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About me

Introduction:

Hi, my name is Jacan.

I am a Theoretical Physics third year at the University of York. I have had a real excitement for gaining an understanding in science at all levels through my education. This was, however, not always facilitated by the attitudes and teaching styles of my teachers. I therefore truly understand the need for trying to create a tailored experience for individuals, generating a rapport and respecting that there are many ways of expressing the same idea.

How I'd like to help

I feel that it should be my role, as a tutor, to try to present the ideas, on each syllabus that I teach, in a multitude of ways, with a strong emphasis on understanding leading to problem solving, rather than the other way around (as is so often the case in secondary education).

During my A-Levels, I became known in my subjects for having a deep understanding, and so gained a lot of experience in informal tutoring. Notably, there were a few occasions during lessons in which teachers asked me to explain and clarify a topic that they were struggling with, themselves, to the class.

Despite this, I understand what it means to be struggling during studies. At AS, I was extremely disappointed to achieve an E grade in Mathematics. With little external support and belief in my abilities, I set about improving this grade by finding techniques to learning that suited me. This occasionally flew in the face of traditional learning styles, but I ended up achieving an A* in Mathematics, achieving 97% of the available UMS marks across AS & A2. I feel that this demonstrates an understanding in the difficulties of struggling students, but also in the steps in devising a system which can even overcome the limiting notion of a student’s “potential” as put upon them by their school.

While mentoring students in learning strategies as individuals is not formally the aim of my tutorials, I hope to help passively establish the tools required to independently approach problems in novel ways, allowing each student to find the techniques in problem-solving for Maths, Physics and Chemistry which suit them.

Schedule

My schedule is often flexible. During term-time I would be comfortable giving up to 5-6 tutorials per week, and over the holidays up to 16. The majority of my available time for tutorials will be over the afternoon and evening.

If you have any questions, please don’t hesitate to contact me or set up a free “Meet the Tutor Session” and I’ll get back to you as soon as I can and be happy to clarify.

I very much look forward to hearing from you,

Have a lovely day,

Jacan.

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £30 /hr
Maths A Level £30 /hr
Physics A Level £30 /hr
Chemistry GCSE £30 /hr
Further Mathematics GCSE £30 /hr
Maths GCSE £30 /hr
Physics GCSE £30 /hr
Science GCSE £30 /hr

Qualifications

QualificationLevelGrade
PhysicsA-LevelA*
MathematicsA-LevelA*
ChemistryA-LevelA
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

General Availability

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Please get in touch for more detailed availability

Ratings and reviews

4.9from 41 customer reviews

Aziz (Parent) May 29 2016

Jacan is very perceptive and inspirational.

Aziz (Parent) May 22 2016

Jacan thought my son for the first time. His feedback to me was very useful.

Emilie (Student) May 22 2016

Very helpful :)

Declan (Student) May 7 2016

covered everything in good depth and helped answer questions to do with the subject.
See all reviews

Questions Jacan has answered

How do I integrate ln(x)?

There is a subtle, but very neat trick to this when applying the rules of integration by parts. If we take ∫ln(x)dx = ∫1*ln(x)dx, and then let our term to be differentiated, u = ln(x), and our term to be integrated, dv/dx = 1, then it follows that:   du/dx = x⁻¹, v = x   and from the integ...

There is a subtle, but very neat trick to this when applying the rules of integration by parts.

If we take ∫ln(x)dx = ∫1*ln(x)dx, and then let our term to be differentiated, u = ln(x), and our term to be integrated, dv/dx = 1, then it follows that:

 

du/dx = x⁻¹, v = x

 

and from the integration by parts formula:

∫u * (dv/dx) dx = uv - ∫v * (du/dx) dx

 

∴ ∫ln(x)dx = xln(x) - ∫(x⁻¹ * x)dx (+ constant)

∫ln(x)dx = xln(x) - ∫dx (+ constant)

Hence, our results turns out to be:

∫ln(x)dx = xln(x) - x + c

 

NB. While our trick here gives us a very straightforward solution to an integration which could have been very laborious via other methods, integration by parts tends to be a last resort, as more, seemingly contrived steps are required. One should generally try integration by substitution, for non-standard integrations, first when unsure of which method to use, as the steps to a result are often far simpler and quicker.

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2 years ago

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