STEP Maths I, II, III 1993 Solutions

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Reply 220

generalebriety

STEP III Q4:
(iii)

$\displaystyle S = \sum_{r=2}^\infty \frac{r2^{r-2}}{3^{r-1}} = \frac{1}{3}\sum_{r=2}^\infty r\bigg( \frac{2}{3} \bigg)^{r-2}.\\[br]\text{Let } u_n = \sum_{r=n}^\infty \bigg( \frac{2}{3} \bigg)^{r-2} = 3\bigg( \frac{2}{3} \bigg)^{n-2}.\\[br]\text{Then } S = \frac{1}{3}(2u_2+u_3+u_4+u_5+\dots). \\[br]u_2 = 3\\[br]u_2+u_3+u_4+\dots = \frac{3}{1 - \frac{2}{3}} = 9.\\[br]\therefore S = \frac{1}{3}(3+9) = 4.$

Alternative way:

$\frac{1}{2}\displaystyle\sum_{r=2}^{\infty}r(\frac{2}{3})^{r-1}$
Let's study $S=\displaystyle\sum_{r=2}^{\infty}r(\frac{2}{3})^{r-1}=2\frac{2}{3}+3(\frac{2}{3})^2+4(\frac{2}{3})^3+...$
Note that $\frac{2}{3}S=2\frac{2}{3})^2+3(\frac{2}{3})^3+4(\frac{2}{3})^4+...$
Now S-(2/3)S=(1/3)S= $\frac{2}{3}+\frac{2}{3}+(\frac{2}{3})^2+(\frac{2}{3})^3+...$
This is almost a GP, so add (1/3) to this and we have $\frac{1}{3}+\frac{1}{3}S=\displaystyle\sum_{r=0}^{\infty}(\frac{2}{3})^r=\frac{1}{1-\frac{2}{3}}=3$

Thus (1/3)S=3-(1/3)=8/3 i.e. S=8 and we're looking for (1/2)S, which is 4.

Reply 221

Glutamic Acid

I'll post my alternatives for question 4 too.

(i) Note that $\tanh^{-1}x = x + \dfrac{x^3}{3} + \dfrac{x^5}{5} + ... \implies \dfrac{\tanh^{-1}x}{x} = 1 + \dfrac{x^2}{3} + \dfrac{x^4}{5} + ...$

Letting x = 1/2 and using the logarithmic form of arctanh(x) gives $2 \times \dfrac{1}{2} \ln \left( \dfrac{1 + \frac{1}{2}}{1 - \frac{1}{2}} \right) = \ln 3$

(iii) $\displaystyle \sum_{r=0}^{\infty} x^r = (1-x)^{-1} \Rightarrow \sum_{r=0}^{\infty} rx^{r-1} = (1-x)^{-2}$ -- the former is the sum of a geometric series and the latter is its derivative. Let x = 2/3 and we get
$\displaystyle \frac{1}{2} \sum_{r=0}^{\infty} [br]r \left( \frac{2}{3} \right)^{r-1} = \frac{9}{2}$ . To find the sum from r = 2 to infinity we need to subtract the terms for r = 1 and 0 which gives 9/2 - 1/2 = 4.

Reply 222

Glutamic Acid

III/14

Reply 223

Mathletics

Speleo

I/4

I like pencil v :smile:

v

N.B. these are rough solutions, I'd write much more explanation of what was happening in the real exam.

Hey could you explain why, for the second integral, you used limits $2\pi$ and $\pi$ instead of $2\pi$ and 0 as the question asks?

Reply 224

jj193

STEP I Question 8 - gen's solution is STEP III Q8.

edit

STEP I '93 Q8 a.jpg318.9KB

STEP I '93 Q8 b.jpg202.0KB

Reply 225

jj193

I just did STEP II Q7, got the 'infinite descent' part but I don't really understand infinite descent (yet I know/think it is this). Why should it be that if I can keep dividing a,b,c by 5 - forever - that a,b,c are 0. I geuss 0/5=0 etc. and zero is still in the intergers. I've just convinced myself it's ok, I still don't like it.

It seems like a huge jump.

Would this be acceptable as an ending:
If an interger can be divided infinitely many times, then the interger must be zero because if it is not it must be infinite which isn't an interger.

Reply 226

miml

I think a neater way (although I don't see anything wrong with your method) is to say,

Assume a is non zero, and an integer
Then there exists some a that contains infinitely many factors of 5 (this bit is crucial. 0 is definitely in the integers, and has infinitely many factors)
No such integer exists
By contradiction a=0

Similarly for b,c
Therefore only solutions when a=b=c=0

Reply 227

themaths

Original post

by Rabite

I think I did Q2 in the Further Maths A.
But it seems quite easy so I've probably made a mistake. It's still red on the front page, so if no one else has typed it out already, I'll do so~

[edit] Here it is anyway.
$\int \cos{mx}\cos{nx} dx = ½ \int \cos (m+n)x + \cos (m-n)x dx$
By the product/sum formulae that no one remembers.
$= ½[\frac{1}{m+n} \sin(m+n)x + \frac{1}{m-n} \sin(m-n)x]_0 ^{2\pi}$

But if m=±n, one of the fractions explodes. So in that case the question is:

$\int \cos^2 {mx} dx$

$=½ \int 1+ cos{2mx}dx$

$= ½[x+\frac{1}{2m} \sin{2mx}]_0 ^{2\pi}$

$=\pi$

If m=n=0, the integral turns to $2\pi$ .

As for the second bit.
Let x = sinh²t

dx = 2sinhtcosht dt

$I = \int \sqrt{\frac{x+1}{x}}dx$

$= \int \sqrt{\frac{\sinh^2 t +1}{\sinh^2 t}}(2\sinh t\cosh t)dt$

$= \int 2 \frac{\cosh t}{\sinh t} (\sinh t \ cosh t) dt$

$= \int 2 \cosh^2 t dt$

$= \int 1+ \cosh 2t dt = t + ½ \sinh 2t$ Ignoring the +c for now

$= t + \sinh t \cosh t = \sinh ^{-1}(\sqrt{x}) + \sqrt(x(x+x)) +c$
Which you can rewrite using the log form of arsinh.

Alternatively for the second integral:
$Let x=tan^2 {z}$
$dx/dz = 2tan {z} sec^2 {z}$

$=\int \sqrt{sec^2{z}/tan^2{z}}.2sec^2{z}tan{z}dz$
$=\int 2sec^3{z}dz$

Let:
$I= \int sec^3{z}dz$
$I= \int sec{z}(1+tan^2{z})dz$
$I= \int sec{z}dz + \int sec{z}tan^2{z}dz$

Now consider the differential of
$sec{z}tan{z}$
By the product rule
$=sec{z}tan^2{z} + sec^3{z}$

Thus
$\int sec{z}tan^2{z}dz +\int sec^3{z}dz = sec{z}tan{z}$

So we now have two simultaneous equation where we can cancel the ugly integral:
$(1) \int sec^3{z}dz = \int sec{z}dz + \int sec{z}tan^2{z}dz$
$(2) \int sec^3{z}dz = sec{z}tan{z} - \int sec{z}tan^2{z}dz$

Add (1) and (2)
$2\int sec^3{z}dz = \int sec{z}dz + sec{z}tan{z}$
$2\int sec^3{z}dz = \ln(sec{z}+tan{z}) + sec{z}tan{z}$

$x=tan^2{z}$
$x+1=sec^2{z}$
$\int \sqrt{1+\frac{1}{x}}dx = \ln(\sqrt{x} + \sqrt{x+1}) + \sqrt{x(x+1)} + C$

The method with the hyperbolic functions is much nicer and removes a horrific sec cubed term but i thought i would have a go regardless using good old trg functions and i believe this is a good method to do it if you have no hyperbolic background.
I wish my STEP exam this year will have a question like this :biggrin:

Reply 228

Dirac Spinor

Original post

by SimonM

(Updated as far as #213) SimonM - 11.05.2009
...

STEP II Q14
Consider the general collision between a particle of mass m and a fixed surface of coefficient of restitution $a$ :
speed before = $v \Rightarrow$ speed after= $av$
From which it follows that the kinetic energy after such collisions is:
$\frac{1}{2}m(av)^2=a^2\frac{1}{2}mv^2=a^2(original K.E.)$

Applying this to the problem:
let $E_n$ be the kinetic energy of the particle just before the nth impact with the ceiling.
By conservation of energy:
$E_1=\frac{1}{2}m(\sqrt{2kgh})^2-mgh=mgh(k-1)[br]\therefore E_2=a^4mgh(k-1)+a^2mgh-mgh[br]\therefore E_3=a^8mgh(k-1)+a^6mgh-a^4mgh+a^2mgh-mgh$
and, more generally:
$E_{n+1}=a^4E_n+a^2mgh-mgh$
Conjecture $(\gamma)$ :
$E_n=mgh(a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^2+1})$
letting n=1:
$E_1=mgh(a^{4-4}(k-1)+\frac{a^{4-4}-1}{a^2+1})=mgh(k-1)$ and $\gamma$ therefore holds for n=1.
Similarly, let n=2:
$E_2=mgh(a^{4}(k-1)+\frac{a^{4}-1}{a^2+1})=mgh(a^{4}(k-1)+a^2-1)=mgha^4(k-1)+a^2mgh-mgh$ and so $\gamma$ holds for n=2 as well.
Now let's assume the result:
$E_x=mgh(a^{4x-4}(k-1)+\frac{a^{4x-4}-1}{a^2+1})$ .
Using this result, let n=x+1:
$E_{x+1}=a^4E_x+a^2mgh-mgh=a^4mgh(a^{4x-4}(k-1)+\frac{a^{4x-4}-1}{a^2+1})+a^2mgh-mgh[br]=mgh(a^{4x}(k-1)+\frac{a^{4x}-a^4+(a^2-1)(a^2+1)}{a^2+1})=mgh(a^{4x}(k-1)+\frac{a^{4x}-1}{a^2+1})$
Therefore, by mathematical induction, $\gamma$ (the required result) holds $\forall n\in \mathbb{Z^+}$

Next part:
the maximum number of times the ball can hit the ceiling is the n that satisfies:
$E_n=mgh(a^{4n-4}(k-1)+\frac{a^{4n-4}-1}{a^2+1})=0$
so:
$a^{4n-4}(k-1)=-\frac{a^{4n-4}-1}{a^2+1}[br]\Rightarrow a^{4(n-1)}(k-1)(a^2+1)=1-a^{4n-4}[br]\Rightarrow a^{4(n-1)}=\frac{1}{(k-1)(a^2+1)+1}$
So, taking logs of both sides:
$4(n-1)ln(a)=-ln((k-1)(a^2+1)+1)[br]\Rightarrow n-1=-\frac{ln(a^2(k-1)+k)}{4lna}[br]\Rightarrow n=1-\frac{ln(a^2(k-1)+k)}{4lna}$
As required.

Reply 229

brianeverit

STEP number 9
$\text{For first vat, chemical }A\text{ flows in at a rate of }a \text{ litres/sec}$
$\text{If }V\text{ is the volume of}A\text{ in vat at time }T\text{ then proportion of }A\text{ is }\dfrac{V}{K} \text{ so it flows out at }-b\dfrac{V}{K} \text{ litres/sec}$
$\text{Hence, }\dfrac{dV}{dt}=a-\dfrac{bV}{K}=\dfrac{aK-bV}{K}$
$\text{Solving this equation we have }\int\dfrac{dV}{aK-bV}=\int\dfrac{dt}{K}\Rightarrow-\dfrac{1}{b}\ln(aK-bV)=\dfrac{1}{K}-\dfrac{1}{b}\lnaK \text{ since }V=0 \text{ at }t=0$
$\text{i.e. }\ln\left( \dfrac{aK-bV}{aK}\right)=-\dfrac{bt}{K}\RightarrowaK-bV=aK\text{exp}\left(-\dfrac{bt}{K}\right) \text{ or }V=\dfrac{aK}{b}\left(1-\text{e}^{-\frac{bt}{K}}\right)\text{ as required}$
$\text{Considering now the second vat, }A\text{ flows in at a rate of }\dfrac{bV}{K} \text{ litres/sec and out at a rate of }\dfrac{cV}{L}$
$\text{where }V\text{ is now the volume of }A\text{ in the second vat, so }\dfrac{dV}{dt}=\dfrac{bV}{K}-\dfrac{cV}{L}=\dfrac{(bL-cK)}{KL}$
$\text{so }\int\left(\dfrac{dV}{(bL-cK)V}}\right)=\int\dfrac{dt}{KL}\Rightarrow -\dfrac{1}{c}\ln{(bL-cK)V}=\dfrac{t}{KL}-\dfrac{1}{c}\ln{bL} \text{ since }V=0\text{ at }t=0$
$\text{i.e. }\ln\left(\dfrac{(bL-cK)V}{bL}\right)=-\dfrac{ct}{KL}\Rightarrow (bL-cK)V=bL\text{e}^{-\frac{ct}{KL}}\text{ or }V=\dfrac{bt}{bL-cK}\left(1-\text{e}^{-\frac{ct}{KL}}\right)$

Reply 230

brianeverit

Can anyone help with STEP III Number 11. I cannot get the given answer for the direction of the initial axis of rotation.

Reply 231

SimonM

STEP III Question 11

Spoiler

I can't get the given answer either

Reply 232

DFranklin

brianeverit

STEP 1993 Q11: Note - done without a good diagram - might be a minor direction mistake somewhere.

Choose a coord frame s.t. C = (0,0), D = (1,0), A = (0, 1), B = (1, 1).
Then AB has CM (0.5, 1), BC has CM (1, 0.5) and CD has CM (0.5, 0).
It follows that the CM of the entire wire is [ (0.5, 1) + (1, 0.5) + (0.5, 0)] / 3 = (2/3, 1/2).

If we hang from A, then BC lies below the horizontal and makes an angle arctan((1/2)/(2/3)) = arctan(3/4) with the horizontal.
If we hang from B, then BC lies below the horizontal and makes an angle arctan((1/2)/(1/3)) = arctan(3/2) with the horizontal (going the other way).

The tan of the angle between these is

tan(arctan(3/4)+arctan(3/2)) = $\frac{3/4+3/2}{1-(3/4)(3/2)} = \frac{9/4}{-1/8} = -18$ .

Since this is negative, it's the oblique angle between the lines; the acute angle will be arctan(18) as required.

Reply 233

SimonM

Ah, my diagram was crap.

Reply 234

Dirac Spinor

guys, you were confusing me. I think you have mixed up STEP III with STEP I. The papers have been mislabelled.

Reply 235

DFranklin

Original post

by ben-smith

guys, you were confusing me. I think you have mixed up STEP III with STEP I. The papers have been mislabelled.

Thanks. I knew about the mislabelling, but thought (incorrectly) it was 1992 and earlier.

Simon M

Care to try again? Not sure I'll have time.

Reply 236

DFranklin

Original post

by brianeverit

Can anyone help with STEP III Number 11. I cannot get the given answer for the direction of the initial axis of rotation.

OK, for the actual correct question you asked about...

I think you can just take moments about each axis. It's easy to make mistakes, but after 3 attempts, I got their answer using this method.

I think you can also define OA = i, OB = j, OC = k and then simply find $\sum {\bf r} \wedge {\bf F}$ where r is the position of a point on the line of force and F is the force itself. (Note that this tallies with the intuitive result that forces through O are irrelevant).

Reply 237

Dirac Spinor

Original post

by DFranklin

It's reassuring that you had to try 3 times because I keep getting an annoyingly similar answer. I keep getting (1,2,1) as opposed to (1,1,2). Did you get that? If so, where did you go wrong?
EDIT: can I just swap the 2 and the 1 around by just redefining the axes or is that cheating?

(edited 14 years ago)

Reply 238

DFranklin

Original post

by ben-smith

You can't swap the axes because you'd also swap the forces, and nothing would change. What I kept getting was (1, -1, 2) - sign errors.

As I calculate it, taking moments anticlockwise about OB you have: -1 (from the force O'A) and that's it.

Taking moments anticlockwise about OC you have: -1 (from AC'), -2 (from C'B) and 1 (from O'A) for a total of -2.

Taking moments anticlockwise about OA you have -1 (from A'C).

I think.

So I get (-1, -1, -2) which is obviously the same axis as (1,1,2).

Reply 239

brianeverit

1993 STEP III number 13

$\text{When }P \text{ has descended a distance }s \text{, let speed of system be }v \text{ then}$
$\text{k.e. of disc }=\dfrac{1}{2} \times \dfrac{4ma^2}{2} \times \dfrac{v^2}{a^2}=mv^2$
$\text{k.e. of chain }=\dfrac{1}{2}(4+\pi)mv^2$
$\text{k.e. of particle }= \dfrac{1}{2}mv^2 \text{ so gain of k.e. Is }\dfrac{1}{2}(\pi +7)mv^2$
$\text{p.e. of system relative to initial position }=\dfrac{(2a-x)xmg}{2a}-\dfrac{(2a+x)xmg}{2a}-xmg=-\dfrac{2(x^2+ax)mg}{2a}$
$\text{so by conservation of energy, }\dfrac{)x^2+ax)mg}{a}=\dfrac{1}{2}(\pi+7)mv^2$
$\Rightarrow (\pi)+7av^2=2g(ax+x^2) \text{ as required}$
$\text{Now consider the portion of the chain below Q as a variable mass }M \text{including mass P}$
$\text{when Q has fallen a distance }2a \text{, total weight of chainand mass P is }(3m+2m)g=5mg$
$\text{By Newton's second law }Mg-T=\dfrac{d}{dt}(Mv)= \dotM v+M\dotv \text{ where }T \text{ is the tension at Q}$
$\dot{M}= \dfrac{v}{a}m \text{ and }v^2= \dfrac{2g(ax+x^2)}{(\pi+7)a} \Rightarrow 2v \dot{v}= \dfrac{2g(a+2x)v}{(\pi+7)a}$
$\Rightarrow \dot{v}= \dfrac{g(a+2x)}{(\pi+7)}a= \dfrac{5g}{\pi+7} \text{ when }x=2a$

$\text{also }v^2=\dfrac{12g}{\pi+7} \text{ when }x=2a$
$\text{hence, }T=5mg-\dfrac{v^2m}{a}-5m \times \dfrac{5g}{\pi+7}=\left(\dfrac{5\pi+35}{\pi+7}- \dfrac{25}{\pi+7}-\dfrac{12}{\pi+7}\right) mg= \dfrac{5\pi-2}{\pi+7} \text{ as required}$

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