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STEP Maths I, II, III 1995 Solutions

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insparato
I would do this but i really havent got the time now. TeXing them all up would take time, time i dont have :frown:.


Yeah, I don't have any time right now either... Still have coursework (will be done next Friday though), but then I'll have to revise for May exams. Though after the exams I'll have all the time in the world so if it hasn't been done by then I might do it.
Step III, Q8:

Equation of normal to π\pi through x is x+λn\mathbf{x}+\lambda\mathbf{n}, distance is (xr)n|(\mathbf{x} -\mathbf{r}) \cdot \mathbf{n}|.

Consider a single circle C with center r lying in the plane xn=p\mathbf{x} \cdot \mathbf{n} = \mathrm p, and suppose C lies on the surface of a sphere S with center s, radius R.

Claim: s=r+λn\mathbf{s} = \mathbf{r} + \lambda \mathbf{n} for some λ\lambda. In other words, s must lie on the normal to the plane through r.

\I think it's arguable that this is "obvious" and doesn't need to be proved. I will prove it, but it's going to be a little tedious...

Put q=sr\mathbf{q} = \mathbf{s} - \mathbf{r}. Consider e=q(qn)n\mathbf{e} = \mathbf{q} - (\mathbf{q} \cdot \mathbf{n}) \mathbf{n}. Note
Unparseable latex formula:

\mathbf{e}\cdot\mathbf{n} = \mathbf{q}\cdot{\mathbf{n}-(\mathbf{q} \cdot \mathbf{n})\mathbf{n}\cdot\mathbf{n} = 0

. (e is basically what we get if we "remove" the part of q that's parallel to n).

Suppose that e is non-zero. We will (eventually!) show this gives us a contradiction. Let μ=e1\mu = |\mathbf{e}|^{-1}, so that μe|\mu\bf{e}| = 1. Then r±μe\bf{r}\pm\mu\bf{e} are both points on the circle C, so if they lie on the sphere S, they must both be the same distance from s.

Now s(r+μe)=qμe\bf{s} - (\bf{r}+\mu\bf{e}) = \bf{q} - \mu\bf{e}.
So s(r+μe)2=(qμe)(qμe)=q22μqe+μe2|\bf{s} - (\bf{r}+\mu\bf{e})|^2 = (\bf{q} - \mu\bf{e})\cdot(\bf{q} - \mu\bf{e})= |\bf{q}|^2 - 2\mu \bf{q}\cdot\bf{e} + |\mu\bf{e}|^2
Similarly, s(rμe)2=(q+μe)(q+μe)=q2+2μqe+μe2|\bf{s} - (\bf{r}-\mu\bf{e})|^2 = (\bf{q} + \mu\bf{e})\cdot(\bf{q} + \mu\bf{e})= |\bf{q}|^2 + 2\mu \bf{q}\cdot\bf{e} + |\mu\bf{e}|^2

So we must have q.e=0    qq(qn)2\bf{q}.\bf{e} = 0 \implies \bf{q}\cdot\bf{q}-(\bf{q}\cdot\bf{n})^2, which since n=1    q=κn|\bf{n}|=1 \implies \bf{q} = \kappa \bf{n} for some κ\kappa (either argue geometrically, or use the equality case of Cauchy-Schwartz). But then e=κn(κnn)n=0\bf{e} = \kappa \bf{n} - (\kappa \bf{n} \cdot \bf{n})\bf{n} = \bf{0}, giving our contradiction.

So e=0\bf{e} = 0, so q=(qn)n\bf{q} = (\bf{q}\cdot\bf{n})\bf{n} and so q\bf{q} is a multiple of n. Thus s=r+λn\mathbf{s} = \mathbf{r} + \lambda \mathbf{n} as claimed.

Finally, let x be any point on C. xn=rn, and so(xr)n=0\bf{x}\cdot\bf{n} = \bf{r}\cdot\bf{n}, \text{ and so} (\bf{x}-\bf{r})\cdot\bf{n} = 0. Then

Unparseable latex formula:

|\bf{s}-\bf{x}|^2 = |\bf{s}-\bf{r}-(\bf{x}-\bf{r})|^2 = |\lambda \bf{n} - (\bf{x}-\bf{r})|^2 \\[br]= \lambda^2 |\bf{n}|^2 -2 (\bf{x}-\bf{r})\cdot\bf{n}+|\bf{x}-\bf{r}|^2 = \lambda^2 + 1

(Since n is a unit vector, and x lies on a unit circle, center r so |x-r| = 1).

But x lies on the sphere, so sx|\bf{s}-\bf{x}| is just the radius of the sphere, R. So λ2+1=R\lambda^2+1 = R

So for S to contain both circles C1,C2C_1, C_2, we must have s=r1+λ1n1=r2+λ2n2\bf{s}=\bf{r_1}+\lambda_1 \bf{n_1} = \bf{r_2}+\lambda_2 \bf{n_2} for some real λ1,λ2\lambda_1, \lambda_2 satisfying λ12+1=λ22+1    λ1=±λ2\lambda_1^2+1 = \lambda_2^2+1 \implies \lambda_1 = \pm \lambda_2.

This proves that for such S, we can find λ\lambda with r1+λn1=r2±λn2\bf{r_1}+\lambda \bf{n_1} = \bf{r_2}\pm\lambda \bf{n_2} as desired.

But then (r1r2)=λn1±λ2n2 (\bf{r_1} - \bf{r_2}) = -\lambda \bf{n_1} \pm \lambda_2 \bf{n_2}. Since n1(n1×n2)=n2(n1×n2)=0\bf{n_1}\cdot(\bf{n_1} \times \bf{n_2}) = \bf{n_2}\cdot(\bf{n_1} \times \bf{n_2}) = 0 we have (r1r2)(n1×n2)=0(\bf{r_1} - \bf{r_2})\cdot(\bf{n_1} \times \bf{n_2}) = 0 as desired. Geometrically, this means that the line between the two circle centers has no component perpendicular to both plane normals; equivalently that it lies in the plane through the origin generated by the two plane normals.

Finally, if r1+λn1=r2+λn2\bf{r_1}+\lambda \bf{n_1} = \bf{r_2} + \lambda \bf{n_2}. Dotting with n1,n2\bf{n_1}, \bf{n_2} gives:

r1n1+λ=r2n1+λn2n1    p1r2n1=λ+λn2n1\bf{r_1}\cdot\bf{n_1} + \lambda = \bf{r_2}\cdot\bf{n_1} + \lambda\bf{n_2}\cdot\bf{n_1} \implies \mathrm{p_1} - \bf{r_2}\cdot\bf{n_1} = -\lambda + \lambda\bf{n_2}\cdot\bf{n_1}

r1n2+λn2n1=r2n2+λ    p2r1n2=λλn2n1\bf{r_1}\cdot\bf{n_2} + \lambda\bf{n_2}\cdot\bf{n_1} = \bf{r_2}\cdot\bf{n_2} + \lambda \implies \mathrm{p_2} - \bf{r_1}\cdot\bf{n_2} = \lambda - \lambda\bf{n_2}\cdot\bf{n_1}

If instead r1+λn1=r2λn2\bf{r_1}+\lambda \bf{n_1} = \bf{r_2} - \lambda \bf{n_2}. Dotting with n1,n2\bf{n_1}, \bf{n_2} gives:

r1n1+λ=r2n1λn2n1    p1r2n1=λλn2n1\bf{r_1}\cdot\bf{n_1} + \lambda = \bf{r_2}\cdot\bf{n_1} - \lambda\bf{n_2}\cdot\bf{n_1} \implies \mathrm{p_1} - \bf{r_2}\cdot\bf{n_1} = -\lambda - \lambda\bf{n_2}\cdot\bf{n_1}

r1n2+λn2n1=r2n2λ    p2r1n2=λ+λn2n1\bf{r_1}\cdot\bf{n_2} + \lambda\bf{n_2}\cdot\bf{n_1} = \bf{r_2}\cdot\bf{n_2} - \lambda \implies \mathrm{p_2} - \bf{r_1}\cdot\bf{n_2} = \lambda + \lambda\bf{n_2}\cdot\bf{n_1}

In either case we have (p1r2n1)2=(p2r1n2)2(\mathrm{p_1} - \bf{r_2}\cdot\bf{n_1})^2 = (\mathrm{p_2} - \bf{r_1}\cdot\bf{n_2})^2 as desired.

Geometrically, this says that this distance of r1\bf{r_1} from the plane π2\pi_2 must equal the distance of r2\bf{r_2} from the plane π1\pi_1

Comment: Doing questions directly into Latex rather than pen/paper is always tricky; I'd done some roughing out on paper, but there was more work still to be done than I thought, and I hadn't realised just how hard Latex is for questions like this involving vectors. It was not a good plan! Just about every other character needs to be enclosed in \bf{}, and the source text ends up near illegible. I must admit to losing the will to live, let alone answer the question, towards the end.
STEP II, 1995, Q2:

rn+1=snr_{n+1} = s_n because if we have an acceptable colouring of n+1 posts with the first and last posts coloured red, then the next to last post must be coloured other than red, and so removing the last post gives a sequece of n posts with the last post coloured other than red. And vice versa.

rn+snr_n + s_n is the number of ways of colouring n posts so that the first post is red and no two posts have the same colour (with no extra condition about the colour of the last post). Colouring the posts in order, for each post other than the first, there are 2 choices for its colour (either of the colours that differ from the previous post). So rn+sn=rn+1+rn=2n1r_n+s_n = r_{n+1}+r_n = 2^{n-1}.

Then rn+1=2n1rnr_{n+1} = 2^{n-1} -r_{n}. Then if rk=2k1+2(1)k13r_k = \frac{2^{k-1} + 2(-1)^{k-1}}{3} we have:
Unparseable latex formula:

r_{k+1} = 2^{k-1} - \frac{2^{k-1} + 2(-1)^{k-1}}{3} \\[br]\implies r_{k+1} = \frac{2^{k-1}(3-1) - 2(-1)^{k-1}}{3} \\[br]\implies r_{k+1} = \frac{2^k + 2(-1)^k}{3}



So since r1=1=20+2(1)03r_1 = 1 = \frac{2^0 + 2(-1)^0}{3}, result follows by induction.

Of course, there's nothing special about painting the first post red (as opposed to blue or white), so the number of ways of painting n+1 posts so no two posts are the same colour and the first and last posts are coloured the same is just 3 times what we've just calculated. But then by identifying the 1st and last posts (which are the same colour), we are left with the same situation as n posts in a circle. (e.g. if we consider n=13, then we can think of having posts 1,2,...,12 as the hours on a clock, and we consider post 13 to be the same as post 1).

So the required answer is just 3rn+1=2n+2(1)n3r_{n+1} = 2^n+2(-1)^n.

Edit: Mistake spotted by DVS.
Reply 63
STEP II Q2

(Proving that r(n+1) = s(n).)

Well, r(n+1) is the number of ways to paint n+1 posts such that the first and last post are red. If we ignore the last post, this is the same as painting the first post red and the post before last any of the two other colors (because no two adjacent posts can have the same color) - but there are s(n) ways to do this. So r(n+1) = s(n).

(Finding r(n) + s(n).)

There are r(n) + s(n) ways to paint the fence if we start with a red post. Now say we painted the first post red; we're then left with n-1 posts to deal with, subject to the restriction that the first post must either be white or blue. Say the first post is white, then there are r(n-1) + s(n-1) ways to paint the remaining posts. Similarly, if instead the first post is blue, then there are also r(n-1) + s(n-1) ways to paint the remaining posts.

Thus, for n>=1,
r(n) + s(n) = 2(r(n-1) + s(n-1))
= 2^(n-1) (r(1) + s(1))
= 2^(n-1) (1 + 0)
= 2^(n-1)

(r(n+1) + r(n) for n>=1.)

So, for n>=1,
r(n+1) + r(n) = s(n) + r(n) = 2^(n-1)

(Proving the formula given for r(n).)

First we check if it holds for n=1:
r(1) = (2^0 + 2(-1)^0)/3 = 3/3 = 1, as expected.

Now let's assume that it holds for n=k. Then, by what we found r(n+1) + r(n) to be,
r(k+1) = 2^(k-1) - r(k)
= 2^(k-1) - [2^(k-1) + 2(-1)^(k-1)]/3
= [3*2^(k-1) - 2^(k-1) - 2(-1)^(k-1)]/3
= (2^k + 2(-1)^k)/3

Thus, by induction, the formula is correct for positive integral n.

(Posts in a circle bit.)

Suppose you have n+1 posts in a line. There are r(n+1) ways to paint the first post and the last post using the same color. Then if you wrap this line around to form a circle, and disregard the last post, you will get a circle fence painted in the required way. Of course, all such arrangement arise like this. That is, if you have such a circle fence, pick any post and unravel the circle at that bit and form a line; at the end of the line place a post of the same color as the same one. Now to account for the 3 colors, multiply by 3.*

So the answer is: 3r(n+1) ways.
----------------------------------------

* Thanks to DFranklin for noticing this.
Reply 64
Ah, damn it! :smile:

At least we agree on the answers, except for the last bit. Am I undercounting? (Edit: or overcounting?!)
dvs
Ah, damn it! :smile:

At least we agree on the answers, except for the last bit. Am I undercounting? (Edit: or overcounting?!)
Bit of each. I erroneously wrote down the answer as 3r_n (now fixed) instead of 3r_{n+1}, but you're only counting arrangements where the "first post is red", which is where the factor of 3 comes from.

The number of "copying" mistakes I've made by flipping between the window with the STEP .pdf files and the forum window is ridiculous. I remember an examiner telling me in all seriousness that he'd seen more mistakes from peope copying equations from one page to the next than anything else!
dvs
So this leaves the geometry STEP III question. To be honest it doesn't look all that exciting, but if no one is doing it, I'll try to see what I can do later.
I assume you saw I've done this. I wouldn't wish trying to typeset it on anyone else!

Not my finest hour: I thought the approach I took would turn out a bit nicer than it actually did.

In hindsight, it would have been better to choose vectors e,f so that e,f,n form an orthogonal coordinate system. Then show that if s is the center of the sphere then (s-r).e = (s-r).f=0. Creating a co-ordinate system always feels a little like cheating, but it would have made this question quite a bit nicer.
Reply 67
Yeah, I spotted the geometry problem - you truly are a champion for typesetting it! Personally, I wouldn't want to touch it let alone \textbf{} every other letter I type in.

Now as for the last bit of the fence problem, I must admit that I didn't immediately see why it was necessary to multiply by 3, since I hadn't explicitly fixed any color or anything. Although perhaps just by using r(n) one implicitly has a color fixed in the background.
dvs
Now as for the last bit of the fence problem, I must admit that I didn't immediately see why it was necessary to multiply by 3, since I hadn't explicitly fixed any color or anything. Although perhaps just by using r(n) one implicitly has a color fixed in the background.
In terms of the question, it explictly says r_n has the first post fixed as red. But it's also explicit in the formula for r_n+s_n; we assume the colour of the first post, and then say we have 2 choices for each subsequent post. If the question wasn't worded as it was, we might say we have 3 choices for the colour of the first post, and 2 choices for each subsequent post, when we'd get the answer 2^n+2(-1)^n directly.

I guess it's just easier to write the exam question if you fix the colours...
STEP III, 1995, Q10.

Suppose the angle of fire is α\alpha, and form the standard equations for x, y (in the coordinate system where x is horizontal; that is, distance along the slope will be x/cosβx/ \cos\beta). We have:

y(t)=utsinα12gt2,x(t)=utcosαy(t) = ut \sin \alpha - \frac{1}{2}gt^2, x(t) = ut \cos \alpha. Then the height h(t) above the ground will be:

h(t)=y(t)xtanβ=ut(sinαcosαtanβ)12gt2h(t) = y(t) - x \tan \beta = ut(\sin \alpha - \cos \alpha \tan \beta) - \frac{1}{2}gt^2

=t2cosβ(2u(sinαcosβcosαsinβ)gt)=t2cosβ(2usin(αβ)gtcosβ)=\frac{t}{2 \cos \beta} (2u(\sin\alpha \cos \beta-\cos\alpha\sin\beta)-gt) = \frac{t}{2 \cos \beta} (2u \sin(\alpha -\beta)-gt \cos \beta)

So the time when it hits the ground will be (ignoring the root when t=0, which is the launch):

t=2ugcosβsin(αβ)t=\frac{2u}{g\cos\beta}\sin(\alpha-\beta). For this value of t, the horizontal range RxR_x is given by:

Rx=utcosα=2u2gcosβ(sin(αβ))cosαR_x = ut \cos \alpha = \frac{2u^2}{g\cos\beta}(\sin(\alpha-\beta))\cos\alpha

=u2gcosβ(sin(2αβ)sinβ)= \frac{u^2}{g\cos\beta}(\sin(2\alpha-\beta)-\sin\beta).

This is maximized when α=π4+β2\alpha = \frac{\pi}{4} + \frac{\beta}{2}. For this value of α,\alpha,, we find Rx=u2gcosβ(1sinβ)R_x = \frac{u^2}{g \cos \beta}(1-\sin \beta).

Now the maximum range along the slope, R, is given by Rx/cosβR_x / \cos \beta. So:

R=Rxcosβ=u2gcos2β(1sinβ)=u2g(1sin2β)(1sinβ)=u2g(1+sinβ).R = \frac{R_x}{\cos \beta} = \frac{u^2}{g \cos^2 \beta}(1-\sin \beta) = \frac{u^2}{g (1-\sin^2 \beta)}(1-\sin \beta) = \frac{u^2}{g(1+\sin\beta)}.

Comments: This is another nice mechanics question that doesn't really require in depth knowledge - standard equation of a projectile is enough.
STEP I Q10

At equilibrium, the tension in the string equals the weight of the ball, that is
λel=mg\frac{\lambda e}{l} = mg
e=mglλe=\frac{mgl}{\lambda}
The total length of the string is the sum of the natural length and the extension at equilibrium, so the total lenght is
l+mglλ=l(1+mgλ)l + \frac{mgl}{\lambda} = l(1+\frac{mg}{\lambda})

First note that when the ball is put into simple harmonic motion:
md2xdt2=λxlm\frac{d^2x}{dt^2} = -\frac{\lambda x}{l}
d2xdt2=λxml\frac{d^2x}{dt^2} = -\frac{\lambda x}{ml}
So for the SHM:
ω2=λml\omega^2 = \frac{\lambda}{ml}

In general the velocity of the ball as a function of the displacement is given by:
v2=ω2(a2x2)v^2 = \omega^2(a^2 - x^2)
(a is the amplitude). Projected with speed u0, so:
u02=λmla2u_0^2 = \frac{\lambda}{ml} a^2
a2=mlu02λa^2 = \frac{m l u_0^2}{\lambda}

Substitute the value of a2 into the equation for the velocity as a function of displacement:
v2=λml2(mlu02λx2)v^2 = \frac{\lambda}{ml}^2(\frac{m l u_0^2}{\lambda} - x^2)
v2+λmlx2=u02v^2 + \frac{\lambda}{ml}x^2=u_0^2
as required. Call this equation (1).

The inequality on h2 just makes that the ball actually hits the surface in its simple harmonic motion. From the last equation we know that the velocity at a distance h satisfies:
v2=u02λmlh2v^2 =u_0^2 - \frac{\lambda}{ml}h^2
The speed immediately after rebounded with the surface is ve, which we shall call v1, hence:
v12=u02e2λmlh2e2v_1^2 =u_0^2e^2 - \frac{\lambda}{ml}h^2 e^2

From equation (1) we can see u1 (the velocity at equilibrium position after being at h with a velocity v1) is given by:
u12=v12+λmlh2u_1^2 = v_1^2 + \frac{\lambda}{ml}h^2
u12=u02e2+λmlh2(1e2)u_1^2 = u_0^2e^2 + \frac{\lambda}{ml}h^2 (1 - e^2)
as required.

After the second bounce, the velocity v2 = v1e. So
v2=u02e4λmlh2e4v_2 = u_0^2e^4 - \frac{\lambda}{ml}h^2 e^4
Finding u0 using equation (1) again we get
u22=u02e4+λmlh2(1e4)u_2^2 = u_0^2e^4 + \frac{\lambda}{ml}h^2 (1- e^4)

More mechanics coming.
STEP III 1995, Q9.

(i) (See left hand diagram).

Taking moments about O, we have mgrsinβ=mgr2sin(2α+β)mgr \sin \beta = \frac{mgr}{2} \sin(2\alpha+\beta). So
2sinβ=sin2αcosβ+cos2αsinβ2\sin\beta = \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta
    (2cos2α)sinβ=sin2αcosβ\implies (2- \cos 2\alpha)\sin \beta = \sin 2\alpha \cos \beta
    tanβ=sin2α2cos2α\implies \tan \beta = \frac{\sin 2\alpha}{2 - \cos 2\alpha} as required.

(ii) To relate mass and moment of inertia, will divide disk into annulae radius x, width dx, each has area 2πxdx2\pi x dx and so mass 2πρx2rdx\frac{2\pi \rho x^2}{r} dx.

Mass of disk:
Unparseable latex formula:

\displaystyle m = \int_0^r \frac{2\pi \rho x^2}{r} dx = \left[\frac{2\pi\rhox^3}{3r}\right]_0^r = \frac{2\pi}{3}\rho r^2



M.I. of disk about C:
Unparseable latex formula:

\displaystyle I_C = \int_0^r \frac{2\pi \rho x^4}{r}dx = \left[\frac{2\pi\rhox^5}{5r}\right]_0^r = \frac{2\pi}{5} \rho r^4 = \frac{3mr^2}{5}

.

As C is the center of mass, M.I. of disk about O = M.I. of disk about C + m(distance of c.of.m. from O)^2 = 3mr25+mr2=8mr25\frac{3mr^2}{5} + mr^2 = \frac{8mr^2}{5}.

(iii) First note that Q is the center of mass of the system (both disk and particle). Originally Q is at the same height as O. When P is directly below O, Q has fallen a distance 3r2sinα\frac{3r}{2} \sin \alpha (see right hand diagram). Note also that OP = 2rsinα2r \sin \alpha as we'll use this later.

By conservation of energy, the kinetic energy of the system at this point is 3mgrsinα3mgr \sin \alpha (recall the system has total mass 2m).

Now the M.I. of the disk about O is 8mr25\frac{8mr^2}{5}, while the M.I. of P about O is m(OP)2=4mr2sin2αm (OP)^2 = 4mr^2 \sin^2 \alpha. So the M.I. of the system about O is I=4mr2sin2α+8mr25=45mr2(2+5sin2α)I = 4mr^2\sin^2\alpha + \frac{8mr^2}{5} = \frac{4}{5}mr^2(2 + 5 \sin^2 \alpha).

So if ω\omega is the angular velocity of the system about O, then the system has K.E. 12Iω2\frac{1}{2}I\omega^2. So we must have:

12Iω2=3mrgsinα\frac{1}{2}I\omega^2 = 3mrg \sin \alpha
25mr2ω2(2+5sin2α)=3mrgsinα\frac{2}{5}mr^2\omega^2(2+5\sin^2\alpha) = 3mrg \sin \alpha
ω2=15gsinα2r(2+5sin2α)\omega^2 = \frac{15g \sin \alpha}{2r(2+5\sin^2\alpha)}.

Since v2=r2ω2v^2 = r^2 \omega^2, we have v2=15grsinα2(2+5sin2α)v^2 = \frac{15gr \sin \alpha}{2(2+5\sin^2\alpha)} as required.

Write f(t)=(t2+5t2)1=2/t+5t.f(t)=2/t2+5,f(t)=4/t3>0 for t>0f(t) = \left(\frac{t}{2+5t^2}\right)^{-1} = 2/t + 5t. f'(t) = -2/t^2+5, f'''(t) = 4/t^3 > 0 \text{ for } t>0. So setting f'(t)=0 gives us a local minimum, attained for t=2/5t=\sqrt{2/5}. Applying this to v2=15grsinα2(2+5sin2α)v^2 = \frac{15gr \sin \alpha}{2(2+5\sin^2\alpha)}, we see it is maximized for sinα=2/5\sin \alpha = \sqrt{2/5} (we need to also check in case sinα=0\sin \alpha =0 gives a maximum, but it clearly doesn't).

For this value of sinα\sin \alpha, we have v2=3gr102(2+4)=gr104.v^2 = \frac{3gr\sqrt{10}}{2(2+4)} = \frac{gr\sqrt{10}}{4}.

Comment: For a change, thought I'd look at one of the more "advanced" mechanics questions involving moments of inertia. This question seems pretty straightforward; again drawing useful diagrams is half the battle. It does seem very long though - having to maximize v^2 at the end is just a bit much...
I just had a look at STEP II Q 13 and started on it, someone will have to point me in the right direction (if what I've done is correct in the first place).

First, why is it distributed with Poisson Distribution? Because the mean will be near constant, and the variance will be extremely small as ϵ\epsilon is very small.
Therefore the poisson distribution which is:
P(rpassengersdoesnotturnup)=λrr!eλP(r passengers doesnot turn up) = \frac{\lambda^r}{r!}e^{-\lambda}
Where λ=Nϵ\lambda=N\epsilon

Profit=IncomeCosts[br]Income=Priceofticket×numberofticketsi.e.A(N+k)Costs=CosttoRefund×max(0,kr)i.e.Bmax(0,kr)Profit= Income - Costs \newline[br]Income=Price of ticket\times number of tickets\newline i.e. A(N+k) \newline Costs = Cost to Refund\times max(0,k-r)\newline i.e. Bmax(0,k-r)
If the company has sold tickets so r<k they will have to refund otherwise there will be no refund. Therefore the 0 is used when k<=r.

u0=NAu1=A(N+1)B(1r)etc.u_0=NA\newline u_1=A(N+1)-B(1-r)\newline etc.
Here I obviously need to know something about r which I can't seem to figure out right now:s-smilie: A hint is welcome!
I guess that E(N)=λE(N)=\lambda but don't see any use of this...
nota bene: u1 is going to equal A(N+1) - B x P(every passenger turns up). (Since if every passenger turns up, one is bumped).
u2 = A(N+2) - B x [2 x P(every passenger turns up) + 1 x (1 passenger does not turn up)]

Etc...
yes, I know I'm thick:p:

Thanks, will see if I get where I want now:smile:

edit: I still have a bit of an embarrassing problem:redface:

u0=ANu1=A(N+1)B(eλ)u2=A(N+2)B(2eλ+λeλ)u3=A(N+3)B(3eλ+2λeλ+λ22!eλ)u4=A(N+4)B(4eλ+3λeλ+2λ22!eλ+λ33!)etc.u_0=AN \newline u_1 = A(N+1) - B(e^{-\lambda}) \newline u_2 = A(N+2) - B(2e^{-\lambda}+\lambda e^{-\lambda}) \newline u_3 = A(N+3) - B(3e^{-\lambda}+2\lambda e^{-\lambda}+\frac{\lambda^2}{2!}e^-\lambda) \newline u_4 = A(N+4) - B(4e^{-\lambda}+3\lambda e^{-\lambda}+2\frac{\lambda^2}{2!}e^-\lambda+\frac{\lambda^3}{3!}) \newline etc.
This is where I am temporarily stuck, it will probably come to me sooner or later...
un=A(N+n)B0nλnn!eλu_n=A(N+n)-B\displaystyle \sum_0^n\frac{\lambda^n}{n!}e^{-\lambda} But the 1st term in the series needs to be multiplied by n... 2nd by (n-1) and nth by (n-(n+1)).
nota bene
This is where I am temporarily stuck, it will probably come to me sooner or later...Note that the question doesn't actually ask you to find a formula for u_n. I suspect this omission is entirely intentional...
DFranklin
Note that the question doesn't actually ask you to find a formula for u_n. I suspect this omission is entirely intentional...

iteration:tongue: ?
saw that before I went to bed, but forgot to edit:redface:

I guess I'll look at this again.....
STEP III, 1995 Q11

If the plane is to be on a constant NW heading to the ship, then at all times, the velocity of the plane relative to the ship must be of the form (-a, a) when going out, or (a,-a) when coming back (for some value of a). Meanwhile, the velocity of the ship relative to the air is (-V, V).

So on the outward flight, the velocity VoutV_{out} of the plane relative to the air is (aoutV,aout+V)(-a_{out}-V,a_{out}+V). We know Vout2=k2V2|V_{out}|^2 = k^2V^2, so 2(aout+V)2=k2V2    (aout+V)2=kV    aout=(k2)V22(a_{out}+V)^2 = k^2V^2 \implies (a_{out}+V)\sqrt{2} = kV \implies a_{out} = \frac{(k-\sqrt{2})V}{\sqrt{2}}.
Similarly, aback=(k+2)V2a_{back} = \frac{(k+\sqrt{2})V}{\sqrt{2}}.

Now the total journey time is 0.5 hours. Let t be the time spent on the outward journey. Then taout=(0.5t)abackt a_{out} = (0.5 - t)a_{back}. So
2aoutt=aback(12t)2 a_{out} t = a_{back}(1-2t)
2(k2)t=(k+2)(12t)2(k-\sqrt{2})t = (k+\sqrt{2})(1-2t)
2kt=k+2    t=k+22k2kt = k + \sqrt{2} \implies t = \frac{k+\sqrt{2}}{2k}.

Now the air is still, the velocity of the ship relative to the air is (-V, 0). This time let (bout,bout)(-b_{out}, b_{out}) be the velocity of the plane relative to the ship, (bback,bback)(b_{back}, -b_{back}) the velocity coming back.

Then k2V2=bout2+(bout+V)2=2bout2+2boutV+V2k^2V^2 = b_{out}^2+(b_{out}+V)^2 = 2b_{out}^2+2b_{out}V+V^2.
k2V22=(bout+V2)2+V24\frac{k^2V^2}{2} = (b_{out}+\frac{V}{2})^2 + \frac{V^2}{4}
(2k21)V24=(bout+V2)2    bout=(2k21)V24V2\frac{(2k^2-1)V^2}{4} = (b_{out}+\frac{V}{2})^2 \implies b_{out} = \sqrt{\frac{(2k^2-1)V^2}{4}} - \frac{V}{2}
    2bout=V(2k211)\implies 2b_{out} = V(\sqrt{2k^2-1} - 1)

Similarly, 2bback=V(2k21+1)2b_{back} = V(\sqrt{2k^2-1} + 1)

Again, if the outward journey time is t, we have
2boutt=bback(12t)    2t=bbackbout+bback2 b_{out} t = b_{back}(1-2t) \implies 2t = \frac{b_{back}}{b_{out}+b_{back}} and so twice the maximum distance is boutbbackbout+bback\frac{b_{out}b_{back}}{b_{out}+b_{back}}.

Now 2boutbback=V22(2k211)(2k21+1)=V22(2k22)=V2(k21)2b_{out}b_{back} = \frac{V^2}{2}(\sqrt{2k^2-1} - 1)(\sqrt{2k^2-1} + 1) = \frac{V^2}{2}(2k^2 - 2) = V^2(k^2-1)
2(bout+bback)=2V2k212(b_{out}+b_{back}) = 2V\sqrt{2k^2-1}.
So distance travelled dc=V(k21)42k21d_c = \frac{V(k^2-1)}{4\sqrt{2k^2-1}}.

Going back to the original scenario, 2aoutaback=(k22)V22a_{out}a_{back} = (k^2-2)V^2, aout+aback=kV2 a_{out}+a_{back} = kV\sqrt{2}.
So dw=V(k22)(42)kd_w = \frac{V(k^2-2)}{(4\sqrt{2})k}.

Thus dw/dc=V(k22)(22)k42k21V(k21)=(k22)2k(k21)4k22d_w/d_c = \frac{V(k^2-2)}{(2\sqrt{2})k} \frac{4\sqrt{2k^2-1}}{V(k^2-1)} = \frac{(k^2-2)}{2k(k^2-1)} \sqrt{4k^2-2}

Comment: Not the kind of question I'd have touched with a bargepole under exam conditions; thought I'd have a go to see how bad it really is. And it's actually fairly straightforward, if maybe a little on the long and fiddly side.
STEP III, 1995, Q14.

If MT(x)M_T(x) is the moment generating function for T, then MT(x)=extf(t)dt=0extf(t)dtM_T(x) = \int_{-\infty}^\infty e^{xt} f(t) dt = \int_0^\infty e^{xt} f(t) dt since f(t) = 0 for t < 0.

So MT(x)=0te(x1)tdtM_T(x) = \int_0^\infty t e^{(x-1)t}dt. We need x < 1 for convergence, so assume x < 1. Integrating by parts:
MT(x)=[tx1te(x1)t]01x10e(x1)tdt=1(x1)2M_T(x) = [\frac{t}{x-1}te^{(x-1)t}]_0^\infty - \frac{1}{x-1}\int_0^\infty e^{(x-1)t} dt = \frac{1}{(x-1)^2}.

Now if X, Y are any two independently distributed variables, then MX+Y(x)=MX(x)MY(x).M_{X+Y}(x) = M_X(x)M_Y(x).. So MU(x)=1(x1)4M_U(x) = \frac{1}{(x-1)^4}.

Now when we come to find the p.d.f., we're going to find ourselves evaluating 0t3e(x1)tdt\int_0^\infty t^3 e^{(x-1)t}dt, and for the p.d.f. of the last bit, evaluating 0tNe(x1)tdt\int_0^\infty t^N e^{(x-1)t}dt. So to avoid repeating ourselves, we'll prove a little reduction formula integral:

Define In(y)=0tnetydtI_n(y) = \int_0^\infty t^n e^{ty} dt (where y < 0). Then In(y)=[tnyety]0ny0tn1etydt=nyIn1(y)I_n(y) = [\frac{t^n}{y} e^{ty}]_0^\infty - \frac{n}{y} \int_0^\infty t^{n-1} e^{ty} dt = \frac{-n}{y} I_{n-1}(y).
Noting that I0(y)=0etydt=[etyy]0=1yI_0(y) = \int_0^\infty e^{ty} dt = [\frac{e^ty}{y}]_0^\infty = -\frac{1}{y}, we have In(y)=n!(y)n+1I_n(y) = \frac{n!}{(-y)^{n+1}}.

So if g(t)=16t3etg(t) = \frac{1}{6}t^3e^{-t}, then 0extg(t)dt=160t3e(x1)tdt=16I3(x1)=163!(x1)4=1(x1)4=MU(x)\int_0^\infty e^{xt} g(t)dt = \frac{1}{6} \int_0^\infty t^3 e^{(x-1)t} dt = \frac{1}{6} I_3(x-1)= \frac{1}{6} \frac{3!}{(x-1)^4} = \frac{1}{(x-1)^4} = M_U(x). Thus g is the p.d.f. corresponding to the moment generating function MUM_U.

Similarly, in the case of answering n questions, let Mn(x)M_n(x) be the moment generating function of the sum. Then we have Mn=1(1x)2nM_n = \frac{1}{(1-x)^{2n}}. Consider hn(t)=t2n1eth_n(t) = t^{2n-1}e^{-t}. Then 0exth(t)dt=I2n1(x1)=(2n1)!(x1)2n\int_0^\infty e^{xt} h(t)dt = I_{2n-1}(x-1) = \frac{(2n-1)!}{(x-1)^2n}, and so gn(t)=1(2n1)!h(t)=1(2n1)!t2n1etg_n(t) = \frac{1}{(2n-1)!}h(t) = \frac{1}{(2n-1)!}t^{2n-1}e^{-t} is the p.d.f. for the sum of n questions.

Comment: Are moment generating functions on the current syllabus? I had done them, but I couldn't remember a blind thing about them; I had to look them up in Wiki before doing this question.
STEP II Q 13
First, why is it distributed with Poisson Distribution? Because the mean will be near constant, and the variance will be extremely small as ϵ\epsilon is very small.
Therefore the poisson distribution which is:
P(rpassengersdoesnotturnup)=λrr!eλ[br]Whereλ=Nϵ[br][br]Profit=IncomeCostsIncome=Priceofticket×numberofticketsi.e.A(N+k)Costs=CosttoRefund×max(0,kr)i.e.Bmax(0,kr)P(r passengers doesnot turn up) = \frac{\lambda^r}{r!}e^{-\lambda}[br]Where \lambda=N\epsilon[br][br]Profit= Income - Costs \newline Income=Price of ticket\times number of tickets\newline i.e. A(N+k) \newline Costs = Cost to Refund\times max(0,k-r)\newline i.e. Bmax(0,k-r)
If the company has sold tickets so r<k they will have to refund otherwise there will be no refund. Therefore the 0 is used when krk\geq r.
The first terms will be:

[br]u0=ANu1=A(N+1)B(eλ)u2=A(N+2)B(2eλ+λeλ)u3=A(N+3)B(3eλ+2λeλ+λ22!eλ)u4=A(N+4)B(4eλ+3λeλ+2λ22!eλ+λ33!)etc.[br]u_0=AN \newline u_1 = A(N+1) - B(e^{-\lambda}) \newline u_2 = A(N+2) - B(2e^{-\lambda}+\lambda e^{-\lambda}) \newline u_3 = A(N+3) - B(3e^{-\lambda}+2\lambda e^{-\lambda}+\frac{\lambda^2}{2!}e^-\lambda) \newline u_4 = A(N+4) - B(4e^{-\lambda}+3\lambda e^{-\lambda}+2\frac{\lambda^2}{2!}e^-\lambda+\frac{\lambda^3}{3!}) \newline etc.
Generally:
un+1=un+AB0nλnn!eλ[br]u_{n+1}=u_n+A-B\displaystyle \sum_0^n\frac{\lambda^n}{n!}e^{-\lambda}[br]
Which then (since they want it in form vk=un+1ukv_k=u_{n+1} - u_k) rearranges to:
un+1un=AB0nλnn!eλu_{n+1}-u_n=A-B\displaystyle \sum_0^n\frac{\lambda^n}{n!}e^{-\lambda}

Then to "show v_k > v_k+1" should be ovbious since we now are subtracting a larger number...0n+1λnn!eλ\displaystyle \sum_0^{n+1}\frac{\lambda^n}{n!}e^{-\lambda}

As n lim infinity the series will approach 1 as it is the sum of all probabilities 0P(X=x)\displaystyle\sum_0^\infty \mathbb{P}(X=x). This means vk=ABv_k=A-B as n lim infinity, as required.

To maximise expected profit we look at
uk=uk+1vkandvk>0uk+1>ukandvk<0uk+1<uku_k=u_{k+1}-v_k\newline\mathrm{and}v_k>0\rightarrow u_{k+1}>u_k\newline\mathrm{and}v_k<0\rightarrow u_{k+1}<u_k
We know vkv_k is decreasing, and we know v_k -> B-A < 0 as k -> infinity. Therefore there must be a value N so that vk0k<Nandvk<0kNv_k\le 0 \forall k < N \newline\mathrm{and} v_k < 0 \forall k\le N
Then the maximum profit will be when N=k.


Thanks DFranklin for the hints!

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