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STEP 2003 Solutions Thread

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SimonM
STEP III, Question 2

Spoiler



can u explain again how u got the first part ??
i can't see the first part.
Reply 101
Adjective
III/8
Taking (a, b)=(5, 2) (a, \ b) = (-5, \ -2) gives n=2 n = 2 so a general solution to the differential equation is:
(y5x)2(y2x)+C \displaystyle (y - 5x)^2(y-2x) + C

Taking (a, b)=(2, 5) (a, \ b) = (-2, \ -5) gives n=12 n = \frac{1}{2} so another solution is:
(y2x)12(y5x)+C \displaystyle (y - 2x)^{\frac{1}{2}}(y-5x) + C

These two solutions are the same, because (y5x)2(y2x)=C \displaystyle (y - 5x)^2(y-2x) = C is the square of (y2x)12(y5x)=c \displaystyle (y - 2x)^{\frac{1}{2}}(y-5x) = c .
Reply 102
wooper
These two solutions are the same, because (y5x)2(y2x)=C \displaystyle (y - 5x)^2(y-2x) = C is the square of (y2x)12(y5x)=c \displaystyle (y - 2x)^{\frac{1}{2}}(y-5x) = c .

You're right! Out of interest, would someone lose marks if they put that these were two different solutions?
Owl_492
You're right! Out of interest, would someone lose marks if they put that these were two different solutions?

One mark at the most probably
Reply 104
If the list is still updated, heres II Q10, 14.
Original post by TwoTwoOne
If the list is still updated, heres II Q10, 14.


SimonM
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Reply 106
I dunno how active this is, but I'm really confused about question 4 (the inequalities one) on the STEP I paper. I mean, if you just draw the graphs of y= sinx + 1 and y=cos x then the answer is bleedingly obvious, but surely that can't be sufficient to get the marks?
Reply 107
I think for STEP drawing a graph would get you most of the marks (after all, they explicitly say that rigour is not required) but if you can also give an algebraic proof than that's not going to hurt.
Reply 108
Original post by blergh22
I dunno how active this is, but I'm really confused about question 4 (the inequalities one) on the STEP I paper. I mean, if you just draw the graphs of y= sinx + 1 and y=cos x then the answer is bleedingly obvious, but surely that can't be sufficient to get the marks?


yup thats it, ain't STEP scary :smile: although you do have to be careful around the edges of the bounds to see if they are valid in the original inequality and also to not miss 1, but apart from that it's pretty much 20 marks for free.

EDIT: ah, got beaten to it, basically what around said as well :smile:
(edited 13 years ago)
2003 STEP III question 9

OPH=12θ so length of HP is 2acos12θ \angle OPH= \dfrac{1}{2} \theta \text{ so length of }HP \text{ is }2a\cos \dfrac{1}{2} \theta
extension of string is thus 2acos12θa tension T=αg(2acos12θa\text{extension of string is thus }2a\cos \dfrac{1}{2}\theta-a \Rightarrow \text{ tension }T=\dfrac{\alpha g(2a\cos \frac{1}{2}\theta}{a}
Equation of motion tangential to circle gives maθ¨=Tsin12θmgsinθ\text{Equation of motion tangential to circle gives }ma\ddot{ \theta}=T \sin\frac{1}{2}\theta-mg\sin\theta
 i.e. θ¨=1a[ag(2acos12θa)sin12θagsinθa]=ga[(a1)sinθsin12θ]\text{ i.e. }\ddot{\theta}=\dfrac{1}{a} \left[\dfrac{ag(2a\cos\frac{1}{2} \theta-a)\sin\frac{1}{2}\theta-ag\sin\theta}{a}\right]=\dfrac{g}{a}[(a-1)\sin\theta-\\sin\frac{1}{2}\theta]
Equilibrium positions exist when θ¨=0 i.e. (α1)sinθαsin12θ=0\text{Equilibrium positions exist when }\ddot{\theta}=0 \text{ i.e. }(\alpha-1)\sin\theta-\alpha\sin\frac{1}{2}\theta=0
2(α1)sin12θcos12θαsin12θ=0sin12θ=0 or cos12θ=α2(α1)\Rightarrow 2(\alpha-1) \sin \frac{1}{2} \theta \cos \frac{1}{2} \theta-\alpha \sin \frac{1}{2} \theta=0 \Rightarrow \sin \frac{1}{2} \theta=0 \text{ or } \cos \frac{1}{2} \theta= \dfrac{\alpha}{2(\alpha-1)}
From sin12θ=0 we have θ=0 and from cos12θ=α2(α1)\text{From }\sin\frac{1}{2}\theta=0 \text{ we have }\theta=0\text{ and from }\cos\frac{1}{2}\theta=\dfrac{ \alpha}{2( \alpha-1)}
 we have a solution in the range 0<θ<π \text{ we have a solution in the range }0<\theta<\pi
if θ<α2(α1)<1 then 2(α1)>αα>2\text{if }\theta<\dfrac{\alpha}{2(\alpha-1)}<1 \text{ then }2(\alpha-1)>\alpha \Rightarrow \alpha>2
if α=2+2 then the above condition is satisfied and θ¨=ga[(1+2)sinθ(2+2)s12θ]\text{if }\alpha=2+\sqrt2 \text{ then the above condition is satisfied and }\ddot{\theta}=\dfrac{g}{a}[(1+\sqrt2)\sin\theta-(2+\sqrt2)s\in\frac{1}{2}\theta]
Unparseable latex formula:

\text{Putting }\theta=\dfrac{\pi}{2}+\phi; \ddot{\theta}=\dfrrac{g}{a}(1+ \sqrt2)[\sin\left(\frac{\pi}{2}+\phi \right)-\sqrt2\left(\sin\frac{1}{2}\left(\frac{\pi}{2}+\phi\right)\right]


ga(1+2)[(sinπ2cosϕ+cosπ2sinϕ)2(sinπ4cos12ϕ+cosπ4sin12ϕ)]\dfrac{g}{a}(1+\sqrt2)\left[\left(\sin\dfrac{\pi}{2}\cos\phi+\cos\dfrac{\pi}{2}\sin\phi \right)-\sqrt2\left(\sin\dfrac{\pi}{4} \cos \dfrac{1}{2}\phi+\cos\dfrac{\pi}{4}\sin\dfrac{1}{2}\phi\right) \right]
=ga(1+2)[cosϕ(cos12ϕ+sin12ϕ)]=\dfrac{g}{a}(1+\sqrt2)\left[\cos\phi-\left(\cos\dfrac{1}{2}\phi+\sin \dfrac{1}{2}\phi\right)\right]
(edited 9 years ago)
2003 STEP II*I question 11

Let particles collide at X after a time t then AX=vt and BX=12t2 \text{Let particles collide at }X \text{ after a time }t \text{ then }AX=vt \text{ and }BX= \dfrac{1}{2}t^2
 By cosine rule in ΔABX v2t2=d2+14a2t4adt2cosβ \text{ By cosine rule in } \Delta ABX \text{ } v^2t^2=d^2+ \dfrac{1}{4}a^2t^4-adt^2 \cos \beta
i.e. a2t44(v2+adcosβ)t2+4d2=0\text{i.e. }a^2t^4-4(v^2+ad \cos \beta)t^2+4d^2=0
which has solutions given by \text{which has solutions given by }
t2=12a2[4(v2+adcosβ)±4(v2+adcosβ)2a2d2]t^2=\dfrac{1}{2a^2}\left[4(v^2+ad \cos \beta) \pm 4 \sqrt{(v^2+ad \cos \beta)^2-a^2d^2} \right]
and solutions exist if (v2+adcosβ)2>a2d2\text{and solutions exist if }(v^2+ad \cos \beta)^2>a^2d^2
Unparseable latex formula:

\Rightarrowv^2+ad \cos \beta>ad \Rightarrow v^2>ad(1- \cos \beta)


Taking the smaller of the two values BX=12at2=1a[v2+adcosβ(v2+adcosβ)2a2d2]\text{Taking the smaller of the two values }BX=\dfrac{1}{2}at^2=\dfrac{1}{a}\left[v^2+ad \cos \beta-\sqrt{(v^2+ad \cos \beta)^2-a^2d^2} \right]
BXd=(cosβ+v2ad)(cosβ+v2ad)21<1 since xx21<1 for all x \Rightarrow \dfrac{BX}{d}=\left(\cos \beta+\dfrac{v^2}{ad} \right)- \sqrt {\left(\cos \beta+ \dfrac{v^2}{ad} \right)^2-1}<1 \text{ since }x-\sqrt{x^2-1}<1 \text{ for all }x
 hence BXd<1BX<d so P has moved a distance less than d\text{ hence } \dfrac{BX}{d}<1 \Rightarrow BX<d \text{ so }P \text{ has moved a distance less than } d
2003 STEP III questyion 12.
I believe thi9s is the last unanswered question for 2003

The number of interruptions in a 10 minute period has a Poisson distribution with mean110\text{The number of interruptions in a 10 minute period has a Poisson distribution with mean}\dfrac{1}{10}
Probability of less than 10 minutes to next interruption is  \text{Probability of less than 10 minutes to next interruption is }
1Pr(no interruption in next 10 minutes1-Pr(\text{no interruption in next 10 minutes}
i.e. 1e0.1 so p.d.f of T is f(t)=0.1e0.1t\text{i.e. }1-\text{e}^{-0.1} \text{ so p.d.f of }T \text{ is f}(t)=0.1\text{e}^{-0.1t}
If first uninterrupted period of 30 minutes is the Nth then the next N1\text{If first uninterrupted period of 30 minutes is the }N^{th} \text{ then the next }N-1
 uninterrupted periods must each be less than 30 minutes, so \text{ uninterrupted periods must each be less than 30 minutes, so }
Pr(N=n)=(030f(t)dt)n1(30f(t)dt)=(1e3)n1×e3Pr(N=n)=\left( \displaystyle \int_0^{30} \text{f}(t)dt \right)^{n-1} \left( \displaystyle \int_{30}^{\infty} \text{f}(t)dt \right)=(1-\text{e}^{-3})^{n-1} \times \text{e}^{-3}
We see that this is the general term of a geometric probability distribution with p=e3\text{We see that this is the general term of a geometric probability distribution with }p=\text{e}^{-3}
Hence, E(N)=1p=e3\text{Hence, E}(N)= \dfrac{1}{p}=\text{e}^3
The number of interruptions before thye first of 30 minutes or more is N1\text{The number of interruptions before thye first of 30 minutes or more is }N-1
so it has expectation e31\text{so it has expectation }\text{e}^3-1
Density function for periods that are less than 30 minutews apart is f(t)k where k is a constant \text{Density function for periods that are less than 30 minutews apart is } \dfrac{\text{f}(t)}{k} \text{ where }k \text{ is a constant }
to scale the area under the graph to unit area, i.e. k=030f(t)dt\text{to scale the area under the graph to unit area, i.e. }k=\displaystyle \int_0^{30} \text{f}(t)dt
The density function for time between interruiptions that are less than 30 minutes apart\text{The density function for time between interruiptions that are less than 30 minutes apart}
Unparseable latex formula:

\text{will be given by }\dfrac{\text{f}(t)}{\int_0^{30} \text{f}(t)dt}\text{ for }0 \leqt \leq30 \text{ the denominator being required to rescale the area under the curve}


since we are given that t<30. Hence, expected wait before first uninterrupted period of 30 minutes or more is\text{since we are given that }t<30. \text{ Hence, expected wait before first uninterrupted period of 30 minutes or more is}
0300.1e0.1tdt1e3=[(t+10)e0.1t1e3]030=40e3+101e3=10(e34)e31\dfrac{\int_0^{30} 0.1\text{e}^{-0.1t}dt}{1-\text{e}^{-3}}=\left[ \dfrac{-(t+10) \text{e}^{-0.1t}}{1-\text{e}^{-3}} \right]_0^{30}= \dfrac{-40 \text{e}^{-3}+10}{1- \text{e}^{-3}}= \dfrac{10( \text{e}^3-4)}{ \text{e}^3-1}
hence, expected wait before first uninterrupted period of 30 minutes or more is \text{hence, expected wait before first uninterrupted period of 30 minutes or more is }
(e31)10(e34)e31=10(e34)161 minutes (\text{e}^3-1) \dfrac{10(\text{e}^3-4)}{\text{e}^3-1}=10( \text{e}^3-4) \approx{161} \text{ minutes}
I don't quite understand the last part. Does anyone know why the sequence is increasing?
Original post by number23
I don't quite understand the last part. Does anyone know why the sequence is increasing?


Haven't actually read the question but by the looks of the working he's binomially expanded (1+x3)12(1+\frac{x}{3})^{\frac{1}{2}} so you get increasing powers of x.
Reply 114
Original post by number23
I don't quite understand the last part. Does anyone know why the sequence is increasing?


In the first bit he expanded:

x^3 * (1+3/x)^1/2

In the second part he's took a factor of 3/x out:

x^3 * (3/x( 1+ x/3))^1/2 = x^3 * (3/x)^1/2 * (1+x/3)^1/2, then expand.
Original post by number23
I don't quite understand the last part. Does anyone know why the sequence is increasing?

He binomially expanded (1 + ax^b)^n where a, b, n were positive; so all the terms are positive and the powers of x grow positively.
Original post by hassi94
Haven't actually read the question but by the looks of the working he's binomially expanded (1+x3)12(1+\frac{x}{3})^{\frac{1}{2}} so you get increasing powers of x.




Original post by f1mad
In the first bit he expanded:

x^3 * (1+3/x)^1/2

In the second part he's took a factor of 3/x out:

x^3 * (3/x( 1+ x/3))^1/2 = x^3 * (3/x)^1/2 * (1+x/3)^1/2, then expand.




Original post by Physics Enemy
He binomially expanded (1 + ax^b)^n where a, b, n were positive; so all the terms are positive and the powers of x grow positively.



I'm just not quite sure about the second from last line of working, since there is a negative term?
Reply 117
Original post by number23
I'm just not quite sure about the second from last line of working, since there is a negative term?


That's because of the binomial expansion.

The point is, is that the powers of x are increasing.
Reply 118
Original post by tommm
STEP II 2003 Q8

Spoiler




the ratio of the stationary values are incorrect, in fact the required value is
y(2)y(1)=(3/2)6ke52k\frac{y(2)}{y(1)} = (3/2)^{6k}e^{-\frac{5}{2}k}

this is just caused by some minor arithmetic mistakes in the step before. The whole thing is generally correct
Reply 119
Original post by tommm
STEP II 2003 Q4

Spoiler



Upon using the substitution y=Rcosuy = R\cos u, and some manipulation with the cosine double-angle formula, we get

R2(u0.5sin(2u)0arccos(d/R)R^2(u - 0.5\sin(2u)|^{\arccos(d/R)}_0 (N.B. that represents limits of integration)

The difficulty in evaluating this lies in the evaluation of 0.5sin(2arccos(d/R))0.5\sin(2\arccos(d/R)). First, using the sine double angle formula, we get

0.5sin(2arccos(d/R))=sin(arccos(d/R))cos(arccos(d/R))=(d/R)sin(arccos(d/R))0.5\sin(2\arccos(d/R)) = \sin(\arccos(d/R))\cos(\arccos(d/R)) = (d/R)\sin(\arccos(d/R))

By drawing a suitable right-angled triangle, we find that
Unparseable latex formula:

\sin(2 \arccos (d/R)) = \frac{\sqrt{R^2 - d^2}{R}}

and from this we can find the required result.


To find the area in the three listed cases, we note that i) and ii) are both specific versions of iii), therefore we need only find the answer to iii) and i) and ii) follow easily.

From a diagram, we see that we must replace d with the perpendicular distance from the line to the furthest point on the circle from the line. This gives us:

d+D=Rd + D = R, where D is the shortest distance from the origin to the line. The distance from a point (x, y) to the centre is given by

D2=(xa)2+(yb)2D^2 = (x - a)^2 + (y - b)^2

But the point (x, y) lies on the line y = mx + c, therefore

D2=(xa)2+(mx+cb)2D^2 = (x - a)^2 + (mx + c - b)^2

To find the minimum value of D, we must differentiate this expression and set it equal to 0. This gives us

xa+m2x+mcmb=0x - a + m^2x + mc - mb = 0

    x=mbmc+am2+1\implies x = \displaystyle\frac{mb - mc + a}{m^2 + 1}

substituting this back into our expression for D^2, we get

D2=(mbmcm2a)2+(ma+cb)2(m2+1)2D^2 = \displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}

and therefore, to find the new area, we must modify (*) such that

d=R(mbmcm2a)2+(ma+cb)2(m2+1)2d = R - \sqrt{\displaystyle\frac{(mb - mc - m^2a)^2 + (ma + c - b)^2}{(m^2 + 1)^2}}

The answer to i) follows by letting m = 0; the answer to ii) follows by letting a = b = 0.



Instead of the integration, could you just use the formulae for an area of a triangle and area of a sector for the first part?

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