We want to find the value of x for which the angular velocity is greatest, so we need to differentiate and set equal to 0. Writing this as y = f(x)/g(x) and applying the quotient rule, we know that this is 0 when f'(x)g(x)-g'(x)f(x) = 0, so let's do some differentiating:

f(x) = 6u(1+e)(a+x)

f'(x) = 6u(1+e)

g(x) = a^2+3x^2 + 6(a+x)^2

g'(x) = 6x + 12(a+x)

So

6u(1+e)[a^2+3x^2 + 6(a+x)^2] - 6u(1+e)(a+x)[6x + 12(a+x)] = 0

a^2 + 3x^2 + 6(a+x)^2 - 6x(a+x) - 12(a+x)^2 = 0

6(a+x)^2 + 6x(a+x) - 3x^2 - a^2 = 0

9x^2 + 18ax + 5a^2 = 0

(3x+a)(3x+5a) = 0

Giving solutions of x = -a/3, x = -5a/3. We reject the latter as it lies off of the rod and take x = -a/3 which plugs into the equation for y to give the angular velocity as u(1+e)/a, as required. In order to show that this is a maximum, we simply plug in a value of x close (e.g. x = 0 and note that the resultant velocity is smaller such that this angular velocity must be greatest.

(edited 10 years ago)

Reply 145

tinto99

1

Original post by GG.TintiN

I forgot what I wrote during the exam...nonetheless I think I got the answer because i was very focused after I failed the 1st attempt by y=ux and used y=ux^2 for second trial

good luck if you are taking III

Do you mean you got the answer with the e^ in it? I still fail to see how using the substitution ux^2 (which I used, and am pretty sure is correct) leads to any ln's or e's

Reply 146

mikelbird

8

I think this is OK for Step 1 Q6

Step2013Paper1Question6.pdf73.7KB

Reply 147

username937760

16

Q7 - STEP III

Part i):

Spoiler

Part ii):

Spoiler

Part iii):

Spoiler

(edited 10 years ago)

Reply 148

bananarama2

Q9 STEP III

This seems very dubious

Spoiler

(edited 10 years ago)

Reply 149

Pyoro

8

STEP III 2013 – Q1
If you spot an error, let me know!
Right-o! This is the tan half-angle substitution taught in further pure maths. The formulae below can be deduced by differentiating and by using formulae for the sine and cosine of an inverse tangent – justify these using right-angled triangles (labelled with angle x/2) or algebra, remembering that x is restricted to -π < x < π.

Unparseable latex formula:

x = 2\tan ^{-1}\left( t\right) \Rightarrow t=\tan \left( x / 2\right) \\[br]\dfrac {dx} {dt}=\dfrac {2} {1+t^{2}}\Rightarrow \dfrac {dt} {dx}=\dfrac {1} {2}\left( 1+t^{2}\right) \\[br]\sin \left( x\right) =2\sin \left( x / 2\right) \cos \left( x / 2\right) =2\dfrac {t} {\sqrt {1+t^{2}}}\dfrac {1} {\sqrt {1+t^{2}}}=\dfrac {2t} {1+t^{2}}[br]

The first result then transpires using this substitution. Two points to note:

•

The integrand turns into a rational function with quadratic denominator. After completing the square and substituting the squared term, it becomes a standard inverse tan integral, because 1 – a² is positive (since 0 < a < 1).

•

The result of integration is the difference between two inverse tan functions, A and B. This is simplified by considering A – B as tan^-1(tan(A – B)) (notice A – B is within the range of tan^-1(x)) and simplifying using the multiple angle formula on the inner bracket.

Unparseable latex formula:

\displaystyle\int _{0}^{\pi/2}\dfrac {dx} {1+a\sin \left( x\right) } = \displaystyle\int _{\tan \left( 0\right) }^{\tan \left(\pi/4\right) }\dfrac {\dfrac {2dt} {1+t^{2}}} {1+\dfrac {2at} {1+t^{2}}} = \displaystyle\int _{0}^{1}\dfrac {2dt} {t^{2}+2at+1} \\[br]= \displaystyle\int _{0}^{1}\dfrac {2dt} {\left( t+a\right) ^{2}+1-a^{2}} = 2\displaystyle\int _{a}^{1+a}\dfrac {du} {u^{2}+\left( 1-a^{2}\right) } \\[br]=\dfrac {2} {\sqrt {1-a^{2}}}\left( \tan ^{-1}\left(\dfrac {1+a} {\sqrt {1-a^{2}}}\right) - \tan ^{-1}\left( \dfrac {a} {\sqrt {1-a^{2}}}\right) \right) \\[br]= \dfrac {2} {\sqrt {1-a^{2}}} \tan ^{-1}\left( \dfrac {\dfrac {1+a-a} {\sqrt {1-a^{2}}}} {1+\dfrac {a\left( 1+a\right) } {1-a^{2}}}\right) = \dfrac {2} {\sqrt {1-a^{2}}} \tan ^{-1}\left( \dfrac {\sqrt {1-a^{2}}} {1-a^{2}+a+a^{2}}\right) \\[br]= \dfrac {2} {\sqrt {1-a^{2}}} \tan ^{-1}\left( \dfrac {\sqrt {1-a}} {\sqrt {1+a}}\right)

The second part turns into a reduction formula whose 0^th term can be evaluated using the above result.

Unparseable latex formula:

I_{n+1}+2I_{n}=\displaystyle\int _{0}^{\pi/2}\dfrac {\sin ^{n+1}\left( x\right) +2\sin ^{n}\left( x\right) } {2+\sin \left( x\right) }dx = \displaystyle\int _{0}^{\pi/2}\dfrac {\sin ^{n}\left( x\right)\left(\sin(x) + 2\right) } {2+\sin \left( x\right) }dx \\[br]= \displaystyle\int _{0}^{\pi/2}\sin ^{n}\left( x\right) dx \Rightarrow I_{n+1} = \displaystyle\int _{0}^{\dfrac {\pi } {2}}\sin ^{n}\left( x\right)dx - 2I_n

Unparseable latex formula:

I_3 = \displaystyle\int _{0}^{\pi/2}\sin ^{2}\left( x\right)dx - 2\left( \displaystyle\int _{0}^{\pi/2}\sin\left( x\right)dx - 2\left(\displaystyle\int _{0}^{\pi/2}dx-2I_{0}\right)\right) \\[br]= \dfrac{1}{2}\displaystyle\int _{0}^{\pi/2} \left(1-\cos \left( 2x\right)\right) dx - 2\displaystyle\int _{0}^{\pi/2}\sin\left( x\right)dx +4\displaystyle\int _{0}^{\pi/2}dx-8I_{0} \\[br]= \dfrac {1} {2}\left( \dfrac {\pi } {2}-\dfrac {1} {2}\left[ \sin \left( 2x\right) \right]_0^{\pi / 2}\right) + 2\left[ \cos \left( x\right) \right] _{0}^{\pi / 2} + 2\pi - 8I_0 \\[br]= \dfrac {\pi } {4} - 2 + 2\pi -\dfrac {8} {2}\displaystyle\int _{0}^{\pi / 2}\dfrac {dx} {1+\dfrac {1} {2}\sin \left( x\right) } \\[br]= \dfrac {9\pi } {4} - 2 - 4\left( \dfrac {2} {\sqrt {1-\frac{1}{4}}} \tan ^{-1}\left( \dfrac {\sqrt {1-\frac{1}{2}}} {\sqrt {1+\frac{1}{2}}}\right)\right) = \dfrac {9\pi } {4} -2 - \dfrac {16} {\sqrt {3}}\tan ^{-1}\left(\dfrac {1} {\sqrt {3}}\right) \\[br]= \dfrac {9\pi } {4} - 2 - \dfrac {8\pi} {3\sqrt {3}}= \left(\dfrac {9} {4} - \dfrac {8\sqrt {3}} {9}\right)\pi - 2[br][br]

(edited 10 years ago)

Reply 150

username1088632

2

Original post by mikelbird

I think this is OK for Q5

I could have sworn it said 45 degrees clockwise about the origin, I checked this twice.

Reply 151

borealis72

STEP III 2013, question 2 solutions:

Original post by mikelbird

I think this is OK for Step 1 Q6

Essentially what I got except I concluded that the series was 'one term off' the Fibonacci sequence as

\binom{0}{0}=1

, does it not? :P

(edited 10 years ago)

Reply 153

Pyoro

8

Original post by borealis72

Yeah there are definitely a couple of errors that I can't quite spot yet, you may want to have a thorough check:

I_3 = \left(\dfrac {9} {4} - \dfrac {8\sqrt {3}} {9}\right)\pi - 2

http://www.wolframalpha.com/input/?i=integrate%28sin%28x%29%5E3%2F%282+%2B+sin%28x%29%29+from+x+%3D+0+to+pi%2F2%29

http://www.wolframalpha.com/input/?i=%289%2F4+-+8%2F%283sqrt%283%29%29%29pi+-+2

Edit:
The errors are:

1) The integral of sin(x) is -cos(x)
2) 8*I_0 is subtracted and not added on, your sign suddenly changes

In the deepest depths of code... SIGN-CHANGE NINJUTSU!!!

Thank you, much obliged :biggrin:

Reply 154

DFranklin

18

STEP III, Q13:

Spoiler

Reply 155

msmith2512

1

Original post by FJacob

3.
i)

Spoiler

ii)

Spoiler

iii)

Spoiler

iv)

Spoiler

Here's my solution to this neglected question, which happened to be my favorite. Please let me know of any mistakes.

This might be picky or the question might be badly worded but in (ii) if x = z then the identities are not distinct. The question doesn't say it but I think it should assume that X, Y and Z are distinct.

Reply 156

Ab3rry

Would love to hear an approximate mark for what I got in Step 1:

Spoiler

(edited 10 years ago)

Reply 157

FJacob

Original post by msmith2512

This might be picky or the question might be badly worded but in (ii) if x = z then the identities are not distinct. The question doesn't say it but I think it should assume that X, Y and Z are distinct.