I dunno how active this is, but I'm really confused about question 4 (the inequalities one) on the STEP I paper. I mean, if you just draw the graphs of y= sinx + 1 and y=cos x then the answer is bleedingly obvious, but surely that can't be sufficient to get the marks?
I think for STEP drawing a graph would get you most of the marks (after all, they explicitly say that rigour is not required) but if you can also give an algebraic proof than that's not going to hurt.
I dunno how active this is, but I'm really confused about question 4 (the inequalities one) on the STEP I paper. I mean, if you just draw the graphs of y= sinx + 1 and y=cos x then the answer is bleedingly obvious, but surely that can't be sufficient to get the marks?
yup thats it, ain't STEP scary although you do have to be careful around the edges of the bounds to see if they are valid in the original inequality and also to not miss 1, but apart from that it's pretty much 20 marks for free.
EDIT: ah, got beaten to it, basically what around said as well
∠OPH=21θ so length of HP is 2acos21θ extension of string is thus 2acos21θ−a⇒ tension T=aαg(2acos21θ Equation of motion tangential to circle gives maθ¨=Tsin21θ−mgsinθ i.e. θ¨=a1[aag(2acos21θ−a)sin21θ−agsinθ]=ag[(a−1)sinθ−sin21θ] Equilibrium positions exist when θ¨=0 i.e. (α−1)sinθ−αsin21θ=0 ⇒2(α−1)sin21θcos21θ−αsin21θ=0⇒sin21θ=0 or cos21θ=2(α−1)α From sin21θ=0 we have θ=0 and from cos21θ=2(α−1)α we have a solution in the range 0<θ<π if θ<2(α−1)α<1 then 2(α−1)>α⇒α>2 if α=2+2 then the above condition is satisfied and θ¨=ag[(1+2)sinθ−(2+2)s∈21θ]
Let particles collide at X after a time t then AX=vt and BX=21t2 By cosine rule in ΔABXv2t2=d2+41a2t4−adt2cosβ i.e. a2t4−4(v2+adcosβ)t2+4d2=0 which has solutions given by t2=2a21[4(v2+adcosβ)±4(v2+adcosβ)2−a2d2] and solutions exist if (v2+adcosβ)2>a2d2
Taking the smaller of the two values BX=21at2=a1[v2+adcosβ−(v2+adcosβ)2−a2d2] ⇒dBX=(cosβ+adv2)−(cosβ+adv2)2−1<1 since x−x2−1<1 for all x hence dBX<1⇒BX<d so P has moved a distance less than d
2003 STEP III questyion 12. I believe thi9s is the last unanswered question for 2003
The number of interruptions in a 10 minute period has a Poisson distribution with mean101 Probability of less than 10 minutes to next interruption is 1−Pr(no interruption in next 10 minutes i.e. 1−e−0.1 so p.d.f of T is f(t)=0.1e−0.1t If first uninterrupted period of 30 minutes is the Nth then the next N−1 uninterrupted periods must each be less than 30 minutes, so Pr(N=n)=(∫030f(t)dt)n−1(∫30∞f(t)dt)=(1−e−3)n−1×e−3 We see that this is the general term of a geometric probability distribution with p=e−3 Hence, E(N)=p1=e3 The number of interruptions before thye first of 30 minutes or more is N−1 so it has expectation e3−1 Density function for periods that are less than 30 minutews apart is kf(t) where k is a constant to scale the area under the graph to unit area, i.e. k=∫030f(t)dt The density function for time between interruiptions that are less than 30 minutes apart
Unparseable latex formula:
\text{will be given by }\dfrac{\text{f}(t)}{\int_0^{30} \text{f}(t)dt}\text{ for }0 \leqt \leq30 \text{ the denominator being required to rescale the area under the curve}
since we are given that t<30. Hence, expected wait before first uninterrupted period of 30 minutes or more is 1−e−3∫0300.1e−0.1tdt=[1−e−3−(t+10)e−0.1t]030=1−e−3−40e−3+10=e3−110(e3−4) hence, expected wait before first uninterrupted period of 30 minutes or more is (e3−1)e3−110(e3−4)=10(e3−4)≈161 minutes
To find the area in the three listed cases, we note that i) and ii) are both specific versions of iii), therefore we need only find the answer to iii) and i) and ii) follow easily.
From a diagram, we see that we must replace d with the perpendicular distance from the line to the furthest point on the circle from the line. This gives us:
d+D=R, where D is the shortest distance from the origin to the line. The distance from a point (x, y) to the centre is given by
D2=(x−a)2+(y−b)2
But the point (x, y) lies on the line y = mx + c, therefore
D2=(x−a)2+(mx+c−b)2
To find the minimum value of D, we must differentiate this expression and set it equal to 0. This gives us
x−a+m2x+mc−mb=0
⟹x=m2+1mb−mc+a
substituting this back into our expression for D^2, we get
D2=(m2+1)2(mb−mc−m2a)2+(ma+c−b)2
and therefore, to find the new area, we must modify (*) such that
d=R−(m2+1)2(mb−mc−m2a)2+(ma+c−b)2
The answer to i) follows by letting m = 0; the answer to ii) follows by letting a = b = 0.
Instead of the integration, could you just use the formulae for an area of a triangle and area of a sector for the first part?