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STEP I, II, III 1999 solutions

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    STEP I, Question 5

    (Please someone check this, I'm not very confident, because I did last year)

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    (i) d = R\thetaTriangle with x on the adjacent, x on the opposite, and h+R on the hypotenuse, implies \sin{\theta} = \frac{x}{R} = \frac{\sqrt{(h+R)^2-R^2}}{R}\frac{\sqrt{h^2+2Rh}}{R}, so d = R\arcsin\left(\frac{\sqrt{h^2+2R  h}}{R}\right).

    (ii) h is small relative to h, so distance tends to d_\text{NEW} = R\arcsin\left(\frac{\sqrt{2Rh}}{  R}\right), but \sin\theta \approx \theta (for obviously small \theta), so d_\text{NEW}\approx \sqrt{2}(Rh)^{1/2}, so k=\sqrt{2}.

    (iii) We draw some strange diagram... and then... !

    d = R(\frac{\pi}{2} - \phi), where \phi is an angle ZOK, where K is a side of the earth, as well as that angle being the angle XZO. So two triangles are congruent, and the opposite is R, hypotenuse h+R and x the adjacent.

    \sin\phi = \frac{R}{\sqrt{h^2+2Rh+R^2}}, and since h gets large compared to R, \phi \approx \frac{R}{\sqrt{h^2+2Rh+R^2}}\app  rox{R}{h^2}. Then, d_\text{NEW} \approx \frac{\pi}{2}R-R\phi\approx \frac{\pi}{2}R-\frac{R^2}{h}, so a=\frac{\pi}{2} and b=-1.
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    STEP I 1999 Question 7

    Could somebody verify the method used in the third part? It really would not hold up as any kind of a proof in my eyes. I did this one last year, too.

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    First part:

    y=\sin(k\arcsin{x})
    \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{k}{\sqrt{1-x^2}}\cos(k\arcsin{x})
    \frac{\mathrm{d}^2y}{\mathrm{d}x  ^2} = -\frac{k^2}{1-x^2}\sin(k\arcsin{x}) + \cos(k\arcsin{x})x(1-x^2)^{-3/2}

    Substituting into the differential equation:
    (1-x^2)(k^2x(1-x^2)^{-3/2}\cos(k\arcsin{x}) - \frac{k^2}{1-x^2}\sin(k\arcsin{x}))-x\frac{k^2}{\sqrt{1-x^2}}\cos(k\arcsin{x})+k^2sin(k\  arcsin{x}) =

    = k^2x(1-x^2)^{-1/2}\cos(k\arcsin{x})-k^2\sin(k\arcsin{x}) - \frac{kx}{\sqrt{1-x^2}}\cos(k\arcsin{x}) + k^2\sin(k\arcsin{x}) =

     = k^2x(1-x^2)^{-1/2}\cos(k\arcsin{x})-k^2x(1-x^2)^{-1/2}\cos(k\arcsin{x}) = 0.

    As required, it is satisfied.

    Second part:
    y=Ax^3+Bx^2+Cx+d, so \frac{\mathrm{d}y}{\mathrm{d}x} = 3Ax^2+2Bx+C, and \frac{\mathrm{d}^2y}{\mathrm{d}x  ^2} = 6Ax+2B

    y'(0) = 3, y(0) = 0, so C = 3, D = 0. At x = 0, also 2B = 0, so B = 0. At the differential equation, also, (1-x^2)(6Ax+2B)-x(3Ax^2+2Bx+3)+9(Ax^2+Bx^2+3x)=0. Equating the x terms, we get 6A-3+27=0=6A+24\implies A = -4. So, A=4, B=0, C = 3, D = 0. Thus, y = 3x-4x^3 is a polynomial solution to the differential equation.

    Third part:
    If we let x = \sin\theta for -1\leq x \leq 1, then we get the equivalent solutions y = 3\sin\theta-4\sin^3\theta = \sin{(3\arcsin{(\sin{x})})} = \sin{3\theta} . This holds because the initial conditions hold for the polynomial solution and for the trigonometric solution. Thus, they must equal.
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    (Original post by Aurel-Aqua)
    STEP I 1999 Question 7

    Could somebody verify the method used in the third part? It really would not hold up as any kind of a proof in my eyes. I did this one last year, too.
    I think you need to show the initial conditions are satisfied, but it's otherwise OK (if a little indirect).

    The following isn't really any different to what you wrote, but it feels a little bit better, somehow:

    My version
    By the first part, we know Y(x) = sin(3 arcsin x) is a solution to the DE, and we can see Y(0) = 0 and Y'(0) = 3. So this must equal our other solution and so

    sin(3 arcsin x) = 3x - 4x^3.

    Then for -1 <= x <= 1 we can write x = sin y to get

    sin 3y = 3 sin y - 4 sin^3 y.


    I don't think you'd lose more than 2 marks for what you did though (and that would basically be because of not checking the initial conditions).
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    STEP I 1999 Question 10

    (Someone please verify that my assumption is correct).


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    Diagram: points A, P and B in one vertical line (A at the top), l being the length of AND, and showing \ddot{x} is going up.
    The spring can be assumed as two separate [equal-length] ones.

    At equilibrium, m\ddot{x} = 0 = -mg + k_Ax_\text{A-eq} - k_Bx_\text{B-eq} = 0, so mg = k_Ax_\text{A-eq} - k_Bx_\text{B-eq} .

    We also know that l=a+x_A+b+x_B and l=a+x_\text{A-eq} + b + x_\text{B-eq}.

    Equating the forces in the general case, m\ddot{x} = mg + k_Ax_A-k_Bx_B. Also x_A = x_\text{A-equilibrium} - x and x_B = x_\text{B-eq} + x. Thus,

    m\ddot{x} = -mg + k_Ax_\text{A-eq} - k_Bx_\text{B-eq} = -k_Ax-k_Bx = -(k_A+k_B)x. We know that the spring constant is the same for both parts , so:
    k_A+k_B = \frac{\lambda}{2l_0}+\frac{\lamb  da}{2l_0} = \frac{\lambda}{l_0} = k and then
    \ddot{x}= -\frac{k}{m} x, where k is the spring constant, which clearly is SHM.

    T = \frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{\kappa}}. From this, the period will only be dependent on the mass, and nothing else (since the spring constant is unique to the spring itself and its natural length, not deformed length).

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    (Original post by Aurel-Aqua)
    STEP I, Question 5

    (Please someone check this, I'm not very confident, because I did last year)

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    (i) d = R\thetaTriangle with x on the adjacent, x on the opposite, and h+R on the hypotenuse, implies \sin{\theta} = \frac{x}{R} = \frac{\sqrt{(h+R)^2-R^2}}{R}\frac{\sqrt{h^2+2Rh}}{R}, so d = R\arcsin\left(\frac{\sqrt{h^2+2R  h}}{R}\right).

    (ii) h is small relative to h, so distance tends to d_\text{NEW} = R\arcsin\left(\frac{\sqrt{2Rh}}{  R}\right), but \sin\theta \approx \theta (for obviously small \theta), so d_\text{NEW}\approx \sqrt{2}(Rh)^{1/2}, so k=\sqrt{2}.

    (iii) We draw some strange diagram... and then... !

    d = R(\frac{\pi}{2} - \phi), where \phi is an angle ZOK, where K is a side of the earth, as well as that angle being the angle XZO. So two triangles are congruent, and the opposite is R, hypotenuse h+R and x the adjacent.

    \sin\phi = \frac{R}{\sqrt{h^2+2Rh+R^2}}, and since h gets large compared to R, \phi \approx \frac{R}{\sqrt{h^2+2Rh+R^2}}\app  rox{R}{h^2}. Then, d_\text{NEW} \approx \frac{\pi}{2}R-R\phi\approx \frac{\pi}{2}R-\frac{R^2}{h}, so a=\frac{\pi}{2} and b=-1.
    I think that's right, although I used cos instead of sin.

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    i) cos{\theta} = \frac{R}{h+R}

    ii) use the approximation and ignore h.

    iii) cos{\theta}=sin{\phi}, ignore R.
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    I don't see how you can assume the string to be the same as two separate equal length ones, when we're given there are two lengths a and b.
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    (Original post by DFranklin)
    I don't see how you can assume the string to be the same as two separate equal length ones, when we're given there are two lengths a and b.
    Oh, my mistake... I think I misread my working from long time ago.
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    STEP I 1999 Question 14

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    f(x) = \begin{cases} 2x & 0 \leq x \leq 1 \\ 0 & \text{otherwise}\\ \end{cases}

    In the valid range, F(x) = \int f(x)\, \mathrm{d}x = x^2+c.
    (i) E(X) = 4(F(0.25)-F(0)) + 3(F(0.5)-F(0.25))+2(F(0.75)-F(0.5)) + (F(1)-F(0.75)) =

    = 1+\frac{9}{16}+\frac{4}{16}+\fra  c{1}{16} = \frac{1}{16}(16+9+4+1) = \frac{30}{16} = \frac{15}{8}.
    (ii) To win, first we have to calculate the probability of winning: P((1,1,1)\cup(2,1)\cup (3)) = P(Z).
    (Note: two ways to get (2,1))
    P(\text{1 point}) = F(1) - F(0.75) = \frac{7}{16}
    P(\text{2 points}) = F(0.75) - F(0.5) = \frac{5}{16}
    P(\text{3 points}) = F(0.5) - F(0.25) = \frac{3}{16}.

    So, P(Z) = \left(\frac{7}{16}\right)^3 + 2\frac{7}{16}\frac{5}{16} = \frac{3}{16}
     = \frac{7^3}{16^3} + \frac{2\cdot 35\cdot 16}{16^3}+\frac{3\cdot 16^2}{16^3} = \frac{2231}{16^3}.

    Using the conditional probability rule, P(\text{Win with } (1,1,1)}) = \frac{P((1,1,1))}{P(Z)} = \frac{\frac{7^3}{16^3}}{\frac{22  31}{16^3}} = \frac{343}{2231}.
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    STEP I, Question 2

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    \sum \left ( (x-x_i)^2 + (y-y_i)^2 \right ) = a^2

    Therefore

    3x^2+3y^2 - 2x(\sum x_i) - 2y(\sum y_i)+(\sum x_i^2)+(\sum y_i^2) = a^2

    (x- \frac{\sum x_i}{3})^2 + (y - \frac{\sum y_i}{3} )^2 = \frac{a^2}{3} - \frac{2}{9} ( \sum x_i^2 - \sum x_i x_j + \sum y_i^2 - \sum y_i y_j )

    Is a circle, centre the centroid.

    Therefore if a^2 is smaller than the distance to the centroid the circle would have to have a negative radius
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    STEP II, Question 10
    Please check it
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    First part
    Forming the basic equations:
    v_2-v_1=e(u_1-u_2)\implies v_2=v_1+eV
    u_1m_1+u_2m_2 = v_1m_1+v_2m_2\implies mV = mv_1+mqv_2\implies V = v_1+qv^2
    \implies v_2 = V-qv_2+eV \implies v_2(1+q) = V(1+e) \implies v_2 = \frac{V(1+e)}{1+q}
    Thus, v_q = V - q\frac{V(1+e)}{1+q} = V\frac{1+q-q-qw}{1+q} = \frac{1-qe}{1+q}V.

    Second part

    For the last hit particle not to be hit by the one before that, the condition: v_n\leq v_{n+1} (This means: final velocity of particle n must be less than the initial particle's (n+1) velocity, so there would be N-1 impacts), which, after factoring, gives:
     \implies \frac{1-eq}{1+q}\leq\left(\frac{1-eq}{1+q}\right)\left(\frac{1+e}{  1+q}\right)\implies 1+q \leq 1 + e \implies q\leq e.

    q=e. We also know that v_{n\text{-final}} = (1-e)V, m_n = e^{n-1}m, and so E_n = \frac{1}{2}(1-e)^2V^2e^{n-1}m. Summing,
    \displaystyle S_N = \sum_{i=0}^{N-1} E_i = \frac{1}{2}(1-e)V^2m\sum_{i=0}^{N-1} e^{i=1} = \frac{1}{2}V^2m(1-e^{N-1}(1-e). Initial K.E. E_0=\frac{1}{2}mV^2.

    S_n - E_0 =  \frac{1}{2}mV^2(1-(1-e^{N-1})(1-e))) = \frac{1}{2}mV^2(e-e^N) = \frac{1}{2}me(1-e^{N-1})V^2, as required.
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    STEP II, Question 13

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    Wlog our stick is of unit length and by symmetry we can choose the length of the shorter piece. Let X \sim U(0, \frac{1}{2})

    The probability that R &lt; r is

    \displaystyle \frac{a}{1-a} &lt; r \Longleftrightarrow \boxed{a &lt; 1- \frac{1}{r+1}}

    Since we're choosing from a uniform distribution, this probability is \displaystyle 2-\frac{2}{r+1}

    This is the cumulative distribution. Therefore the probability density function is

    \displaystyle p(r)=\frac{2}{(r+1)^2}

    The expected value will be

    \displaystyle \int_0^1 r p(r) \, dr = \boxed{2\ln 2 - 1}}

    and the variance

    \displaystyle \int_0^1 r^2 p(r) \, dr - \left ( \int_0^1 r p(r) \, dr \right)^2= \boxed{2-(\ln 4)^2} }
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    STEP II, Question 13
    Don't even quote me on this question, there is absolutely no guidance within it.
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    Ratios for positions \displaystyle x&lt;\frac{1}{2}L and \displaystyle x\geq\frac{1}{2}L are \displaystyle \frac{x}{L-x} and \displaystyle \frac{L-x}{x} respectively.
    Draw a diagram, and then you have a simple probability density:
    \displaystyle f(x) = \begin{cases} \frac{1}{L} & 0 \leq x \leq L \\ 0 & \text{otherwise}\\ \end{cases}

    \displaystyle E(R) = \int_{\frac{1}{2}L}^{L} \frac{L-x}{x}\frac{1}{L}\,\mathrm{d}x + \int_{0}^{\frac{1}{2}L} \frac{x}{L-x}\frac{1}{L}\,\mathrm{d}x =
    \displaystyle  = \frac{1}{L} \left[L\ln|x|-x\right]_{L/2}^L + \left[-L\ln|L-x|-x\right]_{0}^{L/2} =
    \displaystyle  = 2ln2 - 1


    \displaystyle E(R^2) = \int_{\frac{1}{2}L}^{L} \left(\frac{L-x}{x}\right)^2\frac{1}{L}\,\math  rm{d}x + \int_{0}^{\frac{1}{2}L} \left(\frac{x}{L-x}\right)^2\frac{1}{L}\,\mathrm{  d}x = \frac{3}{2} - 4\ln2.

    So \displaystyle Var(R) = \frac{3}{2} - 3\ln2 - (2\ln2-1)^2 = \frac{1}{2} - 4(\ln2)^2.
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    Heh. Different method, different answers. Not great
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    Simon: Surely you really want to have p(R < r) = 1 in the case where r=1?

    Edit: I don't think you're taking into account that your uniform variable is on [0,1/2], not [0,1], (and so has density 2 instead of density 1).
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    (Original post by DFranklin)
    Simon: Surely you really want to have p(R < r) = 1 in the case where r=1?
    This is true.. Better check my logic again

    Ah... spotted the problem. Not too bad to fix

    Edit: Yeah, that was what I ascertained. Stupid mistake
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    STEP II, Question 14

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    Your tactic will essentially be "stop rolling if greater than a given number", this is because each result is independent on the previous ones, so you must take a decision based on what you have rolled. The probability of obtaining any of the scores in the "stop" range is the same (symmetry, relabelling, etc). Therefore, considering cases we have:

    Stop after first:
    \displaystyle \frac{1}{6}(1+2+3+4+5+6) = \frac{7}{2}

    Stop if larger than 2
    \displaystyle \frac{1}{5}(1+3+4+5+6) = \frac{19}{5}

    Stop if larger than 3
    \displaystyle \frac{1}{4}(1+4+5+6) = 4

    Stop if larger than 4
    \displaystyle \frac{1}{3}(1+5+6) = 4

    Stop if larger than 5
    \displaystyle \frac{1}{2}(1+6) = \frac{7}{2}

    Therefore we should stop if we get greater than 3. (or 4, I guess we should consider variances)
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    (Original post by squeezebox)
    Ok, so heres my working:

    I get the integral down to;
    

 \frac{3a^{2}}{8} \displaystyle\int^{2\pi}_{0} \sin^{2}(2t)  \, dt

    which becomes:

    \frac{3a^{2}}{16} \displaystyle\int^{2\pi}_{0} 1-cos(4t) \, dt

    which is equal to \frac{3a^{2}\pi}{8}?
    I think i got the same answer as yours
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    (Original post by lilman91)
    The hare finishes 1.5hrs. This means she has travelled 2X in 1.5hrs.
    So, \frac{2X}{V}=1.5 \Rightarrow 4X=3V
    Substituting expression for V,
    4X=\frac{30}{4X-9} \Rightarrow 16X^2-36X-30=0 \Rightarrow 8X^2-18x-15=0.
    Solving the quadratic but since X&gt;0, X=\frac{18+\sqrt{804}}{16}=\frac  {18+2\sqrt{201}}{16}=\frac{9+\sq  rt{201}}{8}.
    So, finally X=\frac{9+\sqrt{201}}{8}.
    I think you've forgotten to add on the half hour rest at each end, so we end up with:

     \dfrac{2X}{V} + 1 = \dfrac{3}{2}

    And we end up with a somwhat neater final answer:
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     2X + V = 1.5V \iff 4X = V

     \iff 4X = \dfrac{10}{4X-9} \iff 16X^2 - 36X - 10 = 0

     \iff 8X^2 - 18X - 5 = 0

     \iff (4X+1)(2X-5) = 0

     \iff X = \dfrac{5}{2}, -\dfrac{1}{4} \implies X = \boxed{\dfrac{5}{2} \mathrm{km}}
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    For paper III Q2. PDF attached.
    Attached Files
  1. File Type: pdf STEP III 1999 Q2.pdf (267.8 KB, 598 views)
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    STEP III Question 13

    I am not certain that this method is correct, so it would be very much appreciated if somebody were to follow my working and check I have not made any wrong assumptions.

    Firstly,  \int_0^1 f(x) dx = 1 \rightarrow A = 2

    so  \int_0^1 x.f(x) dx = \left[ 2 x^3/3 \right]_0^1  = 2/3

    ie expected proportion of cake.

    Currents are randomly placed so the expected number of currents = 4 * 2/3 = 8/3

    Now taking a proportion X of the cake gives probability X^4 that all 4 currents will be in that portion.

    So taking a portion X and getting all 4 currents in that portion has pdf:

     f(x) = 2x.x^4 = 2x^5

    Hence,  P(all4inportion) = \int_0^1 2x^5 dx = 1/3

    and,  P(all4inportionbiggerthanhalf) = \int_{1/2}^1 2x^5 dx = 1/3 - 1/192 = 63/192

    So finally  P(portionbiggerthanhalf \mid all4inportion) = (63/192)/(1/3) = 189/192

    I have spent a whole day thinking about pdfs and probabilities and venn diagrams and integrals and have got about 100 wrong answers to this question, let alone a lot of messing about with excel formulas to convert rand() into the continuous random variable you want, so now my brain hurts! But, I think I finally understand how to do this question. If anybody has any comments, that would be great, and also if anybody could point me in the direction of similar questions to practice this, that would be good too, as I'm hoping that probability will be one of my specialist subjects in STEP. Now to attempt number 14...

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