Simon T. GCSE Further Mathematics  tutor, A Level Further Mathematics...

Simon T.

£22 - £24 /hr

Natural Sciences Tripos (Masters) - Cambridge University

5.0
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12 reviews

Trusted by schools

This tutor is also part of our Schools Programme. They are trusted by teachers to deliver high-quality 1:1 tuition that complements the school curriculum.

102 completed lessons

About me

Hey, I'm a third year undergraduate at Cambridge studying Natural Science, specialising in Astrophysics. 

I hope my tutoring sessions will be useful and fun.

I enjoy teaching and have been involved in outreach events for my university, demonstrating physics to lower secondary age pupils. Last summer I worked as a researcher in seismology


Hey, I'm a third year undergraduate at Cambridge studying Natural Science, specialising in Astrophysics. 

I hope my tutoring sessions will be useful and fun.

I enjoy teaching and have been involved in outreach events for my university, demonstrating physics to lower secondary age pupils. Last summer I worked as a researcher in seismology


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About my sessions

The 55 minute supervisions will be catered entirely for for the students needs: whether that carefully establishing a firm grasp of the key concepts, or working thorugh the rigours of problems that might be set in exams.

The 55 minute supervisions will be catered entirely for for the students needs: whether that carefully establishing a firm grasp of the key concepts, or working thorugh the rigours of problems that might be set in exams.

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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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28/06/2017

Ratings & Reviews

5from 12 customer reviews
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Joanna (Student)

April 4 2018

Great

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Holly (Student)

August 12 2016

Thanks that was a really good lesson, you have explained things very clearly and it makes it easy to understand and Chemistry is starting to make sense.

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Holly (Student)

August 16 2016

Thanks so much for your help, I really think I am beginning to understand a bit better :).

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Betina (Student)

June 26 2016

The tutorial was great! I'm looking forward to the next one!

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Qualifications

SubjectQualificationGrade
Physics Higher LevelInternational Baccalaureate (IB) (HL)7
Chemistry Higher LevelInternational Baccalaureate (IB) (HL)7
Mathematics Higher LevelInternational Baccalaureate (IB) (HL)6
English Standard LevelInternational Baccalaureate (IB) (SL)6
German Standard LevelInternational Baccalaureate (IB) (SL)6
Geography Standard LevelInternational Baccalaureate (IB) (SL)7

General Availability

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

Subjects offered

SubjectQualificationPrices
PhysicsA Level£24 /hr
Further MathematicsGCSE£22 /hr
PhysicsGCSE£22 /hr
ChemistryIB£24 /hr
MathsIB£24 /hr
PhysicsIB£24 /hr
ScienceIB£24 /hr
-Personal Statements-Mentoring£24 /hr

Questions Simon has answered

Based on Newton's 3 laws of motion why is linear momentum always conserved?

This is a common IB question- fairly tricky the first time you com across it!

This is a common IB question- fairly tricky the first time you com across it!

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2 years ago

811 views

Take the square root of 2i

As with much of complex number the trick here is to change forms to polar representation.

If you think of an argand diagram the number i will be represented as a point straight up on the imaginary axis a distance 2 from the origin.

It can therefore be represented as 2i = 2*e^(iπ/2)

From here it's easy! Just  apply the same indices rules that you have grown so familiar with. 

2 goes to the square root of 2, e^(iπ/2) goes to e^(iπ/4).

so we have the expression (2i)^(1/2) = (2)^(1/2)*(iπ/4)

And now convert back to standard form!

We know the magnitude is square root 2, and the arguement is π/4. Imagined on the argand diagram this is a line slanting at 45 degrees to the horizontal.

We can use the identity e^(iθ) =cos(θ) + i*sin(θ)

cos(π/4)=sin(π/4)= 2^(-1/2)

Thankfully the square roots of 2 cancel (Careful! they will not allways do this!) Therefore we reach the answer:

(2i)^(1/2) = 1 + i

which is satisfyingly elegant

As with much of complex number the trick here is to change forms to polar representation.

If you think of an argand diagram the number i will be represented as a point straight up on the imaginary axis a distance 2 from the origin.

It can therefore be represented as 2i = 2*e^(iπ/2)

From here it's easy! Just  apply the same indices rules that you have grown so familiar with. 

2 goes to the square root of 2, e^(iπ/2) goes to e^(iπ/4).

so we have the expression (2i)^(1/2) = (2)^(1/2)*(iπ/4)

And now convert back to standard form!

We know the magnitude is square root 2, and the arguement is π/4. Imagined on the argand diagram this is a line slanting at 45 degrees to the horizontal.

We can use the identity e^(iθ) =cos(θ) + i*sin(θ)

cos(π/4)=sin(π/4)= 2^(-1/2)

Thankfully the square roots of 2 cancel (Careful! they will not allways do this!) Therefore we reach the answer:

(2i)^(1/2) = 1 + i

which is satisfyingly elegant

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2 years ago

1727 views

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