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STEP maths I, II, III 2007 solutions

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Avatar for Chewwy
Chewwy
17
II 2. thought i had to do my bit. note my cool watermark, lol.
II-2.jpg97.7KB
STEP III, Q5: y=ln(x2−1),coth⁡θ=x,r=x2−1y = ln(x^2-1), \coth \theta = x, r = \sqrt{x^2-1}

First find expressions for cosh⁡θ,sinh⁡θ\cosh \theta, \sinh \theta: coth⁡2−1=cosh⁡2−sinh⁡2sinh⁡2=1sinh⁡2\coth^2-1 = \frac{\cosh^2-\sinh^2}{\sinh^2} = \frac{1}{\sinh^2}, so

sinh⁡θ=1coth⁡2θ−1=1r\displaystyle \sinh \theta = \frac{1}{\sqrt{\coth^2 \theta -1}}=\frac{1}{r}.

Then cosh⁡θ=coth⁡θsinh⁡θ=xr\displaystyle \cosh \theta = \coth \theta \sinh \theta = \frac{x}{r}

Now dydx=2xx2−1=2xr2=2(x/r)r=2cosh⁡θr\displaystyle \frac{dy}{dx} = \frac{2x}{x^2-1} = 2\frac{x}{r^2} = \frac{2(x/r)}{r} = \frac{2\cosh \theta}{r}

Next, find drdx=xx2−1=xr=cosh⁡θ\displaystyle \frac{dr}{dx} = \frac{x}{\sqrt{x^2-1}} = \frac{x}{r} = \cosh \theta.

dθdx\displaystyle \frac{d\theta}{dx} is a little harder, but we have cosh⁡θ=xsinh⁡θ\cosh \theta = x \sinh \theta, so sinh⁡θdθdx=sinh⁡θ+xcosh⁡θdθdx\displaystyle \sinh \theta \frac{d\theta}{dx} = \sinh \theta + x \cosh \theta \frac{d\theta}{dx} and so (1−xcoth⁡θ)dθdx=1\displaystyle (1 - x \coth \theta)\frac{d\theta}{dx} = 1.

Since x=coth⁡θx = \coth \theta we deduce dθdx=−1r2=−1rsinh⁡θ\frac{d\theta}{dx} = -\frac{1}{r^2} = -\frac{1}{r}\sinh \theta

Now with the benefit of having looked at this question before, we can do the rest in one big jump:

ddxcosh⁡nθrn=nsinh⁡nθdθdx1rn−ncosh⁡nθrn+1drdx\displaystyle \frac{d}{dx} \frac{\cosh n\theta}{r^n} = n \sinh n\theta \frac{d\theta}{dx} \frac{1}{r^n} -n \frac{\cosh n\theta}{r^{n+1}}\frac{dr}{dx}

=−nrn+1(sinh⁡nθsinh⁡θ+cosh⁡nθcosh⁡θ)\displaystyle = -\frac{n}{r^{n+1}} (\sinh n\theta \sinh\theta + \cosh n\theta \cosh \theta)

=−ncosh⁡(n+1)θrn+1\displaystyle = -\frac{n \cosh(n+1)\theta}{r^{n+1}}

So d2ydx2=2ddxcosh⁡θr=−2cosh⁡2θr2\displaystyle \frac{d^2y}{dx^2} = 2 \frac{d}{dx} \frac{\cosh \theta}{r} = -\frac{2 \cosh 2\theta}{r^2}

d3ydx3=4cosh⁡3θr3\displaystyle \frac{d^3y}{dx^3} = \frac{4 \cosh 3\theta}{r^3} and generally dnydxn=2(−1)n−1(n−1)!cosh⁡nθrn\displaystyle \frac{d^ny}{dx^n} = 2\frac{(-1)^{n-1}(n-1)!\cosh n\theta}{r^n}.
Ive completed STEP III Question 4 on paper. Bit tricky but wasnt too bad actually.

STEP III Question 4

x=acost+alntant2\displaystyle x = acost + aln tan \frac{t}{2}

y=asint\displaystyle y = asint

dydx=dydt.dtdx\displaystyle \frac{dy}{dx} = \frac{dy}{dt}.\frac{dt}{dx}

dydt=acost\displaystyle \frac{dy}{dt} = acost

dxdt=−asint+asec2t22tant2=−asint+acos2t22sint2cost2\displaystyle \frac{dx}{dt} = - asint + \frac{asec^2\frac{t}{2}}{2tan\frac{t}{2}} = -asint + \frac{\frac{a}{cos^2\frac{t}{2}}}{\frac{2sin\frac{t}{2}}{cos\frac{t}{2}}}
=−asint+a2sint2cost2=−asint+asint=a(1−sin2t)sint=acos2tsint = -asint + \frac{a}{2sin\frac{t}{2}cos\frac{t}{2}} = -asint + \frac{a}{sint}= \frac{a(1-sin^2t)}{sint} = \frac{acos^2t}{sint}

dtdx=sintacos2t\displaystyle \frac{dt}{dx} = \frac{sint}{acos^2t}

dydx=acost.sintacos2t\displaystyle \frac{dy}{dx} = acost.\frac{sint}{acos^2t}

dydx=tant\displaystyle \frac{dy}{dx} = tant

As required.

I might provide a sketch tomorrow but its not exactly a taxing thing to do. (As sketching is the next part to the question)

It is actually best if make a sketch for the next part

Tangent to curve at P is

y−asint=tant(x−acost−alntant2)\displaystyle y - asint = tant (x - acost - aln tan \frac{t}{2})

y = 0

−asint=tantx+tant(−acost−alntant2)\displaystyle - asint = tan t x + tant (-acost - aln tan \frac{t}{2})

tantx=−asint−tant(−acost−alntant2)\displaystyle tant x = -asint - tant(-acost - aln tan \frac{t}{2})

x=−acost+acost+alntant2\displaystyle x = -acost + acost + aln tan \frac{t}{2}

x=alntant2\displaystyle x = a ln tan \frac{t}{2}

PQ2=(asint)2+(alntant2−(acost+alntant2))2\displaystyle PQ^2 = (asint)^2 + (aln tan \frac{t}{2} - (acost + a ln tan \frac{t}{2}))^2

PQ2=a2sin2t+(−acost)2\displaystyle PQ^2 = a^2sin^2t + (-acost)^2

PQ2=a2\displaystyle PQ^2 = a^2

PQ=a\displaystyle PQ = a

As required.

Next part.

y=asint\displaystyle y = asint

dydt=acost\displaystyle \frac{dy}{dt} = acost

d2ydt2=−asint\displaystyle \frac{d^2y}{dt^2} = -asint

x=acost+alntant2\displaystyle x = acost + a ln tan \frac{t}{2}

dxdt=−asint+asint\displaystyle \frac{dx}{dt} = -asint + \frac{a}{sint}

d2xdt2=−acost−acottcosect\displaystyle \frac{d^2x}{dt^2} = - acost - acot t cosect

Right so

ρ=[(−asint+asint)2+a2cos2t]32∣[(−asint+asint).(−asint)−acost.(−acost−acottcosect)]∣\displaystyle \rho = \frac{ [(-asint + \frac{a}{sint})^2 + a^2cos^2t]^{\frac{3}{2}}}{|[(-asint + \frac{a}{sint}).(-asint) -acost.(-acost - acottcosect)]|}

ρ=[a2sin2t+a2sin2t−2a2+a2cos2t]∣32∣a2sin2t−a2+a2cos2t+a2cot2t∣\displaystyle \rho = \frac{[a^2sin^2t + \frac{a^2}{sin^2t} - 2a^2 + a^2cos^2t]^{|\frac{3}{2}}}{|a^2sin^2t-a^2 + a^2cos^2t + a^2 cot^2t|}

ρ=[a2−a2sin2tsin2t]32a2cot2t\displaystyle \rho = \frac{[\frac{a^2-a^2sin^2t}{sin^2t}]^{\frac{3}{2}}}{a^2cot^2t}

ρ=[a2cot2t]32a2cot2t\displaystyle \rho = \frac{[a^2cot^2t]^{\frac{3}{2}}}{a^2cot^2t}

ρ=a3cot3ta2cot2t\displaystyle \rho = \frac{a^3cot^3t}{a^2cot^2t}

ρ=acott\displaystyle \rho = acott

As required.

Normal to P

y−asint=−1tant(x−acost−alntant2)\displaystyle y - asint = -\frac{1}{tant} ( x - acost - aln tan\frac{t}{2})

You really need to look at a diagram here. If CQ is parallel to y axis the equation x=alntant2 x = a ln tan \frac{t}{2} should intersect the normal to P at C. (The prove of this comes to the fact that we can confirm the distance between P and C is rho which will be explained)

Let x=alntant2\displaystyle x = a ln tan \frac{t}{2}

y−asint=−costsint(−acost)\displaystyle y - asint = -\frac{cost}{sint}(-acost)

y=acos2tsint+asint\displaystyle y = \frac{acos^2t}{sint} + asint

y=acos2t+asin2tsint\displaystyle y = \frac{acos^2t+asin^2t}{sint}

y=asint\displaystyle y = \frac{a}{sint}

Now We are told that the distance between P and C is rho = acott

Distance2=(acost)2+(asint−asint)2\displaystyle Distance^2 = (acost)^2 + (\frac{a}{sint} - asint)^2

Distance2=a2cos2t+[(a−asin2tsint]2\displaystyle Distance^2 = a^2cos^2t + [(\frac{a-asin^2t}{sint}]^2

Distance2=a2cos2t+[acos2tsint]2\displaystyle Distance^2 = a^2cos^2t + [\frac{acos^2t}{sint}]^2

Distance2=a2cos2t+a2cos4tsin2t\displaystyle Distance^2 = a^2cos^2t + \frac{a^2cos^4t}{sin^2t}

Distance2=a2cos2tsin2t+a2cos4tsin2t\displaystyle Distance^2 = \frac{a^2cos^2tsin^2t + a^2cos^4t}{sin^2t}

Distance2=a2cos2t(sin2t+cos2t)sin2t\displaystyle Distance^2 = \frac{a^2cos^2t(sin^2t + cos^2t)}{sin^2t}

Distance2=a2cos2tsin2t\displaystyle Distance^2 = \frac{a^2cos^2t}{sin^2t}

Distance=acott=ρ\displaystyle Distance = acott = \rho

So if the coordinates of C are (alntant2,asint)\displaystyle (alntan\frac{t}{2},\frac{a}{sint}) and i.e it intersects withx=alntant2\displaystyle x = a ln tan \frac{t}{2}

The distance between P and C is \rho. Which we are told it is in the question. This means that the coordinates of C are (alntant2,asint)\displaystyle (alntan\frac{t}{2},\frac{a}{sint}) and intersects with x=alntant2 x = aln tan \frac{t}{2} at C. Therefore CQ must be parallel to the y axis.

I'll post a few diagrams up tomorrow hopefully that will make it alot clearer.
Right, I/8 here. Sorry I had to use webcam to "scan" my answers in (includes graphs).
The quality is quite poor - apologies for that!
2007 I8 1.jpg39.5KB
2007 I8 2.jpg35.8KB
2007 I8 3.jpg31.9KB
I'm surprised nobody's had a bash at STEP II Question 3. I'll have a go now.
insparato
Im surprised nobodys had a bash at STEP II Question 3. I'll have a go now.

I have, but I'm missing something (probably swapping for sin/cos over the limits...)
Reply 46
Avatar for Rabite
Rabite
8
Ooh I did that one. Want me to check any part for you? Wouldn't be surprised if a sine sprouted legs and started walking into the next line.
STEP II Question 3(except last part so far)

Unparseable latex formula:

\displaystyle \int \frac{1}{a^2 + x^2} \hspace5 dx



x=atanθ x = atan\theta

dxdθ=asec2θ \frac{dx}{d\theta} = asec^2\theta

dx=asec2θ.dθ dx = asec^2\theta.d\theta

Unparseable latex formula:

\displaystyle \int \frac{1}{a^2 + a^2tan^2\theta}.asec^2\theta \hspace5 d\theta



Unparseable latex formula:

\displaystyle \int \frac{a(1+tan^2\theta)}{a^2(1+tan^2\theta)} \hspace5 d\theta



Unparseable latex formula:

\displaystyle \int \frac{1}{a} \hspace5 d\theta



=1aθ+C = \frac{1}{a}\theta + C

=1aarctanxa+C = \frac{1}{a}arctan \frac{x}{a} + C
i) a)
Unparseable latex formula:

I = \displaystyle \int_0^{\frac{\pi}{2}} \frac{cosx}{1+sin^2x} \hspace5 dx



u=sinx u = sinx

dudx=cosx \frac{du}{dx} = cosx

ducosx=dx \frac{du}{cosx} = dx

∫01cosx1+u2.ducosx \displaystyle \int_0^1 \frac{cosx}{1+u^2}.\frac{du}{cosx}

Unparseable latex formula:

\displaystyle \int_0^1 \frac{1}{1+u^2} \hspace5 du



I guess the first part was there for a reason :p:

=[arctanu]01= [arctanu]_0^1

=arctan1=π4 = arctan 1 = \frac{\pi}{4}

Now looking back i fancied taking a bash using the t substitution formula. I'll come back to this once ive attempted the next part.

i) b)

Unparseable latex formula:

\int_0^1 \frac{1-t^2}{1+6t^2+t^4} \hspace5 dt



t=tanx2 t = tan\frac{x}{2}

dtdx=12sec2x2 \frac{dt}{dx} = \frac{1}{2}sec^2\frac{x}{2}

dt=12sec2x2.dx dt = \frac{1}{2}sec^2\frac{x}{2}.dx

Unparseable latex formula:

\displaystyle \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{(1 -tan^2\frac{x}{2})(1+tan^2\frac{x}{2})}{1 + 6tan^2\frac{x}{2} + tan^4\frac{x}{2}} \hspace5 dx



Now this is a bloody leap of faith because i actually did the t substitution with the last part on paper.

sint=2tanx21+tan2x2 sint = \frac{2tan\frac{x}{2}}{1+tan^2\frac{x}{2}}

cost=1−tan2x21+tan2x2 cost = \frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}

cosx1+sin2x=1−tan2x21+tan2x21+4tan2x2(1+tan2x2)2 \frac{cosx}{1+sin^2x} = \frac{\frac{1-tan^2\frac{x}{2}}{1+tan^2\frac{x}{2}}}{1+\frac{4tan^2\frac{x}{2}}{(1+tan^2\frac{x}{2})^2}}

=(1−tan2x2)(1+tan2x2)(1+tan2x2)2+4tan2x2 = \frac{(1-tan^2\frac{x}{2})(1+tan^2\frac{x}{2})}{(1+tan^2 \frac{x}{2})^2 + 4tan^2\frac{x}{2}}

=(1−tan2x2)(1+tan2x2)tan4x2+6tan2x2+1 = \frac{(1-tan^2\frac{x}{2})(1+tan^2\frac{x}{2})}{tan^4\frac{x}{2} + 6tan^2\frac{x}{2} + 1}

Therefore

Unparseable latex formula:

\displaystyle \frac{1}{2} \int_0^{\frac{\pi}{2}} \frac{(1 -tan^2\frac{x}{2})(1+tan^2\frac{x}{2})}{1 + 6tan^2\frac{x}{2} + tan^4\frac{x}{2}} \hspace5 dx = \frac{1}{2} \int \frac{cosx}{1+sin^2x} = \frac{1}{2}I



A BIG leap of faith im still alittle perplexed by it myself. It just popped into my head to do t substitution on the original integral. Hmm.

I'm going to have a bash at the next part now.

Okay i think i have this now.

Unparseable latex formula:

\displaystyle \int_0^1 \frac{(1-t^2)}{1+14t^2 + t^4} \hspace5 dt



t=tanx2 t = tan\frac{x}{2}

dtDx=12sec2x2 \frac{dt}{Dx} = \frac{1}{2}sec^2\frac{x}{2}

dt=12sec2x2dx dt = \frac{1}{2}sec^2\frac{x}{2} dx

Unparseable latex formula:

\displaystyle \frac{1}{2}\int_0^{\frac{\pi}{2}} \frac{(1-t^2)(1+t^2)}{1 + 14t^2 + t^4} \hspace5 dx



I've left it in t make it easier at the moment.

Consider

cosx1+3sin2x \frac{cosx}{1 + 3sin^2x}

=1−t21+t21+3(2t)2(1+t2)2 = \frac{\frac{1-t^2}{1+t^2}}{1 + \frac{3(2t)^2}{(1+t^2)^2}}

=(1−t2)(1+t2)(1+t2)2+12t2 = \frac{(1-t^2)(1+t^2)}{(1+t^2)^2 + 12t^2}

=(1−t2)(1+t2)t4+14t2+1 = \frac{(1-t^2)(1+t^2)}{t^4 + 14t^2 + 1}

So

Unparseable latex formula:

\displaystyle \frac{1}{2}\int_0^{\frac{\pi}{2}} \frac{(1-t^2)(1+t^2)}{1 + 14t^2 + t^4} \hspace5 dx = \frac{1}{2}\int_0^{\frac{\pi}{2}} \frac{cosx}{1+3sin^2x}



u=sinx u = sinx

dudx=cosx \frac{du}{dx} = cosx

ducosx=dx \frac{du}{cosx} = dx

12∫0π2cosx1+3sin2x=12∫01cosx1+3u2.ducosx\displaystyle \frac{1}{2}\int_0^{\frac{\pi}{2}} \frac{cosx}{1+3sin^2x} = \frac{1}{2}\int_0^1 \frac{cosx}{1+3u^2}.\frac{du}{cosx}

=12∫0111+3u2\displaystyle = \frac{1}{2} \int_0^1 \frac{1}{1+3u^2}

=12.13∫01113+u2\displaystyle =\frac{1}{2}.\frac{1}{3} \int_0^1 \frac{1}{\frac{1}{3}+u^2}

=16[3arctan3x]01= \frac{1}{6}[\sqrt3 arctan \sqrt3 x]_0^1

=36arctan3=3π18 = \frac{\sqrt{3}}{6} arctan \sqrt{3} = \frac{\sqrt{3}\pi}{18}
I havent a clue whether thats right. But im a bit chuffed with myself on the last two parts.
STEP III, Q11.

(i) Let O be the point of contact with the table edge, and H be the position of the hub. Define θ\theta to be the angle OH makes with the vertical.

To show the frictional force is zero, I think you just need to take moments about H - as the gravitational force and the reaction both act through H, any frictional force will result in a net torque being applied. But since the hub has zero moment of inertia any such torque will result in infinite angular acceleration, which is physically impossible.

For the next bit, conservation of energy gives us 12a2θ˙2=12u2+ga(1−cos⁡θ)\frac{1}{2}a^2\dot{\theta}^2 = \frac{1}{2}u^2+ga(1-\cos\theta) and then differentiating gives us θ¨=gasin⁡θ\ddot{\theta} = \frac{g}{a} \sin \theta

Now consider the y-coordinate of the hub:

y=acos⁡θy = a \cos \theta. Differentiate (twice):

y˙=−aθ˙sin⁡θ\dot{y} = -a \dot{\theta} \sin \theta

y¨=−aθ˙2cos⁡θ−aθ¨sin⁡θ\ddot{y} = - a\dot{\theta}^2\cos \theta -a \ddot{\theta}\sin\theta. Substitute in our values for θ˙2,θ¨\dot{\theta}^2, \ddot{\theta} to get:

ay¨=(−u2+2ga(cos⁡θ−1))cos⁡θ−a2gasin⁡2θa\ddot{y} = (-u^2+2ga(\cos \theta - 1)) \cos \theta - a^2 \frac{g}{a} \sin^2 \theta

=−(u2+2ga)cos⁡θ+2gacos⁡2θ−gasin⁡2θ=-(u^2+2ga)\cos \theta + 2ga \cos^2\theta -ga \sin^2 \theta

=−(u2+2ga)cos⁡θ+3gacos⁡2θ−ga=-(u^2+2ga)\cos \theta + 3ga \cos^2\theta-ga

Now if there was no reaction force from the table, we'd have y¨=−g\ddot{y} = -g (full acceleration due to gravity). So at that point we have:

−ga=−(u2+2ga)cos⁡θ+3gacos⁡2θ−ga-ga = -(u^2+2ga)\cos \theta + 3ga \cos^2\theta-ga

0=cos⁡θ((u2+2ga)−3gacos⁡θ)0= \cos \theta((u^2+2ga)-3ga \cos \theta)

So cos⁡θ=0\cos \theta = 0 or cos⁡θ=u2+2ga3ga\cos \theta = \frac{u^2+2ga}{3ga}. The 2nd condition will occur before we get to cos⁡θ=0\cos \theta = 0, so the wheel loses contact when cos⁡θ=u2+2ga3ga\cos \theta = \frac{u^2+2ga}{3ga}. At this point, it has dropped a distance a(1−cos⁡θ)=3ga−(u2+2ga)3g=ag−u23ga(1-\cos \theta) = \frac{3ga - (u^2+2ga)}{3g} = \frac{ag-u^2}{3g} as desired.

(ii) I don't think any significant calculations are actually needed (or expected?) here. Again, take moments about the center of the wheel. The only torque is from the frictional force, but in this case the wheel has a non-zero moment-of-inertia, so this time we need that torque to provide the angular acceleration. (It's intuitively obvious that θ¨≠0\ddot{\theta} \neq 0, but if you want to prove it, use a similar energy argument to the first part to show θ¨=g2asin⁡θ\ddot{\theta} = \frac{g}{2a}\sin \theta). When the wheel loses contact, there's obviously no reaction force and therefore no frictional force. So an "intermediate value argument" says at some point before losing contact, the frictional force will be insufficient to provide the necessary rotational acceleration and the wheel will slip.
I'm having a go at STEP II Question 1 now.
insparato
I havent a clue whether thats right. But im a bit chuffed with myself on the last two parts.

We've got the same working except that I didn't see how to prove the 'nasty' expressions with t's simplified to something nice (probably because I've never done a t-substitution before and thus didn't think of sinx=1-tanx/1+tanx etc.)


edit: though I'm not sure what you get when integrating cos1+3sin2\frac{cos}{1+3sin^2}; that should surely be 13arctan(3x)\frac{1}{\sqrt{3}}arctan(\sqrt{3}x) ? (or maybe what you have is equivalent, I'm too tired to check:redface:
a = 1/rt3 in this case

so 1/a = rt3
I'm tired ive completed two questions and STEP II Question 1 the binomial one i cant see how to turn this stupidly large fraction into a decimal. Unless theres a much better approach im not seeing.
insparato
a = 1/rt3 in this case

so 1/a = rt3

I'm too tired for this, so I still don't see - this is what I did:

Unparseable latex formula:

\frac{1}{2}\int \frac{cos(x)}{1+3sin^2(x)}dx=\frac{1}{\sqrt{3}}arctan(\sqrt{3}sin(x)}+c

Applying the limits we have 1233π6−0=3π36\frac{1}{2}\frac{\sqrt{3}}{3}\frac{\pi}{6}-0=\frac{\sqrt{3}\pi}{36}

I seem to be a factor of 2 away from your result. Someone want to play detective and find the faulty working? :tongue:
Ok, I/5. Explanatory diagram provided, excuse my master drawing :wink: No problems with i, but my answer to ii (4/3) seems too small a ratio :s: Any thoughts on how to get the proper answer?

Enc.
2007 I5 1 (explanatory).jpg32.1KB
2007 I5 2 (part i).jpg31.5KB
2007 I5 3 (part ii) - shaky.jpg28.7KB
I'm assuming we get the same integral here ?

=12∫0111+3u2\displaystyle = \frac{1}{2} \int_0^1 \frac{1}{1+3u^2}

=12.13∫01113+u2\displaystyle =\frac{1}{2}.\frac{1}{3} \int_0^1 \frac{1}{\frac{1}{3}+u^2}

Heres the integral

∫1a2+x2=1aarctanxa\int \frac{1}{a^2 + x^2} = \frac{1}{a}arctan \frac{x}{a}

in this case

a2=13 a^2 = \frac{1}{3}

a=13 a = \frac{1}{\sqrt3}

1a=3 \frac{1}{a} = \sqrt3

so
=12.13∫01113+u2\displaystyle =\frac{1}{2}.\frac{1}{3} \int_0^1 \frac{1}{\frac{1}{3}+u^2}

=16[3arctan3x]01= \frac{1}{6}[\sqrt3 arctan \sqrt3 x]_0^1

=36arctan3=3π18 = \frac{\sqrt{3}}{6} arctan \sqrt{3} = \frac{\sqrt{3}\pi}{18}
For some reason I thought arctan(sqrt3)=pi/6 was an excellent idea...:rolleyes: otherwise our working is equivalent, just somewhat different (and me integrating by recognition as usual...)
Reply 58
Avatar for ukgea
ukgea
6
calcium878
Ok, I/5. Explanatory diagram provided, excuse my master drawing :wink: No problems with i, but my answer to ii (4/3) seems too small a ratio :s: Any thoughts on how to get the proper answer?


The centres of the faces of the octahedron (i.e. the point where the segment of length L meets the octahedron) are not halfway up the altitudes of the faces, as they are on your picture, but in fact they are only 1/3 of the way up, and so what is 3/4\sqrt 3 / 4 in your diagram should be 3/6\sqrt 3 / 6.

I'm not sure how they wanted you to derive this, but I simply quoted the mechanics section of the OCR formula booklet: The centre of mass of a uniform triangular lamina is 2/3 along median from vertex.

Otherwise the solution looks correct.
Cheers ukgea, I'll update it when I get back from work later :smile:

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