That works
(Though for the first part, you can do it backwards.
Which is a little easier.)
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STEP Maths I, II, III 1994 Solutions

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 41
 30042007 00:22
Last edited by Rabite; 30042007 at 00:25. 
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 30042007 00:33

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 01052007 17:49
(Original post by nota bene)
STEP III Q5
So we have. Meaning: Where k= 0, 1, 2...
So the coefficients of the terms are
etc. so the MacLaurin expansion is
To find the power series expansion for we can see it as for which we can use the McLaurin series for Ln(1+x).
So:
Even terms cancel and leaves
This means the the power series for g(x) is and thus the coefficient is
Comparing coefficients of the series for f(x) and g(x) we can see that g(x)>f(x). For example the coefficients of which are for g(x) and f(x) respectively.
Do I need a better justification for the coefficients of g(x) being larger? I find it a bit hard to express a general term for the power series of f(x). (well I can get it off wikipedia obviously but I have no idea how to get there and especially how to see it during a test).
Where k= 0, 1, 2...
or
Where k= 0, 1, 2...
Then note that (by comparing the maclaurin series of g(x) that you've found with the :
Where k= 0, 1, 2...
Since g(0)=f(0), it soon follows that for k>1,
i.e the coefficients of maclaurin series of g(x) is greater than that of f(x) except for the first term.
And nota bene, you've done a great job for this question.Last edited by khaixiang; 01052007 at 17:52. 
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 44
 01052007 19:20
STEP III
(ds/dt)^2 + 2gy must always be constant as this equation forms the energy equation (incl. kinetic and GP) multiplied by 2 and divided by m, which is a constant. this equation must be a constant as no external forces are applied to the particle so energy is conserved.
(ds/dt)^2 + 2gk^(1)(s/2)^2=c
so:
2(ds/dt)(d^2s/dt^2) + (sg/k)(ds/dt)=0
=> 2(d^2s/dt^2) + (sg/k)=0
=> (d^2/dt^2)= (g/2k)s
so angular speed= (g/2k)^0.5
so period= 2pi(2k/g)^(0.5)
time taken to reach V is 0.25 of period= pi(k/2g)^(0.5) 
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 45
 01052007 19:36
I quite liked this question, it doesn't really have anything complicated in it... Although it turned out to be a bit long, but not too bad.
*bobo* are you going to add the finishing off on that question? (just looked and there is something with "describe the motion" of the m and thingies...). Good job either way 
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 01052007 19:39
yes just had to go for my tea thoguh

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 01052007 19:52
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.

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 01052007 20:53
STEP II
9) centre of mass from B is a distance (2a/3)
let A be the angle between OA and OB
bsinA/sin(900.5A)=2a
=>bcos0.5Asin0.5A=acos0.5A
=>sin0.5A=a/b
O must be directly above centre of mass so:
bsin(900.5A)/costheta + (2a/3)sintheta= bsin(900.5A+ theta)
=>(b^2 a^2)^0.5/costheta +(2a/3)sintheta= b(cos0.5Acostheta + sin0.5Asintheta)
=(b^2a^2)^0.5costheta + asintheta
=> (a/3)sintheta= (b^2a^2)^0.5sin^2theta/costheta
=>tan theta= a/(3(b^2a^2)^0.5) 
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 01052007 21:12
moments aboutB:
mgcos theta= T(b^2 a^2)^0.5/b
sin theta= costheta a/ (3(b^2a^2)^0.5
1cos^2theta= (acos^2theta)^2/ (9(b^2a^2)
=> 1 T^2(b^2a^2)/(mgb)^2= a^2T^2/9(mgb)^2
=>T^2= 9(mbg)^2/(9b^28a^2)
T= 3mgb/(9b^28a^2)^0.5 
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 50
 02052007 17:01
STEP II
questions 9 and 10 
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 02052007 17:05
q 11

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 02052007 17:06
11 cont. couldnt upload full attachment

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 02052007 17:07
and 3rd part to 11

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 19052007 23:57
STEP I Q 12
i)
ii)
iii) 1(probability of failure) Probability of failure: So the answer is
iv)
v) P(Newnham+New HallNewnham)= Then we know P(N) from ii) and
From that follows
vi) Same as above, both probabilities double and hence cancel.
vii) Now we already have the probabilities needed, from the answer in iii) and the from v). So 
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 20052007 00:33
STEP II question 1
i)work mod 10. So the possible residues of n mod10 are 1,3,7,9. It suffices to show that there exists m mod10 such that n*m=1 (mod10) for each residue of n, so taking m=1,7,3,9 respectively gives the required result.
ii)If n ends in 1,3,7 or 9 we have shown we can multiply by some m to give a number ending in 1. So we only have to worry about n ending in 0,2,4,5,6,8,10. Now, clearly our number is divisible by 2 or 5, so suppose it is divisible by 2 and 5 a certain number of times each. All we have do is find an m containing the correct number of 2s or 5s such that the product n*m contains the same number of factors of 2 and 5. This means that n*m is divisble by 10 this number of times, and furthermore it is not divisible by 2 or 5 any further and hence the last digit is either 1,3,7 or 9. Applying the first part again gives the required result.
iii)Suppose n is a kdigit number as described. Using the number m=900..008000....00020..0010, where there are at least k+1 zeros between the adjacent 9,8,..,2,1 will work.Last edited by ad absurdum; 20052007 at 00:56. 
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 20052007 00:55
STEP II question 6
Proof by induction. For n=1:
, as required
Assume the given formula holds. So, for n+1,
, completing the proof by induction.
For the last part, note that as n> infinity, the angle we are taking the cot of gets very small, and as sin(theta)>theta and cos(theta)>1, we can write the RHS for large n as:
, as required.Post rating:2 
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 57
 20052007 00:58
STEP I  Question 11
i)
To ensure the ball hits the ground, it's mass must be greater than the parallel component of the wagon:
By considering both bodies separately, there will be a tension pulling the wagon up the slope and a tension resisting the ball from falling, by applying N2L, firstly to the wagon:
and to the ball:
Adding these gives:
So:
By taking the length of the slope to be:
And applying the fact the system starts from rest,
Suvat can be applied:
As required.
ii)
If the wagon is not to collide with the pulley, then as the ball is on the point of hitting the floor, it must have a velocity that is less than or equal to 0:
hence:
So:
Provided
QED (I hope)
I'm just in two minds about my reasoning in the last part, if somebody could take a quick look at it that'd be great. 
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 20052007 01:38
I would be interested if anybody could show how to do STEP I Q8. I get the impression that you must use the rathertoocoincidental limits to evaluate the integrals, but I cannot see how. Can anybody help?

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 20052007 02:21
Last edited by insparato; 20052007 at 02:24. 
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 20052007 02:54
I = INT (0 to pi/4) ln(1+tanx) dx
let y = pi/4  x
dy/dx = 1
>  INT (pi/4 to 0) ln(1+tan(pi/4  y)) dy
= INT (0 to pi/4) ln(1+ tan(pi/4  y)) dy
= INT ln(1 + [1tany/1+tany]) dy
= INT ln(2/1+tany) dy
I = INT ln(2/1+tanx) dx
Add original integral:
2I = INT (0 to pi/4) ln(1+tanx) + ln(2/1+tanx) dx
= INT ln2 dx
I = (1/2) INT (0 to pi/4) ln2 dx
= (1/2)(pi/4)ln2
= pi.ln2/8
Tell me if you want me to do the other parts too.
EDIT: glancing at the other parts, the second one looks like x = tanu transforms it to the original integral or very close, the last one is a y = pi/2  x substitution, noting that sin and cos can be swapped over with the given limits, and applying a similar trick to get a simple integral (adding the original integral).Last edited by Speleo; 20052007 at 03:01.Post rating:1
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Updated: February 11, 2017
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