I heve retyped this in npost no. 235(Original post by coffeym)
STEP I Q5
LetUnparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\displaystyle \psi=\int^{\beta}_{\alpha}{\frac {1}{\sqrt{(x\alpha)(\betax)}}\,dx
Now if we let then we can say:
and
Also, (best not to write this in a simpler form)
Substituting all of these results into the integral, and changing limits as required, we reach the result:
as required by the first part.
If , then simply
For the next part, let
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\displaystyle \omega=\int^{\beta}_{\alpha}{\fr ac{1}{x\sqrt{(x\alpha)(\betax)}}\,dx
and make the substitution
Hence, after making the substitution, and swapping limits with help from the minus sign, we reach:
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\displaystyle \omega=\int^{\frac{1}{\alpha}}_{ \frac{1}{\beta}}{{\frac{1}{y\sqr t{(\frac{1}{y}\alpha)(\beta\frac{1}{y})}}\,dx
Multiplying the lone inside the square root, we find that
Unparseable or potentially dangerous latex formula. Error 4: no dvi output from LaTeX. It is likely that your formula contains syntax errors or worse.\displaystyle =\int^{\frac{1}{\alpha}}_{\frac{ 1}{\beta}}{{\frac{1}{\sqrt{\alph a\beta (\frac{1}{\alpha}y)(y\frac{1}{\beta})}}}\,dy
Now note that, taking out the factor of , that the integral is simply the same as the one we evaluated in the first part, with certain symbols being inverted. This does not change the value of the integral, as everything is consistent. Hence
as required.
This completes the question.
STEP Maths I,II,III 1987 Solutions
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(Original post by bensmith)
(I've posted some solutions in the 1992 thread if you care)
STEP III Q11
Call the position of the mother A, the point where she enters the river B, and the position of the child C.
So, total time
Differentiating,
set the derivative to zero to find the minimum time. x must satisfy that.
For the next part, is the same but the velocity in has an extra in it's horizontal component. Think of it's velocity in terms of her swimming vector incllned at an angle and then add on the horizontal component due to the river at the end. Because she can only swim in straight lines:
. Using the identity:
and rearranging we get:
which is a quadratic in , so, using the quadratic formula:
. We can discard the 'plus' root as it gives theta to be which makes no sense because the mother would never reach the child if she had no vertical component in her velocity.So:
To find the speed for we use pythagoras:
, substituting in and simplifying:
so, just like in the previous part:
and differentiating we get:
letting to find minima and rearranging:
as required.
Can't really upload the sketches so if anyone wants to do it be my guest
is the vertical distance as she enters the river at time t
is the horizontal distance still.
Then by pythagoras as before (where t is the journy time in the river):
Rearrange for t (the square bits in t cancel)
as desired 
(Original post by themaths)
Hey i have a much easier way in showing the second equation in x. Think of the relative velocity of the river to the position of the girl in the river.
is the vertical distance as she enters the river at time t
is the horizontal distance still.
Then by pythagoras as before (where t is the journy time in the river):
Rearrange for t (the square bits in t cancel)
as desired 
(Original post by bensmith)
Sorry, I don't follow. how is that last result the same as mine? Have you skipped over the calculus? I'm baffled as to how you could find the x that results in the shortest possible total time time without calculus.
I just meant for the part where you were using theta and omitting roots of quadratics and things to find her journey time in the river, when i did it i just used relative velocities to find the time she was in the river. Everything else after that i did the same as you (assuming the diagrams were the same also but they were quite standard after getting the result).
Sorry i didn't really make it clear reading my comment back 
(Original post by themaths)
Oh no i did the calculus part exactly the same as you to find the minimum time you obviously need to take dT/dx as 0 to find the minimum.
I just meant for the part where you were using theta and omitting roots of quadratics and things to find her journey time in the river, when i did it i just used relative velocities to find the time she was in the river. Everything else after that i did the same as you (assuming the diagrams were the same also but they were quite standard after getting the result).
Sorry i didn't really make it clear reading my comment back 
(Original post by brianeverit)
1987 STEP Fma numbers 12 16
(Original post by SimonM)
...
Brian did this by using an energy argument. I have a different way.
let denote the unit tangent vector and the tension in the string respectively.
If the bead can be in static equilibrium fo all positions then:
Multiplying through by an integrating factor of we get:
where C is a constant.
The minimum obviously occurs when
Note that this is the same answer as Brian's. I am taking angles about the x axis whereas he has taken them about the y. 
*Subscribe*

(Original post by Swayum)
STEP I, question 4.
Let
Take each term to logarithm base 2 (change the base rule)
Split the second brackets into two parts.
P = 1 + 1/4 + 1/16...
N = 1/2  1/8  1/32...
So
P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:
N is a geometric series with first term 1/2 and common ratio 1/4. Its sum to infinity is:
Change the base to e, so we have the natural logarithm.
Using the power rule
(multiplying top and bottom by 1/2)
WWWWW. 
(Original post by brianeverit)
1987 Paper 2 no.11 
(Original post by brianeverit)
1987 Paper 2 no.11 
(Original post by nota bene)
I'll post up my attempt to a solution to I/7, using the approach with taking the imaginary parts...
Spoiler:Show
Note that the later is a G.P., where .
Using the sum for a G.P. and we have:
Now, note that [is shown by writing it as cos(a)+isin(a)cos(a)isin(a)=isin(a)isin(a)].
Then we have
Now, taking the imaginary part
And substituting back for u:
Note that, so
To the second part, where the series concerned can be written as
The latter is now a G.P.
Again, introducing will simplify the latexing a bit...
Now, use the fact that [show it by writing cos(a)+isin(a)+cos(a)+isin(a)=cos(a)+cos(a)]
Therefore
Now, taking the imaginary part (and substituting back for u):
By earlier argument and by symmetry follows that
Putting k=10 into my above general formula we have = [using the product formula for cos(a)sin(b)] And the top of that fraction is equal to and thus the required answer in
Tell me if I've made some silly mistakes, quite likely...
edit: The solution should now be correct.
I have compressed things slightly and I think you have the wrong sign for the last part...see document.... 
(Original post by mikelbird)
I have compressed things slightly and I think you have the wrong sign for the last part...see document.... 
(Original post by Seernb)
Follow up post. I'm now not sure you have the right answer. Suppose theta is zero. Then there doesn't need to be any frictional force to keep rod in place. Hence coefficient of friction can be zero, so "mu" = tan(alpha)sin(theta). 
(Original post by mikelbird)
Slightly more succinct and using the given hint.....
to:
I have already done this question, but I can't see the link between the two without having to write a couple terms out. 
(Original post by Blazy)
Sorry for being a bit daft, but could you please tell me how you got from:
to:
I have already done this question, but I can't see the link between the two without having to write a couple terms out. 
(Original post by gff)
GP? . The good old formula should work, then think about where this infinity symbol goes.
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