STEP Maths I, II, III 1989 solutions

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Reply 60

II/7:

$t = x^{\alpha} \Rightarrow \dfrac{\text{d}t}{\text{d}y} = \alpha x^{\alpha - 1}\dfrac{\text{d}x}{\text{d}y} \Rightarrow \dfrac{\text{d}y}{\text{d}x} = \alpha x^{\alpha - 1} \dfrac{\text{d}y}{\text{d}t}$

$\Rightarrow \dfrac{\text{d}^2y}{\text{d}x^2} = \alpha (\alpha - 1)x^{\alpha - 2} \dfrac{\text{d}y}{\text{d}t} + \alpha x^{\alpha - 1} \dfrac{\text{d}}{\text{d}x} \left( \dfrac{\text{d}y}{\text{d}t} \right)$

Note that $\dfrac{\text{d}}{\text{d}x} \left( \dfrac{\text{d}y}{\text{d}t} \right) = \dfrac{\text{d}}{\text{d}t} \left( \dfrac{\text{d}y}{\text{d}t} \right) \dfrac{\text{d}t}{\text{d}x} = \dfrac{\text{d}^2y}{\text{d}t^2} \dfrac{\text{d}t}{\text{d}x}$ .

$\Rightarrow \dfrac{\text{d}^2y}{\text{d}x^2} = \alpha (\alpha - 1)x^{\alpha - 2} \dfrac{\text{d}y}{\text{d}t} + \alpha^2 x^{2 \alpha - 2} \dfrac{\text{d}^2y}{\text{d}t^2}$ .

Substituting this horrendous crap leads to: $\alpha^2 x^{2 \alpha} \dfrac{\text{d}^2y}{\text{d}t^2} + (\alpha(\alpha - 1) -b \alpha)x^{\alpha} \dfrac{\text{d}y}{\text{d}t} + x^{2b + 2}y = 0$ .

Now choosing $\alpha = b + 1$ and substituting will smash up dy/dt one time bro', leaving

$(b+1)^2 t^2 \dfrac{\text{d}^2y}{\text{d}t^2} + t^2y = 0 \Rightarrow (b+1)^2 \dfrac{\text{d}^2y}{\text{d}t^2} + y = 0 \Rightarrow y = A \cos \left( \dfrac{t}{b+1} \right) + B \sin \left( \dfrac{t}{b + 1} \right)$

$= A \cos \left( \dfrac{x^{b+1}}{b + 1} \right) + B \sin \left( \dfrac{x^{b+1}}{b + 1} \right)$ .

If A = 1, then as x -> 0 then y -> and dy/dx -> 0, but as B can be anything, this is not a unique solution.

Reply 61

Glutamic Acid

II/4:

$\text{f}(x) = \dfrac{(x-a)(x-b)}{(x-c)(x-d)} \Rightarrow \ln \text{f}(x) = \ln(x - a) + \ln(x - b) - \ln (x-c) - \ln(x - d)$
$\Rightarrow \dfrac{\text{f'}(x)}{\text{f}(x)} = \dfrac{1}{x - a} + \dfrac{1}{x - b} - \dfrac{1}{x - c} - \dfrac{1}{x - d}$
$\Rightarrow \dfrac{x \text{f'}(x)}{\text{f}(x)} = \left( \dfrac{x-a}{x} \right)^{-1} + \left( \dfrac{x - b}{x} \right)^{-1} - \left( \dfrac{x - c}{x} \right)^{-1} - \left( \dfrac{x - d}{x} \right)^{-1}$
$= \left( 1 - \dfrac{a}{x} \right)^{-1} + \left( 1 - \dfrac{b}{x} \right)^{-1} - \left( 1 - \dfrac{c}{x} \right)^{-1} - \left( 1 - \dfrac{d}{x} \right)^{-1}$

If |x| is much greater than |a|, |b|, |c| and |d| we can use the first two terms of the binomial expansion with sufficient* accuracy.

$\dfrac{x \text{f'}(x)}{\text{f}(x)} \approx 1 + \dfrac{a}{x} + 1 + \dfrac{b}{x} - 1 - \dfrac{c}{x} - 1 - \dfrac{d}{x}$

$\Rightarrow \dfrac{x^2 \text{f'}(x)}{\text{f}(x)} \approx a + b - c - d$ .

As |x| gets very large, f(x) tends to 1 (can be seen by writing f(x) as $\dfrac{(1 - a/x)(1 - b/x)}{(1 - c/x)(1 - d/x)}$ , and x^2 is always positive, so x^2/f(x) is positive. So f'(x) (the gradient of f(x)) has the same sign as a + b - c - d.

Let $z = \dfrac{(x -a)(x -b)}{(x-c)(x-d)} \Rightarrow z(x-c)(x-d) = (x-a)(x-b)$ . Now, the value of z much be such that there is a root of this equation (i.e. a value of x). As it's a quadratic in x, the discriminant of this quadratic, the expression shown, will be greater than or equal to zero.

Reply 62

brianeverit

1989 Paper 1 numbers 10,12-15

1989PAPER1.no.10.pdf30.9KB

1989PAPER1.no.12.pdf38.8KB

1989PAPER1.no.13.pdf45.8KB

1989PAPER1.no.14.pdf32.0KB

1989PAPER1.no.15.pdf24.4KB

Reply 63

brianeverit

1989 Paper 1 number 16

1989PAPER1.no.16.pdf22.5KB

Reply 64

brianeverit

1989 Paper 2 nos. 4,7,8,9,10

1989PAPER2.no.4.pdf41.8KB

1989PAPER2.no.7.pdf37.4KB

1989PAPER2.no.8.pdf35.4KB

1989PAPER2.no.9.pdf35.1KB

1989PAPER2.no.10.pdf39.1KB

Reply 65

brianeverit

1989 Paper 2 nos. 12 & 13

1989PAPER2.no.12.pdf39.4KB

1989PAPER2.no.13.pdf55.0KB

Reply 66

brianeverit

1989 Paper 2 nos. 15 & 16

1989PAPER2.no.15.pdf48.9KB

1989PAPER2.no.16.pdf33.6KB

Reply 67

brianeverit

1989 Paper 3 nos. 3-7

1989PAPER III no.3.pdf39.3KB

1989PAPER III no.4.pdf46.9KB

1989PAPER III no.5.pdf32.4KB

1989PAPER III no.6.pdf38.7KB

1989PAPER III no.7.pdf41.4KB

Reply 68

brianeverit

1989 Paper 3 nos. 13-16
Now can ANYONE do numbers 11 and 12?

1989PAPER III no.13.pdf32.6KB

1989PAPER III no.14.pdf40.5KB

1989PAPER III no.15.pdf40.5KB

1989PAPER III no.16.pdf32.8KB

Reply 69

brianeverit

1989 Paper III
Sorry, there is an error in my solutoion for number 13 so here is an amended version.

1989PAPER III fmb.13.pdf32.6KB

Reply 70

Glutamic Acid

brianeverit

1989 Paper 3 nos. 13-16
Now can ANYONE do numbers 11 and 12?

I have quite a different answer to the end of 15.

I get E(Z) = 0 as z(c^2 - z^2)^n is an odd function.

For Var(Z):

Note that $\displaystyle \int_{-c}^c (c^2 - z^2)^n \, \text{d}z = \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!}$ . Now, let $z = c \sin \theta$ and we get $\displaystyle \int_{-\pi/2}^{\pi/2} c^{2n+1} \cos^{2n+1} \theta \, \text{d} \theta = \frac{(n!)^2 (2c)^{2n+1}}{(2n+1)!}$

For Var(Z), we want to evaluate $\displaystyle \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int z^2(c^2 - z^2)^n \, \text{d}z$ . Letting $z = c \sin \theta$ as before transforms this to

$\displaystyle c^2 \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-\pi/2}^{\pi/2} c^{2n+1} \cos^{2n+1} \theta \, \text{d} \theta - \frac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \int_{-\pi/2}^{\pi/2} c^{2n+3} \cos^{2n+3} \theta \, \text{d}\theta$ . Using what we know for the first integral, and replacing n by n + 1 for the second, we get

$c^2 - \dfrac{(2n+1)!}{(n!)^2 (2c)^{2n+1}} \cdot \dfrac{((n+1)!)^2 (2c)^{2n+3}}{(2n+3)!} = \dfrac{c^2}{2n+3}$ .
Quite a nice answer, makes me wonder if there's an easier way.

Reply 71

brianeverit

Having looked at number Paper III 15 again I agree with your solution.

Reply 72

brianeverit

Dystopia

STEP 1, Q3

$\textbf{x} = x\textbf{a}, \; \textbf{y} = (1-x)\textbf{b}$

$\textbf{c} = \frac{1}{3}(2\textbf{a} + \textbf{b}), \textbf{d} = \frac{1}{3}(\textbf{a} + 2\textbf{b})$

(By the ratio theorem.)

$\overrightarrow{CY} = (\frac{2}{3} - x)\textbf{b} - \frac{2}{3}\textbf{a}$
$\overrightarrow{DX} = (x - \frac{1}{3})\textbf{a} - \frac{2}{3}\textbf{b}$

These vectors are perpendicular, so the scalar product is zero.

$(\frac{2}{9} + x - x^{2})\textbf{a}.\textbf{b} - \frac{2}{9} = 0$

(Note that the scalar product is commutative and distributive.)

$\textbf{a}.\textbf{b} = \frac{-2}{9x^{2} - 9x - 2}$

Let $f(x) = \frac{-2}{9x^{2} - 9x - 2}, \; 0 \leq x \leq 1$

$f'(x) = \frac{2(18x - 9)}{(9x^{2}-9x-2)^{2}}$

Minimum and maximum values occur at the endpoint of a range or at a turning point. There is a turning point when x = 1/2.

$f(0) = 1, \; f(\frac{1}{2}) = \frac{8}{17}, \; f(1) = 1$

So $\frac{8}{17} \leq \textbf{a}.\textbf{b} \leq 1$

Note that $\textbf{a}.\textbf{b} = |a||b|\cos\theta = \cos\theta$

Where $\theta = \angle AOB, \; 0 < \theta < \pi$

The maximum value is $\theta$ occurs when $\cos\theta = \frac{8}{17}$

Suppose that $\lambda \overrightarrow{CY} = \overrightarrow{CE},\; \mu \overrightarrow{DX} = \overrightarrow{DE}$

Then $\lambda (\frac{1}{6} \textbf{b} - \frac{2}{3} \textbf{a}) = \frac{1}{3} ( \textbf{b} - \textbf{a} ) + \mu ( \frac{1}{6} \textbf{a} - \frac{2}{3} \textbf{b} )$

So $\frac{1}{6} \lambda = \frac{1}{3} - \frac{2}{3}\mu$

$-\frac{2}{3}\lambda = -\frac{1}{3} + \frac{1}{6}\mu$

$\lambda = \mu = \frac{2}{5}$

$\textbf{e} = \textbf{c} + \frac{2}{5}\overrightarrow{CY} = \frac{2}{5}\textbf{a} + \frac{2}{5}\textbf{b}$

By the cosine rule, $AB^{2} = \frac{18}{17}$
Let F be the midpoint of AB. By Pythagoras, $OF^{2} = 1 - \frac{9}{34} = \frac{25}{34}, \; OF = \frac{5}{\sqrt{34}}$

$OE = \frac{3}{5} \times \frac{5}{\sqrt{34}} = \frac{3}{\sqrt{34}}$

OF=1/2(a+b) and OE=2/5 (a+b) so surely OE=4/5(a+b)=4/5OF not 3/5?

Reply 73

username219321

I made a sign error at one point (accidently multiplied by A sinh t instead of -A sinh t) so there's a bit of a mess on the last little bit (I'd have wrote it out again in a real exam) but it should be understandable - STEP II 1989 Q9 is attatched (It'd probably take a very long time to latex)

step21989q9part1.jpg276.2KB

step21989q9part2.jpg207.8KB

step21989q9part3.jpg179.9KB

Reply 74

r2enigma

brianeverit

1989 Paper 3 nos. 13-16
Now can ANYONE do numbers 11 and 12?

I did 12 and did the first part of 11 but have no idea about the last part

Reply 75

r2enigma

brianeverit

1989 Paper III
Sorry, there is an error in my solutoion for number 13 so here is an amended version.

holy **** how did you think of that? :eek3:

Reply 76

Physics Enemy

Haha 1989 ... I bet the STEP papers were monstrous back then! :eek:

Reply 77

Farhan.Hanif93

Original post

by squeezebox

...

'SimonM'

I wasn't sure if the OP was still active.

STEP I - Q1

solution

Reply 78

Dirac Spinor

Original post

by brianeverit

1989 Paper 1 numbers 10,12-15

Hi, I´ve been trying to do Q10 and I just saw your solution which is different to my method.

Couldn´t you just calculate the loss in kinetic energy and then equate that to the work done by the force? i.e. :
$\Delta K.E = \int^d_0 \frac{M\omega (v^2 +V^2)}{v}dx$
where x is the distance from the start of the field and $v=\dot{x}$

I was having a bit of trouble with the integration so I was wondering if anyone could give me a hand :colondollar:

It might just all be wrong but any help would be appreciated.

Reply 79

DFranklin

Original post

by ben-smith

Well, you can't really integrate that without ending up doing what Brian did.

Let's write v(x) for v, to emphasise that v is a function of x.

Then you have

$\frac{1}{2}v(0)^2 - \frac{1}{2}v(d)^2 = \int_0^d \frac{\omega (v(x)^2 +V^2)}{v(x)}dx$

So, here's the problem - at this point you have absolutely no idea what v(x) is, which is going to make finding the integral pretty tricky.

So, let's relabel some of the variables: I'm going to write 'x' instead of d, and 't' instead of 'x'.

$\frac{1}{2}v(0)^2 - \frac{1}{2}v(x)^2 = \int_0^d \frac{\omega (v(t)^2 +V^2)}{v(t)}dt$

Now diff both sides w.r.t. x:

$v(x)\frac{dv}{dx} = \frac{\omega (v(x)^2 +V^2)}{v(x)}$

Which is Brian's first line. So from here you can solve for v as a function of x, which will give you the desired result.

Obviously it's quicker to just use the v dv/dx form of acceleration to start with.

(edited 15 years ago)

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