PremiumBen J. A Level Maths tutor, A Level Further Mathematics  tutor

Ben J.

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Mathematics (Integrated Masters) - Durham University

5.0
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157 reviews

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204 completed lessons

About me

Hello there! My names Ben, I'm originally from Sheffield and this year I'm doing an integrated Master's in Mathematics at Durham University :) I spent my first two years at Durham and then studied abroad in Switzerland last year, attempting to do some maths in French! 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. Although I do really enjoy studying, I'm not just an introverted bookworm! I play lots of sports and music 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 lots 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! :)

Hello there! My names Ben, I'm originally from Sheffield and this year I'm doing an integrated Master's in Mathematics at Durham University :) I spent my first two years at Durham and then studied abroad in Switzerland last year, attempting to do some maths in French! 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. Although I do really enjoy studying, I'm not just an introverted bookworm! I play lots of sports and music 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 lots 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! :)

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About my sessions

I like to let the student take the lead, I'll try to listen and understand your aims and goals and discuss with you what the most effective way of learning would be. We're all different, and everybody has slightly different ways of communicating and learning! I consider myself to be quite flexible in this regard; I'm happy to go into depth explaining the theory, or focusing on questions and exam technique. I enjoy preparing for tutorials and making revision guides, or if you'd prefer to fire unseen questions at me and we work through them together that works to! The whole point of tutoring is to listen to what your goals are, and to find ways together of making sure you can achieve them! Thanks, Ben

I like to let the student take the lead, I'll try to listen and understand your aims and goals and discuss with you what the most effective way of learning would be. We're all different, and everybody has slightly different ways of communicating and learning! I consider myself to be quite flexible in this regard; I'm happy to go into depth explaining the theory, or focusing on questions and exam technique. I enjoy preparing for tutorials and making revision guides, or if you'd prefer to fire unseen questions at me and we work through them together that works to! The whole point of tutoring is to listen to what your goals are, and to find ways together of making sure you can achieve them! Thanks, Ben

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Ratings & Reviews

5from 157 customer reviews
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Hanna (Parent from Bristol)

September 27 2017

Good tutorial :)

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V (Parent from London)

September 16 2017

First session today but my son felt really positive about working with Ben. We will be arranging weekly sessions! Thank you Ben!

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Jung (Student)

March 16 2017

Super friendly and encouraging! Thank you for being patient and for making integration less daunting :)

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Sulayman (Student)

December 1 2016

Enjoyed the lesson. Thanks!

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
PhysicsA-level (A2)A*
STEP IIUni admission test1
STEP IIIUni admission test3

General Availability

Pre 12pm12-5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further MathematicsA Level£24 /hr
MathsA Level£24 /hr

Questions Ben has answered

how to find flight time/distance and greatest hight of projectiles?

is acceleration, v is velocity, is position

vertically:

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 s= 0)

horizontally:

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

greatest hight:

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


:)

is acceleration, v is velocity, is position

vertically:

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 s= 0)

horizontally:

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

greatest hight:

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


:)

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3 years ago

950 views

What's the deal with Integration by Parts?

EXPLANATION:

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

METHOD:

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

:)

EXPLANATION:

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

METHOD:

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

:)

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3 years ago

962 views

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