STEP Maths I, II, III 1997 Solutions

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STEP II Q12
If there are three people the game can be won if everyone hears everyone else correctly or if two people hear incorrectly.

\displaystyle \left(\frac{2}{3}\right)^3 = \frac{8}{27}

\displaystyle \binom{3}{2} \times \frac{2}{3} \times \left(\frac{1}{3}\right)^2 = \frac{6}{27}

\displaystyle P(Winning) = \frac{8}{27} + \frac{6}{27} = \frac{14}{27}

(a + b)^n = a^n + \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + ... + \binom{n}{n-1}ab^{n-1} + b^n

(a - b)^n = a^n - \binom{n}{1}a^{n-1}b + \binom{n}{2}a^{n-2}b^2 + ... + \binom{n}{n-1}a(-b)^{n-1} + (-b)^n

(a+b)^n + (a-b)^n = 2a^n + 2\binom{n}{2}a^{n-2}b^2 + 2\binom{n}{4}a^{n-4}b^4 + 2\binom{n}{6}a^{n-6}b^6 + ...

Similarly, the game will be won when everyone hears correctly or an even number of people hear incorectly. The above formula can be used to model this if a represents hearing corretly and b incorrectly.

\displaystyle P(Winning) = \frac{(p + q)^n + (p - q)^n}{2}

But

\displaystyle q = 1 - p

\displaystyle P(Winning) = \frac{1 + (2p - 1)^n}{2}

To find the expected number of round consider the first few possibilities.

\displaystyle P(Rounds = 1) = \frac{1 + (2p - 1)^n}{2}

\displaystyle P(Rounds = 2) = \left(\frac{1 + (2p - 1)^n}{2}\right)\left(1-\frac{1 + (2p - 1)^n}{2}\right) = \left(\frac{1 + (2p - 1)^n}{2}\right)\left(\frac{1 - (2p - 1)^n}{2}\right)

\displaystyle P(Rounds = 3) = \left(\frac{1 + (2p - 1)^n}{2}\right)\left(\frac{1 - (2p - 1)^n}{2}\right)^2

The pattern is obvious so using the formula for expectation:

\displaystyle \mu = \sum_{k=1}^\infty k\left(\frac{1 + (2p - 1)^n}{2}\right)\left(\frac{1 - (2p - 1)^n}{2}\right)^{k-1} = \frac{1 + (2p - 1)^n}{2} \times \sum_{k=1}^\infty k\left(\frac{1 - (2p - 1)^n}{2}\right)^{k-1}

Now use the formula provided at the bottom of the question.

\displaystyle \mu = \left(\frac{1 + (2p - 1)^n}{2}\right)\left(1 - \frac{1 - (2p - 1)^n}{2}\right)^{-2} = \frac{2}{1+(2p-1)^n}

Reply 61

datr

STEP II Q14

u = 30 s= 9600 t = 320
v = 32 s= 9600 t = 300

So only cars leaving within 20 seconds after me will have time to catch up and form a queue. If the expected number of cars to enter the tunnel in a minute is 6, the expected number of cars to enter in 20 seconds is 2.

The probability that exactly two cars enter the tunnel within 20 seconds of my doing so is equal to:

P(X=2|X\sim Po(2)) = \frac{e^{-2}\times 2^2}{2!} = 2e^{-2}

For them to catch me up each must be travelling at 32ms^-1 the probability of this is 0.5*0.5 = 0.25

P = 2e^{-2} \times \frac{1}{4} = \frac{e^{-2}}{2}

I think for this last part of the question it wants you to consider the possibility that more than 2 cars enter the tunnel within the 20 second time frame but only two manage to catch up with you.

P(X>2) = 1 - P(X\le 2) = 1 - e^{-2}\left(1 + 2 + \frac{2^2}{2!}\right) = 1-5e^{-2}

For two cars to catch up and no more the first two must be travelling at 32ms^-1 and the third must be travelling at 30ms^-1.

\left(\frac{1}{2}\right)^3 = \frac{1}{8}

\frac{1-5e^{-2}}{8}

P = \frac{e^{-2}}{2} + \frac{1-5e^{-2}}{8} = \frac{1 - e^{-2}}{8}

Hmm... this seems much too easy, in fact I'm sure I'm missing something.

Reply 62

DFranklin

STEP III, Q12.

(i)

\mathbb{P}(K=k) = p(1-p)^{k-1} = pq^{k-1}.

.

So

G(s) = \sum_1^\infty pq^{k-1}s^k = ps \sum_1^\infty (qs)^{k-1} = \frac{ps}{1-qs}

.

Now

G'(s) = \sum_1^\infty k pq^{k-1}s^{k-1}

and so

G'(1) = \sum_1^\infty k pq^{k-1} = \mathbb{E}[K]

.

Take logs:

\ln G(s) = \ln ps - \ln(1-qs) \implies \frac{G'(s)}{G(s)} = \frac{1}{s} + \frac {q}{1-qs}

.

So

G'(s) = \left[\frac{1}{s} + \frac {q}{1-qs}\right]G(s)

. Putting s=1, and noting that G(1) = 1, we get:

\mathbb{E}[K] = G'(1) = 1 + \frac {q}{1-q} = \frac{1}{1-q} = \frac{1}{p}

Similarly

G''(s) = \sum_1^\infty k(k-1) pq^{k-1}s^{k-2}

and so

G''(1) = \sum_1^\infty k(k-1) pq^{k-1} = \mathbb{E}[K(K-1)]

Unparseable latex formula:

G''(s) = \frac{d}{ds} \left\{\left[\frac{1}{s} + \frac {q}{1-qs}\right]G(s)\right\} \\[br]= \left\{ G(s) \frac{d}{ds}\left[\frac{1}{s} + \frac {q}{1-qs}\right]\right\} + \left[\frac{1}{s} + \frac {q}{1-qs}\right] G'(s)

= G(s) \left[\frac{-1}{s^2} + \frac {q^2}{(1-qs)^2}\right] + \left[\frac{1}{s} + \frac {q}{1-qs}\right] G'(s)

We don't bother tidying up much, because we only want to find G''(1). Recall

G(1) = 1, G'(1) = \frac{1}{p}

. So

G''(1) = -1 + \frac{q^2}{(1-q)^2}+ \left[1 + \frac {q}{1-q}\right]\frac{1}{p} = -1 + \frac{q^2}{p^2} + \frac{1}{p^2}

=\frac{-p^2+(1-p)^2+1}{p^2} = \frac{2(1-p)}{p^2}=\frac{2q}{p^2}.

Then

\mathbb{E}[K^2] = \mathbb{E}[K(K-1)] + \mathbb{E}[K] = \frac{2q}{p^2} + \frac{1}{p} = \frac{2q+p}{p^2} = \frac{2-p}{p^2}

.

Finally,

Var[K] = \mathbb{E}[K^2] - \mathbb{E}[K]^2 = \frac{2-p}{p^2}-\frac{1}{p^2} = \frac{1-p}{p^2}

.

(ii) Suppose I have already sampled i-1 cards. Then the probability that I will sample a new card on any particular draw is (53-i)/52 . Define

N_i

to be the number of draws needed to sample the new card. Then by part (i),

\mathbb{E}[N_i] = \frac{52}{53-i}

.

Clearly

N = \sum_1^{52}N_i = 52 \sum_1^{52} \frac{1}{53-i} = 52 \sum_1^{52} \frac{1}{i}

.

The approximation given implies

N \approx 52 (0.5 + \ln 52)

.

The fact that they give an approximate value of 'e' implies we're supposed to estimate ln 52.

2.7^2 = 7.29, 7.29^2 =49 + 14 \times 0.29 + 0.29^2 \approx 49 + 4.2 - 0.14 + .09 \approx 53.1

. Deduce

\ln 52 \approx 4

and so

N \approx 52 \times 4.5 = 234

Comment: Both parts of this question are fairly classic; part (ii) in particular comes up lots and lots in STEP questions. Part (i) is surprisingly fiddly: I think there are better ways of finding the mean and variance (e.g. if E is the expected number of trials, then by considering the result of 1 trial we see E = 1 + (1-p)E and so E=1/p).

All in all a somewhat long question; needing to use a (bad) approximation for e was a bit the final straw.

Reply 63

DFranklin

STEP III, Q14.

Let L be the length and B the breadth. Then

E[L] =\mu_1, E[b] =\mu_2, Var[L] = \sigma_1^2, Var[b] = \sigma_2^2

.

The perimeter

P = 2L + 2B

. By linearity of expectation,

E[P] = 2(\mu_1+\mu_2)

, while for independent variables

Var[aX+bY]=a^2Var[X]+b^2Var[Y]

so we have

Var[P] = 4( \sigma_1^2 + \sigma_2^2

, so P has standard deviation

2\sqrt{\sigma_1^2 + \sigma_2^2}

.

The area

A = LB

. Now for independent variables, E[XY]=E[X]E[Y], so

E[A] = E[L]E[b] = \mu_1\mu_2

. To find the variance, we consider

E[A^2] = E[L^2B^2] = E[L^2]E[B^2]

(since

L^2, B^2

are also independent).

Now

E[L^2] = (E[L^2]-E[L]^2] + E[L]^2) = Var[L] + \mu_1^2 = (\mu_1^2+\sigma_1^2)

; similarly

E[B^2] = (\mu_2^2+\sigma_2^2)

.

So

E[A^2] = E[L^2]E[B^2](\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2)

.

So

Var[A] = E[A^2] - E[A]^2 = (\mu_1^2+\sigma_1^2)(\mu_2^2+\sigma_2^2) - \mu_1^2\mu_2^2 = \mu_1^2\sigma_2^2+\mu_2^2\sigma_1^2 + \sigma_1^2\sigma_2^2

So A has standard deviation

Unparseable latex formula:

\sqrt{\mu_1^2\sigma_2^2+\mu_2^2\sigma_1^2 + \sigma_1^2\sigma_2^2

.

Finally,

E[P \times A] = 2E[(L+B) \times LB]=2 E[L^2B +B^2L] = 2E[L^2] E[b]+2E[B^2]E[L]

= 2(\mu_1^2+\sigma_1^2)\mu_2 + 2(\mu_2^2+\sigma_2^2)\mu_1 =2(\mu_1+\mu_2)\mu_1\mu_2 + 2\sigma_1^2\mu_2+2\sigma_2^2\mu_1

.

On the other hand,

E[P]E[A] = 2(\mu_1+\mu_2)\mu_1\mu_2

. So

E[P \times A] \ne E[P]E[A]

and so P, A are not independent.

Reply 64

DFranklin

STEP III, Q13:

(i)

Unparseable latex formula:

\displaystyle\mathbb{E}(e^{\theta X}) = \int_{-\infty}^\infty e^{\theta x} (2\pi)^{-1/2} e^{-x^2/2} dx = (2\pi)^{-1/2}\int_{-\infty}^\infty e^{-(x^2-2\theta x)/2} dx\\[br]=(2\pi)^{-1/2}\int_{-\infty}^\infty e^{-(x-\theta)^2/2}\,e^{\theta^2/2} dx =(2\pi)^{-1/2} e^{\theta^2/2} \int_{-\infty}^\infty e^{-(x-\theta)^2/2}dx

Set

y=x-\theta

to get

\displaystyle\mathbb{E}(e^{\theta X}) = e^{\theta^2/2} \int_{-\infty}^\infty (2\pi)^{-1/2} e^{-y^2/2} dy = e^{\theta^2/2}

(ii) General results: if

G(\theta)

is the MGF of X then

G(a\theta)

is the MGF of aX. If

G(\theta), H(\theta)

are the MGFs of X,Y then

G(\theta)H(\theta)

is the MGF of X+Y.

So the MGF of aX+bY is

e^{(a^2+b^2)\theta^2/2}

. Thus aX+bY has the same distribution as X iff

a^2+b^2 = 1

.

(iii)

Z^{\theta} = e^{\mu \theta + \sigma \theta X}.

. So we have:

\displaystyle\mathbb{E}(Z^\theta) = e^ (2\pi)^{-1/2} \int_{-\infty}^\infty e^{\mu \theta + \sigma \theta x} e^{-x^2/2} dx \\=[br](2\pi)^{-1/2} e^{\mu \theta} \int_{-\infty}^\infty e^{\sigma \theta x} e^{-x^2/2} dx

.

This last integral is the same as the integral we evaluated in part (i), only with

\sigma \theta

instead of

\theta

. So

\displaystyle(2\pi)^{-1/2} \int_{-\infty}^\infty e^{\sigma \theta x} e^{-x^2/2} dx = e^{\sigma^2 \theta^2/2}

\mathbb{E}(Z^\theta) = e^{\mu \theta} e^{\sigma^2 \theta^2/2} = e^{\mu \theta + \sigma^2 \theta^2/2}

Putting

\theta = 1

, we see that the expectation of Z is

\mathbb{E}(Z) = e^{\mu + \sigma^2 / 2}

For the variance, we put

\theta = 2

and find that

\mathbb{E}(Z^2) = e^{2\mu+2\sigma^2}

. So

Var(Z) = \mathbb{E}(Z^2) - \mathbb{E}(Z)^2 = e^{2\mu+2\sigma^2} - e^{2\mu + \sigma^2} = e^{2\mu +\sigma^2}(e^{\sigma^2} - 1)

Reply 65

DFranklin

STEP III, Q13: This question is absurdly easy; so much so I won't even bother with a diagram:

Consider the motion of the clapper and bell separately; at the point where they separate, they must have equal rates of angular acceleration. Let

\theta

be the angle of the bell handle from vertical, so

\alpha = \theta + \beta

at the time of separation.

For the clapper we have:

mgl \sin(\theta+\beta) = ml^2\ddot{\theta} \implies \ddot{\theta} = \frac{g}{l} \sin(\theta+\beta)

For the bell we have:

gMh \sin \theta = Mk^2 \ddot{\theta} \implies \ddot{\theta} = \frac{gh}{k^2} \sin \theta

.

Equating the two expressions for

\ddot{\theta}

we get:

\sin(\theta+\beta) = \frac{hl}{k^2}\sin \theta

. Use

\alpha = \theta + \beta

to get:

\frac{k^2}{hl} \sin \alpha = \sin(\alpha-\beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta

. Divide by

\sin \alpha sin \beta

\frac{k^2}{hl} \text{cosec} \beta = \cot \beta - \cot \alpha \implies \cot \alpha = \cot \beta - \frac{k^2}{hl} \text{cosec} \beta

.

Comment: What on earth happened here? This can't be the full solution, surely?

Reply 66

DFranklin

STEP III, Q10:

The M.I. of a uniform sphere is

\frac{2}{5}ma^2

. (Don't know if you're expected to prove it, but if so it's completely standard bookwork).

Let the frictional force be F, the angular velocity be

\omega

and the linear velocity be v.

Then

\frac{2}{5}ma^2 \dot{\omega} = -Fa,\quad m\dot{v} = -F

, so

\frac{d}{dt}(\frac{2}{5}a\omega-v) = 0

\frac{2}{5}a\omega-v

is constant.

(i) When marble is at a complete stop,

\omega = v = 0

\frac{2}{5}a\omega-v = 0

, so we must have

\frac{2}{5}a\omega_0 = v_0 \implies v_0/(a\omega_0) = 2/5

.

(ii) When marble is rolling backwards with velocity

v_0/7, v=-v_0/7, a\omega = v_0/7

. So we know

Unparseable latex formula:

\displaystyle \frac{2}{5}a\omega-v = \frac{2}{5}\ctimes \frac{v_0}{7} + \frac{v_0}{7} = \frac{v_0}{5}

\displaystyle \frac{2}{5}a\omega_0 - v_0 = \frac{v_0}{5} \implies \frac{2}{5}a\omega_0 = \frac{6}{5} v_0 \implies v_0/(a\omega_0) = 1/3

.

Have I missed something here? Seems ridiculously easy again.

Reply 67

Kolya

Those solutions do seem extremely easy David; they almost look like something you might find in the A level textbook. (I'm afraid I cannot check them as I do not know STEP III Mechanics theory (it's on the "to-do" list) but *hmmm*....)

Reply 68

DFranklin

Lusus Naturae

Actually, if memory serves, Q10 used to be a completely standard A-level Further Maths questions back when I did the A-level. Only thing is, I don't remember anything about how you were supposed to do these questions. I'm really doing them from first principles, so I could well be making "gotcha" errors; I'm a bit nervous about whether anything odd goes on with a change in the direction of friction in Q10, for example.

Reply 69

Kolya

DFranklin

I will have a little look at the Mechanics textbooks tomorrow, to see if it is indeed "routine". Unfortunately, I do not have the old Mechanics 6 textbook, but moments of inertia is covered in Mechanics 5, so it may be an exercise in there; I shall have to get back to you on it.

Reply 70

DFranklin

STEP II 1997, Q10:

For most of this question we only need to consider the velocities perpendicular to the cylinder axis. Let

P_1

be the particle of mass qM and

P_2

the other . Let

v_1,v_2

be the respective velocities perpendicular to the axis. (Note that v_2 will be measured in the opposite direction to v_1).

Then conservation of momentum gives us

qMv_1 = (1-q)Mv_2 \implies \frac{q}{1-q} = \frac{v_2}{v_1}

. Note that since

q\le\frac{1}{2}

we must have

v_1\le v_2

But at the time when the particles coalesce, one particle has travelled a distance 5L/2 (it has gone to the cylinder boundary, back to the axis, and a further L/2), and the other 3L/2. So

\frac{v2}{v1} = 3/5

5q=3(1-q) \implies q = 3/8

.

If the collision occurs at time

\frac{5L}{v}

then P1 has travelled a distance 5L/2 in time 5L/v, so

v_1 = v/2

. Since

v_2/v_1 = 3/5

we find

v_2=3v/10

.

For the total energy, we need to consider the velocities

u_1, u_2

parallel to the axis. (Comment: Actually, of course, we don't. You'll see that the energy parallel to the axis is unchanged. It's fairly obvious this will be the case, but I think it is easier to do a few extra lines of maths than to try to explain why convincingly).

By conservation of momentum we have

Mqu_1 + M(1-q)u_2 = Mv

, while the fact that the particles collide

\implies u_1 = u_2

. So

u_1 = u_2 = v

.

So the total energy of the 2 particles is

Unparseable latex formula:

\displaystyle \frac{1}{2}[qM(u_1^2+v_1^2)+(1-q)M(u_2^2+v_2^2)]\\[br]=\frac{M}{2}[qv^2 +(1-q)v^2 + q(v/2)^2 +(1-q)(3v/10)^2]\\[br]=\frac{Mv^2}{2}\left[1 + \frac{3}{8}\times \frac {1}{4} + \frac{5}{8} \times \frac{9}{100}c] \\[br]=\frac{Mv^2}{2} + \frac{Mv^2}{2} \left[\frac{3}{32} + \frac{9}{160}\right]

As the initial K.E. was Mv^2/2, the K.E. gained is

\frac{Mv^2}{2} [\frac{15}{160} + \frac{9}{160}] = Mv^2\frac{24}{320} = Mv^2\frac{3}{40}

.

Finally, after coalescence, the velocity is again v down the axis of the cylinder (by conservation of momentum).

Reply 71

DFranklin

STEP II, Q1997, Q13: (although I'm fairly sure I've misunderstood part (ii) or gone wrong somewhere)

(i) Let

\theta

be the acute angle the needle makes with the lines. Then

\theta

is uniformly distributed on

[0,\pi/2]

. Then the needle crosses the line if

\sin \theta > x

, or equivalently if

\theta > \sin^{-1} x

. Since

\mathbb{P}(\theta \le sin^{-1} x) = \frac{2}{\pi} \sin^{-1}x

, the desired probability is

1 - \frac{2}{\pi} \sin^{-1}x = \frac{2}{\pi}(\pi/2 - sin^{-1}x) = \frac{2}{\pi} cos^{-1} x

.

(ii) The length of needle (measured from the center) required to reach the line is

\frac{x}{\sin \theta}

, so the length

c

of the segment cut off by the line is given by

c = 1 - \frac{x}{\sin \theta}

. Note we have

0 \le c \le 1-x

.

Now

c>=0 \Leftrightarrow \sin \theta \ge x

, and

c \le C \Leftrightarrow 1-\frac{x}{\sin \theta} \le C \Leftrightarrow 1-C \le \frac{x}{\sin \theta} \Leftrightarrow \sin \theta \le \frac{x}{1-C}

Then

\mathbb{P}(0 \le c \le C) = \mathbb{P}(x \le \sin \theta \le \frac{x}{1-C} ) = \frac{2}{\pi}\left[\sin^{-1}\left(\frac{x}{1-C}\right) - \sin^{-1} x \right]

= \frac{2}{\pi}\left[\cos^{-1} x - \cos^{-1}\left(\frac{x}{1-C}\right)\right]

Then the conditional probability given the needle hits the line is just this divided by the result from (i). That is:

\mathbb{P}(0 \le c \le C | \text{ needle hits}) = \frac{2}{\pi}\left[\cos^{-1} x - \cos^{-1}\left(\frac{x}{1-C}\right)\right] \Big/ \left(\frac{2}{\pi} \cos^{-1}x \right)

= 1 - \frac{\cos^{-1}\left(\frac{x}{1-C}\right)}{\cos^{-1} x}

(where

0 \le C \le 1-x

).

(iii) For a fixed x, the probability that the needle misses all the lines is

\frac{2}{\pi}\sin^{-1}x

(from (i)).

Since x is uniformly distributed on [0,1], the general probability will be

\frac{2}{\pi} \int_0^1 \sin^{-1}x dx = \frac{2}{\pi} [x \sin^{-1} x]_0^1 - \frac{2}{\pi}\int_0^1 \frac{x}{\sqrt{1-x^2}}

= 1 + \frac{2}{\pi}[\sqrt{1-x^2}]_0^1 = 1 - \frac{2}{\pi}

Slightly easier, albeit making less use of earlier results:

The "width" spanned by a needle at a fixed angle

\theta

2 \sin \theta

. Since this has to fit in the gap between the lines (of width = 2), the probability it will do so is

1 - \sin \theta

. Since

\theta

is uniformly distributed on

[0,\pi/2]

, the general probability is:

\frac{2}{\pi} \int_0^{\pi/2} 1 - \sin \theta d\theta = \frac{2}{\pi} [\theta + \cos \theta]_0^{\pi/2} = \frac{2}{\pi}(\pi/2 - 1) = 1 - \frac{2}{\pi}

Reply 72

DFranklin

STEP I 1997, Q10 (not my finest hour)

Let T be the tension in the string. Resolving vertically and horizontally, we get

T \cos \theta - T \cos \phi = mg \implies 2T \sin \frac{\theta+\phi}{2} \sin \frac{\phi-\theta}{2} = mg

T \sin \theta + T \sin \phi = mg \implies 2T \sin \frac{\theta+\phi}{2} \cos \frac{\phi-\theta}{2} = mr\omega^2

Dividing the first by the second gives

\tan \frac{\phi-\theta}{2} = \frac{g}{r\omega^2}

as required.

Now square the 2nd form of the first 2 equations:

4T^2 \sin^2 \frac{\theta+\phi}{2} \sin^2 \frac{\phi-\theta}{2} = m^2g^2

4T^2 \sin^2 \frac{\theta+\phi}{2} \cos^2 \frac{\phi-\theta}{2} = m^2r^2\omega^4

Adding, we get:

4T^2 \sin^2 \frac{\theta+\phi}{2} = m^2(g^2+r^2\omega^4) \implies 2T \sin \frac{\theta+\phi}{2} = m \sqrt{g^2+r^2\omega^4}

T = \frac{mg}{2} \left(\sin \frac{\theta+\phi}{2}\right)^{-1} \sqrt{1+\frac{r^2\omega^4}{g^2}}

Don't really know what is expected for the last bit. I don't see any easy of getting the result. Here's one approach:

\tan \frac{\phi-\theta}{2} = \frac{g}{r\omega^2} \implies \tan \frac{\pi - (\phi - \theta)}{2} = \frac{r\omega^2}{g}

.

Write

\delta = \frac{r\omega^2}{g}

, and ignore terms of order

\delta^2

and higher. Then we have

\frac{\pi - ( \phi - \theta)}{2} = \delta

(using

\tan x \approx x

).

So

\pi - \phi + \theta = 2\delta

. Since

0\le \theta, \phi \le \pi

we deduce

\phi >= \pi - 2\delta , \theta \le 2\delta

and so

\pi/2-\delta \le \frac{\theta+\phi}{2} \le \pi /2

\sin(\pi/2 - \delta) \le \sin \frac{\theta+\phi}{2} \le 1 \implies \cos \delta \le \sin \frac{\theta+\phi}{2} \le 1

Since

\cos(\delta) \approx 1- \delta^2/2

we deduce

\sin \frac{\theta+\phi}{2} = 1

ignoring terms of

\delta^2

and higher.

Meanwhile

\sqrt{1+\delta^2} \approx 1 + \delta^2/2 = 1

(ignoring

\delta^2

and higher).

Thus

T = \frac{mg}{2}

ignoring terms in

\left(\frac{r\omega^2}{g}\right)^2

and higher.

Comment: I found this very hard for STEP I. I don't know if I missed some useful trig identity for the last part, or if they expected you just to say "

\tan \frac{\phi-\theta}{2}

is large, so

\phi-\theta

is near pi, so

\phi+\theta

is also near pi, so

\sin \frac{\theta+\phi}{2} \approx 1

. Which seems a bit dubious, even for STEP I. Ah well...

Reply 73

justinsh

I don't agree with dvs on post 51.

It shouldn't be v = pi/2 + kpi but v = pi/2 + 2kpi. (beware the signs!)

A simpler argument may be to say that arg(z)=pi/2=v (mod 2pi) and |z|=e=exp(u).

Reply 74

speedy_s

ukgea

0 < a < 1

, then

\ln a < 0

and so

f'(x)

is strictly positive (again for x>0). This means that there can be at most 1 solution to the equation. Consider now

f(a)

f(a) = \ln a - a\ln a = (1 - a)\ln a

This may be a slightly dumb question...

Ok, so I understand that the function has to be negative once, to prove that it has 1 real root. But, how did you figure out you had to find f(a). Why not some other x value?

Reply 75

nota bene

I think it is the intuitive thing to do, you have f(x)=\ln(x)-x\ln(a) and want to find something that has useful information, and we know something about ln(a) and a, then we want f(x) to be in terms of a.

I do not think there is a better answer to your question :frown:

Reply 76

DFranklin

speedy_s

No reason. f(a^2) works, for example.

Or even an argument along the lines of "

x\ln a \to 0

\ln x \to -\infty

x \to 0

, so

f(x)\to -\infty

x \to 0

. Since f(1) > 0 there must be a sign change in (0,1]".

Reply 77

ukgea

speedy_s

Well, it was a year ago, so I have no clue about what I actually was thinking when I wrote that, but I think generally the train of thought ought to have been something like this:

First, you have to note that what you want to do is find an x where f(x) < 0, just as you have already seen. Having realised that, what you would probably do is try the simplest thing that you can think of, say, f(0.001). This won't work though if a is very very small, because ln a can then be arbitrarily low. So it is clear that you need an x that depends on a. The simplest expression that depends on a is just the expression 'a', and so you try f(a), and voilà, it works!

Reply 78

brianeverit

DFranklin

STEP III, Q10:

The M.I. of a uniform sphere is

\frac{2}{5}ma^2

. (Don't know if you're expected to prove it, but if so it's completely standard bookwork).

Let the frictional force be F, the angular velocity be

\omega

and the linear velocity be v.

Then

\frac{2}{5}ma^2 \dot{\omega} = -Fa,\quad m\dot{v} = -F

, so

\frac{d}{dt}(\frac{2}{5}a\omega-v) = 0

\frac{2}{5}a\omega-v

is constant.

(i) When marble is at a complete stop,

\omega = v = 0

\frac{2}{5}a\omega-v = 0

, so we must have

\frac{2}{5}a\omega_0 = v_0 \implies v_0/(a\omega_0) = 2/5

.

(ii) When marble is rolling backwards with velocity

v_0/7, v=-v_0/7, a\omega = v_0/7

. So we know

Unparseable latex formula:

\displaystyle \frac{2}{5}a\omega-v = \frac{2}{5}\ctimes \frac{v_0}{7} + \frac{v_0}{7} = \frac{v_0}{5}

\displaystyle \frac{2}{5}a\omega_0 - v_0 = \frac{v_0}{5} \implies \frac{2}{5}a\omega_0 = \frac{6}{5} v_0 \implies v_0/(a\omega_0) = 1/3

.

Have I missed something here? Seems ridiculously easy again.

According to the published hints and solutions the answer to the last part should be 14/37

Reply 79

brianeverit

DFranklin

STEP III, Q12.

(i)

\mathbb{P}(K=k) = p(1-p)^{k-1} = pq^{k-1}.

.

So

G(s) = \sum_1^\infty pq^{k-1}s^k = ps \sum_1^\infty (qs)^{k-1} = \frac{ps}{1-qs}

.

Now

G'(s) = \sum_1^\infty k pq^{k-1}s^{k-1}

and so

G'(1) = \sum_1^\infty k pq^{k-1} = \mathbb{E}[K]

.

Take logs:

\ln G(s) = \ln ps - \ln(1-qs) \implies \frac{G'(s)}{G(s)} = \frac{1}{s} + \frac {q}{1-qs}

.

So

G'(s) = \left[\frac{1}{s} + \frac {q}{1-qs}\right]G(s)

. Putting s=1, and noting that G(1) = 1, we get:

\mathbb{E}[K] = G'(1) = 1 + \frac {q}{1-q} = \frac{1}{1-q} = \frac{1}{p}

Similarly

G''(s) = \sum_1^\infty k(k-1) pq^{k-1}s^{k-2}

and so

G''(1) = \sum_1^\infty k(k-1) pq^{k-1} = \mathbb{E}[K(K-1)]

Unparseable latex formula:

= G(s) \left[\frac{-1}{s^2} + \frac {q^2}{(1-qs)^2}\right] + \left[\frac{1}{s} + \frac {q}{1-qs}\right] G'(s)

We don't bother tidying up much, because we only want to find G''(1). Recall

G(1) = 1, G'(1) = \frac{1}{p}

. So

G''(1) = -1 + \frac{q^2}{(1-q)^2}+ \left[1 + \frac {q}{1-q}\right]\frac{1}{p} = -1 + \frac{q^2}{p^2} + \frac{1}{p^2}

=\frac{-p^2+(1-p)^2+1}{p^2} = \frac{2(1-p)}{p^2}=\frac{2q}{p^2}.

Then

\mathbb{E}[K^2] = \mathbb{E}[K(K-1)] + \mathbb{E}[K] = \frac{2q}{p^2} + \frac{1}{p} = \frac{2q+p}{p^2} = \frac{2-p}{p^2}

.

Finally,

Var[K] = \mathbb{E}[K^2] - \mathbb{E}[K]^2 = \frac{2-p}{p^2}-\frac{1}{p^2} = \frac{1-p}{p^2}

.

(ii) Suppose I have already sampled i-1 cards. Then the probability that I will sample a new card on any particular draw is (53-i)/52 . Define

N_i

to be the number of draws needed to sample the new card. Then by part (i),

\mathbb{E}[N_i] = \frac{52}{53-i}

.

Clearly

N = \sum_1^{52}N_i = 52 \sum_1^{52} \frac{1}{53-i} = 52 \sum_1^{52} \frac{1}{i}

.

The approximation given implies

N \approx 52 (0.5 + \ln 52)

.

The fact that they give an approximate value of 'e' implies we're supposed to estimate ln 52.

2.7^2 = 7.29, 7.29^2 =49 + 14 \times 0.29 + 0.29^2 \approx 49 + 4.2 - 0.14 + .09 \approx 53.1

. Deduce

\ln 52 \approx 4

and so

N \approx 52 \times 4.5 = 234

It need not be such a long question if you simply note that G(s) is a probability generating function for K and then quote standard results E(K)=G'(1) etc.