(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").
STEP 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 x2+y2=r2.
So differentiating implicitly:
2x+2ydxdy=0
⇒dxdy=y−x.
So using the formula for surface area (∫2πy1+(dxdy)2dx) the area of the zone between the lines x = a and x=b (r > b > a > -r), is : (∫ab2πy1+(yx)2dx=∫ab2πx2+y2dx=∫ab2πrdx,=2×π×r×(b−a)(∗)as required.
This next bit is really hard to explain without a diagram....
let θ be the angle between the tangent and the x-axis. Then, since angle OPC (c = centre of circle) = 90 degrees, tan(θ)=c2−a2a.
So the equation of the tangent is: y=c2−a2ax. And the x co-ord of P = c2−a2cos(θ)=c2−a2×cc2−a2=cc2−a2.
Using (*), the surface area of the sphere bit is: 2π×a×[(a+c)−(cc2−a2)]=c2πa2(a+c).
And the surface area formed by rotating the tangent (y=c2−a2ax) part (the bit from the orgin to P) about the x-axis is : ∫0cc2−a22πy1+(dxdy)2dx=∫0cc2−a22π×c2−a2ax×1+(c2−a2a2dx=∫0cc2−a2c2−a22πacx=[c2−a2πacx2]0cc2−a2=cπa(c2−a2).
So the total area = cπa(c2−a2)+c2πa2(a+c)=cπa(a+c)2
EDIT: I'm not even sure if this is the correct method.....
Let the tension in the string when it has rotated through θ degrees be T, and the velocity of the particle at this point be V.
Then using conservation of energy:
amgsin(θ)=21mV2(1)
and Newtons 2nd law:
amV2=T−mgsin(θ)(2)
(1) -> (2): a2amgsin(θ)=T−mgsin(θ)
⇒T=3mgsin(θ).
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:
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.
Is there any way of solving this using, the complementary function + particular integral method.
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?
The normal to the curve will therefore have gradient -1/f'(x). The equation of the normal would be
y−f(x)=f′(x)−1(x′−x) (where x' is the general x coordinate of the normal).
At the point Q, x' = 0 as it cuts the y axis.
y−f(x)=f′(x)x
y=f′(x)x+f(x)
The distance PQ can be worked out using Pythagoras' theorem.
PQ2=(f(x)−(f′(x)x+f(x)))2+x2
=f′(x)2x2+x2
It's given that PQ2=ex2+x2
So
f′(x)2x2+x2=ex2+x2
f′(x)2x2=ex2
ex2x2=f′(x)2
e0.5x2x=f′(x)
∫1dy=∫e0.5x2xdx
y=∫xe−0.5x2dx
y=−e−0.5x2+c (use a substitution of u = x^2 if you can't see why)
−2=−1+c
c=−1
y=−e−0.5x2−1
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?
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?
I 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.