STEP maths I, II, III 1991 solutions

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  1. DFranklin's Avatar
    181 12-05-2008  12:28
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by insparato)
    I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.
    Exactly.

    When i attempted this question, this is what plagued me... I don't think you can assume that the two spheres are going to amount to half the initial height and be half the initial radii.
    Form a differential equation for R as a function of time. Solve it for both spheres. Add the answers to get a formula for height as a function of time. Etc...
  2. nota bene's Avatar
    182 12-05-2008  12:30
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by insparato)
    I think the problem is, the head could melt slower or faster than the body, which means you could reach half the initial height without the head or body necessarily being exactly half the respective radius.
    Ah, of course! Thanks, this should set me off
  3. nota bene's Avatar
    183 12-05-2008  13:17
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    Re: STEP maths I, II, III 1991 solutions
    \frac{dV_U}{dt}=k16\pi R^2=k16\pi(\frac{3V_U}{4\pi})^{\ frac{2}{3}} and
    \frac{dV_L}{dt}=k36\pi R^2=k36\pi(\frac{3V_L}{4\pi})^{\ frac{2}{3}}
    Separating variables \int dt=\int \frac{1}{16k\pi} (\frac{3V_U}{4\pi})^{ -\frac{2}{3}}dV_U \Rightarrow t=\frac{1}{4k}(\frac{V_U}{4\pi}) ^{\frac{1}{3}}+c_1= \frac{1}{4k}(\frac{R^3}{3})^{\fr ac{1}{3}}+c_1= \frac{1}{4k \sqrt[3]{3}}R+c_1
    Separating variables on the other one \int dt=\int \frac{1}{36k\pi}(\frac{3V_L}{4\p i})^{-\frac{2}{3}}dV_L \Rightarrow t=\frac{1}{9k}(\frac{3V_L}{4\pi} )^{\frac{1}{3}}+c_2=\frac{1}{9k} R+c_2

    What do I do with the constants now? They're bugging me.

    2\times4k\sqrt[3]{3}(t-c_1) is the diameter (height) of the upper sphere and 3\times9k(t-c_2) is the diameter of the lower sphere. So the height is h(t)=27k(t-c_2)+8k\sqrt[3]{3}(t-c_1)

    So 0.5=h(t) but don't see how this helps. I'm giving up, I have never been able to do rates of change:/
  4. DFranklin's Avatar
    184 12-05-2008  13:42
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    Re: STEP maths I, II, III 1991 solutions
    I'd do something like:

    Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

    v \varpropto r^3

    So \frac{dv}{dt} \varpropto 3r^2 \frac{dr}{dt} \varpropto r^2 \frac{dr}{dt}

    And s \varpropto r^2, \frac{dv}{dt} \varpropto -s

    So r^2 \frac{dr}{dt} \varpropto -r^2, so \frac{dr}{dt} = -k for some constant k.

    Subbing in for the initial values, we get R_u(t) = 2R-kt, R_l(t)=3R-kt.

    Now find kt when R_u+R_l = 5R/2, etc...
    Last edited by DFranklin; 12-05-2008 at 13:48.
  5. brianeverit's Avatar
    185 16-05-2008  08:31
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by DFranklin)
    I'd do something like:

    Consider the rate of evaporation of a sphere of radius r, volume v, surface area s.

    v \varpropto r^3

    So \frac{dv}{dt} \varpropto 3r^2 \frac{dr}{dt} \varpropto r^2 \frac{dr}{dt}

    And s \varpropto r^2, \frac{dv}{dt} \varpropto -s

    So r^2 \frac{dr}{dt} \varpropto -r^2, so \frac{dr}{dt} = -k for some constant k.

    Subbing in for the initial values, we get R_u(t) = 2R-kt, R_l(t)=3R-kt.

    Now find kt when R_u+R_l = 5R/2, etc...
    This question is number 7 in Advanced Problems in Mathematics
  6. Glutamic Acid's Avatar
    186 27-04-2009  18:33
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    Re: STEP maths I, II, III 1991 solutions
    II/6:


    The inequality 0 \le \sin x \le x is satisfied at x = 0. Differentiating: 0 \le \cos x \le 1, which is true. So x increases faster than sin x which increases faster than 0, so the inequality will still be satisfied: so 0 \le \sin x \le x.

    The inequality \cos x \ge 1 - \dfrac{x^2}{2} is satisfied at x = 0. Differentiating: - \sin x \ge - x \Leftrightarrow \sin x \le x, which has been shown previous. So cos x decreases slower than 1 - x^2/2, so the inequality will still be satisfied: so \cos x \ge 1 - \dfrac{x^2}{2}.

    3 \ge 2 + \cos x \ge 3 - \dfrac{x^2}{2} \implies \dfrac{1}{3} \le \dfrac{1}{2 + \cos x} \le \dfrac{1}{3 - x^2/2}

    \implies\dfrac{1}{9} \le \dfrac{1}{(2 + \cos x)^2} \le \dfrac{1}{(3 - x^2/2)} \implies \dfrac{x}{9} \le \dfrac{x}{(2 + \cos x)^2} \le \dfrac{x}{(3 - x^2/2)^2}

    \displaystyle \implies \int_0^{1/10} \frac{x}{9} \, \text{d}x} \le \int_0^{1/10} \frac{x}{(2 + \cos x)^2} \, \text{d}x \le \int_0^{1/10} \frac{x}{(3 - x^2/2)^2} \, \text{d}x \implies \left[ \frac{x^2}{18} \right]^{1/10}_0 \le \int_0^{1/10} \frac{x}{(2 + \cos x)^2} \, \text{d}x \le \left[ \frac{1}{3 - x^2/2} \right]^{1/10}_0

    \displaystyle \implies \frac{1}{1800} \le \int_0^{1/10} \frac{x}{(2 + \cos x)^2} \, \text{d}x \le \frac{1}{1797}

    \sin x \ge x - x^3/6 is satisfied at x = 0. Differentiating: \cos \ge 1 - x^2/2, which has been previously shown. sin x increases faster than x - x^3/6, so the inequality holds.

    Similarly:
    \displaystyle \int_0^{1/10} x^2 \, \text{d}x \le \int_0^{1/10} \frac{x^2}{(1 - x + \sin x)^2} \, \text{d}x \le \int_0^{1/10} \frac{x^2}{(1 - x^3/6)^2} \, \text{d}x

    \displaystyle \implies \left[ \frac{x^3}{3} \right]^{1/10}_0 \le \int_0^{1/10} \frac{x^2}{(1 - x + \sin x)^2} \, \text{d}x \le \left[ \frac{2}{1 - x^3/6} \right]^{1/10}_0

    \displaystyle \implies \frac{1}{3000} \le \int_0^{1/10} \frac{x^2}{(1 - x + \sin x)^2} \, \text{d}x \le \frac{2}{5999}
  7. toasted-lion's Avatar
    187 27-05-2009  17:21
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    Re: STEP maths I, II, III 1991 solutions
    STEP II Q16

    Part 1a

    Obviously it needs to rain on both days. The next few arguments assume it does. Note also it is necessary that V>u. If this is not the case, then Q=0, and the most specific expression for u is u \geq V. If Q \not= 0 then the following applies.

    At midnight on the first day the volume will be u below the top.

    We require that on the second day it rains at a time t such that V>(1 + \frac{t}{24})u ie t < 24(\frac{V}{u} - 1) .

    The probability of this being the case is \frac{V}{u} - 1.

    So Q = p^2(\frac{V}{u} - 1) and u = \frac{Vp^2}{p^2 +Q}.

    Part 1b

    We require P(R>k)<\frac{1}{10} where R is a random variable representing the number of times it rains, R \sim B(18,\frac{1}{3}).

    R \sim B(18,\frac{1}{3}) can be approximated by S \sim N(6,4).

    P(R>k) \approx P(S > k + \frac{1}{2}) = P(Z > \frac{k - 5.5}{2}) where Z \sim N(0,1).

    We require P(R>k)<\frac{1}{10} so we look for a value of k such that P(Z > \frac{k - 5.5}{2}) \approx \frac{1}{10}.

    Now P(Z>1.28) = 0.1003 so \frac{k-5.5}{2} \approx 1.28 \implies k \approx 8.06.

    The probability must be strictly less than 0.1 so he should set k=9.

    Part 2

    The variance in the total amount of rain will be lower, allowing for a lower value of k.
  8. toasted-lion's Avatar
    188 27-05-2009  17:24
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    Re: STEP maths I, II, III 1991 solutions
    On the last part of question 15 of STEP II I got \frac{d(m+1)}{6m}[2N(m-1) + 3m] instead of \frac{d(m-1)}{6m}[2N(m+1) + 3m] . Did anybody else get this? What are the chances they made a typo? I may type up my solution in a bit, but I just typed up 16 and I'm latexed out.
  9. toasted-lion's Avatar
    189 27-05-2009  18:32
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by justinsh)
    I may have done mistakes but here it goes:

    \ln(1+x+x^2+x^3)\\
    =\ln(1+x)+\ln(1+x^2)\\
    =(x-\frac{1}{2}x^2+\frac{1}{3}x^3+.. .+\frac{(-1)^{n+1}}{n}x^n)+(x^2-\frac{1}{2}x^4+\frac{1}{3}x^6+.. .+\frac{(-1)^{n+1}}{n}x^{2n})\\
    =x+\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{3}{4}x^4+...+\left[\frac{(-1)^{n+1}}{n}-\frac{1}{2n}\right]x^{2n}+\frac{x^{2n+1}}{2n+1}
    In my personal opinion, both solutions to STEP III Q1 part b on here fail to find a general term (although this second one is quite nice with two fairly simple general terms), so I am going to post the way I did it, which is quite similar to the above method except that I actually put a general term, rather than a set of general terms. I don't know whether they expect you to do this, but before looking at these I presumed you had to. Anyway, here goes:

    ln(1+x+x^2+x^3) = ln(1+x)+ln(1+x^2) \\ \\ = (x-\frac{1}{2}x^2+\frac{1}{3}x^3+.. .+\frac{(-1)^{n+1}}{n}x^n+...)+(x^2-\frac{1}{2}x^4+\frac{1}{3}x^6+.. .+\frac{(-1)^{n+1}}{n}x^{2n}+...) \\ \\ = (x-\frac{1}{2}x^2+\frac{1}{3}x^3+.. .+\frac{(-1)^{n+1}}{n}x^n+...)+(x^2-\frac{1}{2}x^4+\frac{1}{3}x^6+.. .+\frac{(-1)^{\frac{n}{2}+1}}{n/2}\frac{1+(-1)^n}{2}x^{n}+...) \\ \\ = x+\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{3}{4}x^4+\frac{1}{5}x^5+.. .+\big[\frac{(-1)^{n+1}}{n}+ \frac{(-1)^{\frac{n}{2}+1}}{n}(1+(-1)^n)\big]x^n+...

    It's just that this question would be too easy without this, I think that what they're testing is your ability to think of a way to merge the xn and x2n terms, and finding a fairly simple function that returns 1 for even n and 0 for odd n.
  10. toasted-lion's Avatar
    190 27-05-2009  22:27
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Rabite)
    I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...


    Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors. :p:
    I updated the attachment.

    Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.
    1. There is a typo early on, you put 11 instead of 1.
    2. No minimum at (1, -4) of any sort.
    3. But there is a minimum at (2, -5), which is also necessary for the graph.
    4. In the last part, you are defining f-1(x) so the x we are talking about is a different x to before if that makes sense, ie it is what was on the y-axis before not the x-axis. So your table thing doesn't really make sense.
    5. If we ignore point 4, the right-most box is wrong anyway, because the positive root is only valid to the right of 2.

    So in the last part you need to consider x<-5,-5 \leq x \leq -4, -4 \leq x \leq 1, 1 \leq x \leq 4, x \geq 4 . The first is undefined, all the others have two possibilities (one of which is the same for all four). It's seriously boring stuff. This question is not particularly challenging, but loads of cases to consider, and loads of room for little errors.
  11. toasted-lion's Avatar
    191 28-05-2009  12:56
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    Re: STEP maths I, II, III 1991 solutions
    STEP III Q16
    Part 1
    P(X=x)=\displaystyle\sum_{y=1}^{ 2(a-x)} c(2x+y) \\ =2(a-x)c2x + c(a-x)(2a-2x+1) \\ =c(a-x)(2a+2x+1) \\ =c\left[a(2a+1) - x - 2x^2 \right]

    Part 2
    \displaystyle\sum_{x=1}^{a-1} P(X=x) = 1 \\ \implies c \displaystyle\sum_{x=1}^{a-1} \left[a(2a+1) - x - 2x^2 \right] = 1 \\ \implies c\left[(a-1)a(2a+1) - \frac{1}{2}(a-1)a - \frac{1}{3}(a-1)a(2a-1)\right] = 1 \\ \implies c(a-1)a(12a +6-3-4a+2) = 6 \\ \implies c = \frac{6}{a(a-1)(8a+5)}

    Part 3
    P(Y=y) = \displaystyle\sum_{y=1}^{a-\frac{y}{2}} c(2x +y) \\ =c\left[(a-\frac{y}{2})y + (a - \frac{y}{2})(a - \frac{y}{2} + 1)\right] \\ =\frac{3(2a-y)(2a+2+y)}{2a(a-1)(8a+5)}
    for even y.

    Part 4
    P(Y=y) = \displaystyle\sum_{y=1}^{a-\frac{y+1}{2}} c(2x +y) \\ =c\left[(a-\frac{y+1}{2})y + (a - \frac{y+1}{2})(a - \frac{y+1}{2} + 1)\right] \\ =\frac{12a^2 - 3(y+1)^2}{2a(a-1)(8a+5)}
    for odd y.

    Part 5
    E(Y)=\displaystyle\sum_{y=1}^{a-1} \frac{6y(2a-2y)(2a+2+2y) + (2y-1)\left( 12a^2 - 3(2y)^2\right)}{2a(a-1)(8a+5)} \\ = \frac{3}{2a(a-1)(8a+5)}\displaystyle\sum_{y=1} ^{a-1}\left(-16y^3 - 4y^2 +(16a^2 +8a)y - 4a^2 \right) \\ = \frac{3}{2a(a-1)(8a+5)} \left( -4a^2 (a-1)^2 - \frac{2}{3}(a-1)a(2a-1) + \frac{1}{2} (16a^2 +8a)a(a-1)- 4a^2(a-1) \right)
    \\ \\ =\frac{6a^2 +4a +1}{(8a+5)}
    Woooooooo! This has taken me almost all day and I've finally done it! I'm almost certain the answer is right as it works for a=2 and a=3, and everything follows nicely.
    Last edited by toasted-lion; 28-05-2009 at 17:32.
  12. Glutamic Acid's Avatar
    192 11-06-2009  01:19
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    Re: STEP maths I, II, III 1991 solutions
    This alternative to III/1 part (c) is quite cute.

    e^{x \ln(1+x)} = (1+x)^x = 1 + x^2 + \dfrac{x(x-1)}{2!}x^2 + \dfrac{x(x-1)(x-2)}{3!}x^3 + ... = 1 + x^2 - \dfrac{x^3}{2} + \dfrac{5x^4}{6} + ... from the binomial expansion.
  13. themaths's Avatar
    193 17-05-2011  10:17
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    Re: STEP maths I, II, III 1991 solutions
    (Original post by Rabite)
    I should hope not. Not sure what I've done to make someone want to neg me though. Guess it's because I sound like a moron or something. Oh wait, I am a moron...


    Anyway - I did III Q3. Thanks to Khai for pointing out my stupid errors. :p:
    I updated the attachment.

    Also, has anyone done the one in STEP III about some particle and two observers? Q4 I think it was...I thought I had the answer, but the direction vector has a 22 in it, so I'm guessing it's wrong.
    Hi there just wanted to point out one final thing i noticed when doing this question about the inverse functions is you have to check the functions are still valid in the intervals you gave them:
    For instance for -1<x<0
    \sqrt\frac{x-1}{3}
    is undefined
    Also, in the interval x<-1
    2- \sqrt{x+5}
    is undefined once x<-5 so the interval should be -5<x<-1
    and the inverse functions are undefined for the intervals -1<x<0 and x<-5
  14. mikelbird's Avatar
    194 14-06-2011  08:29
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    Re: STEP maths I, II, III 1991 solutions
    I think these solve 1991 Step 3 Q4 and Q9
    Attached Files
  15. File Type: pdf Step1991Paper3Question4.pdf (56.5 KB, 46 views)
  16. File Type: pdf Step1991Paper3Question9.pdf (77.4 KB, 35 views)
  17. brianeverit's Avatar
    195 18-07-2011  12:05
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    Re: STEP maths I, II, III 1991 solutions
    STEP 1991 Paper II question 9

    \text {Closure } \begin{pmatrix}a&b\\0&c \end{pmatrix}\begin{pmatrix}x & y \\ 0 & z \end{pmatrix}= \begin{pmatrix}ax&ay+bz\\0&cz \end{pmatrix}
    \text{Associativity may be assumed}
    \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ is an obvious identity element}
    A= \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \implies A^{-1}= \dfrac {1}{ac} \begin{pmatrix}c & -b \\ 0 & a \end{pmatrix} \text{ and }ac\not=0 \implies \text{ existence of inverses}
    \text{The order of G is the number of its elements}
    a \text{ and }c \text{ can each take 4 values, 1,2,3, or 4 whilst }b \text{ can take 5, 0,1,2,3, or 4}
    \text{so there are }4 \times 4 \times 5=80 \text{ distinct elements so the order of the group is }80
    \text{G is not commutative, } \begin{pmatrix} x & y \\ 0 & z \end{pmatrix} \begin{pmatrix} a & b \\ 0 & c \end{pmatrix}= \begin{pmatrix} ax & bx+cy \\ 0 & cz \end{pmatrix} \text{ and } bx+cy \text { need not be the same as }ay+bz
    \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \text{ is the only element of order 1, if } \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \text{ is an element of order 2 then}
    \begin{pmatrix} a & b \\ 0 & c \end{pmatrix}\begin{pmatrix} a & b \\ 0 & c \end{pmatrix}= \begin{pmatrix} a2 & ab+bc \\ 0 & c \end{pmatrix}= \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\implies a^2 = c^2 =1 \text{ and } b(a+c)=0
    \text{possibilities are } b=0,a=\pm 1,c=\pm 1, \text{ or } a+c=0 \implies a=1, c=-1,b=1,2,3 \text{ or }4
    \text{or }a=-1,c=1,b=1,2,3 \text{ or }4 \text { giving a total of 12 elements}
    \text{so there are 13 elements of order 1 or 2 hence this subset cannot be a subgroup since 13 is not a factor of 80}
  18. brianeverit's Avatar
    196 18-07-2011  14:12
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    Re: STEP maths I, II, III 1991 solutions
    \text {STEP 1991 paper II question 10}

    An absolute swine to Latex but here goes

    \text{Let the sun be S, the pole AB and the shadow AP with origin O}
    \mathbf {OS} = \begin{pmatrix} R \cos \theta \\ R \sin \theta \\ 0 \end{pmatrix}, \mathbf {OB}= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix} \text{ so } \mathbf {SB} = \begin{pmatrix} -R \cos \theta \\ b \cos \phi-R \sin \theta \\ b \sin \phi \end{pmatrix}}
    \text{horizontal plane through A has equation } (\cos \phi )y+( \sin \phi)z=a
    \text{so, taking equation of SB as }\mathbf r= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix}+ \lambda \begin{pmatrix}-R \cos \theta \\ b \cos \phi-R \sin \theta \\ b \sin \phi \end{pmatrix}}
    \text{P is the point where } \cos \phi (b \cos \phi + \lambda( b \cos \phi - R \sin \theta ))+b \sin^2 \phi + \lambda b \sin^2 \phi=a
    \implies b+ \lambda b-\lambda R \sin \theta \cos \phi=a \implies \lambda= \dfrac{b-a}{R \sin \theta \cos \phi-b}= \dfrac{h} {R \sin \theta \cos \phi -b}
    \mathbf{OP}= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix}+ \left(\dfrac{h}{R \sin \theta \cos \phi -b}\right) \begin{pmatrix} -R \cos \theta \\ b \cos \phi - R \sin \theta \\ b \sin \phi \end{pmatrix}
    \text {and so } \mathbf {AP}= \begin{pmatrix} 0 \\ b \cos \phi \\ b \sin \phi \end{pmatrix} +\left( \dfrac{h}{R \sin \theta \cos \phi -b}\right) \begin{pmatrix} -R \cos \theta \\ b \cos \phi - R \sin \theta \\ b \sin \phi \end{pmatrix}- \begin{pmatrix} 0 \\ a \cos \phi \\ a \sin \phi \end {pmatrix}
    = \dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix} -hR \cos \theta \\ b \cos \phi(R \sin \theta \cos \phi -b)+bh \cos \phi - hR \sin \theta - a \cos \phi (R \sin \theta \cos \phi -b) \\b \sin \phi (R \sin \theta \cos \phi -b)+bh \sin \phi-a \sin \phi (R \sin \theta \cos \phi -b) \end{pmatrix}
    = \dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix}-hR \cos \theta \\ bR\sin \theta \cos^2 \phi -b^2 \cos \phi+bh \cos \phi - hR \sin \theta -aR\sin \theta \cos^2 \phi +ab \cos \phi \\ bRsin \theta \sin \phi \cos \phi -b^2 \sin \phi+bh \sin \phi -aR\sin \phi \sin \theta \cos \phi +ab \sin \phi \end{pmatrix}
    =\dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix}-hR \cos \theta \\ hR\sin \theta \cos^2 \phi -b^2 \cos \phi+(b^2-ab) \cos \phi - hR \sin \theta +ab \cos \phi \\ hRsin \theta \sin \phi \cos \phi -ab \sin \phi+ab \sin \phi \end{pmatrix}
    =\dfrac{1}{R \sin \theta \cos \phi-b} \begin{pmatrix}-hR \cos \theta \\ hR\sin \theta \cos^2 \phi - hR \sin \theta \\ hRsin \theta \sin \phi \cos \phi \end{pmatrix}
    = \dfrac {Rh}{R \sin \theta \cos \phi-b} \begin{pmatrix}- \cos \theta \\ -\sin \theta \sin^2 \phi \\ sin \theta \sin \phi \cos \phi \end{pmatrix}
  19. brianeverit's Avatar
    197 18-07-2011  14:56
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    Re: STEP maths I, II, III 1991 solutions
    \text{STEP 1991 paper II question 12}

    \text{When hanging in equilibrium we have }mg= \dfrac{\lambda c}{a}
    \text {At time }t, mg= \dfrac{\lambda (c+x-b \sin pt)}{a}=m \dfrac{ \text{d}^2x}{ \text{d}t^2} \implies \dfrac{ \text{d}^2x}{ \text{d}t^2}=- \dfrac {g}{c}(x-b \sin pt)=n^2(x-b \sin pt) \text { as required}
    \text{rearranging, }\dfrac{ \text{d}^2x}{ \text{d}t^2}+n^2x=n^2 \sin pt \text{ which has C.F. } A \cos nt+B \sin nt \text { and a P.I. } x=C \sin pt}
    \text {hence }-Cp^2 \sin pt+Cn^2 \sin pt=bn^2 \sin pt \implies C= \dfrac{bn^2}{n^2-p^2}
    \text {so }x=A \cos nt+B \sin nt+ \dfrac{bn^2}{n^2-p^2} \sin pt \implies \dfrac{ \text{d}^2x}{ \text{d}t^2}=-n^2A \cos nt-n^2 B \sin nt- \dfrac{bn^2p^2}{n^2-p^2} \sin pt
    \text{so }-n^2A \cos nt-n^2 B \sin nt- \dfrac {bn^2p^2}{n^2-p^2} \sin pt+A \cos nt + B \sin nt+ \dfrac{bn^2}{n^2-p^2} \sin pt=b \sin pt
    \text{hence, }-n^2A+A=0 \implies A=0 \text{ and }-n^2B- \dfrac{bn^2p^2}{n^2-p^2}+B+ \dfrac{bn^2}{n^2-p^2}=b
    \text{i.e. }B(1-n^2)+ \dfrac{bn^2}{n^2-p^2}(1-p^2)=b \implies B= \dfrac{b- \frac{bn^2}{n^2-p^2}(1-p^2)}{1-n^2}= \dfrac{b(n^2-p^2)-bn^2+bn^2p^2}{(1-n^2)(n^2-p^2)}= \dfrac{bp^2(n^2-1)}{(1-n^2)(n^2-p^2)}
    \text{so solution is }x=- \dfrac{bp^2}{n^2-p^2} \sin nt+ \dfrac{bn^2}{n^2-p^2} \sin pt= \dfrac{bn}{n^2-p^2}(n \sin pt-p \sin nt)
    \text{The condition }0&lt;p&lt;n \text{ is required as otherwise the initial movement of the particle is upwards}
    \text{which is clearly impossible}
    \text{String remains taut throughout the motion if }x-b \sin pt&gt;-c
    \text{i.e. }c+ \dfrac{bn}{n^2-p^2}(n \sin pt-p \sin nt)&gt;b \sin pt \implies cn^2-cp^2+bn^2 \sin pt-bnp \sin nt &gt; bn^2 \sin pt-bp^2 \sin pt
    \implies cn^2-cp^2-bnp \sin nt+bp^2 \sin pt&gt;0 \text { but } \sin nt&lt; \sin pt
    \text{so } cn^2-cp^2+(bnp+bp^2) \sin pt&gt;0 \implies c(n^2-p^2)+bp(n+p) \sin pt&gt;0 \implies c(n-p)+bp \sin pt&gt;0
    \implies \sin pt&gt; \dfrac{c(p-n)}{bp} \text { which is certainly true if } \dfrac{c(p-n)}{bp}&lt;-1 \implies bp&lt;c(n-p)
    Last edited by brianeverit; 18-07-2011 at 14:58.
  20. brianeverit's Avatar
    198 18-07-2011  19:26
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP maths I, II, III 1991 solutions
    \text{STEP 1991 Paper II question 13}

    \text {For particle we have (1) } R-amg \sin \theta =amx \ddot \theta
    \text{ and (2) }F-amg \cos \theta=amx \dot \theta^2
    \text{For rod, we have (3) }mk^2 \ddot \theta =-Rx-dmg \sin \theta
    \text{From (1) and (3) } mk^2 \ddot \theta+dmg \sin \theta=-ax^2 \ddot \theta-amgx \sin \theta
    \text{ i.e. }m(k^2+ax^2) \ddot \theta=-mg \sin \theta(ax+d)
    \text{so } \ddot \theta= -\dfrac{g \sin \theta(ax+d)}{k^2+ax^2} \implies 2 \dot \theta \ddot \theta =- \dfrac{2g \sin \theta(ax+d)}{k^2+ax^2 \dot \theta}
    \implies \dot \theta^2= \dfrac {2g(ax+d)}{k^2+ax^2} \cos \theta}
    \text {substituting in (2) gives }F=amg \cos \theta+ \dfrac{2amxg(ax+d)}{k^2+ax^2} \cos \theta
    =amg \cos \theta \left( \dfrac{k^2+ax^2+2x(ax+d)}{k^2+ax ^2} \right) \text { and } R=amg \sin \theta- \dfrac{amxg \sin \theta(ax+d)}{k^2+ax^2}
    =amg \sin \theta \left( \dfrac{k^2+ax^2-x(ax+d)}{k^2+ax^2} \right) = amg \sin \theta \left( \dfrac{k^2-dx}{k^2+ax^2}\right)
    \text{hence, } \mu= \dfrac{F}{R}= \dfrac{k^2+ax^2+2x(ax+d)}{(k^2-dx) \tan \theta} \implies \mu \tan \theta= \dfrac{3ax^2+k^2+2xd}{k^2-dx} \text { as required}
    \text {(i) If }k^2=xd \text { then } \ddot \theta=- \dfrac{g \soin \theta}{x} \text { and particle moves as if it is a simple pendulum of length }x
    \text{(ii) If }k^2&lt;xd \text { then particle willm slide upwards towards }O.
    Attached Thumbnails
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  21. brianeverit's Avatar
    199 19-07-2011  08:50
    • Exalted Member
    • Location: Lincoln
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    Re: STEP maths I, II, III 1991 solutions
    \text{1991 STEP Paper II question 14}

    \text{Velocity vector of boat is } \begin{pmatrix} -v \cos \theta \\ av-v \sin \theta \end{pmatrix}
    \text{Using polar coordinates, }x=r \cos \theta, y=r \ sin \theta
    \text{we have } \dfrac{ \text{d}x}{\text{d}t}=-r \sin \theta \dot \theta+ \dot r \cos \theta \dfrac{ \text{d}y}{ \text{d}t}=r \cos \theta \dot \theta+ \dot r \sin \theta
    \terxt{so } \tan \theta \dfrac{ \text{d}x}{ \text{d}t} + x \sec^2 \theta \dfrac{ \text{d}y}{ \text{d}t} = \dfrac{\sin \theta}{\cos \theta} (-r \sin \theta \dot \theta+ \dot r \cos \theta)+ \dffrac{r \cos \theta}{ \cos^2 \theta} \dfrac { \text{d} \theta}{ \text{d}t}=\dot r \sin \theta- \lefty( \dfrac{r \sin^2 \theta}{ \cos \theta}- \dfrac{r}{ \cos \theta} \right) \dfrac{ \text{d} \theta}{ \text{d}t}
    = \dot r \sin \theta+r \cos \theta \dfrac { \text{d} \thewta}{ \text{d}t}=\dfrac{ \text{d}y}{ \text{d}t}
    \text{so velocity vector of boat is } \begin{pmatrix} \dfrac { \text{d}x}{ \text{d}t} \\ \tan \theta \dfrac {\tyext{d}x}{ \text{d}t}+ x \sec^2 \theta \dfrac{ \text{d} \theta}{ \text{d}t} \end{pmatrix}
    \text {comparing with previous expression we have } \dfrac{ \text{d}x}{ \text{d}t}=-v \cos \theta \text{ and } \tan \theta \dfrac{ \text{d}x}{ \text{d}t}+ x \sec^2 \theta \dfrac{ \text{d} \theta}{ \text{d}t}=av-v \sin \theta
    \implies \dfrac{ \cos \theta}{a- \sin \theta}= \dfrac{ \frac{ \text{d}x}{ \text6{d}t}}{ \tan \theta \frac{ \text{d}x}{ \text{d}t}+ x \sec^2 \theta \frac { \text{d} \theta}{} \text{d}t}}= \dfrac{1}{ \tan \thewta+x \sec^2 \theta \frac { \text{d} \theta}{} \text{d}x}}
    \implies \dfrac{ \text{d}x}{ \text{d} \theta}= \dfrac{x}{a \cos \theta} \text{ or } a \dfrac{ \text{d}x}{ \text{d} \theta}=-x \sec \theta
    \text{integrating we have } \int \dfrac{a}{x} \terxt{d}x=-\int \sec \theta \text{d} \theta \implies a \ln x=- \ln( \sec \theta+ \tan \theta)+C
    x=h \text{ when } \theta=0 \text{ so } C=a \ln h \implies a \ln \left( \dfrac{x}{h} \right) = \ln \left( \dfrac {1}{ \sec \theta+ \tan \theta} \right) \implies \left( \dfrac{x}{h} \right)^a =( \sec \theta+ \tan \theta)^{-1}
    x=h \text{e}^{-s} \implies \dfrac{ \text{d}x}{ \text{d}t}=-h \text{e}^{-s} \dfrac{ \text{d}s}{ \text{d}t} \implies-h \text{e}^{-s} \dfrac{ \text{d}s}{ \text{d}t}=-v \cos \theta \implies \dfrac{h \text{e}^{-s}}{ \cos \theta} \dfrac { \text{d}s}{ \text{d}t}=v
    \text{but } \dfrac{1}{ \cos \theta}= \sqrt{\sec^2 \theta} = \sqrt{1+ \tan^2 \theta}= \sqrt{1+ \sinh^2(as)}= \cosh(as)
    \text{hence, }h \text{e}^{-s} \cosh(as) \dfrac { \text{d}s}{ \text{d}t} =v \text { as required}
    \text{integrating } t= \dfrac{h}{v} \displaystyle\int_0^\infty \text{e}^{-s} \cosh(as) \text{d}s \impliest= \dfrac{h}{2v} \displaystyle\int_0^\infty \texy{e}^{-s}( \text{e}^{as}+ \text{e}^{-as}) \text{d}s= \dfrac{h}{2v} \left[ \dfrac{ \text{e}^{-(1-a)s}}{a-1}- \dfrac{ \text{e}^{-(a+1)s}}{a+1} \right]_0^\infty
    \text{i.e. } t= \dfrac{h}{2v} \left(0- \left( \dfrac{1}{a-1}- \dfrac{1}{a+1} \right) \right)= \dfrac{h}{v(1-a^2)}=hv^{-1}(1-a^2)^{-1} \text { as required}
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  22. brianeverit's Avatar
    200 19-07-2011  10:02
    • Exalted Member
    • Location: Lincoln
    • Posts: 293
    Re: STEP maths I, II, III 1991 solutions
    \text{1991 STEP Paper II question 16}

    \text{The dam cannot overflow during the first day since it starts with spare capacity }V
    \text{and starts emptying immediately}
    \text{Suppose it rains on both days, the time, in days, at which the second rainfall occurs is uniform in } [1,2]
    \text{the amount of water drained out by then is uniform in }[u,2u]
    \text {and so is the spare capacity, hence, the dam overflows if this value is less than }V
    \text {which happens with probability } \dfrac{V-u}{u}= \dfrac{V}{u}-1
    .\text {so probability of overflow is }p^2 \left( \dfrac{V}{u}-1 \right) \text{ and for a given value of }Q
    \text {the required }u \text { satisfies }p^2 \left( \dfrac{V}{u} \right)=Q \implies u= \dfrac{V}{(Q/p^2)+1}
    \text {(i) For the engineers' holiday, then only thing that matters is how often it rains.}
    \text {If it's }k \text { or less, there will be no overflow.}
    \text {so we must choose }k \text { so that P}(n \geq k) \leq \dfrac{1}{10} \text { where }n \text { has a B}\left(18, \dfrac{1}{3} \right) \text { distribution}
    \text {approximating th8is by N}(6,4) \text { we have P}(n \geq k) \approx \text {P} \left( z \geq \dfrac{k+0.5-6}{2} \right) \leq 0
    \text { i.e. if } \dfrac{k-5.5}{2} \geq 1.282 \implies k \geq 5.5+2.56=8.06 \text { so he must choose }k=9
    \text {(ii) Mean number of rainfalls is still the same, but with half the variance so we can afford to use a similar value of }k
    Last edited by brianeverit; 19-07-2011 at 10:04.
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