2004 a STEP thread...

OP

STEP I, Question 7

Spoiler

This question is about ordinary generating functions. If this whetted your appetite, consider reading generatingfunctionology

Reply 41

Aurel-Aqua

5

2004 III Question 10
Someone could check the last part.
Diagram:


 --o A -----------
 |       | 6a    |
 |       |       |
x|     ---       |
 |       | x_AE  |
 |       -       | 9a
 |       | x_A   |
 --o P ---       |
         | x_B   |
       ---       |
         | x_BE  |
       ---       |
         | 2a    |
   o B -----------

First things first

x_{Ae}

denotes the extension of AP at equilibrium, and

x_{A}

denotes the extra extension when not in equilibrium, and same for PB.

k_A = \frac{mg}{a}

,

k_B = \frac{3mg}{a}

.

Adding all the lengths, we know that

9a = 8a + x_{Be} + x_{Ae} + x_B + x_A \iff a = \_{Be}+x_{Ae}+x_B+x_A

. Also

x_B+x_A = 0

(*)

At equilibrium,

x_A = x_B = 0

\implies a = x_{Be}+x_{Ae}

and also

mg + k_{B}x_{Be} = k_Ax_{Ae}

(**)

\iff x_Ae(k_A+k_B) = mg + k_Ba \iff x_{Ae} = \frac{mg+k_Ba}{k_A+k_B} = a \iff x_{Be} = 0

.

First part of the motion
Writing the main equation of motion when both strings are in some sort of tension, for

6a < x < 7a

:

m\ddot{x_A} = mg + k_B(x_B+x_{Be})-k_A(x_A+x_{Ae})

. By (**), this is equal to

=k_Bx_B-k_Ax_A

, and by (*), it equals

= -x_A(k_B+k_A) = - \frac{4mg}{a}x_A \iff \ddot{x_A} = -\frac{4g}{a}x_A

. From before, we know that

x = 7a + x_A

\iff \ddot{x_A} = \frac{4g}{a}(x-7a)

. Since accelerations

\ddot{x_A} = \ddot{x}

, result

\ddot{x} = \boxed{\frac{4g}{a}(x-7a)}

follows.

Second part of the motion
When

7a < x < 9a

, PB exerts no force, so

m\ddot{x_A} = mg - k_A(x_a+x_{Ae}) = -k_Ax_A

\iff \ddot{x_A} = \ddot{x} = -\frac{g}{a}(7a-x) = \boxed{\frac{g}{a}(x-7a)}

, as required.

Finding the time to reach B
Considering the differential equation

\ddot{x_A} = - \frac{4}{a}x_A

, we deduce that

x_A = A\cos\left(\frac{2}{\sqrt{a}}t\right)+B\sin \left( \frac{2}{\sqrt{a}} t \right)

\iff x = 7a+A\cos\left(\frac{2}{\sqrt{a}}t\right)+B\sin \left( \frac{2}{\sqrt{a}} t \right)

. Initial conditions:

x(0) = 6a \iff A = -a

,

\dot{x}(0) = 0 \iff \frac{2B}{\sqrt{a}} = 0 \implies x = 7a - a\cos\left(\frac{2}{\sqrt{a}}t\right)

. Finding the lowest point, we get

x(t_1) = 7a \implies t_1 = \frac{\pi\sqrt{a}}{4}

.

Using the second equation found, and using t=0 as the point where x = 7a, we come to

x = 7a+2a\sin\left(\frac{1}{\sqrt{a}}t\right)

.

x(t_2) = 9a \implies \frac{1}{\sqrt{a}}t_2 = \frac{\pi}{2} \iff \frac{\pi\sqrt{a}}{2}

.

Adding the two times, we get

t_T = t_1+t_2 = \boxed{\frac{3\pi\sqrt{a}}{4}}

seconds.

Edit: see sonofdot's post: http://www.thestudentroom.co.uk/showpost.php?p=19452840&postcount=59

Reply 42

Aurel-Aqua

5

2004 II, Question 12

Finding

k

:
Firstly:

\displaystyle 1 = \int_0^1 kxe^{-ax^2}\,\mathrm{d}x = \left[-\frac{k}{2a}e^{-ax^2}\right]_0^1 = \frac{k}{2a}\left(1-e^{-a}\right) = 1 \iff k = \frac{2a}{1-e^{-a}}

.
This gives

\boxed{F(x) =\frac{1}{1-e^{-a}}\left(1-e^{-ax^2}\right)}

.

Finding the mode:
To get the mode, we differentiate to find the maximum point (since

f(0) = 0

and

f(x)>0

for other

x

:

f'(m) = 0 = ke^{-am^2}(1-axm^2) \iff 2am^2 = 1

. In case

a<\frac{1}{2}

,

m^2

will be less than 1, so this case has no mode (since it would be outside the range

0\leq m \leq 1

. In case

a > \frac{1}{2}

,

\boxed{m=\frac{1}{2a}}

.

Reply 43

Aurel-Aqua

5

2004 III, Question 14

If

\sigma_0 < \sigma_1

, then

\bar{X}

lies in a set [smaller] range, then

\sigma_0

should be the variance, but if the deviation is larger, then

\sigma = \sigma_1

must be used as naturally it would be expected to be out of the range of

\sigma_0

. This is acceptable also since only one of the two variances have to be chosen.

Accept

H_0

when true:

\displaystyle = P(\mu-c<\bar{X}<\mu+c) = 2P(\bar{X}-\mu<c) - 1 = 2\Phi\left(\frac{c}{\sigma_0/\sqrt{n}}\right)-1

.

Accept

H_1

when false:

\displaystyle = 1 - (2\Phi\left(\frac{c}{\sigma_0/\sqrt{n}}\right)-1) = 2-2\Phi(z_\alpha) = \alpha \iff c = \frac{\sigma_0z_\alpha}{\sqrt{n}}

.

Accept

H_0

when false:

\displaystyle = \beta = 2\Phi\left(\frac{c}{\sigma_1/\sqrt{n}}\right)-1 = 2\Phi\left(\frac{\sigma_0z_\alpha}{\sigma_1}\right)-1

.

\displaystyle\beta < 0.05 \iff \Phi\left(\frac{\sigma_0z_\alpha}{\sigma_1}\right) < 0.525\iff \frac{\sigma_0z_\alpha}{\sigma_1} < 0.063 \iff \frac{z_\alpha}{0.063} < \cdot\frac{\sigma_1}{\sigma_0}

\displaystyle\alpha < 0.05 \iff 2 - 2\Phi(z_\alpha) < 0.05 \iff \Phi(z_\alpha) >0.0975 \iff z_\alpha > 1.960

Combining the two, we get

\displaystyle\frac{\sigma_1}{\sigma_0}>\frac{1.960}{0.063} \approx 30 \,\,\square

.

Reply 44

Aurel-Aqua

5

2004 II, Question 4

Part (i)
Take the maximum stretch case, and assume that L exactly touches the corner at some point. At that point, the angle of L against the horizontal is equal to

\phi

. By similarity, the angle from the horizontal on the side of b is also

\phi

. The lower part of L in the diagram satisfies

b = L_1\cos{\phi}

and the other part

a = L_2\sin{\phi}

. Since

L = L_1+L_2

,

L = a\mathrm{cosec}\phi + b\sec{\phi}

.

Differentiating w.r.t. L:

\dot{L} = -\frac{a\cos{\phi}}{\sin^2{\phi}} + \frac{b\sin{\phi}}{\cos^2{\phi}}

Since we're working at a maximum problem, we're going to be maximising the value of L. At the maximum,

\displaystyle \dot{L} = 0 \iff a\cos^3{\alpha} = b\sin^3{\alpha} \iff \frac{a}{b}\tan^3{\alpha}

(

\iff

are valid since

\phi

is not divisible by

\pi/2

; and the change between

\phi

and

\alpha

is valid because the latter is now a constant). Substituting backwards, we obtain that

\displaystyle L \leq a\mathrm{cosec}{\alpha} + b\sec{\alpha}

for

\tan^3{\alpha} = \frac{a}{b}

, as required.

Part (ii)
The best way to imagine this problem is as if we're simply extending the sides a and b by a certain bit. We know that the vertical angle of the perpendicular from L intersecting the corner is equal to

\phi

as well, so we can say that

a' = a - w\cos{\phi}

and

b' = b-w\sin{\phi}

.
Thus,

\displaystyle L = a\mathrm{cosec}{\phi}+b\sec{\phi} - w\left(\frac{\cos{\phi}}{\sin{\phi}}+\frac{\sin{ \phi }}{\cos{\phi}}\right)

\displaystyle = a\mathrm{cosec}{\phi}+b\sec{\phi}-w\left(\frac{\sin^2{\phi}+\cos^2{\phi}}{\frac{1}{2}\sin{2\phi}}\right)=

\displaystyle =a\mathrm{cosec}{\phi}+b\sec{\phi}-2w\mathrm{cosec}{2\phi}

.

Differentiating as previously, we get

\displaystyle \dot{L} =-\frac{a\cos{\phi}}{\sin^2{\phi}} + \frac{b\sin{\phi}}{\cos^2{\phi}} +\frac{4w\cos{2\phi}}{\sin^2{2\phi}}

.
Then,

\displaystyle \dot{L} = 0 \iff \frac{1}{\cos^2{\phi}\sin^2{\phi}}\left(-a\cos^3{\phi}+b\sin^3{\phi}+w\cos{2\phi}\right) = 0

\displaystyle w = \frac{a\cos^2{\phi}-b\sin^3{\phi}}{\cos{2\phi}} =

(did I make a mistake here?)

\displaystyle = \left(\frac{a-b}{1-\tan^2{\phi}}\right)\cos{\phi}

, as required. Using the previous method and conditions, the result follows.

Reply 45

Aurel-Aqua

5

Anybody want to do II Q6?

Reply 46

Aurel-Aqua

5

STEP 2004, III, Question 13
(Believe me, it's a mess and I probably did something wrong, so check for yourselves and inform me!)

Representing A, B, C as the peoples' wins and A', B', C' as their dropping outs respectively, we can continue onto the question:

Part (i)
For all to drop out at once, the following are the possibilities:

A'B'C'
ABC.A'B'C'
ABC.ABC.A'B'C'
....

Summing these, we get

\displaystyle P(\text{all out at once}) = (1-p)^3(1+(p^3)+(p^3)^2+....) = \frac{(1-p)^3}{1-p^3}

.

Part (ii)
For two drop out at rth round and one earlier, we must have a case such that:

ABC.ABC.ABC. .... A'BC(r-1).B'C'(r)
ABC.ABC.ABC. .... A'BC(r-2).BC(r-1).B'C'(r)
ABC.ABC.ABC. .... A'BC(r-3).BC(r-2).BC(r-1).B'C'(r)
...

For case r = 5, we sum and multiply by three the probabilities of these cases:

ABC.ABC.ABC.A'BC.B'C'
ABC.ABC.A'BC.BC.B'C'
ABC.A'BC.BC.BC.B'C'
A'BC.BC.BC.BC.B'C'

\iff

(not in horizontal sequence)

A'B'C'.ABC.ABC.ABC.BC
A'B'C'.ABC.ABC.BC.BC
A'B'C'.ABC.BC.BC.BC
A'B'C'.BC.BC.BC.BC

\iff

A'B'C'.(BC)^4.A^3
A'B'C'.(BC)^4.A^2
A'B'C'.(BC)^4.A
A'B'C'.(BC)^4

which for any r equals:

\displaystyle P(Y) = 3(1-p)^3(p^2)^{r-2}\sum_{i=0}^{r-1}p^i =

\displaystyle = 3(1-p)^3p^{2(r-2)}\frac{(1-p^r)}{1-p} = 3(1-p)^2p^{2(r-2)}(1-p^r)

Part (iii)
Probability that exactly one wins is opposite of the probability that (all three drop out at once) plus (two drop out at once with one in any other round).

Thus, it is:

Unparseable latex formula:

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

(which I will not bother to factorise).

Reply 47

SimonM

OP

18

STEP II, Question 6

Spoiler

(edited 10 years ago)

Reply 48

SimonM

OP

18

STEP III, Question 11

Spoiler

Spoiler

Aurel-Aqua

STEP 2004, III, Question 13
(Believe me, it's a mess and I probably did something wrong, so check for yourselves and inform me!)

Spoiler

Don't you want to simplify this? I got:

i)

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

ii)

(p^{r-1} - p^r)^2(1 - p^{r-1})

iii)

1 - \frac{1-p}{1 + p + p^2}

Let me know if it's the same.

Reply 51

maltodextrin

5

DeanK22

TEP III, Question 5

Spoiler

I think the answer to part i) is w + pi and w + 3pi/2

Notice that when cosx = 4/5 sinx = -3/5 and so x cannot equal pi/2 - w as then sinx would be positive

Reply 52

sonofdot

8

Aurel-Aqua

2004 III Question 10

For the last part, I get

t = \frac{3\pi}{4} \sqrt{\frac{a}{g}}

I can't see why your g has disappeared from your DE, so maybe you made a small error there (although of couse its perfectly plausible that I have made a mistake) :smile:

Reply 53

Machodog

I think there is a small mistake in the solution to STEP II 2004 question 1. For part (ii), x=4 does not satisfy the equation given.

Reply 54

Daniel Freedman

7

STEP II 2004, Question 7

Spoiler

Reply 55

Daniel Freedman

7

STEP II 2004, Question 8

Spoiler

Reply 56

Daniel Freedman

7

STEP I 2004, Question 8

Spoiler

Reply 57

SimonM

OP

18

\mathbf{R} \cdot \mathbf{a} = ((1-4k+3l)\mathbf{a} + (1-l)\mathbf{b} + \mathbf{c} ) \cdot \mathbf{a} = ((1-4k+3l)\mathbf{a} \cdot \mathbf{a} + (1-l)\mathbf{b} \cdot \mathbf{a} + \mathbf{c} \cdot \mathbf{a} ) = (1-4k+3l) +(1-l) 3 +(-2)

Reply 58

around

13

Yeah, I thought I'd overlooked something really basic. Also I didn't know the dot product distributed like that :s-smilie:

learn something every day

Reply 59

abstraction98

It's actually a joke, how much of a cakewalk the Oxford admission papers are in comparison.

Although, to be fair, they are sat before the end of the A level curriculum.

2004 a STEP thread...

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