I tend to avoid second order differential equations because they can become quite nasty. In this case solving the first order DE was somewhat neater and very STEPlike with the discriminant. Energies is nice too, although it's usually the last thing I think of.

und
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Last edited by und; 14042013 at 00:01. 
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(Original post by und)
Solution 50
Using as the initial velocity.
Since the function of is continuous, if for some , then at that instant the frictional force must be greater than the weight, thus the rope will remain stationery. It remains to find a range of values of for which will become .
Creating and solving the differential equation, we obtain , so if is to be for some positive , then there must exist a positive root , so a necessary and sufficient condition (assuming all the constant terms are positive) is as required.
If the rope has mass , then the impulse applied to accelerate the rope to a speed is given by the change of momentum . Hence , giving the maximum impulse as . 
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 14042013 00:05
(Original post by und)
Sorry to be blunt but this is all rather unconvincing. Consider what happens if we decide to make theta smaller. We can then increase ab.
If it wasn't clear enough in my original post I am editing it now.
Edit: I'll admit that my original post didn't have the best explanation, but the idea was there that you couldn't get longer shortest side lengths than of 2 and 3.Last edited by metaltron; 14042013 at 00:12. 
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 14042013 00:23
Last edited by Indeterminate; 14042013 at 00:51. 
und
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 14042013 00:27
(Original post by metaltron)
I said that if you increase the side length of 3, then the other side can becomes at most 3. This means that you have this new side of length 3 and the old one of length 2. If you increase the one of length 2 the other side must become at most 2, hence you still have sides of at most length 3 and 2. If you increase both then you have sides lengths of at most 2, just over 3 and just over 3. But you can have side lengths of 2 and 3 and one larger so this still gives the most area as the area depends only on the two sides 2 and 3 and the angle between them.
If it wasn't clear enough in my original post I am editing it now.
Edit: I'll admit that my original post didn't have the best explanation, but the idea was there that you couldn't get longer shortest side lengths than of 2 and 3.
Spoiler:ShowLabel the three sides of the triangle , and and assume without loss of generality that . Then , and . Now consider the two shorter sides of the triangle, and . Label the angle between them , and consider the formula for the area of the triangle given by . Clearly this is a maximum when , meaning we have a right triangle. We check that the condition is satisfied when and take their respective maximum values, and it is because . Hence the maximum area is given by .
I'm very disappointed I never gave this question greater thought in the actual exam. This is a nice solution.Last edited by und; 14042013 at 02:42. 
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 14042013 00:30
Spoiler:Showuse the t sub? 
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 14042013 00:32
(Original post by und)
I see what you mean, but it could have been explained better in the solution. So your argument is as follows if I understood correctly.
Spoiler:ShowLabel the three sides of the triangle , and and assume without loss of generality that . Then , and . Now consider the two shorter sides of the triangle, and . Label the angle between them , and consider the formula for the area of the triangle given by . Clearly this is a maximum when , meaning we have a right triangle. We check that the condition is satisfied, and it is because . Hence the maximum area is given by .
I'm very disappointed I never gave this question greater thought in the actual exam. This is a nice solution. 
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 14042013 00:35
(Original post by und)
I see what you mean, but it could have been explained better in the solution. So your argument is as follows if I understood correctly.
Spoiler:ShowLabel the three sides of the triangle , and and assume without loss of generality that . Then , and . Now consider the two shorter sides of the triangle, and . Label the angle between them , and consider the formula for the area of the triangle given by . Clearly this is a maximum when , meaning we have a right triangle. We check that the condition is satisfied, and it is because . Hence the maximum area is given by .
I'm very disappointed I never gave this question greater thought in the actual exam. This is a nice solution.
"We cannot use a side of length 4 as one of the legs of the triangle, as the hypotenuse would be larger than 4, which isn't allowed. Thus take the legs of the triangle to be 2 and 3; by Pythagoras, the hypotenuse is 3^2 + 2^2 = 13 < 4^2 (so satisfies the restrictions). Hence the maximum area of the triangle is (1/2)*2*3=3."
(This is obviously the same as what you've written, except I think you should make the point about the a leg of length 4 explicit.)
Are all BMO1 problems this short?Last edited by shamika; 14042013 at 00:39. 
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 14042013 00:40
(Original post by shamika)
The better way of explaining this I think is once you get to the formula for the area, to say something like:
"We cannot use a side of length 4 as one of the legs of the triangle, as the hypotenuse would be larger than 4, which isn't allowed. Thus take the legs of the triangle to be 2 and 3; by Pythagoras, the hypotenuse is 3^2 + 2^2 = 13 < 4^2 (so satisfies the restrictions). Hence the maximum area of the triangle is (1/2)*2*3=3."
Are all BMO1 problems this short?
This year the paper was much easier than usual. The only reasonably difficult question was Q6  I wasted a lot of time trying to solve that one without any success. 
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 14042013 00:41
(Original post by shamika)
The better way of explaining this I think is once you get to the formula for the area, to say something like:
"We cannot use a side of length 4 as one of the legs of the triangle, as the hypotenuse would be larger than 4, which isn't allowed. Thus take the legs of the triangle to be 2 and 3; by Pythagoras, the hypotenuse is 3^2 + 2^2 = 13 < 4^2 (so satisfies the restrictions). Hence the maximum area of the triangle is (1/2)*2*3=3."
(This is obviously the same as what you've written, except I think you should make the point about the a leg of length 4 explicit.)
Are all BMO1 problems this short? 
und
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 14042013 00:47
(Original post by shamika)
Isn't there a really easy way of doing this, only needing ALevel knowledge? (Don't read the spoiler if you want to do the question)
Spoiler:Showuse the t sub?Spoiler:ShowThe bit in bold misled me into NOT wanting to try Weierstrass! 
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 14042013 00:49
(Almost complete) Solution 27
This representation is all well and good, but as noted by my earlier attempts, any attempt to evaluate in any coordinate system with the integrand in this form leads to a very cumbersome/impossible integral. The aim is thus to represent it in an equivalent form that will make the integration easier.
Notice that if are the vectors such that , then:
If we take the unit vector in the direction of and call it , then
. In order to be able to express any point in , we need two more unit vectors which are all perpendicular to each other, say . As the three vectors are mutually orthogonal and of unit length in their directions, then they can now form an alternative way of expressing points in and so form an orthonormal basis.
Let these three unit vectors have associated coordinates . Then since they are orthonormal, the region we are integrating over can still be written in the same form, just with the coordinates substituted for . In our new orthonormal basis:
and so we can rewrite as:
, with the region in that we are integrating over defined by . However, this could be potentially a troublesome integral because of the limits and so this integral would better be evaluated in the spherical coordinate system rather than the cartesian. So changing the variables over redefines the region in that we are integrating over as , multiplying by the Jacobian and picking our order of integrating carefully so as not to run into trouble gives:
One iteration of integration by parts gives
The second part of this question is still to be completed.Last edited by Stargirl; 14042013 at 09:27. Reason: LaTeX fails as usual... 
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 14042013 00:53
(Original post by DJMayes)
No, that's fair enough. I tend to opt for Differential Equations because I like transforming a Mechanics question into a Pure question (And one of the nicer Pure topics in my opinion) but some of the energy solutions are really nice. 
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 14042013 00:55
Solution 57
Let the integral be . Apply to get (by the oddity of tan), then add the two forms of together and divide by 2 to obtain . 
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(Original post by Farhan.Hanif93)
Solution 57
Let the integral be . Apply to get (by the oddity of tan), then add the two forms of together and divide by 2 to obtain . 
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(Original post by Farhan.Hanif93)
Solution 57
Let the integral be . Apply to get (by the oddity of tan), then add the two forms of together and divide by 2 to obtain . 
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 14042013 01:06
Well I guess I'll provide another physicsy mathsy problem
Problem 58 **/***
A double pendulum consists of a mass m_{2} suspended by a rod of length l_{2} from a mass m_{1}, which is itself suspended by a rod of length l_{1} from a fixed pivot, as shown below.
Show that the equations of motion for small displacements can be written as
where,
and 
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 14042013 01:06
Solution 57
Spoiler:Show
Why I am so slow with latex T_T
I may give the residue theorem a go laterLast edited by Llewellyn; 14042013 at 01:11.
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