STEP Maths I, II, III 1989 solutions
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Re: STEP Maths I, II, III 1989 solutionsI think so, yes. (Modulo whether it's e^At or e^-At; I haven't looked closely enough to be sure).
Spoiler:ShowI think it's a slightly weird question: the earlier parts give you all the pieces to basically just write down the answer, but it's most definitely not the approach you'd come to naturally. -
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Re: STEP Maths I, II, III 1989 solutions(Original post by DFranklin)
I think so, yes. (Modulo whether it's e^At or e^-At; I haven't looked closely enough to be sure).
Spoiler:ShowI think it's a slightly weird question: the earlier parts give you all the pieces to basically just write down the answer, but it's most definitely not the approach you'd come to naturally.Spoiler:ShowDoesn't multiplying by F work just as well?
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Re: STEP Maths I, II, III 1989 solutions(Original post by DFranklin)
Spoiler:ShowThey're the same thing, surely?Spoiler:ShowI'm not quite sure what I was thinking. I did it by just noticing that multiplying by F would work rather than doing it 'properly'. -
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Re: STEP Maths I, II, III 1989 solutionsWould this be correct for the third part?
Expected winnings =
(So X = 2b_3 - b_2, Y = b_1 - b_3, Z = b_2 - 2b_1)
And for the next part, is it enough if I just state
b_1 = 0.25
b_2 = 0.5
b_3 = 0.25?
With these numbers, X, Y and Z all turn out 0 which is non-negative. And so the expectation is 0 as well, regardless of a_1, a_2 and a_3, which means that he won't LOSE any money (in the "long run"). -
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Re: STEP Maths I, II, III 1989 solutionsSTEP III - Question 9
i)
![1 + \frac{2}{2} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + ..... = [ 1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{2}} + ......] + [\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{2}} + \frac{1}{2^{4}} +......] + [ \frac{1}{2^{2}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ......] + ....... = \frac{1}{1-\frac{1}{2}} + \frac{\frac{1}{2}}{1-\frac{1}{2}} + \frac{\frac{1}{4}}{1-\frac{1}{2}} + \frac{\frac{1}{8}}{1-\frac{1}{2}} + ....... = 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ..... = 2 + \frac{1}{1 - \frac{1}{2}} = 4](http://www.thestudentroom.co.uk/latexrender/pictures/13/13dd1c9f233c60a422e38f68c01d981a.png "1 + \frac{2}{2} + \frac{3}{2^{2}} + \frac{4}{2^{3}} + ..... = [ 1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{2}} + ......] + [\frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{2}} + \frac{1}{2^{4}} +......] + [ \frac{1}{2^{2}} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + ......] + ....... = \frac{1}{1-\frac{1}{2}} + \frac{\frac{1}{2}}{1-\frac{1}{2}} + \frac{\frac{1}{4}}{1-\frac{1}{2}} + \frac{\frac{1}{8}}{1-\frac{1}{2}} + ....... = 2 + 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ..... = 2 + \frac{1}{1 - \frac{1}{2}} = 4")
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ii) Consider the Maclaurins series of
.
for 
.
.
So putting x = 0.5 (its ok since it's in the required range.), we get:
iii) consider
so putting
:
Last edited by squeezebox; 30-05-2008 at 01:16. -
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Re: STEP Maths I, II, III 1989 solutionsSTEP III - Question 1 - I'm not really sure if I've done this right
We form the sphere by rotating a circle 360 degrees about the x-axis. WLOG, let the equation of the circle be
.
So differentiating implicitly:
.
So using the formula for surface area (
) the area of the zone between the lines x = a and x=b (r > b > a > -r), is : (
as required.
This next bit is really hard to explain without a diagram....
let
be the angle between the tangent and the x-axis. Then, since angle OPC (c = centre of circle) = 90 degrees, 
.
So the equation of the tangent is:
. And the x co-ord of P = 
.
Using (*), the surface area of the sphere bit is:![2\pi \times a \times [ (a+c) - (\frac{c^{2}-a^{2}}{c})] = \frac{2\pi a^{2}(a+c)}{c}](http://www.thestudentroom.co.uk/latexrender/pictures/40/40eb92124f3ad5ffee6d837212ff86dc.png "2\pi \times a \times [ (a+c) - (\frac{c^{2}-a^{2}}{c})] = \frac{2\pi a^{2}(a+c)}{c}")
.
And the surface area formed by rotating the tangent () part (the bit from the orgin to P) about the x-axis is : ![\displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} 2 \pi y \sqrt{1+(\frac{dy}{dx})^{2}} \, dx = \displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} 2 \pi \times \frac{ax}{\sqrt{c^{2}-a^{2}}} \times \sqrt{1+(\frac{a^{2}}{c^{2}-a^{2}}} \, dx = \displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} \frac{2 \pi a c x}{c^{2}-a^{2}} = [\frac{\pi a c x^{2}}{c^{2}-a^{2}}]^{\frac{c^{2}-a^{2}}{c}}_{0} = \frac{\pi a (c^{2}-a^{2})}{c}](http://www.thestudentroom.co.uk/latexrender/pictures/b6/b6a9c45d67863b50fa6e737e9b022163.png "\displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} 2 \pi y \sqrt{1+(\frac{dy}{dx})^{2}} \, dx = \displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} 2 \pi \times \frac{ax}{\sqrt{c^{2}-a^{2}}} \times \sqrt{1+(\frac{a^{2}}{c^{2}-a^{2}}} \, dx = \displaystyle\int^{\frac{c^{2}-a^{2}}{c}}_{0} \frac{2 \pi a c x}{c^{2}-a^{2}} = [\frac{\pi a c x^{2}}{c^{2}-a^{2}}]^{\frac{c^{2}-a^{2}}{c}}_{0} = \frac{\pi a (c^{2}-a^{2})}{c}")
.
So the total area =
EDIT: I'm not even sure if this is the correct method.....
Last edited by squeezebox; 31-05-2008 at 01:17. -
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Re: STEP Maths I, II, III 1989 solutionsSTEP II - Question 14
Let the tension in the string when it has rotated through degrees be T, and the velocity of the particle at this point be V.
Then using conservation of energy:
and Newtons 2nd law:
(1) -> (2):
.
This next bit is hard to describe without a diagram, sorry . EDIT: forget it, I've attached a rubbish paint diagram.
To find the force at the edge (let this be called R), you need to note that each section of the string exerts a force on the edge. Drawing a triangle of forces (see attachment) and using the cosine rule we get:
.
Then
We find that R is a maximum when
(justify with 2nd derivative, that I don't wanna type...)
When
and 
.
Which gives the maximum value of R as
Last edited by squeezebox; 01-06-2008 at 01:39. -
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Re: STEP Maths I, II, III 1989 solutionsIs there any way of solving this using, the complementary function + particular integral method.(Original post by Dystopia)
STEP III, Q8
Let
Then
Using an integrating factor,
, we get
Applying boundary conditions:
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I think that I shall restrict myself to a maximum of three solutions per day (at least at the start), so that everyone has a chance to answer some if they want.
I got to x'' - 4x' + 4x = 48(e^2t + e^-2t) and used complementary function x = Ae^2t + Bte^2t
Particular integral (most likely where i've gone wrong) x = Q(t^2)e^2t + R(t^2)e^-2t. (multiplying by t^2 because Bte^2t is in the complementary function)
It ends up giving the wrong answer.
I've never tried using a particular integral like that before and so it was basically a guess. Could anyone give me some kind of guidelines for when to not bother looking for a particular integral and just use another method instead? -
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Re: STEP Maths I, II, III 1989 solutionsI see, thanks for your help. Just so i'm totally sure on this, say you had(Original post by DFranklin)
You only need to multiply e^2t by t^2. So you'd look for a PI of form x = Q(t^2)e^2t + Re^-2t.
ax'' + bx' + cx = e^4x(2cos3x + 6sin2x)
and the complementary function was e^4x(Acos3x + Bsin3x)
Would the PI be Qx(e^4x)cos3x + R(e^4x)sin2x? -
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Re: STEP Maths I, II, III 1989 solutionsMy answer differs slightly from yours in that I thought you had to take both the positive and negative square root of (f'(x))^2 = x^2/e^(x^2) doing this gives(Original post by Swayum)
STEP I Question 6
The normal to the curve will therefore have gradient -1/f'(x). The equation of the normal would be
(where x' is the general x coordinate of the normal).
At the point Q, x' = 0 as it cuts the y axis.
The distance PQ can be worked out using Pythagoras' theorem.
It's given that
So
(use a substitution of u = x^2 if you can't see why)
y = +/-(e^(-x^2/2) + C
If you then differentiate twice you get f''(x) = +/-e^(-x^2/2)*(1 - x^2).
f''(0) = +/-1 but we are given that f(0) is a minimum and so f''(0) must equal +1. Consequently (after plugging in the initial conditions) we get y = -(e^(-x^/2) + 1).
I figured this must be what they were expecting because i couldn't figure out why else they would mention that (0, -2) is a minimum. Could anyone confirm? -
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Re: STEP Maths I, II, III 1989 solutionsI haven't looked at the question again, but what you're saying makes sense. I was careless when writing that solution. I'll edit it soon - thanks.(Original post by maltodextrin)
My answer differs slightly from yours in that I thought you had to take both the positive and negative square root of (f'(x))^2 = x^2/e^(x^2) doing this gives
y = +/-(e^(-x^2/2) + C
If you then differentiate twice you get f''(x) = +/-e^(-x^2/2)*(1 - x^2).
f''(0) = +/-1 but we are given that f(0) is a minimum and so f''(0) must equal +1. Consequently (after plugging in the initial conditions) we get y = -(e^(-x^/2) + 1).
I figured this must be what they were expecting because i couldn't figure out why else they would mention that (0, -2) is a minimum. Could anyone confirm?
