STEP Maths I, II, III 1993 Solutions

generalebriety

STEP III Q4:
(iii)

Unparseable latex formula:

\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

17

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.

III/14

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

14

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

14

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

17

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

10

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.

(edited 12 years ago)

Reply 229

brianeverit

9

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}

Unparseable latex formula:

\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

Unparseable latex formula:

\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}

Unparseable latex formula:

\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

9

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

OP

18

STEP III Question 11

Spoiler

I can't get the given answer either

(edited 12 years ago)

Reply 232

DFranklin

18

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.

(edited 12 years ago)

Reply 233

SimonM

OP

18

Ah, my diagram was crap.

Reply 234

Dirac Spinor

10

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

Reply 235

DFranklin

18

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

18

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

10

Original post by DFranklin

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).

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 12 years ago)

Reply 238

DFranklin

18

Original post by ben-smith

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?

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).