Thanks! It's been so long now I can't reember what my final answer was, buI didn't do it the same way as you: I found the point of unstable equilibrium and calculated the change in GPE needed to get there.

Reply 26

mikelbird

8

Solution to Question 10 of step II

Step2015Paper2Question10.pdf58.8KB

Reply 27

mikelbird

8

for what it is worth regarding Question 9, I do think the answer given makes rather heavy weather of the first two parts....but let it be said the answer is right!!

Step2015Paper2Question9.pdf62.6KB

Reply 28

mikelbird

8

I think this is OK for question 11....any faults please let me know!

Step2015Paper2Question11.pdf80.9KB

Reply 29

brianeverit

9

Original post by Gawain

STEP II 2015

Q1) team-f-maths (attached to first post)
Q2) Gawain
Q3) Gawain
Q4) team-f-maths (attached to first post)
Q5) Gawain
Q6)
Q7) Goose Lane
Q8) Gawain
Q9) [url="http://www.thestudentroom.co.uk/showpost.php?p=57555953&
postcount=25"]Brammer

Q10)
Q11)
Q12)

I'll upload Q2 once I've managed to upload my diagram.

In 4(ii) surely the curve tends to the same horizontal line as x tends to plus or minus infinity.

Reply 30

brianeverit

9

Here are my solutions for questions 12 and 13.

step 2015 II question 12..pdf44.9KB

Step 2015 II question 13..pdf127.6KB

Reply 31

lambda98

2

Original post by Gawain

Q3)Extra triangles gained adding a rod length 8:
(8,7,1-6) = 6 Triangles
(8,6,2,5) = 4 Triangles
(8,5,3-4)= 2 Triangles

So

\displaystyle T_8-T_7=6+4+2 =12

Through a similar method

\displaystyle T_7-T_6=5+3+1=9

Therefore

\displaystyle T_8-T_6=21

Trivially:

\displaystyle T_{2m}-T_{2m-1}=m(m-1)

and

\displaystyle T_{2m}-T{2m-2}=(2m-1)(m-1)

Induction:

Base: With 4 rods we can make 3 triangles, also

\displaystyle T_4=3

so our base case works.

Assumption:

\displaystyle T_{2k}=\frac{k(k-1)(4k+1)}{6}

Induction:

\displaystyle T_{2(k+1)}=(2k+1)(k) + T_{2k}

Simplifying we get:

\displaystyle T_{2k}=\frac{(k+1((k+1)-1)(4(k+1)+1)}{6}

Using the second difference formula we get:

\displaystyle T_{2m-1}=\frac{m(m-1)(4m-5)}{6}

I know it's meant to be the easiest part of the question but can anyone explain why 4 rods make 3 triangles, given that you can onl have one of each length isn't it just 2 triangles?

Reply 32

Alisdair

1

Why are

\tan(\alpha) = -2n^2

and

\tan(\alpha) = \frac{1}{2n^2}

equivalent statements ?

(edited 7 years ago)

Reply 33

DFranklin

18

Original post by Alisdair

Why are $\tan(\alpha) = -2n^2$ and $\tan(\alpha) = \frac{1}{2n^2}$ equivalent statements ?

Obviously, they're not equivalent (barring some other conditions that seem unlikely to be true). Some context, please?

Reply 34

Alisdair

1

Original post by DFranklin

Obviously, they're not equivalent (barring some other conditions that seem unlikely to be true). Some context, please?

Sorry, that was supposed to be a reply to this post :
https://www.thestudentroom.co.uk/showthread.php?t=3411533#post57155041

Reply 35

DFranklin

18

Original post by Alisdair

Sorry, that was supposed to be a reply to this post :
https://www.thestudentroom.co.uk/showthread.php?t=3411533#post57155041

OK, well without having the paper or done the question, I think what they're saying is that when you solve for

\tan \alpha

those are the two possible solutions (i.e. it's NOT saying they're equivalent).

I assume that the question context lets you rule out

\tan \alpha = -2n^2

as a valid solution.

(It should have been explained, I guess, but you should note that there's no guarantee that posted solutions are perfect, or even correct!)

Edit: found the question; since it's an angle in a triangle,

0 \leq 2\alpha \leq \pi

and so we can rule out the -ve solution.

(edited 7 years ago)

Reply 36

Alisdair

1

Original post by DFranklin

OK, well without having the paper or done the question, I think what they're saying is that when you solve for

\tan \alpha

those are the two possible solutions (i.e. it's NOT saying they're equivalent).

I assume that the question context lets you rule out

\tan \alpha = -2n^2

as a valid solution.

(It should have been explained, I guess, but you should note that there's no guarantee that posted solutions are perfect, or even correct!)

Edit: found the question; since it's an angle in a triangle,

0 \leq 2\alpha \leq \pi

and so we can rule out the -ve solution.

Right ! I was a bit stuck there and didn't realise there were two solutions. Thanks !

Reply 37

amd87

3

Original post by Gawain

STEP II 2015

Q1) team-f-maths (attached to first post)
Q2) Gawain
Q3) Gawain
Q4) team-f-maths (attached to first post)
Q5) Gawain
Q6)
Q7) Goose Lane
Q8) Gawain
Q9) Brammer ; Alt - mikelbird
Q10) mikelbird
Q11) mikelbird
Q12) brianeverit
Q13) brianeverit

Paper (Google Drive link courtesy of jneill)

Q4)

\arctan\left (\frac{1}{2} \right ) \neq \frac{\pi}{4}

Reply 38

Alisdair

1

Q6) (almost a complete solution)
I apologise in advance for any typographic errors, it’s late and the post preview isn’t showing me the LaΤεΧ. I’ll fix it as soon as I can.

i)

\sec^2 \left( \frac{\pi}{4} - \frac{x}{2} \right) = \dfrac{1}{\cos^2 \left( \frac{\pi}{4} - \frac{x}{2} \right)}

Now,

\cos 2a = 2\cos^2 a - 1

, therefore

\cos^2 a = \dfrac{\cos 2a +1}{2}

.

Therefore,

\dfrac{1}{\cos^2 \left( \frac{\pi}{4} - \frac{x}{2} \right)} = \dfrac{2}{\cos \left( \frac{\pi}{2} - x \right) + 1} = \dfrac{2}{1 + \cos \left( \frac{\pi}{2} - x \right)} = \dfrac{2}{1 + \sin x }

For the integral:

\displaystyle\frac{1}{2} \int \dfrac{2}{1+\sin x} dx = \frac{1}{2} \int \sec^2 \left(\frac{\pi}{4} - \frac{x}{2} \right) = \tan\left( \frac{x}{2} - \frac{\pi}{4} \right) + C

(From trying

-2\tan \left( \frac{\pi}{4} - \frac{x}{2} \right)

as an antiderivative, and carrying the negative from inside the tangent function to outside, as the tangent function is odd.)

ii)

\displaystyle\int^\pi_0 x f\left(\sin x \right) dx \longrightarrow \int^0_\pi (\pi - y) f \left( \sin(\pi-y) \right) \, d(\pi - y)

\displaystyle = \int^0_\pi \left[ (\pi f \left( \sin(\pi-y) \right) - \pi f \left( \sin(\pi-y) \right) \right] \times -1 \, dy

\displaystyle \int^\pi_0 x f\left(\sin x \right) dx = \pi \int^\pi_0 f\left(\sin y\right) dy - \int^\pi_0 y f\left(\sin y\right) dy

Because this is now exactly in the form originally given, we can pretend all the y-variables are x-variables or, more rigorously, perform the substitution

x=y

.

Therefore

\displaystyle 2 \int^\pi_0 x f\left(\sin x \right) dx = \pi \int^\pi_0 x f\left(\sin x \right) dx

,
and

\displaystyle \int^\pi_0 x f\left(\sin x \right) dx = \frac{\pi}{2} \int^\pi_0 x f\left(\sin x \right) dx

,
as required.

Evaluating

\displaystyle\int^\pi_0 \dfrac{x}{1 + \sin x} dx = \frac{\pi}{2} \int^\pi_0 \dfrac{1}{1 + \sin x} dx

= \left. \dfrac{\pi}{2} \tan \left(\frac{x}{2} - \frac{\pi}{4} \right) \right|_0^\pi

Unparseable latex formula:

= \dfrac{\pi}{2} \left( \tan \left( \frac{\pi}{4} \right) - \tan \left( -\frac{\pi}{4} \right) = \frac{\pi}{2} (1--1) = \pi

iii)

\displaystyle\int_0^\pi \dfrac{2x^3 - 3\pi x^2}{\left(1 + \sin x\right)^2} dx

Try applying the same trick as earlier with

\displaystyle\int^\pi_0 x f\left(\sin x \right) dx

on a higher power of x. I first tried on

x^2

but it turns out we should use

x^3

, substituting

u = \pi - x

.

\displaystyle\int^\pi_0 x^3 f\left(\sin x \right) dx \longrightarrow \int^0_\pi (\pi - u)^3 f \left( \sin(\pi-u) \right) \times -1 du

\displaystyle = \int^\pi_0 (\pi^3 - 3\pi^2 u + 3\pi u^2 - u^3) f \left( \sin u \right) du

Rearranging and applying the rule calculated in part ii on the second term of the binomial expansion gives

\displaystyle\int^\pi_0 \left(\pi^3 - 3\pi^2 \times \frac{\pi}{2} \right) f \left( \sin u \right) + (3\pi u^2 - u^3) f \left( \sin u \right) du

.

Therefore

\displaystyle\int^\pi_0 x^3 f\left(\sin x \right) dx = \int^\pi_0 \left(\pi^3 - 3\pi^2 \times \frac{\pi}{2} \right)f \left( \sin u \right) + (3\pi u^2 - u^3) f \left( \sin u \right) du

.

Replacing u with x and rearranging:

\displaystyle\int^\pi_0 (2x^3 -3\pi x^2) f\left(\sin x \right) dx = -\frac{\pi}{2} \int^\pi_0 f\left(\sin x \right) dx

And therefore,

\displaystyle\int_0^\pi \dfrac{2x^3 - 3\pi x^2}{\left(1 + \sin x\right)^2} dx = -\frac{\pi}{2} \int^\pi_0 \dfrac{1}{\left(1 + \sin x \right)^2} dx

And that’s as far as I got. I haven’t yet been able to work out the integral of

\dfrac{1}{\left(1 + \sin x \right)^2}

.

{Help-request edit: can anyone tell me why the LaΤεΧ formatting isn’t working?} Nevermind, I'm an idiot.

(edited 6 years ago)

Reply 39

etothepiiplusone

15

@DFranklin in Q8 here it's fine to take

\bf{x_1}=\bf{0}

for part ii) and then

\bf{x_2}=\bf{0}

for iii) right, since no origin was specified?

I think this makes the computation significantly easier (than both here, in integral maths and the MS), and leads to an almost elegant solution to a dreaded vectors question :smile:

(edited 5 years ago)

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