STEP maths I, II, III 1990 solutions Watch

DFranklin
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#21
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#21
(Original post by Rabite)
Oh. When does this work?
Is that just something that textbooks happened to forget to tell us? I hardly know what the determinant is... (Except for how to calculate it)
Whenever you apply a matrix transformation, the area is multiplied by the determinant. (Also in 3D, except it's the volume).

I do confess, sometimes I get these thngs the wrong way round, so I checked the result for this question by writing a proglet to find the actual area.
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Speleo
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#22
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So in general, if a curve, A, is transformed by a matrix, M, into another curve, B, then the area of B is |M|*the area of A?

(thanks Rabite and DFranklin )

EDIT: ok, great
There are so many simple things I don't know about matrices... were they covered in that much more depth at A-level back in the early 90's?
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insparato
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#23
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 \int y^2 = \frac{1}{3}\int y^4 - Yep!

Yeah i noticed that we couldnt handle  \int x(t) So i was purposely where possible using u = x(t) when an integral popped up. Yeah i was thinking the exact same thing, theres no way i could do the last two parts in an exam.
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DFranklin
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(Original post by Speleo)
So in general, if a curve, A, is transformed by a matrix, M, into another curve, B, then the area of B is |M|*the area of A?

(thanks Rabite and DFranklin )

EDIT: ok, great
There are so many simple things I don't know about matrices... were they covered in that much more depth at A-level back in the early 90's?
Yes. At any rate, the bit about areas under transformations was in my FM A-level (1986), even though it was really only shown for rectangles rather than general curves.

Do you guys do eigenvalues/eigenvectors?
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insparato
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Yeah ive tried to look up the definition of the determinant and it didnt shed any light what so ever. I get the impression it was just some expression that happens to useful.
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Speleo
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#26
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(Original post by DFranklin)
Do you guys do eigenvalues/eigenvectors?
Yes, and that's pretty much it. There's also inverses of 2 or 3 rowed matrices and diagonalising a matrix. Nothing beyond that.
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Speleo
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#27
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OK, I've completed III/6 now, I might get round to latexing/typing it up it in a bit, but until then here are hints and answers I got:
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First bit is easy. Next bit, with the curve, write x = cost and y = sint to make the transformation easier. Then write out what 8X^2 + ... would be and show that it's 80. For the area, see DFranklin's posts on this page and the last. For the maximum value, complete the square and find Y in terms of X. There will be a square root of a number which you need to ensure stays positive. For the tangent, use the transformation on the co-ordinates given and then do as normal, differentiate to find the gradient etc. etc., I get: Y = (-4/5)X + (332/125) but I'm not all that confident about the accuracy of that solution because I was working it out in the corner and margins of the sheet of paper

EDIT: the equation of the tangent I gave seems to be wrong, see below (my mental arithmetic has failed me)
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khaixiang
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#28
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(Original post by Speleo)
OK, I've completed III/6 now, I might get round to latexing/typing it up it in a bit, but until then here are hints and answers I got:
Spoiler:
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First bit is easy. Next bit, with the curve, write x = cost and y = sint to make the transformation easier. Then write out what 8X^2 + ... would be and show that it's 80. For the area, see DFranklin's posts on this page and the last. For the maximum value, complete the square and find Y in terms of X. There will be a square root of a number which you need to ensure stays positive. For the tangent, use the transformation on the co-ordinates given and then do as normal, differentiate to find the gradient etc. etc., I get: Y = (-4/5)X + (332/125) but I'm not all that confident about the accuracy of that solution because I was working it out in the corner and margins of the sheet of paper
Err.. some comments on this question
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You could just find the equation of tangent at the original curve then apply transformation T It's simpler this way since you don't have to deal with the very messy fraction, maybe this is the reason why a hideous coordinate (3/5, 4/5) was given. The equation of tangent is 21y+13x-50=0, I did using your method too, and got the same equation.

But can anyone tell me if it's always true that if a curve undergoes a transformation, then its' equation of tangent, will be transformed to an equation of tangent of the transformed curve at the transformed coordinate?
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Speleo
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Clever
I didn't notice that since it seemed so straight-forward (if arithmetically intensive) to do it the obvious way.
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Rabite
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#30
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Any transformation representing a stretch or a rotation (linear?) should preserve the tangent line, because it's just like changing the axes isn't it?

I suppose a translation would too.
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DFranklin
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#31
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#31
(Original post by khaixiang)
But can anyone tell me if it's always true that if a curve undergoes a transformation, then its' equation of tangent, will be transformed to an equation of tangent of the transformed curve at the transformed coordinate? [/SPOILER]
Should be OK for the tangent. It is not true for the normal!
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Kolya
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As I touched on in the other thread, the reason for the confusion is that transformations and rotations using matrices are still on the Edexcel syllabus, but the popular Heinemann textbook gives no questions at all on them.
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*bobo*
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#33
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STEP I
10) i)total resistive forces= 50+2(v-3)= 2(22 +v) where v is pos. in direction of turning point from starting point.
using P=Fv, where F=resistive forces due to no acceleration
300=2v(22+v)
2v^2 + 44v- 300=0
v=0.25 (in forward journey), v=1.25(backward journey) to 2dp
using v=s/t:
total time for race= 5000/0.25 + 5000/1.25
= 24000 seconds (supposed to answer to nearest second but don't have a calc on me)


ii) total energy expended using P=W/t
W=300x24000=7200000

x(1)t(1)+x(2)t(2)= 7200000

x(1)= v(1)(50+2(v(1)-3))= v(1)(44 + 2v)
t(1)= 5000/x(1)

x(2)= v(2)(50+2(v(2) +3))=v(2)(56+2v(2))
t(2)=5000/v(2)

=> 5000(44+2v(1))+5000(56+2v(2))=72 00000

=> 100+ 2v(1) + 2v(2)= 1440
=> v(1) +v(2)=670 so is independant of x(1) and x(2) values.
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Dystopia
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#34
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STEP 1, Q8.

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\cos \frac{\alpha}{2} \cos \frac{\alpha}{4} = \frac{\sin \alpha}{4\sin \frac{\alpha}{4}}, \; \alpha \not= k\pi \Leftrightarrow 2 \cos \frac{\alpha}{2} \left(2 \cos \frac{\alpha}{4} \sin \frac{\alpha}{4} \right) = \sin\alpha
(k \in ℤ)

Which is obvious by the half angle formula.

\cos\frac{\alpha}{2} \cdots \cos\frac{\alpha}{2^{n}} = \frac{\sin\alpha}{2^{n} \sin\frac{\alpha}{2^{n}}}, \; \alpha \not = k\pi

 \Leftrightarrow 2\cos\frac{\alpha}{2}\left(2\cos  \frac{\alpha}{4} \left(\cdots \left(2\sin\frac{\alpha}{2^{n}} \cos\frac{\alpha}{2^{n}}\right)\  cdots\right)\right) = \sin\alpha

Which is again immediate by n applications of the half angle formula.

It is then necessary to show that \displaystyle \lim_{n \to \infty} 2^{n}\sin\frac{\alpha}{2^{n}} = \alpha

I initially considered L'Hopital's rule, but I haven't used it before and wasn't particularly sure if my arguments were justified (although it gave the correct result). Then I realised it was much simpler to use the power expansion.

2^{n} \sin\frac{\alpha}{2^{n}} = 2^{n} \left( \frac{\alpha}{2^{n}} - \frac{\alpha ^{3}}{2^{3n} \times 3!} + \cdots \right)

From which the limit is evident.

Rearranging, and letting n \to \infty, \; \alpha = \frac{\sin\alpha}{\cos\frac{\alp  ha}{2}\cos \frac{ \alpha}{4} \cos\frac{\alpha}{8} \cdots}

Substituting \alpha = \frac{\pi}{2} and using \cos\frac{\pi}{4} = \sqrt{\frac{1}{2}}, \; \cos\frac{x}{2} = \sqrt{\frac{1}{2} + \frac{1}{2} \cos x} gives the required expression.
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Rabite
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#35
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Quick question about II/8 (the one with x(t)=int y dt or something).
After multiplying by x, you get
xy'' + (y^2-1)y'x + xy=0

To integrate the big thing in the middle, can you not just do
 u=x \rightarrow u'=y
 v=\frac{1}{3} y^3 - y \leftarrow v'=(y^2-1)y'
Without the messy business?

Since it's a function and a derivative "recognition" type thing.

Also, does the last part not just pop out if you times the given DE by y and then integrate?
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yy'' + y(y^2-1)y' + y^2=0
Integrate each little bit. yy'' by parts:
u=y . . . u'=y'
v=y' . . . v'=y''
\int yy'' dt = [yy'] - \int [y']^2 dt
Put limits in and the thing in []s is zero because of the y1=y0 etc...
Middle bit. Consider
\frac{d}{dt} (\frac{1}{4} y^4-\frac{1}{2} y^2)=(y^3-y)*\frac{dy}{dt}
 = y(y^2-1)y' which is what we wanted to integrate anyway...
If you put limits on it, it goes to zero from the given conditions.

So you're left with  -\int [y']^2 dt + \int y^2 dt=0?

Or can you not just use y' as a dy/dt so casually...
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DFranklin
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#36
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#36
(Original post by Rabite)
To integrate the big thing in the middle, can you not just do
 u=x \rightarrow u'=y
 v=\frac{1}{3} y^3 - y \leftarrow v'=(y^2-1)y'
Without the messy business?
Yes, I think so. That is absolutely brilliant. Your argument for the last bit seems fine as well. First class.
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cuddy
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#37
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are you guys still looking for pre 1990 papers? I was surfing through maths sites earlier and I found the 1987 papers for STEP I, II and III. If you want the link pm me.
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Speleo
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#38
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#38
1987 III/6

\frac{dx}{dt} + 2x - 5y = 0
\frac{dy}{dt} + x - 2y = 2\cos t

at t=0; x=0, \frac{dy}{dt}=0

Can anyone get me started?
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insparato
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#39
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Are these differential equations connected ?
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Speleo
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#40
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Yeah, they're simultaneous, sorry.
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