The Student Room Group

STEP Maths I, II, III 1995 Solutions

Scroll to see replies

1995 STEP I number 11 Alternative solution
The system is completely symmetrical so we need only consider one half  \text{The system is completely symmetrical so we need only consider one half }
Since cylinders are tending to move apart S=0 \text {Since cylinders are tending to move apart }S=0
Resolving vertically R=3mg2 \text{Resolving vertically }R=\dfrac{3mg}{2}
Resolving horizontally for A Xsin30Ycos30=F \text{Resolving horizontally for A }X\sin30^\circ-Y\cos30^\circ=F
Resolving vertically for A R=Xcos30+Ysin30 \text{Resolving vertically for A }R=X\cos30^\circ+Y\sin30^\circ
eliminating Xwe haveY(sin230+cos230)=Rsin30Fcos30Y=Rsin30Fcos30 \text{eliminating }X \text{we have} Y(\sin^230^\circ+\cos^230^\circ)=R\sin30^\circ-F\cos30^\circ \Rightarrow Y=R\sin30^\circ-F\cos30^\circ
Taking moments about centre of A F=Y \text{Taking moments about centre of A }F=Y
hence, F(1+cos30)=Rsin30FR=sin301+cos30=12+3=231 \text{hence, } F(1+ \cos30^ \circ)=R \sin30^ \circ \Rightarrow \dfrac{F}{R}=\dfrac{ \sin30^ \circ}{1+ \cos30^ \circ}= \dfrac{1}{2+ \sqrt3}= \dfrac{2- \sqrt3}{1}
 and so μ(23) as required] \text{ and so } \mu \geq (2-\sqrt3) \text{ as required} ]
STEP II/10
Let velocity of firs sphere after first collision be v and that of the second sphere w \text{Let velocity of firs sphere after first collision be }v \text{ and that of the second sphere }w
After second collision let velocity of third sphere be x \text {After second collision let velocity of third sphere be }x
 then by conservation of momentum for first collision m1u=m1v+m2w (1) \text{ then by conservation of momentum for first collision }m_1u=m_1v+m_2w \text{ (1)}
for second collision m2w=m2v+m3x (2) \text{for second collision }m_2w=m_2v+m_3x \text{ (2)}
By law of restitution at first collision wv=eu (3) \text{By law of restitution at first collision }w-v=eu \text{ (3)}
and at second collision xu=ew (4) \text{and at second collision }x-u=ew \text{ (4)}
from (1) and (3) we have m1u=m1v+m2(v+eu)vu=m1em2m1+m2\text{from (1) and (3) we have } m_1u=m_1v+m_2(v+eu)\Rightarrow \dfrac{v}{u}=\dfrac{m_1-em_2}{m_1+m_2}
Unparseable latex formula:

\text{from (2),(3) and (4) }m_2(v+eu)=m_2u+m_3(u+e(v+eu))\Rightarrow \frac{v}{u}= \dfrac{m_2+m_3+e^2m_3-em_2}{m_2-em_3}}


Unparseable latex formula:

\text{so } }=\dfrac{m_1-em_2}{m_1+m_2}=\dfrac{m_2+m_3+e^2m_3-em_2}{m_2-em_3}}}}


Unparseable latex formula:

[br]\Rightarrow (m_1-em_2)(m_2-em_3)=(m_2+m_3+e^2m_3-em_2)(m_1+m_2) }}


Unparseable latex formula:

\Rightarrow m_1m_2-em_2^2-em_1m_3+e^2m_2m_3 }}


em1m3=m1m3+e2m1m3em1m2+m22+m2m3\Rightarrow -em_1m_3=m_1m_3+e^2m_1m_3-em_1m_2+m_2^2+m_2m_3
m1m3(1+e+e2)=em1m2m2(m2+m3)em2=m3(1+e+e2)+m1(m2+m3)m1 \Rightarrow m_1m_3(1+e+e^2)=em_1m_2-m_2(m_2+m_3) \Rightarrow em_2=m_3(1+e+e^2)+ \dfrac{m_1(m_2+m_3)}{m_1}
Clearly m2(m2+m3)m1>0em2>m3(1+e+e2) \text{Clearly } \dfrac{m_2(m_2+m_3)}{m_1}>0 \Rightarrow em_2>m_3(1+e+e^2)
In the new situation, consider the motion relative to m2 and m3 initially stationary \text{In the new situation, consider the motion relative to }m_2 \text{ and }m_3 \text{ initially stationary}
Equations (1) to (4) then still hold with u replaced by uv, so after second collision, velocity of second sphere is  \text{Equations (1) to (4) then still hold with }u \text{ replaced by }u-v \text{, so after second collision, velocity of second sphere is }
uv relative to this moving frame of reference. Hence it is traveling with veloicity u in absolute termsu-v \text{ relative to this moving frame of reference. Hence it is traveling with veloicity }u \text{ in absolute terms}
1995 STEP II/14

Unparseable latex formula:

\text{If f}(x)=Ax^2\text{exp}\left(-\dfrac{x^2}{2}\right) \text{ is a p.d.f. Then we must have } \displaystyle \int_{-\infty}^\inftyAx^2\text{exp}\left(-\dfrac{x^2}{2}\right)dx=1[br]


consider Ax2exp(x22)dx=[Axexp(x22)]+ inftyAxexp(x22)dx\text{consider }\displaystyle \int_{-\infty}^\infty Ax^2\text{exp}\left(-\dfrac{x^2}{2}\right)dx=\left[-Ax\text{exp}\left(-\dfrac{x^2}{2}\right)\right]_{-\infty}^\infty+\displaystyle \int_{-\infty}^\ infty Ax\text{exp}\left(-\dfrac{x^2}{2}\right)dx
=0+[Aexp(x22)]=A2π=1A=12π =0+ \left[-A\text{exp}\left(-\dfrac{x^2}{2}\right)\right]_{-\infty}^ \infty =A\sqrt{2\pi}=1\Rightarrow A=\dfrac{1}{\sqrt{2\pi}}
My group think the outcome has a mean of 0 and so the probability of a single result as large or\text{My group think the outcome has a mean of 0 and so the probability of a single result as large or}
Unparseable latex formula:

\text{larger than 87.3 is given by }\left[-\dfrac{1}{\sqrt{2\pi}}\text{exp}\left(-\dfrac{x^2}{2}\right]_{87.3}^\infty


which will be infinitessimally small and my faith in the hypothesis would certainly be shaken\text{which will be infinitessimally small and my faith in the hypothesis would certainly be shaken}
The rival hypothesis however can be made to seem quite reasonable by a suitably large value of μ \text{The rival hypothesis however can be made to seem quite reasonable by a suitably large value of } \mu
For 1000 trials, by the central limit theorem, my hypothesis would still suggest a mean of 0 but a  \text{For }1000 \text{ trials, by the central limit theorem, my hypothesis would still suggest a mean of 0 but a }
variance equal to E[X2]=12πx4exp(x22)dx= E[X4] where X is standard normal\text{variance equal to E}[X^2]=\dfrac{1}{\sqrt{2\pi}} \displaystyle \int_{-\infty}^\infty x^4\text{exp}\left(-\dfrac{x^2}{2}\right)dx=\text{ E}[X^4] \text{ where }X\text{ is standard normal}
i.e. E[X2]=3 since we are given that E[X4]=3 \text{i.e. E}[X^2]=3 \text{ since we are given that E}[X^4]=3
Variance of the average of 1000 trials is thus 0.003 or s.d. is 0.055\text{Variance of the average of 1000 trials is thus 0.003 or s.d. is 0.055}
a result of 0.23 is thus approx 4 s.d above the mean and so would again cause some doubt as to the \text{a result of 0.23 is thus approx 4 s.d above the mean and so would again cause some doubt as to the}
validity of my hypothesis thopugh not so much as the previous case. \text{validity of my hypothesis thopugh not so much as the previous case.}
Again the other scientists could make their hypothesis fit by choosing a value of μ close to 0.23 \text{Again the other scientists could make their hypothesis fit by choosing a value of }\mu \text{ close to 0.23}
Reply 123
Here is my solution for STEP I 1995 Q11:

Edit: I was so pleased to fit it onto one page :smile: that I forgot to mention that one should explicitly check which contact is limiting: here the contact between the cylinders would slip first because the contact force there (mg/2) is smaller than that with the plane (3mg/2). Although it's fairly obvious it should definitely be mentioned in a good answer.
(edited 12 years ago)
Reply 124
STEP II Q9
The current link on the first page is the answer for STEP II Q9 1996; STEP II Q9 1995 hasn't yet been posted. So here's mine.
(edited 12 years ago)
Reply 125
STEP II Q11.

I'm posting this because it's a bit more economical/easier to read than the previous answer in the thread.
Reply 126
Regarding STEP 1995 III Q4

Original post by DFranklin
Very nice! I was aware of the method but didn't realise how much shorter it was than the one I chose of summing eikθe^{ik\theta} and taking real/imaginary parts.

Looking at the Siklos booklet, this particular trick seems to come up quite a bit. So it's worth making note of, particularly given how many people don't bother with the product forms of the trig identities.

(This was referring to Speleo's posted answer, post 32 in thread)


I actually think the complex number way is much clearer, as long as you know the right way of doing it (to be honest, it comes to the same thing really). See my answer below; this was one of the easiest questions I've done from the 1995 set so far:
(edited 12 years ago)
Reply 127
Here is an alternative answer for STEP III 1995 Q6.
I always try to do locus problems without resorting to Cartesian form, this is a good example :smile:
Reply 128
Original post by DFranklin
Step III, Q8:

Equation of normal to π\pi through x is x+λn\mathbf{x}+\lambda\mathbf{n}, distance is (xr)n|(\mathbf{x} -\mathbf{r}) \cdot \mathbf{n}|.

Consider a single circle C with center r lying in the plane xn=p\mathbf{x} \cdot \mathbf{n} = \mathrm p, and suppose C lies on the surface of a sphere S with center s, radius R.

Claim: s=r+λn\mathbf{s} = \mathbf{r} + \lambda \mathbf{n} for some λ\lambda. In other words, s must lie on the normal to the plane through r.

\I think it's arguable that this is "obvious" and doesn't need to be proved. I will prove it, but it's going to be a little tedious...

Put q=sr\mathbf{q} = \mathbf{s} - \mathbf{r}. Consider e=q(qn)n\mathbf{e} = \mathbf{q} - (\mathbf{q} \cdot \mathbf{n}) \mathbf{n}. Note
Unparseable latex formula:

\mathbf{e}\cdot\mathbf{n} = \mathbf{q}\cdot{\mathbf{n}-(\mathbf{q} \cdot \mathbf{n})\mathbf{n}\cdot\mathbf{n} = 0

. (e is basically what we get if we "remove" the part of q that's parallel to n).

Suppose that e is non-zero. We will (eventually!) show this gives us a contradiction. Let μ=e1\mu = |\mathbf{e}|^{-1}, so that μe|\mu\bf{e}| = 1. Then r±μe\bf{r}\pm\mu\bf{e} are both points on the circle C, so if they lie on the sphere S, they must both be the same distance from s.

Now s(r+μe)=qμe\bf{s} - (\bf{r}+\mu\bf{e}) = \bf{q} - \mu\bf{e}.
So s(r+μe)2=(qμe)(qμe)=q22μqe+μe2|\bf{s} - (\bf{r}+\mu\bf{e})|^2 = (\bf{q} - \mu\bf{e})\cdot(\bf{q} - \mu\bf{e})= |\bf{q}|^2 - 2\mu \bf{q}\cdot\bf{e} + |\mu\bf{e}|^2
Similarly, s(rμe)2=(q+μe)(q+μe)=q2+2μqe+μe2|\bf{s} - (\bf{r}-\mu\bf{e})|^2 = (\bf{q} + \mu\bf{e})\cdot(\bf{q} + \mu\bf{e})= |\bf{q}|^2 + 2\mu \bf{q}\cdot\bf{e} + |\mu\bf{e}|^2

So we must have q.e=0    qq(qn)2\bf{q}.\bf{e} = 0 \implies \bf{q}\cdot\bf{q}-(\bf{q}\cdot\bf{n})^2, which since n=1    q=κn|\bf{n}|=1 \implies \bf{q} = \kappa \bf{n} for some κ\kappa (either argue geometrically, or use the equality case of Cauchy-Schwartz). But then e=κn(κnn)n=0\bf{e} = \kappa \bf{n} - (\kappa \bf{n} \cdot \bf{n})\bf{n} = \bf{0}, giving our contradiction.

So e=0\bf{e} = 0, so q=(qn)n\bf{q} = (\bf{q}\cdot\bf{n})\bf{n} and so q\bf{q} is a multiple of n. Thus s=r+λn\mathbf{s} = \mathbf{r} + \lambda \mathbf{n} as claimed.

Finally, let x be any point on C. xn=rn, and so(xr)n=0\bf{x}\cdot\bf{n} = \bf{r}\cdot\bf{n}, \text{ and so} (\bf{x}-\bf{r})\cdot\bf{n} = 0. Then

Unparseable latex formula:

|\bf{s}-\bf{x}|^2 = |\bf{s}-\bf{r}-(\bf{x}-\bf{r})|^2 = |\lambda \bf{n} - (\bf{x}-\bf{r})|^2 \\[br]= \lambda^2 |\bf{n}|^2 -2 (\bf{x}-\bf{r})\cdot\bf{n}+|\bf{x}-\bf{r}|^2 = \lambda^2 + 1

(Since n is a unit vector, and x lies on a unit circle, center r so |x-r| = 1).

But x lies on the sphere, so sx|\bf{s}-\bf{x}| is just the radius of the sphere, R. So λ2+1=R\lambda^2+1 = R

So for S to contain both circles C1,C2C_1, C_2, we must have s=r1+λ1n1=r2+λ2n2\bf{s}=\bf{r_1}+\lambda_1 \bf{n_1} = \bf{r_2}+\lambda_2 \bf{n_2} for some real λ1,λ2\lambda_1, \lambda_2 satisfying λ12+1=λ22+1    λ1=±λ2\lambda_1^2+1 = \lambda_2^2+1 \implies \lambda_1 = \pm \lambda_2.

This proves that for such S, we can find λ\lambda with r1+λn1=r2±λn2\bf{r_1}+\lambda \bf{n_1} = \bf{r_2}\pm\lambda \bf{n_2} as desired.

But then (r1r2)=λn1±λ2n2 (\bf{r_1} - \bf{r_2}) = -\lambda \bf{n_1} \pm \lambda_2 \bf{n_2}. Since n1(n1×n2)=n2(n1×n2)=0\bf{n_1}\cdot(\bf{n_1} \times \bf{n_2}) = \bf{n_2}\cdot(\bf{n_1} \times \bf{n_2}) = 0 we have (r1r2)(n1×n2)=0(\bf{r_1} - \bf{r_2})\cdot(\bf{n_1} \times \bf{n_2}) = 0 as desired. Geometrically, this means that the line between the two circle centers has no component perpendicular to both plane normals; equivalently that it lies in the plane through the origin generated by the two plane normals.

Finally, if r1+λn1=r2+λn2\bf{r_1}+\lambda \bf{n_1} = \bf{r_2} + \lambda \bf{n_2}. Dotting with n1,n2\bf{n_1}, \bf{n_2} gives:

r1n1+λ=r2n1+λn2n1    p1r2n1=λ+λn2n1\bf{r_1}\cdot\bf{n_1} + \lambda = \bf{r_2}\cdot\bf{n_1} + \lambda\bf{n_2}\cdot\bf{n_1} \implies \mathrm{p_1} - \bf{r_2}\cdot\bf{n_1} = -\lambda + \lambda\bf{n_2}\cdot\bf{n_1}

r1n2+λn2n1=r2n2+λ    p2r1n2=λλn2n1\bf{r_1}\cdot\bf{n_2} + \lambda\bf{n_2}\cdot\bf{n_1} = \bf{r_2}\cdot\bf{n_2} + \lambda \implies \mathrm{p_2} - \bf{r_1}\cdot\bf{n_2} = \lambda - \lambda\bf{n_2}\cdot\bf{n_1}

If instead r1+λn1=r2λn2\bf{r_1}+\lambda \bf{n_1} = \bf{r_2} - \lambda \bf{n_2}. Dotting with n1,n2\bf{n_1}, \bf{n_2} gives:

r1n1+λ=r2n1λn2n1    p1r2n1=λλn2n1\bf{r_1}\cdot\bf{n_1} + \lambda = \bf{r_2}\cdot\bf{n_1} - \lambda\bf{n_2}\cdot\bf{n_1} \implies \mathrm{p_1} - \bf{r_2}\cdot\bf{n_1} = -\lambda - \lambda\bf{n_2}\cdot\bf{n_1}

r1n2+λn2n1=r2n2λ    p2r1n2=λ+λn2n1\bf{r_1}\cdot\bf{n_2} + \lambda\bf{n_2}\cdot\bf{n_1} = \bf{r_2}\cdot\bf{n_2} - \lambda \implies \mathrm{p_2} - \bf{r_1}\cdot\bf{n_2} = \lambda + \lambda\bf{n_2}\cdot\bf{n_1}

In either case we have (p1r2n1)2=(p2r1n2)2(\mathrm{p_1} - \bf{r_2}\cdot\bf{n_1})^2 = (\mathrm{p_2} - \bf{r_1}\cdot\bf{n_2})^2 as desired.

Geometrically, this says that this distance of r1\bf{r_1} from the plane π2\pi_2 must equal the distance of r2\bf{r_2} from the plane π1\pi_1

Comment: Doing questions directly into Latex rather than pen/paper is always tricky; I'd done some roughing out on paper, but there was more work still to be done than I thought, and I hadn't realised just how hard Latex is for questions like this involving vectors. It was not a good plan! Just about every other character needs to be enclosed in \bf{}, and the source text ends up near illegible. I must admit to losing the will to live, let alone answer the question, towards the end.


On the question of whether it needs to be proved that the normal line passes through the centre of the sphere:
I would think it sufficient to observe that the line from the centre of a circle to a chord must be perpendicular to the chord, because the chord forms an isosceles triangle combined with the centre. (i.e. "cut" the problem into 2D).
Overall I found this a really tricky question. Here's my effort, for what it's worth. I hope I have interpreted the "deduce necessary condition" part correctly - it seems to be saying that you should prove that the existence of the sphere is a necessary condition that the given equations hold.

Oh and one minor point, you have λ2+1=R \lambda^2+1=R , should be λ2+1=R2\lambda^2+1 = R^2
(edited 12 years ago)
Reply 129
Original post by DFranklin
STEP III 1995, Q9.

(i) (See left hand diagram).

Taking moments about O, we have mgrsinβ=mgr2sin(2α+β)mgr \sin \beta = \frac{mgr}{2} \sin(2\alpha+\beta). So
2sinβ=sin2αcosβ+cos2αsinβ2\sin\beta = \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta
    (2cos2α)sinβ=sin2αcosβ\implies (2- \cos 2\alpha)\sin \beta = \sin 2\alpha \cos \beta
    tanβ=sin2α2cos2α\implies \tan \beta = \frac{\sin 2\alpha}{2 - \cos 2\alpha} as required.

(ii) To relate mass and moment of inertia, will divide disk into annulae radius x, width dx, each has area 2πxdx2\pi x dx and so mass 2πρx2rdx\frac{2\pi \rho x^2}{r} dx.

Mass of disk:
Unparseable latex formula:

\displaystyle m = \int_0^r \frac{2\pi \rho x^2}{r} dx = \left[\frac{2\pi\rhox^3}{3r}\right]_0^r = \frac{2\pi}{3}\rho r^2



M.I. of disk about C:
Unparseable latex formula:

\displaystyle I_C = \int_0^r \frac{2\pi \rho x^4}{r}dx = \left[\frac{2\pi\rhox^5}{5r}\right]_0^r = \frac{2\pi}{5} \rho r^4 = \frac{3mr^2}{5}

.

As C is the center of mass, M.I. of disk about O = M.I. of disk about C + m(distance of c.of.m. from O)^2 = 3mr25+mr2=8mr25\frac{3mr^2}{5} + mr^2 = \frac{8mr^2}{5}.

(iii) First note that Q is the center of mass of the system (both disk and particle). Originally Q is at the same height as O. When P is directly below O, Q has fallen a distance 3r2sinα\frac{3r}{2} \sin \alpha (see right hand diagram). Note also that OP = 2rsinα2r \sin \alpha as we'll use this later.

By conservation of energy, the kinetic energy of the system at this point is 3mgrsinα3mgr \sin \alpha (recall the system has total mass 2m).

Now the M.I. of the disk about O is 8mr25\frac{8mr^2}{5}, while the M.I. of P about O is m(OP)2=4mr2sin2αm (OP)^2 = 4mr^2 \sin^2 \alpha. So the M.I. of the system about O is I=4mr2sin2α+8mr25=45mr2(2+5sin2α)I = 4mr^2\sin^2\alpha + \frac{8mr^2}{5} = \frac{4}{5}mr^2(2 + 5 \sin^2 \alpha).

So if ω\omega is the angular velocity of the system about O, then the system has K.E. 12Iω2\frac{1}{2}I\omega^2. So we must have:

12Iω2=3mrgsinα\frac{1}{2}I\omega^2 = 3mrg \sin \alpha
25mr2ω2(2+5sin2α)=3mrgsinα\frac{2}{5}mr^2\omega^2(2+5\sin^2\alpha) = 3mrg \sin \alpha
ω2=15gsinα2r(2+5sin2α)\omega^2 = \frac{15g \sin \alpha}{2r(2+5\sin^2\alpha)}.

Since v2=r2ω2v^2 = r^2 \omega^2, we have v2=15grsinα2(2+5sin2α)v^2 = \frac{15gr \sin \alpha}{2(2+5\sin^2\alpha)} as required.

Write f(t)=(t2+5t2)1=2/t+5t.f(t)=2/t2+5,f(t)=4/t3>0 for t>0f(t) = \left(\frac{t}{2+5t^2}\right)^{-1} = 2/t + 5t. f'(t) = -2/t^2+5, f'''(t) = 4/t^3 > 0 \text{ for } t>0. So setting f'(t)=0 gives us a local minimum, attained for t=2/5t=\sqrt{2/5}. Applying this to v2=15grsinα2(2+5sin2α)v^2 = \frac{15gr \sin \alpha}{2(2+5\sin^2\alpha)}, we see it is maximized for sinα=2/5\sin \alpha = \sqrt{2/5} (we need to also check in case sinα=0\sin \alpha =0 gives a maximum, but it clearly doesn't).

For this value of sinα\sin \alpha, we have v2=3gr102(2+4)=gr104.v^2 = \frac{3gr\sqrt{10}}{2(2+4)} = \frac{gr\sqrt{10}}{4}.

Comment: For a change, thought I'd look at one of the more "advanced" mechanics questions involving moments of inertia. This question seems pretty straightforward; again drawing useful diagrams is half the battle. It does seem very long though - having to maximize v^2 at the end is just a bit much...


STEP 1995 III Q9

I agree with the above except for the very last line: it should surely be
v2=310gr8 v^{2} = \frac{3\sqrt{10}gr}{8}
Also, I think you need to check cosα=0\cos{\alpha}=0, which yields v2=15gr14v^{2} = \frac{15gr}{14}, rather than sinα=0\sin{\alpha}=0

Anyway, for comparison, here's my answer. This question was ridiculously long, I totally agree about that.
Original post by waxwing
STEP 1995 III Q9

I agree with the above except for the very last line: it should surely be
v2=310gr8 v^{2} = \frac{3\sqrt{10}gr}{8} Yes, that's a fairly obvious LaTeX fail, I think.

Also, I think you need to check cosα=0\cos{\alpha}=0, which yields v2=15gr14v^{2} = \frac{15gr}{14}, rather than sinα=0\sin{\alpha}=0
Yes, probably.
Reply 131
Original post by DFranklin
Yes, that's a fairly obvious LaTeX fail, I think.


Yeah, sorry, that was a pretty pointless "correction" wasn't it :smile:

Anyway, on to STEP 1995 III Q10.

This is a very routine question using a normal approach. But I'm posting this because I recalled a very snazzy and unique way of approaching this question, which I'd read on some guy's internet page a year or two back. The answer below contains both approaches, see what you think..
1995 STEP III question 12 \text {1995 STEP III question 12}

XN(0,1)    ΦX(x)=12πexp(12x2)     Pr(xXx+δx)=δx2π exp(12x2) X \sim \text {N}(0,1) \implies \Phi_X(x)= \dfrac{1}{\sqrt{2 \pi}}\text {exp} \left( -\dfrac{1}{2}x^2 \right) \implies \text{ Pr}(x \leq X \leq x+ \delta x)=- \dfrac{ \delta x}{ \sqrt{2 \pi}} \text{ exp} \left( - \dfrac{1}{2}x^2 \right)
similarly YN(0,1)    Pr(yYy+δy)=δy2πexp(12y2)\text {similarly }Y \sim \text{N}(0,1) \implies \text {Pr}(y \leq Y \leq y+ \delta y)= \dfrac{ \delta y}{ \sqrt{2 \pi}} \text {exp} \left(- \dfrac{1}{2}y^2 \right)
     Pr(xXx+δx and yYy+δy)=δxδy2πexp(12x212y2) \implies \text { Pr}(x \leq X \leq x+ \delta x \text { and } y \leq Y \leq y+ \delta y)= \dfrac{ \delta x \delta y}{2 \pi} \text {exp} \left( - \dfrac {1}{2}x^2- \dfrac{1}{2}y^2 \right)
    f)x,y)=12πe12(x2+y2) as required\implies \text {f})x,y)= \dfrac{1}{2 \pi} \text{e}^{- \frac{1}{2}(x^2+y^2)} \text { as required}
Unparseable latex formula:

\text {if the polar coordinates of the point }(x,y) \text { are }(r, \theta) \text{ then f}(x,y)= \phi(r, \theta) \text{ say, where } \Phi(r, \theta)= \dfrac{1}{2\pi} \text{e}^_{-r^2}


 radial symmetry means that this expression is independent of θ\text{ radial symmetry means that this expression is independent of }\theta
Let the density function be ϕ(r) \text {Let the density function be }\phi(r)
Pr(UV<k)= Pr(U<kV,V>0)+ Pr(U>kV,V<0)\text {Pr} \left( \dfrac{U}{V}<k \right)= \text{ Pr}(U<kV,V>0)+ \text { Pr}(U>kV,V<0)
radial symmetry guarantees that Pr(U<kV,V>)= Pr(U>kV,V<0) \text {radial symmetry guarantees that Pr}(U<kV,V>)= \text { Pr}(U>kV,V<0)
hence, Pr(UV<k)=2Pr(U<kV,V>0) so Pr(kUVk+δk)=2 Pr(kV<U<(k+δk)V,V>0)\text {hence, Pr} \left( \dfrac{U}{V}<k \right)=2 \text{Pr}(U<kV,V>0) \text { so Pr} \left(k \leq \dfrac{U}{V} \leq k+ \delta k \right)=2 \text { Pr}(kV<U<(k+ \delta k)V,V>0)
By radial symmetry the area required is δα2π \text {By radial symmetry the area required is }\dfrac{ \delta \alpha}{2 \pi}
Unparseable latex formula:

\delta \alpha= \tan^{-1}(k+ \delta k)- \tan^{-1}k= \sdelta k \dfrac{ \text{d}}{ \text{d}k} \tan^{-1} k=\dfrac{ \delta k}{1+k^2}


 hence, Pr(kUVk+δk)=2δα2π=δαπ=δkπ(1+k2)\text{ hence, Pr} \left(k \leq \dfrac{U}{V} \leq k+ \delta k \right)=2 \dfrac{ \delta \alpha}{2 \pi}= \dfrac{ \delta \alpha}{ \pi}= \dfrac{\delta k}{ \pi(1+k^2)}
and so g(k)=1π(1+k2) as required \text {and so g}(k)= \dfrac{1}{ \pi (1+k^2)} \text { as required}
Pr(message received correctly)=[αp1+(1α)p0]10k\text {Pr(message received correctly)}=[\alpha p_1+(1-\alpha)p_0]^{10^k}
β=αp1+(1α)(1p0)=α(p11+p0)+1p0=α(1q11+1q0)+q0=α(0q1)+q0\beta=\alpha p_1+(1-\alpha)(1-p_0)=\alpha(p_1-1+p_0)+1-p_0=\alpha(1-q_1-1+1-q_0)+q_0=\alpha(0-q_1)+q_0
Unparseable latex formula:

\text {now if } \alpha= \beta \text{ we have }\alpha=\alpha(1-q_0-q_1)+q_0 \Leftrightarrow \alpha(q_0+q_1)=q_0 \Leftrigfhtarrow \alpha= \dfrac{q_0}{q_0+q_1}


Pr(message is correct | all zeros received)=Pr(message is all zeros and is correct)Pr(all zeros received) \text {Pr(message is correct | all zeros received)}= \dfrac{ \text{Pr(message is all zeros and is correct)}}{\text{Pr(all zeros received)}}
=(1p0)10k(1α)10k since α=β in this case =\dfrac{(1-p_0)^{10^k}}{(1- \alpha)^{10^k}} \text{ since }\alpha = \beta \text{ in this case}
if α=0.5 we have 0.5=[12(1q)+12)(1q)]10k\text {if } \alpha=0.5 \text { we have }0.5= \left[ \dfrac{1}{2}(1-q)+ \dfrac{1}{2})(1-q) \right]^{10^k}
    ln(1q)=1106    1q=exp(1106ln0.5) \implies \ln(1-q)= \dfrac{1}{10^6} \implies1-q= \text{exp} \left( \dfrac{1}{10^6} \ln 0.5 \right)
Unparseable latex formula:

\implies q=1- \text{exp} \left( \dfrac{1}{10^6} \ln o=0.5 \right)=1- \text{exp} \lefdt( -\dfrac {\ln 2}{10^6} \right)= \dfrac{\ln2}{10^6}


i.e. q=6.93×107=6.9×107 to 2 d.p. \text {i.e. }q=6.93 \dots \times 10^{-7}=6.9 \times 10^{-7} \text { to 2 d.p.}
q=kL2 where L is the cable lengthq=kL^2 \text { where }L \text { is the cable length}
Unparseable latex formula:

\text { when }L=L_0 \text { say }q= \dfrac{ \ln2}{10^6} \text { or } \dfrac{\ln2}{10^6}=kL_0 \implies q= \dfrac{\ln2}{10^6L_0^2}L^2 \text { so when }L= \drac{1}{2}L_0, q= \dfrac{\ln2}{4 \times 10^6}


(1q)106=(1ln24×106)106114ln2(1-q)^{10^6}= \left(1- \dfrac{\ln2}{4 \times 10^6} \right)^{10^6} \approx 1- \dfrac{1}{4} \ln2
so Pr(message is correct over two such stages)=(114ln2)2=0.683 to 3 d.p \text {so Pr(message is correct over two such stages)}= \left(1-\dfrac{1}{4}\ln2 \right)^2=0.683 \text { to 3 d.p}
Original post by insparato
STEP I Question 2
i)

ST=cosxsinxcosx+sinx=ln(cosx+sinx) S-T = \frac{cosx-sinx}{cosx+sinx} = ln (cosx+sinx)


ahh what would happen if you didnt realise the f'(x)/f(x) to integrate straight into a ln?

i did this, i know its a fairly stupid way but can anyone say if the final answer can actually be obtained this way:

ST=cosxsinxcosx+sinx=(cosxsinx)(cosxsinx)(cosxsinx)(cosx+sinx) S-T = \frac{cosx-sinx}{cosx+sinx} = \frac{(cosx-sinx)(cosx-sinx)}{(cosx-sinx)(cosx+sinx)}

which goes to

=(1sin2x)cos2x=1cos2xtan2x= \frac{(1-sin2x)}{cos2x} = \frac{1}{cos2x}-tan2x

from here i integrated to get

12[(lnsec2x+tan2x)lnsec2x]\frac{1}{2}[(ln|sec2x+tan2x|) - ln|sec2x|]

which simplified to

=12ln1+sin2x=\frac{1}{2}ln|1+sin2x|

which i guess isnt the same as cosx + sinx
Reply 135
Original post by 8inchestall
ahh what would happen if you didnt realise the f'(x)/f(x) to integrate straight into a ln?

i did this, i know its a fairly stupid way but can anyone say if the final answer can actually be obtained this way:

ST=cosxsinxcosx+sinx=(cosxsinx)(cosxsinx)(cosxsinx)(cosx+sinx) S-T = \frac{cosx-sinx}{cosx+sinx} = \frac{(cosx-sinx)(cosx-sinx)}{(cosx-sinx)(cosx+sinx)}

which goes to

=(1sin2x)cos2x=1cos2xtan2x= \frac{(1-sin2x)}{cos2x} = \frac{1}{cos2x}-tan2x

from here i integrated to get

12[(lnsec2x+tan2x)lnsec2x]\frac{1}{2}[(ln|sec2x+tan2x|) - ln|sec2x|]

which simplified to

=12ln1+sin2x=\frac{1}{2}ln|1+sin2x|

which i guess isnt the same as cosx + sinx


Almost there:

1/2ln(1+sin2x) = 1/2ln(cos^2x+sin^2x+2sinxcosx) = 1/2ln(cosx+sinx)^2 = ln(cosx+sinx).
Original post by f1mad
Almost there:

1/2ln(1+sin2x) = 1/2ln(cos^2x+sin^2x+2sinxcosx) = 1/2ln(cosx+sinx)^2 = ln(cosx+sinx).


thanks a lot :smile:

in terms of step marking, do you know what leaving it in a form where i cant see an obvious simplification would affect?

and another Q for anyone, in question 1 the final part where it says 'shade in the region,' where do I shade? i got the lines y=x, y=-x, 4y=x but where one is >0 another one isnt...
Original post by 8inchestall
thanks a lot :smile:

in terms of step marking, do you know what leaving it in a form where i cant see an obvious simplification would affect? If you do the question in an "expected" way and miss a simple simplification, I'd only expect you to lost a mark or so.

If you do it an unexpected way, you're more likely to get punished for not getting the expected answer.

Also, in this case I wouldn't say the simplification was that obvious.

So I'd guess at 3-4 marks. It's just a guess, though.
Original post by 8inchestall
ahh what would happen if you didnt realise the f'(x)/f(x) to integrate straight into a ln?

i did this, i know its a fairly stupid way but can anyone say if the final answer can actually be obtained this way:

ST=cosxsinxcosx+sinx=(cosxsinx)(cosxsinx)(cosxsinx)(cosx+sinx) S-T = \frac{cosx-sinx}{cosx+sinx} = \frac{(cosx-sinx)(cosx-sinx)}{(cosx-sinx)(cosx+sinx)}

which goes to

=(1sin2x)cos2x=1cos2xtan2x= \frac{(1-sin2x)}{cos2x} = \frac{1}{cos2x}-tan2x

from here i integrated to get

12[(lnsec2x+tan2x)lnsec2x]\frac{1}{2}[(ln|sec2x+tan2x|) - ln|sec2x|]

which simplified to

=12ln1+sin2x=\frac{1}{2}ln|1+sin2x|

which i guess isnt the same as cosx + sinx


But it is because 1+sin2x=sin^2(x)+cos^2(x)+2sinxcosx = (sinx+cosx)^2
(edited 11 years ago)
Reply 139
Original post by brianeverit
But it is because 1+sin2x=sin^2(x)+cos^2(x)+2sinxcosx = (sinx+cosx)^2


http://www.thestudentroom.co.uk/showthread.php?t=359567&page=7&p=36970234#post36970234

Quick Reply

Latest