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STEP Maths I,II,III 1987 Solutions

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Original post by coffeym
STEP I Q5

Let
Unparseable latex formula:

\displaystyle \psi=\int^{\beta}_{\alpha}{\frac{1}{\sqrt{(x-\alpha)(\beta-x)}}\,dx



Now if we let x=αcos2θ+βsin2θ\displaystyle x=\alpha\cos^2{\theta}+\beta\sin^2{\theta} then we can say:

(xα)=(βα)sin2θ\displaystyle (x-\alpha)=(\beta-\alpha)\sin^2{\theta} and

(βx)=(βα)cos2θ\displaystyle (\beta-x)=(\beta-\alpha)cos^2{\theta}

Also, dx=2(βα)cosθsinθ\displaystyle dx=2(\beta-\alpha)\cos{\theta}\sin{\theta} (best not to write this in a simpler form)

Substituting all of these results into the integral, and changing limits as required, we reach the result:

ψ=0π22(βα)sinθcosθ(βα)2sin2θcos2θdθ\displaystyle \psi=\int^{\frac{\pi}{2}}_0{\frac{2(\beta-\alpha)\sin{\theta}\cos{\theta}}{\sqrt{(\beta-\alpha)^2\sin^2{\theta}\cos^2{\theta}}}}\,d\theta

    ψ=20π2dθ=π\displaystyle \implies \psi=2\int^{\frac{\pi}{2}}_0{}\,d\theta=\pi as required by the first part.

If α>β\displaystyle \alpha>\beta, then simply ψ=π\psi=-\pi

For the next part, let

Unparseable latex formula:

\displaystyle \omega=\int^{\beta}_{\alpha}{\frac{1}{x\sqrt{(x-\alpha)(\beta-x)}}\,dx



and make the substitution x=1y    dx=1y2dy\displaystyle x=\frac{1}{y}\implies dx=-\frac{1}{y^2}dy

Hence, after making the substitution, and swapping limits with help from the minus sign, we reach:

Unparseable latex formula:

\displaystyle \omega=\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{y\sqrt{(\frac{1}{y}-\alpha)(\beta-\frac{1}{y})}}\,dx



Multiplying the lone yy inside the square root, we find that

ω=1β1α1(1αy)(βy1)dy\displaystyle \omega=\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{\sqrt{(1-\alpha y)(\beta y-1)}}}}\,dy

Unparseable latex formula:

\displaystyle =\int^{\frac{1}{\alpha}}_{\frac{1}{\beta}}{{\frac{1}{\sqrt{\alpha\beta (\frac{1}{\alpha}-y)(y-\frac{1}{\beta})}}}\,dy



Now note that, taking out the factor of 1αβ\displaystyle \frac{1}{\sqrt{\alpha\beta}}, that the integral is simply the same as the one we evaluated in the first part, with certain symbols being inverted. This does not change the value of the integral, as everything is consistent. Hence

ω=1αβψ\displaystyle \omega=\frac{1}{\sqrt{\alpha\beta}}\psi

    ω=παβ\displaystyle \implies \omega=\frac{\pi}{\sqrt{\alpha\beta}} as required.

This completes the question.


I heve retyped this in npost no. 235
1987 STEP II question 14

Revised solution

Let y be the distance of an element of the band from the vertex of the cone. See diagram  \text{Let }y \text{ be the distance of an element of the band from the vertex of the cone. See diagram }
component of force on element along slant face of cone is Tδθsinα \text{component of force on element along slant face of cone is }T\delta\theta \sin \alpha
 and T=λ(xl)l so force =λ(xl)sinαlδθ \text{ and }T=\dfrac{\lambda (x-l)}{l} \text{ so force }=\dfrac{\lambda(x-l)\sin \alpha}{l}\delta \theta
Hence, ignoring gravitational effects, equation of motion is \text{Hence, ignoring gravitational effects, equation of motion is }
mδθ2πd2ydt2=λ(xl)sinαlδθ - \dfrac{m \delta\theta}{2\pi}\dfrac{d^2y}{dt^2}=\dfrac{\lambda(x-l)\sin\alpha}{l}\delta\theta
y=rsinα=x2πsinα so d2ydt2=12πsinαd2xdt2y=\dfrac{r}{\sin\alpha}=\dfrac{x}{2\pi\sin\alpha} \text{ so }\dfrac{d^2y}{dt^2}=\dfrac{1}{2 \pi \sin\alpha}\dfrac{d^2x}{dt^2}
Hence, m4π2sinαd2xdt2=λ(xl)sinαld2xdt2+4π2λ(xl)sin2αml=0\text{Hence, }-\dfrac{m}{4\pi^2\sin\alpha} \dfrac{d^2x}{dt^2}= \dfrac{\lambda(x-l)\sin\alpha}{l} \Rightarrow \dfrac{d^2x}{dt^2}+\dfrac{4\pi^2\lambda(x-l)\sin^2\alpha}{ml}=0
writing this as d2xdt2+4π2λxsin2αml=4π2λlsin2αml we see that it is a standard differential equation  \text{writing this as } \dfrac{d^2x}{dt^2}+ \dfrac{4 \pi^2 \lambda x \sin^2\alpha}{ml}= \dfrac{4\pi^2 \lambda l \sin^2 \alpha}{ml} \text{ we see that it is a standard differential equation }
 with C.F. Acoskt+Bsinkt where k=2πsinαλml\text{ with C.F. }A\cos{kt}+B\sin{kt} \text{ where }k=2\pi sin \alpha\sqrt \dfrac{\lambda}{ml}
x=l is an obvious P.I. so general solution is x=Acoskt+Bsinkt+lx=l \text{ is an obvious P.I. so general solution is }x=A \cos{kt}+B \sin{kt}+l
dxdt=0 when t=0B=0 so x=Acoskt \dfrac{dx}{dt}=0 \text{ when }t=0 \Rightarrow B=0 \text{ so } x=A\cos{kt}
d2xdt2=0 when x=lA=1k2 so x=1k2coskt+l\dfrac{d^2x}{dt^2}=0 \text{ when }x=l \Rightarrow A=- \dfrac{1}{k^2} \text{ so }x=-\dfrac{1}{k^2}\cos{kt}+l
Band will become slack when x=lcoskt=0t0=π2k=14sinαmlλ \text{Band will become slack when }x=l \Rightarrow \cos{kt}=0 \Rightarrow t_0= \dfrac{ \pi}{2k}= \dfrac{1}{4 \sin \alpha } \sqrt{ \dfrac{ml}{ \lambda}}
1987 STEP II question 16

Revised solution

Let X=[There is a body in the fridge],H=[ P says there is] andK=[Q says there is]\text{Let }X =[\text{There is a body in the fridge}],H=[\text{ P says there is}] \text{ and} K=[\text{Q says there is}]
Then we require Pr(XHK)=Pr(XHK)Pr(HK)=12pq12pq+12(1p)(1q)=pq1pq+2pq\text{Then we require }Pr(X|H \cup K)= \dfrac{Pr(X \cup H \cup K)}{Pr(H \cup K)}= \dfrac{\frac{1}{2}pq}{\frac{1}2pq+\frac{1}{2}(1-p)(1-q)}=\dfrac{pq}{1-p-q+2pq}

Now let S=[A is in the fridge] and T=[B is in the fridge],L=[P says A is],M=[Q says A is] \text{Now let }S=[\text{A is in the fridge}] \text{ and }T=[\text{B is in the fridge}], L=[\text{P says A is}],M=[\text{Q says A is}]
Unparseable latex formula:

\text{Then we require }Pr(S \cup T | L \cup M)= \dfrac{\frac{1}{3}pq}{_\frac{2}{3}pq+\frac{1}{3}(1-p)(1-q)}=\dfrac{pq}{1-p-q+3pq}}

Original post by SimonM
STEP III, Question 8

Spoiler



This is actually question 10 NOT 8.
1987 STEP III question 8

Consider the parallelogram ABCD where the angle between the diagonals is θ \text{Consider the parallelogram }ABCD \text{ where the angle between the diagonals is }\theta
Area of ABCD=2×12(AD×BD2sinθ)=12AC.BDsinθ \text{Area of }ABCD=2 \times \dfrac{1}{2} \left(AD \times\dfrac{BD}{2} \sin \theta \right) =\dfrac{1}{2}AC.BD\sin \theta
which is clearly a maximum when sinθ=1 diagonals perpendicular\text{which is clearly a maximum when }\sin \theta=1 \Rightarrow \text{ diagonals perpendicular}

if ACBD then 12AC.BDsinθAC2sinθ12AC2\text {if }AC \geq BD \text{ then } \dfrac{1}{2}AC.BD \sin \theta \leq AC^2 \sin \theta \leq \dfrac{1}{2}AC^2

A is the parallelogram with sides defined by  A \text{ is the parallelogram with sides defined by }
a1x+b1y=±δ and a2x+b2y+±δ a_1x+b_1y=\pm \delta \text{ and }a_2x+b_2y+\pm \delta
and is the inverse image under the linear map by the matrix M=(a1 b1a2 b2)\text{and is the inverse image under the linear map by the matrix }M=\begin{pmatrix} a_1\ b_1\\a_2\ b_2 \end{pmatrix}
of the square with vertices (±δ,±δ) and hence of area 4δ2 so A has area 4δ2a1b2a2b1\text{of the square with vertices }(\pm\delta,\pm\delta) \text{ and hence of area }4\delta^2 \text{ so }A\text{ has area }\dfrac{4\delta^2}{|a_1b_2-a_2b_1|} ]
Reply 245
Original post by ben-smith
(I've posted some solutions in the 1992 thread if you care)
STEP III Q11

Call the position of the mother A, the point where she enters the river B, and the position of the child C.
AB=a2+(bx)2[br]BC=x2+c2AB=\sqrt{a^2+(b-x)^2}[br]BC=\sqrt{x^2+c^2}
So, total time T=a2+(bx)2u+x2+c2vT=\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{x^2+c^2}}{v}
Differentiating, T=xbua2+(bx)2+xvx2+c2T'=\frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{x}{v\sqrt{x^2+c^2}}
set the derivative to zero to find the minimum time. x must satisfy that.
For the next part, ABAB is the same but the velocity in BCBC has an extra vv in it's horizontal component. Think of it's velocity in terms of her swimming vector incllned at an angle θ\theta and then add on the horizontal component due to the river at the end. Because she can only swim in straight lines:
xc=vcosθ+vvsinθ\frac{x}{c}=\frac{vcos\theta +v}{vsin\theta}. Using the identity:
Unparseable latex formula:

1+Tan^2\theta =sec^2\theata

and rearranging we get:
(c2x2)sec2θ+2c2secθ+c2+x2=0(c^2-x^2)sec^2\theta+2c^2sec\theta+c^2+x^2=0 which is a quadratic in secθsec\theta, so, using the quadratic formula:
secθ=2c2±4c44(c2x2)(c2+x2)2(c2x2)=c2±x2c2x2sec\theta=\frac{-2c^2\pm\sqrt{4c^4-4(c^2-x^2)(c^2+x^2)}}{2(c^2-x^2)}=\frac{-c^2 \pm x^2}{c^2-x^2}. We can discard the 'plus' root as it gives theta to be π\pi which makes no sense because the mother would never reach the child if she had no vertical component in her velocity.So:
secθ=c2x2c2x2cosθ=c2x2c2x2 sec\theta=\frac{-c^2 - x^2}{c^2-x^2} \Rightarrow cos\theta=\frac{c^2-x^2}{-c^2-x^2}
To find the speed for BCBC we use pythagoras:
speedBC=v2sin2θ+(vcosθ+v)2=v2(2cosθ+2)speed_{BC}=\sqrt{v^2sin^2\theta+(vcos\theta+v)^2}=\sqrt{v^2(2cos \theta+2)}, substituting in and simplifying:
speedBC=2vxc2+x2speed_{BC}=\frac{2vx}{\sqrt{c^2+x^2}}
so, just like in the previous part:
T=a2+(bx)2u+c2+x22vxc2+x2=a2+(bx)2u+c2+x22vxT=\frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{\sqrt{c^2+x^2}}{\frac{2vx}{\sqrt{c^2+x^2}}}= \frac{\sqrt{a^2+(b-x)^2}}{u}+\frac{c^2+x^2}{2vx} and differentiating we get:
T=xbua2+(bx)2+4vx22vx22vc24v2x2=xbua2+(bx)2+x2c22vx2T'= \frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{4vx^2-2vx^2-2vc^2}{4v^2x^2}= \frac{x-b}{u\sqrt{a^2+(b-x)^2}}+\frac{x^2-c^2}{2vx^2}
letting T=0T'=0 to find minima and rearranging:
bxua2+(bx)2=x2c22vx22vx2(bx)=u(x2c2)[a2+(bx)2]1/2\frac{b-x}{u\sqrt{a^2+(b-x)^2}}= \frac{x^2-c^2}{2vx^2} \Rightarrow 2vx^2(b-x)=u(x^2-c^2)[a^2+(b-x)^2]^{1/2} as required.
Can't really upload the sketches so if anyone wants to do it be my guest:biggrin:


Hey i have a much easier way in showing the second equation in x. Think of the relative velocity of the river to the position of the girl in the river.
(xvt)(x-vt) is the vertical distance as she enters the river at time t
cc is the horizontal distance still.
Then by pythagoras as before (where t is the journy time in the river):
t=c2+(xvt)2vt=\frac{\sqrt{c^2+(x-vt)^2}}{v}
t2v2=c2+x22vtx+v2t2t^2v^2=c^2+x^2-2vtx+v^2t^2
Rearrange for t (the square bits in t cancel)
t=x2+c22vxt=\frac{x^2+c^2}{2vx}
as desired :biggrin:
Original post by themaths
Hey i have a much easier way in showing the second equation in x. Think of the relative velocity of the river to the position of the girl in the river.
(xvt)(x-vt) is the vertical distance as she enters the river at time t
cc is the horizontal distance still.
Then by pythagoras as before (where t is the journy time in the river):
t=c2+(xvt)2vt=\frac{\sqrt{c^2+(x-vt)^2}}{v}
t2v2=c2+x22vtx+v2t2t^2v^2=c^2+x^2-2vtx+v^2t^2
Rearrange for t (the square bits in t cancel)
t=x2+c22vxt=\frac{x^2+c^2}{2vx}
as desired :biggrin:


Sorry, I don't follow. how is that last result the same as mine? Have you skipped over the calculus? I'm baffled as to how you could find the x that results in the shortest possible total time time without calculus.
Reply 247
Original post by ben-smith
Sorry, I don't follow. how is that last result the same as mine? Have you skipped over the calculus? I'm baffled as to how you could find the x that results in the shortest possible total time time without calculus.


Oh no i did the calculus part exactly the same as you to find the minimum time you obviously need to take dT/dx as 0 to find the minimum.
I just meant for the part where you were using theta and omitting roots of quadratics and things to find her journey time in the river, when i did it i just used relative velocities to find the time she was in the river. Everything else after that i did the same as you (assuming the diagrams were the same also but they were quite standard after getting the result).
Sorry i didn't really make it clear reading my comment back :biggrin:
Original post by themaths
Oh no i did the calculus part exactly the same as you to find the minimum time you obviously need to take dT/dx as 0 to find the minimum.
I just meant for the part where you were using theta and omitting roots of quadratics and things to find her journey time in the river, when i did it i just used relative velocities to find the time she was in the river. Everything else after that i did the same as you (assuming the diagrams were the same also but they were quite standard after getting the result).
Sorry i didn't really make it clear reading my comment back :biggrin:


Oh, I see. Well, your way is definitely simpler, it's good to show both methods I guess.
Original post by brianeverit
1987 STEP Fma numbers 12 -16


For question 12, aren't polar coordinates supposed to be taken anticlockwise from the +ve x axis? It probably doesn't matter, it just took me a while to realise why my answer differed from yours.


Original post by SimonM
...

Alternative solution to the first bit of question 12 STEP II.
Brian did this by using an energy argument. I have a different way.
let γ,T\mathbf{\gamma},\mathbf{T} denote the unit tangent vector and the tension in the string respectively.
If the bead can be in static equilibrium fo all positions then:
Unparseable latex formula:

m \mathbf{g}. \mathbf{\gamma}+ \mathbf{T}.\mathbf{\gamma}=0[br]\Rightarrow (m \mathbf{g}+\mathbf{t}).\mathbf{ \gamma}=0[br]\Rightarrow-mg\begin{pmatrix} cos\theta \\ 1+sin\theta\end{pmatrix}.\frac{ \mathbf{ \dot{r}}}{|\mathbf{\dot{r}}|}=0 \Rightarrow -mg\begin{pmatrix} cos\theta \\ 1+sin\theta\end{pmatrix}. \begin{pmatrix} \dot{r}cos\theta-rsin\theta \\ \dot{r} sin\theta+rcos\theta \end{pmatrix} \frac{1}{\sqrt{\dot{r}^2+r^2}}=0[br]\Rightarrow \dot{r}cos^2\theta-rsin\thetacos\theta+\dot{r} sin\theta+rcos\theta+\dot{r}sin^2\theta+rsin\thetacos\theta=0[br]\Rightarrow \dot{r}(1+sin\theta)+rcos\theta=0[br]\Rightarrow \dot{r}+r\frac{cos\theta}{1+sin\theta}=0


Multiplying through by an integrating factor of eln(1+sinθ)e^{ln(1+sin\theta)} we get:
ddθ[reln(1+sinθ)]=0r=Csinθ+1\frac{d}{d \theta}[re^{ln(1+sin\theta)}]=0 \Rightarrow r=\frac{C}{sin\theta+1} where C is a constant.
The minimum obviously occurs when sinθ=1r=2dsinθ+1sin\theta=1 \Rightarrow r=\frac{2d}{sin\theta+1}
Note that this is the same answer as Brian's. I am taking angles about the x axis whereas he has taken them about the y.
(edited 11 years ago)
*Subscribe*
Original post by Swayum
STEP I, question 4.

Let S=log2elog4e+log16e...S = \mathrm{log}_2 e - \mathrm{log}_4 e + \mathrm{log}_{16} e...

Take each term to logarithm base 2 (change the base rule)

S=log2e1log2e2+log2e4...\displaystyle S = \frac{\mathrm{log}_2 e}{1} - \frac{\mathrm{log}_2 e}{2} + \frac{\mathrm{log}_2 e}{4}...

Unparseable latex formula:

S = (\mathrm{log}_2 e})(1 - 1/2 + 1/4 - 1/8 + 1/16 - 1/32...)



Split the second brackets into two parts.

P = 1 + 1/4 + 1/16...
N = -1/2 - 1/8 - 1/32...

So

Unparseable latex formula:

S = (\mathrm{log}_2 e})(P + N)



P is a geometric series with first term 1 and common ratio 1/4. Its sum to infinity is:

P=1/(11/4)=4/3P = 1/(1-1/4) = 4/3

N is a geometric series with first term -1/2 and common ratio 1/4. Its sum to infinity is:

N=(1/2)/(11/4)=2/3N = (-1/2)/(1-1/4) = -2/3

Unparseable latex formula:

S = (\mathrm{log}_2 e})(P + N) = (\mathrm{log}_2 e})(4/3 - 2/3)



Unparseable latex formula:

S = 2/3(\mathrm{log}_2 e})



Change the base to e, so we have the natural logarithm.

Unparseable latex formula:

\displaystyle S = 2/3(\mathrm{log}_2 e}) = 2/3(\frac{\mathrm{log}_e e}{\mathrm{log}_e 2}) = 2/3(\frac{1}{\mathrm{log}_e 2})


S=23ln2\displaystyle S = \frac{2}{3\mathrm{ln} 2}

Using the power rule

S=23ln2=2ln8\displaystyle S = \frac{2}{3\mathrm{ln} 2} = \frac{2}{\mathrm{ln} 8}


S=21212ln8\displaystyle S = \frac{2*\frac{1}{2}}{\frac{1}{2}\mathrm{ln} 8} (multiplying top and bottom by 1/2)

S=112ln8=1ln8\displaystyle S = \frac{1}{\frac{1}{2}\mathrm{ln} 8} = \frac{1}{\mathrm{ln} \sqrt 8}

S=1ln22\displaystyle S = \frac{1}{\mathrm{ln} 2\sqrt 2}

WWWWW.


Slightly more succinct and using the given hint.....
Reply 252
Original post by brianeverit
1987 Paper 2 no.11


Your vector product seems to be wrong, it should be sin(theta) in the first term rather than cos(theta), though it get's the right answer at the end.
Reply 253
Original post by brianeverit
1987 Paper 2 no.11


Follow up post. I'm now not sure you have the right answer. Suppose theta is zero. Then there doesn't need to be any frictional force to keep rod in place. Hence coefficient of friction can be zero, so "mu" = tan(alpha)sin(theta).
Original post by nota bene
I'll post up my attempt to a solution to I/7, using the approach with taking the imaginary parts...

Spoiler



Tell me if I've made some silly mistakes, quite likely...

edit: The solution should now be correct.



I have compressed things slightly and I think you have the wrong sign for the last part...see document....
(edited 11 years ago)
Original post by mikelbird
I have compressed things slightly and I think you have the wrong sign for the last part...see document....


corrected document
Original post by Seernb
Follow up post. I'm now not sure you have the right answer. Suppose theta is zero. Then there doesn't need to be any frictional force to keep rod in place. Hence coefficient of friction can be zero, so "mu" = tan(alpha)sin(theta).

Thanks for pointing out the error. I agree with your answer.
Reply 257
Original post by mikelbird
Slightly more succinct and using the given hint.....


Sorry for being a bit daft, but could you please tell me how you got from:

r=0(1)r12r\sum_{r=0}^{\infty}(-1)^{r}\frac{1}{2^{r}}

to:

11.5\frac{1}{1.5}

I have already done this question, but I can't see the link between the two without having to write a couple terms out.
(edited 11 years ago)
Reply 258
Original post by Blazy
Sorry for being a bit daft, but could you please tell me how you got from:

n=012r(1)r \sum_{n=0}^{\infty}\frac{1}{2^{r}}(-1)^{r}

to:

11.5 \frac{1}{1.5}

I have already done this question, but I can't see the link between the two without having to write a couple terms out.


GP? (1)r12r=(12)r(-1)^r \frac{1}{2^r} = \left(-\frac{1}{2}\right)^r. The good old formula should work, then think about where this infinity symbol goes.
Reply 259
Original post by gff
GP? (1)r12r=(12)r(-1)^r \frac{1}{2^r} = \left(-\frac{1}{2}\right)^r. The good old formula should work, then think about where this infinity symbol goes.


Yeah I got there eventually...D:. Thanks though :smile:

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