brianeverit
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(Original post by squeezebox)
It is good that you are pointing out mistakes in some of the solutions, but if you are going to you could atleast suggest improvement. This comment is not the slightest bit helpful. :rolleyes:
Sorry about that. The point is that the total energy is not conserved. It is not a "perfectly" elastic collision so the coefficient of restitution needs to be taken into account I assume. I think thet motion will continue with collisions always occurring at V until the particles finally come to rest, but I don't know how much explanation the examiners would require.
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Swayum
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Need help with STEP I question 5.

I've found that the coordinates of R are ((p-q)/2, (p^2 - q^2)/2)

So x = (p-q)/2
y = (p-q)(p+q)/2 = x(p+q)

Also, using the information about POQ being a right angle, I found pq = -1 (I won't bother posting my working as it won't really help anything - it's just using Pythagoras several times to find the distance PQ). But I can't deduce from this what the locus of R would be.

For the second part, I got

T (x, y)

x = (p+q)/2
y = 2pq

xy = pq(p+q) = -(p+q)

And again I don't know what to do from here.
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