Hello there! My names Ben, I'm a third year maths student at Durham University, but this year I'm actually studying at the University of Neuchâtel in Switzerland, attempting to do some maths in French! I'm originally from Sheffield, and next year I'll be returning to Durham to do a Masters :)
I have a real passion for Mathematics, and I'm really enthusiastic about helping and encouraging others to enjoy Maths, whether that's purely exam based or a more just-for-fun attitude. I've been quite successful academically, I was fortunate enough to achieve 14A* at GCSE, including an A* with Distinction at Further Maths, I got 3 A* in Maths, Further Maths and Physics at A Level and I got two AS levels at A grades in Music and French. I also sat some extra maths exams, I got a merit in AEA, a grade 1 in STEP II and grade 3 in STEP III (the latter meaning I just missed out on a Cambridge offer).
Although I do really enjoy studying, I'm not just an introverted bookworm! I play loads of sports and enjoying going out with friends a lot, I think I can be a good communicator and I don't take myself too seriously :) I've had a bit of tutoring experience so if you need a hand with Maths, I'd be more than happy to give you any help I can, and hopefully we can have a laugh and enjoy it along the way! :)
|Further Mathematics||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
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Sulayman (Student) December 1 2016
Sulayman (Student) November 27 2016
Marcin (Parent) November 20 2016
Ben (Student) November 6 2016
a is acceleration, v is velocity, s is position
a = - g
v = u - gt (by integrating with respect to t and setting initial speed u as constant of integration)
s = ut -gt2/2 + s0 (by integrating again and setting initial position s0 as constant of integration, although object if usually projected from origin so s0 = 0)
a = 0
v = u (constant velocity as acceleration is zero)
s = ut + s0 (again s0 usually 0)
so for an object projected from the origin at speed u at an angle θ from the horizontal, the initial speed in the x direction is ucos(θ) and usin(θ) in the y direction
thus sx = utcos(θ) and sy = utsin(θ) - gt2/2
flight time: this is the value of t when sy returns to zero
0 = utsin(θ) - gt2/2
so either t = 0 (at launch) or usin(θ) - gt/2 = 0
=> flight time is t = 2usin(θ)/g
flight distance: this is the value of sx when sy = 0
when sy = 0, t = 2usin(θ)/g
so sx = utcos(θ) = 2u2sin(θ)cos(θ)/g = u2sin(2θ)/g (as sin(2θ) = 2sin(θ)cos(θ))
=> flight distance is x = u2sin(2θ)/g
this is when sy is at a stationary point, ie. when dsy/dt = 0. This is also when vy = 0.
vy = usin(θ) - gt = 0
=> t = usin(θ)/g
=> sy = utsin(θ) - gt2/2 = u2sin2(θ)/g - u2sin2(θ)/2g = u2sin2(θ)/2g
=> greatest height is h = u2sin2(θ)/2g
Integration by parts states that ∫uv' dx = uv - ∫u'v dx, where u & v are funcions of x and the notation u' means du/dx.
to do integration by parts given an integral ∫f(x) dx, it involves writing f(x) as f(x) = u(x)v'(x), and then following the formula by determining u' and v.
The whole point of IBP is that ∫u'v dx is hopefully easier to integrate than ∫uv' dx.
eg(1) take ∫ xsin(x) dx. This is quite hard to integrate directly, so we use integration by parts. When choosing which is u(x) and v'(x), remember that you will have to integrate v' and differentiate u later. Often what happens is when you differentiate u, u'(x) turns out to be 1, which leaves you with a simple integration.
so take u(x) =x and v'(x)=sin(x)
=>u' =1, v = -cos(x) (don't worry about the "+c", it's included at the end)
so following the formula: ∫xsin(x)dx = -xcos(x) - ∫-cos(x) = xcos(x) + ∫cos(x)dx = xcos(x) + sin(x) + c
this is the final answer to that particular question, and we see that integration by parts gives us another method of integtration
1) for ∫ f(x) dx, choose suitable functions u(x) and v'(x) such that f(x) = u(x) * v'(x).
2) determine u'(x) and v(x) by differentiating and integrating respectively
3) use the formula ∫uv' dx = uv - ∫u'v dx to find the answer!
DERIVATION: (not usually necessary for exam but interesting to see!)
differentiation by parts works like this, for, u & v as functions of x,
where u' = du/dx etc.
d/dx(uv) = (uv)' = uv' + u'v (proof omitted)
if we integrate both sides wrt x
=> uv = ∫uv' dx + ∫u'v dx
=> ∫uv' dx = uv - ∫u'v dx