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    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 STEP-like with the discriminant. Energies is nice too, although it's usually the last thing I think of.
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    (Original post by und)
    Solution 50

    Using u as the initial velocity.

    Since the function of v is continuous, if v \to 0^{+} for some x, 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  u for which v will become  0 .

    Creating and solving the differential equation, we obtain v^2=\frac{g(\mu +1)x^2}{l}-2\mu g x+u^2, so if v is to be  0 for some positive x, then there must exist a positive root x=\frac{\mu l \pm \sqrt{(\mu gl)^2-u^{2}gl(\mu +1)}}{g(\mu +1)}, so a necessary and sufficient condition (assuming all the constant terms are positive) is u^2 \leq \frac{{\mu}^{2}gl}{(\mu +1)} as required.

    If the rope has mass m, then the impulse applied to accelerate the rope to a speed u is given by the change of momentum mu. Hence \text{Impulse}^2 \leq  \frac{{m^{2}\mu}^{2}gl}{\mu +1}, giving the maximum impulse as m\mu \sqrt{\frac{gl}{\mu+1}}.
    I think this is the most natural solution to the question. But it is very STEP-esque to have multiple possible solutions.
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    (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.
    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.
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    Problem 56 **/***

    Evaluate

    \displaystyle \int_0^{2\pi} \dfrac{1}{\cos x + 5} \ dx
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    (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.
    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:
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    Label the three sides of the triangle a, b and c and assume without loss of generality that a\leq b\leq c. Then a\leq 2, b\leq 3 and c\leq 4. Now consider the two shorter sides of the triangle, a and b. Label the angle between them \theta, and consider the formula for the area of the triangle given by A=\frac{1}{2}ab\sin{\theta}. Clearly this is a maximum when \theta=\frac{\pi}{2}, meaning we have a right triangle. We check that the condition c\leq 4 \Leftrightarrow c^2\leq 16 is satisfied when a and b take their respective maximum values, and it is because c^2=a^2+b^2=13\leq 16. Hence the maximum area is given by A=\frac{1}{2}ab\sin{\frac{\pi}{2  }}=3.


    I'm very disappointed I never gave this question greater thought in the actual exam. This is a nice solution.
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    (Original post by Indeterminate)
    Problem 56 ***

    Evaluate

    \displaystyle \int_0^{2\pi} \dfrac{1}{\cos x + 5} \ dx
    Isn't there a really easy way of doing this, only needing A-Level knowledge? (Don't read the spoiler if you want to do the question)

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    use the t sub?
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    (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:
    Show
    Label the three sides of the triangle a, b and c and assume without loss of generality that a\leq b\leq c. Then a\leq 2, b\leq 3 and c\leq 4. Now consider the two shorter sides of the triangle, a and b. Label the angle between them \theta, and consider the formula for the area of the triangle given by A=\frac{1}{2}ab\sin{\theta}. Clearly this is a maximum when \theta=\frac{\pi}{2}, meaning we have a right triangle. We check that the condition c<4 \Leftrightarrow c^2<16 is satisfied, and it is because c^2=a^2+b^2=13<16. Hence the maximum area is given by A=\frac{1}{2}ab\sin{\frac{\pi}{2  }}=3.


    I'm very disappointed I never gave this question greater thought in the actual exam. This is a nice solution.
    Yeah that's it! It's a really easy question if you can word your answer right, but for some reason in that exam I had so many mental blocks. Was pretty upset after that paper, especially since it was the easiest in ages. Anyway, bring on next year's!
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    (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:
    Show
    Label the three sides of the triangle a, b and c and assume without loss of generality that a\leq b\leq c. Then a\leq 2, b\leq 3 and c\leq 4. Now consider the two shorter sides of the triangle, a and b. Label the angle between them \theta, and consider the formula for the area of the triangle given by A=\frac{1}{2}ab\sin{\theta}. Clearly this is a maximum when \theta=\frac{\pi}{2}, meaning we have a right triangle. We check that the condition c<4 \Leftrightarrow c^2<16 is satisfied, and it is because c^2=a^2+b^2=13<16. Hence the maximum area is given by A=\frac{1}{2}ab\sin{\frac{\pi}{2  }}=3.


    I'm very disappointed I never gave this question greater thought in the actual exam. This is a nice solution.
    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?
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    (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?
    Isn't it OK simply to consider the two shorter sides which are less than 2 and 3 respectively?

    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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    Problem 57 *

    find

    \displaystyle\int^{1}_{-1} \frac{1+x^4 tan(x)}{1+x^2}\ dx
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    (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?
    The completing the square one was even shorter surely?
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    (Original post by shamika)
    Isn't there a really easy way of doing this, only needing A-Level knowledge? (Don't read the spoiler if you want to do the question)

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    use the t sub?
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    The bit in bold misled me into NOT wanting to try Weierstrass!
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    (Almost complete) Solution 27

    I =  \displaystyle\quad \iiint\limits_{x^2+y^2+z^2 \leq 1} \cos(ax+by+cz) \ \, \mathrm{d} x\,\mathrm{d} y\,\mathrm{d}z

    This representation is all well and good, but as noted by my earlier attempts, any attempt to evaluate I 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 \textbf{a} ,\ \textbf{r} are the vectors such that \textbf{a} = \begin{pmatrix} a & b & c \end{pmatrix} ,\ \textbf{r} = \begin{pmatrix} x & y & z \\ \end{pmatrix} , then:

    \textbf{a} \cdot \textbf{r} = \begin{pmatrix} a & b & c \end{pmatrix} \cdot \begin{pmatrix} x & y & z \end{pmatrix} = ax+by+cz

    If we take the unit vector in the direction of \textbf{a} and call it \textbf{e}_3 , then

    \textbf{e}_3 = \dfrac{\textbf{a}}{|\textbf{a}|} . In order to be able to express any point in \mathbb{R} ^3 , we need two more unit vectors which are all perpendicular to each other, say \textbf{e}_1 ,\ \textbf{e}_2 . 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 \mathbb{R} ^3 and so form an orthonormal basis.

    Let these three unit vectors have associated coordinates X,Y,Z . Then since they are orthonormal, the region we are integrating over can still be written in the same form, just with the x,y,z coordinates substituted for X,Y,Z . In our new orthonormal basis:

    \textbf{a} \cdot \textbf{r} = (0,0,|\textbf{a}|) \cdot (X,Y,Z) = |\textbf{a}|Z and so we can rewrite I as:

    I =  \displaystyle\quad \iiint\limits_{X^2+Y^2+Z^2 \leq 1} \cos(\textbf{a} Z) \ \, \mathrm{d} X\,\mathrm{d} Y\,\mathrm{d}Z , with the region in \mathbb{R} ^3 that we are integrating over defined by -1\leq X \leq 1,\ -1\leq Y\leq 1,\ -\sqrt{1-X^2-Y^2} \leq Z\leq \sqrt{1-X^2-Y^2} . However, this could be potentially a troublesome integral because of the Z 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 \mathbb{R} ^3 that we are integrating over as 0\leq r \leq 1,\ 0\leq \theta \leq 2\pi ,\ 0\leq \phi \leq \pi , multiplying by the Jacobian and picking our order of integrating carefully so as not to run into trouble gives:

    \displaystyle I= \int^1_0 \int^{\pi }_0 \int^{2\pi }_0 \cos (|\textbf{a} |r\cos \phi )r^2\sin \phi \ d\theta \ d\phi \ dr

    \displaystyle = \int^1_0 \int^{\pi }_0 \left[ \cos (|\textbf{a} |r\cos \phi )r^2\sin \phi  \cdot \phi \right]_0^{2\pi} \ d\phi \ dr

    \displaystyle = \int^1_0 \int^{\pi }_0 2\pi r^2\cos (|\textbf{a} |r\cos \phi )\sin \phi \ d\phi \ dr

    \displaystyle = \int^1_0  -\frac{2\pi r}{|\textbf{a} |} \left[\sin (|\textbf{a} |r\cos \phi \right]_0^{\pi} \ dr

    \displaystyle =  -\frac{2\pi }{|\textbf{a} |} \int^1_0 r\left( \sin (-|\textbf{a} |r) -\sin (|\textbf{a} |r) \right) \ dr

    \displaystyle =  \frac{4\pi }{|\textbf{a} |} \int^1_0 r\sin (|\textbf{a} |r) \ dr

    One iteration of integration by parts gives

    \displaystyle I = \frac{4\pi }{|\textbf{a} | ^3 }\left[ \sin |\textbf{a} |- | \textbf{a} | \cos | \textbf{a} | \right]  .

    The second part of this question is still to be completed.
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    (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.
    Yeah - DEs can give really neat solutions for mechanics questions, e.g. SHM. :cute:
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    Solution 57

    Let the integral be I. Apply x\to -x to get  I=\displaystyle\int_{-1}^{1} \dfrac{1-x^4\tan x}{1+x^2} dx (by the oddity of tan), then add the two forms of I together and divide by 2 to obtain I = \displaystyle\int_{-1}^{1} \dfrac{1}{1+x^2} dx = \dfrac{\pi}{2}.
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    (Original post by Farhan.Hanif93)
    Solution 57

    Let the integral be I. Apply x\to -x to get  I=\displaystyle\int_{-1}^{1} \dfrac{1-x^4\tan x}{1+x^2} dx (by the oddity of tan), then add the two forms of I together and divide by 2 to obtain I = \displaystyle\int_{-1}^{1} \dfrac{1}{1+x^2} dx = \dfrac{\pi}{2}.
    Nice, my method was to split it into 2 integrals, 1/(1+x^2) and another where the integrand was odd so it went to 0.
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    (Original post by Farhan.Hanif93)
    Solution 57

    Let the integral be I. Apply x\to -x to get  I=\displaystyle\int_{-1}^{1} \dfrac{1-x^4\tan x}{1+x^2} dx (by the oddity of tan), then add the two forms of I together and divide by 2 to obtain I = \displaystyle\int_{-1}^{1} \dfrac{1}{1+x^2} dx = \dfrac{\pi}{2}.
    Damn beat me to it :/
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    Well I guess I'll provide another physicsy mathsy problem

    Problem 58 **/***

    A double pendulum consists of a mass m2 suspended by a rod of length l2 from a mass m1, which is itself suspended by a rod of length l1 from a fixed pivot, as shown below.

    Name:  Untitled.jpg
Views: 225
Size:  12.1 KB

    Show that the equations of motion for small displacements can be written as

    M \ddot{\Theta} = -K \Theta where,

    M= \begin{pmatrix} {l_1} ^2(m_1 +m_2) & l_1 l_2 m_2 \\l_1 l_2 m_2 & {l_2} ^2 m_2 \end{pmatrix}

     K = \begin{pmatrix} gl_1 (m_1 +m_2) & 0 \\ 0 & gl_2 m_2  \end{pmatrix}

    and  \Theta = \begin{pmatrix} \theta_1 \\ \theta_2 \end{pmatrix}
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    Hint for 56:

    Consider the integral

     \displaystyle \int_0^{2\pi} F(\cos x, \sin x) \ dx

    with the substitution z=e^{ix}

    Residue theorem!
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    Solution 57
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    \int^{1}_{-1} \frac{1+x^4 tan(x)}{1+x^2}\ dx
    \int^{1}_{-1} \frac{1}{1+x^2}\ dx + \int^{1}_{-1} \frac{x^4 tan(x)}{1+x^2}\ dx
    \arctan(1) - \arctan(-1) + \int^{1}_{-1} \frac{x^4 tan(x)}{1+x^2}\ dx
     \frac{\pi}{2} + \int^{1}_{-1} \frac{x^4 tan(x)}{1+x^2}\ dx

    Note that \frac{x^4 tan(x)}{1+x^2} is an odd function in the interval (\frac{\pi}{2}, -\frac{\pi}{2}) (as it is the product of the even function \frac{x^4}{1+x^2} and the odd function tanx) and ergo  \int^{1}_{-1} \frac{x^4 tan(x)}{1+x^2}\ dx = 0

    Therefore the answer is  \frac{\pi}{2}


    Why I am so slow with latex T_T

    I may give the residue theorem a go later
 
 
 
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