Results are out! Find what you need...fast. Get quick advice or join the chat
Hey there Sign in to join this conversationNew here? Join for free

STEP maths I, II, III 1991 solutions

Announcements Posted on
    • 14 followers
    Offline

    ReputationRep:
    (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...
    • 3 followers
    Offline

    ReputationRep:
    (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 followers
    Offline

    ReputationRep:
    \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:/
    • 14 followers
    Offline

    ReputationRep:
    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...
    • 6 followers
    Offline

    ReputationRep:
    (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
    • 10 followers
    Offline

    ReputationRep:
    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}
    • 0 followers
    Offline

    ReputationRep:
    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.
    • 0 followers
    Offline

    ReputationRep:
    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.
    • 0 followers
    Offline

    ReputationRep:
    (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.
    • 0 followers
    Offline

    ReputationRep:
    (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.
    • 0 followers
    Offline

    ReputationRep:
    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.
    • 10 followers
    Offline

    ReputationRep:
    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.
    • 0 followers
    Offline

    ReputationRep:
    (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
    • 1 follower
    Offline

    ReputationRep:
    I think these solve 1991 Step 3 Q4 and Q9
    Attached Files
  1. File Type: pdf Step1991Paper3Question4.pdf (56.5 KB, 74 views)
  2. File Type: pdf Step1991Paper3Question9.pdf (77.4 KB, 54 views)
    • 6 followers
    Offline

    ReputationRep:
    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}
    • 6 followers
    Offline

    ReputationRep:
     \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}
    • 6 followers
    Offline

    ReputationRep:
     \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)
    • 6 followers
    Offline

    ReputationRep:
     \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
    Click image for larger version. 

Name:	1991 II 13(a).jpg 
Views:	26 
Size:	19.4 KB 
ID:	108412   Click image for larger version. 

Name:	1991 II 13(b).jpg 
Views:	24 
Size:	16.0 KB 
ID:	108413  
    • 6 followers
    Offline

    ReputationRep:
     \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}
    Attached Thumbnails
    Click image for larger version. 

Name:	1991 II 14.jpg 
Views:	23 
Size:	6.7 KB 
ID:	108451  
    • 6 followers
    Offline

    ReputationRep:
     \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

Reply

Submit reply

Register

Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: June 10, 2013
New on TSR

THE world university rankings 2014-2015 revealed

Will they affect your uni choices?

Article updates
Reputation gems:
You get these gems as you gain rep from other members for making good contributions and giving helpful advice.