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The Proof is Trivial!

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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.

(edited 11 years ago)

Reply 401

Llewellyn

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.

Reply 402

metaltron

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.

(edited 11 years ago)

Reply 403

Indeterminate

Problem 56 **/***

Evaluate

\displaystyle \int_0^{2\pi} \dfrac{1}{\cos x + 5} \ dx

(edited 11 years ago)

Reply 404

und

Original post by metaltron

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

I'm very disappointed I never gave this question greater thought in the actual exam. This is a nice solution.

(edited 11 years ago)

Reply 405

shamika

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)

Spoiler

Reply 406

metaltron

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

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!

Reply 407

shamika

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

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?

(edited 11 years ago)

Reply 408

und

Original post by shamika

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.

Reply 409

james22

Problem 57 *

find

\displaystyle\int^{1}_{-1} \frac{1+x^4 tan(x)}{1+x^2}\ dx

Reply 410

metaltron

Original post by shamika

The completing the square one was even shorter surely?

Reply 411

und

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)

Spoiler

Reply 412

Star-girl

(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.

(edited 11 years ago)

Reply 413

Star-girl

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:

Reply 414

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}

Reply 415

james22

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.

Reply 416

bananarama2

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 :/

Reply 417

dknt

Well I guess I'll provide another physicsy mathsy problem :colone:

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.