Thanks, I sort of get what you mean. My thinking was that v is the derivative of x so if we integrated something with v in it we would get something in terms of x.(Original post by DFranklin)
Well, you can't really integrate that without ending up doing what Brian did.
Let's write v(x) for v, to emphasise that v is a function of x.
Then you have
So, here's the problem  at this point you have absolutely no idea what v(x) is, which is going to make finding the integral pretty tricky.
So, let's relabel some of the variables: I'm going to write 'x' instead of d, and 't' instead of 'x'.
Now diff both sides w.r.t. x:
Which is Brian's first line. So from here you can solve for v as a function of x, which will give you the desired result.
Obviously it's quicker to just use the v dv/dx form of acceleration to start with.
I´ll think it over.
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STEP Maths I, II, III 1989 solutions

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 81
 13042011 11:08

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 82
 13042011 11:30
You can integrate v w.r.t. x, sure. Integrating 1/v is going to be tricky...

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 83
 13042011 19:43
(Original post by DFranklin)
You can integrate v w.r.t. x, sure. Integrating 1/v is going to be tricky... 
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 84
 14042011 13:06
STEP 1 question 8
A solution has allready been posted using the suggested way but the question let you solve it any other way (apart from newton rapheson et al). This is the way you could do it if , like me, you haven´t done De moivre´s theorem in FP2.
(I am using a spanish keyboard which makes typing in latex a nightmare so, for my own ease, I have let theta=x)
using the identities:
We get:
as required.
Next part:
By standard results:
So, we do the same as in the previous part:
as required.
Next part:
Whenever they give you questions like this, it is worth checking for easy roots. This one has two, x=1 and, noticing that all the powers are even, x=1.
We can then factorise it to get:
is a quadratic in terms of x^2 so all you have to do is use the quadratic formula and you are done. 
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 85
 21042011 17:11
(Original post by Farhan.Hanif93)
STEP I  Q1
solutionfirst partNote that the radius of CDE is . Therefore the length HD is given by .
Considering the area of a circle of this diameter:
.
Consider the areas of the following semicircles
AGH:
ABC and EFG:
CDE:
Note that the area of ABCDEFGH is found by considering the value of . Therefore, since , the area enclosed by ABCDEFGH is equal to the area of a circle of diameter HD, as required.
second partWe now seek to find the equations of the 4 parabolas and then use integration to find the areas.
Consider an (x,y) coordinate system with an origin at A.
The equations of the following parabolas are:
AHG:
ABC:
Considering the area enclosed between these parabolae and the positive xaxis:
Area under parabola AGH = .
Area under parabola ABC = Area under parabola EFG =
Now consider another coordinate system with an origin at the midpoint of the line CE.
Area enclosed by parabola CDE and the xaxis =
Therefore area of ABCDEFGH = . (We subtract because it has a negative area using our coordinate systems and we're looking for an absolute value of the area enclosed by ABCDEFGH)
Therefore, we have the following, where is the area of this shape:
The final answer I got was, . This fitted with my answer for the first part, as , is the length HD.
Slighty of topic, but do you think that we would not lose marks for just stating the equations of the parabolas/ having similar "jumps" in our working for STEP? Is the marking really strict, and do we get our papers back? 
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 86
 11052011 23:30
(Original post by brianeverit)
1989 Paper 3 nos. 37
In it, you say:
"if P=M then
But when k=m not for any value of k which is what you seem to have assumed. Am I reading it wrongly? 
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 87
 07062011 14:54
(Original post by bensmith)
Sorry but I don't understand a line in your solution to STEP III Q3.
In it, you say:
"if P=M then
But when k=m not for any value of k which is what you seem to have assumed. Am I reading it wrongly? 
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 88
 07062011 15:03
Here are my attempts at STEP III numbers 11 and 12.
I'm not at all confident of them so would welcome comments or criticisms. 
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 89
 08062011 19:06
Here is my solution to 1989 Step 3 Question7

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 90
 19062011 17:10
(Original post by squeezebox)
STEP II Q1
Substituting x= ya into the equation:
Notice that the reduced form shown in the question has no term, so we need to find the value of a which will get rid of the term.
Expanding out and collectiing terms we get:
So the value of a to get rid of the is 1.
So we now have:
(*)
Substituting y=z/b into (*);
dividing both sides by 2 and multiplying by ;
.
And so to make this equal to;
we make b = .
Lets take b = and let z =
so;
and (these give distinct values of , and hence, three distinct solutions.)
we know that; x = y  a = (z/b)  a.
Hence the solutions of the equation are:
and
The question also wanted a counterexample to show that not all cubics can be solved by this method and trivially
is such a counterexample 
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 91
 20062011 09:33
(Original post by Dystopia)
STEP III, Q8
Let
Then
Using an integrating factor, , we get
Applying boundary conditions:

I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.

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 92
 27062011 07:51
Here's my solution for Q10, STEP I 1989.
Sorry for dodgy presentation, I am practicing using my new graphics tablet
I realise that LaTeX would be easier to read
If I have understood this question, I don't like it at all... not a good STEP question.
On the other hand I may be wrong in some important way, if so please correct!Post rating:1 
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 93
 02072011 19:03
Here is my solution for Q11 STEP II 1989. (sorry I called it Q10 in the pic!)
I did it because I thought it hadn't been done (because I thought it was Q10, doh!), but anyway you might find my slightly simpler method interesting.
It's a very easy question I think, although it helps a lot to consider relative acceleration.
One last comment: the second part (why the lift comes to rest) I described as "obvious" .. do you see why?Last edited by waxwing; 02072011 at 19:20. Reason: Wrong question number 
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 94
 03072011 07:48
Here is my solution to STEP II 1989 Q12.
Quite easy using Lami's theorem; I suspect the first part might be a bit messy otherwise. 
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 95
 03072011 15:06
Here's my answer for STEP III 1989 Q12.
Note to Brian Everit: I was pleased to see after I did it that we modeled it almost exactly the same way.
The only difference is in the end; my reading of the question suggested they didn't want an explicit solution for but just to deduce from the form of .
But however you look at it, I think you did a great job (That's assuming neither of missed something..) 
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 96
 18072011 07:02
Here is my answer for STEP II 1989 Q4.
After comparing with Brian Everit's solution on post 65, I see that we have exactly the same answer. Only thing different is I have a quicker way of doing the first part.
Not the best question ever  asking students to "tell me how you would prove it, but don't prove it" and then asking them to draw 4 different quadratic/quadratic graphs is pretty evil... 
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 97
 25072011 13:36
(Original post by waxwing)
Here is my answer for STEP II 1989 Q4.
After comparing with Brian Everit's solution on post 65, I see that we have exactly the same answer. Only thing different is I have a quicker way of doing the first part.
Not the best question ever  asking students to "tell me how you would prove it, but don't prove it" and then asking them to draw 4 different quadratic/quadratic graphs is pretty evil... 
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 98
 25072011 16:53
(Original post by brianeverit)
Well done finding that method for the first part. I wish I had thought of it myself.
I can't take if f(x) is negative!
Well, I guess I can if I'm allowing complex answers. 
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 99
 25072011 16:56
(Original post by waxwing)
Yes, it is clever isn't it, except for the tiny detail that I've just realised it doesn't work!
I can't take if f(x) is negative!
Well, I guess I can if I'm allowing complex answers. 
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 100
 25072011 17:15
(Original post by DFranklin)
I don't have the question in front of me, but if you're only interested in the behaviour when x is large then f(x) won't be negative then.
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Updated: November 28, 2015
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