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STEP Maths I, II, III 1989 solutions

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I guess you got to (αβ)=C\begin{pmatrix}\alpha\\ \beta\end{pmatrix}=C and thus (x(t)y(t))=(αβ)eAt\begin{pmatrix}x(t) \\ y(t)\end{pmatrix}=\begin{pmatrix}\alpha\\ \beta\end{pmatrix}e^{-At}

does this help you?

edit: no, that doesn't help because we're still left with a matrix in the exponent, which can't be what is meant!
I can't tell, do you know how to find eAte^{-At} or not?

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Yes I was just about to type that out as a interesting after thought from muddling about with it.

I'll have another little think.
Reply 43
Are you actually meant to multiply by eAte^{\mathbf{A}t}?
I think so, yes. (Modulo whether it's e^At or e^-At; I haven't looked closely enough to be sure).

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Reply 45
DFranklin
I think so, yes. (Modulo whether it's e^At or e^-At; I haven't looked closely enough to be sure).

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Reply 47
DFranklin

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working

Reply 49
nota bene

working


What is drdt(reAt)\frac{\mathrm{d}r}{\mathrm{d}t} (\mathbf{r} e^{ \mathbf{A} t})?
Reply 50
Would this be correct for the third part?

Expected winnings = (2b3b2)a1+(b1b3)a2+(b22b1)a3(2b_3 - b_2)a_1 + (b_1 - b_3)a_2 + (b_2 - 2b_1)a_3

(So X = 2b_3 - b_2, Y = b_1 - b_3, Z = b_2 - 2b_1)

And for the next part, is it enough if I just state

b_1 = 0.25
b_2 = 0.5
b_3 = 0.25?

With these numbers, X, Y and Z all turn out 0 which is non-negative. And so the expectation is 0 as well, regardless of a_1, a_2 and a_3, which means that he won't LOSE any money (in the "long run").
Reply 51
STEP III - Question 9

i)


1+22+322+423+.....=[1+12+122+122+......]+[12+122+122+124+......]+[122+122+123+......]+.......=1112+12112+14112+18112+.......=2+1+12+14+18+.....=2+1112=4 1 + \frac{2}{2} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + ..... = [ 1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{2}} + ......] + [\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{2}} + \frac{1}{2^{4}} +......] + [ \frac{1}{2^{2}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ......] + ....... = \frac{1}{1-\frac{1}{2}} + \frac{\frac{1}{2}}{1-\frac{1}{2}} + \frac{\frac{1}{4}}{1-\frac{1}{2}} + \frac{\frac{1}{8}}{1-\frac{1}{2}} + ....... = 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ..... = 2 + \frac{1}{1 - \frac{1}{2}} = 4 .

ii) Consider the Maclaurins series of ln(1x) \ln(1-x) .

ln(1x)=xx22x33x44.....\ln(1-x) = -x - \frac{x^{2}}{2} - \frac{x^{3}}{3} - \frac{x^{4}}{4} - ..... for 1<x -1 < x \leq .

x+x22+x33+x44+......=ln(11x)\Rightarrow x + \frac{x^{2}}{2} + \frac{x^{3}}{3} + \frac{x^{4}}{4} + ...... = \ln(\frac{1}{1-x}) .

So putting x = 0.5 (its ok since it's in the required range.), we get:

12+122×12+123×13+124×x44+......=ln(1112). \frac{1}{2} + \frac{1}{2^{2}} \times \frac{1}{2} + \frac{1}{2^{3}} \times \frac{1}{3} + \frac{1}{2^{4}} \times \frac{x^{4}}{4} + ...... = \ln(\frac{1}{1-\frac{1}{2}}) .

1+12×12+122×13+123×14+......=2ln(2)=ln(4)\Rightarrow 1 + \frac{1}{2} \times \frac{1}{2} + \frac{1}{2^{2}} \times \frac{1}{3} + \frac{1}{2^{3}} \times \frac{1}{4} + ...... = 2\ln(2) = \ln(4)

iii) consider (1x)12 (1-x)^{\frac{1}{2}}

(1x)12=1+x2+1×3x2222!+1×3×5×x3233!+....... (1-x)^{\frac{-1}{2}} = 1 + \frac{x}{2} + \frac{1 \times 3 x^{2}}{2^{2}2!} + \frac{1 \times 3 \times 5 \times x^{3}}{2^{3}3!} + .......


so putting x=23x=\frac{2}{3}:

1+12×23+1×3222!2232+1×3×5233!2333+......=(123)12=31 + \frac{1}{2} \times \frac{2}{3} + \frac{1 \times 3}{2^{2}2!}\frac{2^{2}}{3^{2}} + \frac{1 \times 3 \times 5 }{2^{3}3!}\frac{2^{3}}{3^{3}} + ...... = (1-\frac{2}{3})^{\frac{-1}{2}} = \sqrt{3}

1×32!132+1×3×53!133+......=3113 \Rightarrow \frac{1 \times 3}{2!}\frac{1}{3^{2}} + \frac{1 \times 3 \times 5 }{3!}\frac{1}{3^{3}}+...... = \sqrt{3} - 1 - \frac{1}{3}

1×32!13+1×3×53!132+......=3331=334\Rightarrow \frac{1 \times 3}{2!}\frac{1}{3} + \frac{1 \times 3 \times 5 }{3!}\frac{1}{3^{2}}+...... = 3\sqrt{3} - 3 -1 = 3\sqrt{3} -4
Reply 52
STEP III - Question 1 - I'm not really sure if I've done this right

We form the sphere by rotating a circle 360 degrees about the x-axis. WLOG, let the equation of the circle be x2+y2=r2x^{2} + y^{2} = r^{2} .

So differentiating implicitly:

2x+2ydydx=02x + 2y\frac{dy}{dx} = 0

dydx=xy\Rightarrow \frac{dy}{dx} = \frac{-x}{y} .

So using the formula for surface area (2πy1+(dydx)2dx\displaystyle\int 2 \pi y \sqrt{1+(\frac{dy}{dx})^{2}} \, dx ) the area of the zone between the lines x = a and x=b (r > b > a > -r), is : (ab2πy1+(xy)2dx=ab2πx2+y2dx=ab2πrdx,=2×π×r×(ba)()\displaystyle\int^{b}_{a} 2 \pi y \sqrt{1+(\frac{x}{y})^{2}} \, dx = \displaystyle\int^{b}_{a} 2 \pi \sqrt{x^{2} + {y}^{2}} \, dx = \displaystyle\int^{b}_{a} 2 \pi r \, dx, = 2 \times \pi \times r \times (b-a) (*)as required.

This next bit is really hard to explain without a diagram....

let θ\theta be the angle between the tangent and the x-axis. Then, since angle OPC (c = centre of circle) = 90 degrees, tan(θ)=ac2a2\tan(\theta) = \frac{a}{\sqrt{c^{2}-a^{2}}}.

So the equation of the tangent is: y=axc2a2y= \frac{ax}{\sqrt{c^{2}-a^{2}}}. And the x co-ord of P = c2a2cos(θ)=c2a2×c2a2c=c2a2c\sqrt{c^{2} - a^{2}}\cos(\theta) = \sqrt{c^{2} - a^{2}} \times \frac{\sqrt{c^{2} - a^{2}}}{c} = \frac{c^{2} - a^{2}}{c}.

Using (*), the surface area of the sphere bit is: 2π×a×[(a+c)(c2a2c)]=2πa2(a+c)c 2\pi \times a \times [ (a+c) - (\frac{c^{2}-a^{2}}{c})] = \frac{2\pi a^{2}(a+c)}{c}.

And the surface area formed by rotating the tangent (y=axc2a2y= \frac{ax}{\sqrt{c^{2}-a^{2}}}) part (the bit from the orgin to P) about the x-axis is : 0c2a2c2πy1+(dydx)2dx=0c2a2c2π×axc2a2×1+(a2c2a2dx=0c2a2c2πacxc2a2=[πacx2c2a2]0c2a2c=πa(c2a2)c\displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} 2 \pi y \sqrt{1+(\frac{dy}{dx})^{2}} \, dx = \displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} 2 \pi \times \frac{ax}{\sqrt{c^{2}-a^{2}}} \times \sqrt{1+(\frac{a^{2}}{c^{2}-a^{2}}} \, dx = \displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} \frac{2 \pi a c x}{c^{2}-a^{2}} = [\frac{\pi a c x^{2}}{c^{2}-a^{2}}]^{\frac{c^{2}-a^{2}}{c}}_{0} = \frac{\pi a (c^{2}-a^{2})}{c} .

So the total area = πa(c2a2)c+2πa2(a+c)c=πa(a+c)2c \frac{\pi a (c^{2}-a^{2})}{c} + \frac{2\pi a^{2}(a+c)}{c} = \frac{\pi a (a+c)^{2}}{c}


EDIT: I'm not even sure if this is the correct method.....:redface:
Reply 53
STEP II - Question 14

Let the tension in the string when it has rotated through θ\theta degrees be T, and the velocity of the particle at this point be V.

Then using conservation of energy:

amgsin(θ)=12mV2(1)a mg \sin(\theta) = \frac{1}{2} m V^{2} (1)

and Newtons 2nd law:

mV2a=Tmgsin(θ)(2) \frac{mV^{2}}{a} = T - mg \sin(\theta) (2)

(1) -> (2): 2amgsin(θ)a=Tmgsin(θ)\frac{2amg \sin(\theta)}{a} = T - mg \sin(\theta)

T=3mgsin(θ)\Rightarrow T = 3mg \sin(\theta) .

This next bit is hard to describe without a diagram, sorry :redface: . EDIT: forget it, I've attached a rubbish paint diagram.

To find the force at the edge (let this be called R), you need to note that each section of the string exerts a force on the edge. Drawing a triangle of forces (see attachment) and using the cosine rule we get:

R=2T22T2cos(θ)=2Tsin(θ2)=6mgsin(θ2)sin(θ)R = \sqrt{2T^{2} - 2T^{2}\cos(\theta)} = 2T\sin(\frac{\theta}{2}) = 6mg\sin(\frac{\theta}{2})\sin(\theta) .

Then
Unparseable latex formula:

\frac{dR}{d\theta} = 3mg( \cos(\frac{theta}{2}) \sin(\theta) + 2\cos(\theta) \sin(\frac{\theta}{2}) ) = 3mg \sin(\frac{\theta}{2})(3\cos^{2}(\frac{\theta}{2}}) - 1)



We find that R is a maximum when cos(θ2)=13\cos(\frac{\theta}{2}) = \frac{1}{\sqrt{3}} (justify with 2nd derivative, that I don't wanna type...)

When cos(θ2)=13,sin(θ2)=23\cos(\frac{\theta}{2}) = \frac{1}{\sqrt{3}}, \sin(\frac{\theta}{2}) = \frac{\sqrt2}{\sqrt3} and sin(θ)=223 \sin(\theta) = \frac{2\sqrt2}{3} .

Which gives the maximum value of R as 8mg3 \frac{8mg}{\sqrt3}
Dystopia
STEP III, Q8

d2xdt2=4dxdt4dydt\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} = 4\frac{\mathrm{d}x}{\mathrm{d}t} - 4\frac{\mathrm{d}y}{\mathrm{d}t}

d2xdt24dxdt+4x=48e2t+48e2t\frac{\mathrm{d}^{2}x}{\mathrm{d}t^{2}} - 4\frac{\mathrm{d}x}{\mathrm{d}t} + 4x = 48e^{2t} + 48e^{-2t}

Let u=dxdt2xu = \frac{\mathrm{d}x}{\mathrm{d}t} - 2x

Then dudt2u=48e2t+48e2t\frac{\mathrm{d}u}{\mathrm{d}t} - 2u = 48e^{2t} + 48e^{-2t}

Using an integrating factor, e2te^{-2t}, we get

ue2t=48+48e4t  dt=48t12e4t+cue^{-2t} = \int 48 + 48e^{-4t} \; \mathrm{d}t = 48t - 12e^{-4t} + c

u=e2t(48t+c)12e2tu = e^{2t}(48t + c) - 12e^{-2t}

dxdt2x=e2t(48t+c)12e2t\frac{\mathrm{d}x}{\mathrm{d}t} - 2x = e^{2t}(48t + c) - 12e^{-2t}

xe2t=48t+c12e4t  dt=24t2+ct+3e4t+dxe^{-2t} = \int 48t + c - 12e^{-4t} \; \mathrm{d}t = 24t^{2} + ct + 3e^{-4t} + d

x=e2t(24t2+ct+d)+3e2tx = e^{2t}(24t^{2} + ct + d) + 3e^{-2t}

Applying boundary conditions:

0=d+3d=30 = d + 3 \Rightarrow d = -3

dxdt=2e2t(24t2+ct+d)+e2t(48t+c)6e2t=4(xy)\frac{\mathrm{d}x}{\mathrm{d}t} = 2e^{2t}(24t^{2} + ct + d) + e^{2t}(48t + c) - 6e^{-2t} = 4(x-y)

2d+c6=0c=62d=6+6=122d + c - 6 = 0 \Rightarrow c = 6 - 2d = 6 + 6 = 12

x=e2t(24t2+12t3)+3e2tx = e^{2t}(24t^{2} + 12t - 3) + 3e^{-2t}

4y=4xdxdt=e2t(48t24t18)+18e2t=4te2t(12t1)36sinh2t4y = 4x - \frac{\mathrm{d}x}{\mathrm{d}t} = e^{2t}(48t^{2} - 4t - 18) + 18e^{-2t} = 4te^{2t}(12t - 1) - 36\sinh 2t

y=te2t(12t1)9sinh2ty = te^{2t}(12t - 1) - 9\sinh 2t

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I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.


Is there any way of solving this using, the complementary function + particular integral method.

I got to x'' - 4x' + 4x = 48(e^2t + e^-2t) and used complementary function x = Ae^2t + Bte^2t

Particular integral (most likely where i've gone wrong) x = Q(t^2)e^2t + R(t^2)e^-2t. (multiplying by t^2 because Bte^2t is in the complementary function)

It ends up giving the wrong answer.

I've never tried using a particular integral like that before and so it was basically a guess. Could anyone give me some kind of guidelines for when to not bother looking for a particular integral and just use another method instead?
You only need to multiply e^2t by t^2. So you'd look for a PI of form x = Q(t^2)e^2t + Re^-2t.
DFranklin
You only need to multiply e^2t by t^2. So you'd look for a PI of form x = Q(t^2)e^2t + Re^-2t.


I see, thanks for your help. Just so i'm totally sure on this, say you had

ax'' + bx' + cx = e^4x(2cos3x + 6sin2x)

and the complementary function was e^4x(Acos3x + Bsin3x)

Would the PI be Qx(e^4x)cos3x + R(e^4x)sin2x?
I'm not 100% sure without doing it. You might need a Sx(e^4x)sin 3x term as well. Try it and see.
Swayum
STEP I Question 6

y=f(x)y = f(x)
dy/dx=f(x)dy/dx = f'(x)

The normal to the curve will therefore have gradient -1/f'(x). The equation of the normal would be

yf(x)=1f(x)(xx)y - f(x) = \frac{-1}{f'(x)}(x' - x) (where x' is the general x coordinate of the normal).

At the point Q, x' = 0 as it cuts the y axis.

yf(x)=xf(x)y - f(x) = \frac{x}{f'(x)}

y=xf(x)+f(x)y = \frac{x}{f'(x)} + f(x)

The distance PQ can be worked out using Pythagoras' theorem.

PQ2=(f(x)(xf(x)+f(x)))2+x2PQ^2 = (f(x) - (\frac{x}{f'(x)} + f(x)))^2 + x^2

=x2f(x)2+x2= \frac{x^2}{f'(x)^2} + x^2

It's given that PQ2=ex2+x2PQ^2 = e^{x^2} + x^2

So

x2f(x)2+x2=ex2+x2\frac{x^2}{f'(x)^2} + x^2 = e^{x^2} + x^2

x2f(x)2=ex2\frac{x^2}{f'(x)^2} = e^{x^2}

x2ex2=f(x)2\frac{x^2}{e^{x^2}} = f'(x)^2

xe0.5x2=f(x)\frac{x}{e^{0.5x^2}} = f'(x)

1dy=xe0.5x2dx\int 1 \mathrm{d}y = \int \frac{x}{e^{0.5x^2}} \mathrm{d}x

y=xe0.5x2dxy = \int xe^{-0.5x^2} \mathrm{d}x

y=e0.5x2+cy = -e^{-0.5x^2} + c (use a substitution of u = x^2 if you can't see why)

2=1+c-2 = -1 + c

c=1c = -1

y=e0.5x21y = -e^{-0.5x^2} - 1


My answer differs slightly from yours in that I thought you had to take both the positive and negative square root of (f'(x))^2 = x^2/e^(x^2) doing this gives

y = +/-(e^(-x^2/2) + C

If you then differentiate twice you get f''(x) = +/-e^(-x^2/2)*(1 - x^2).

f''(0) = +/-1 but we are given that f(0) is a minimum and so f''(0) must equal +1. Consequently (after plugging in the initial conditions) we get y = -(e^(-x^/2) + 1).

I figured this must be what they were expecting because i couldn't figure out why else they would mention that (0, -2) is a minimum. Could anyone confirm?
Reply 59
maltodextrin
My answer differs slightly from yours in that I thought you had to take both the positive and negative square root of (f'(x))^2 = x^2/e^(x^2) doing this gives

y = +/-(e^(-x^2/2) + C

If you then differentiate twice you get f''(x) = +/-e^(-x^2/2)*(1 - x^2).

f''(0) = +/-1 but we are given that f(0) is a minimum and so f''(0) must equal +1. Consequently (after plugging in the initial conditions) we get y = -(e^(-x^/2) + 1).

I figured this must be what they were expecting because i couldn't figure out why else they would mention that (0, -2) is a minimum. Could anyone confirm?


I haven't looked at the question again, but what you're saying makes sense. I was careless when writing that solution. I'll edit it soon - thanks.

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