Jacan C. A Level Maths tutor, A Level Physics tutor, GCSE Physics tut...

Jacan C.

£36 /hr

Studying: Theoretical Physics (Masters) - York University

4.9
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71 reviews| 192 completed tutorials

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About me

At AS, I was extremely disappointed to achieve an E grade in maths. 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 maths, achieving 97% of the available UMS marks across AS and A2. My explanatory skills have been widely celebrated by my tutees, peers (over the course of my secondary and undergraduate education), and my professors - recently scoring 93% on an assessment of my ability to communicate undergraduate level topics.

At AS, I was extremely disappointed to achieve an E grade in maths. 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 maths, achieving 97% of the available UMS marks across AS and A2. My explanatory skills have been widely celebrated by my tutees, peers (over the course of my secondary and undergraduate education), and my professors - recently scoring 93% on an assessment of my ability to communicate undergraduate level topics.

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About my sessions

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. I request that all tutees send me an email at least 18 hours prior to our tutorials on the given topic to cover, in order that I can tailor information to the needs of individual exam boards/syllabuses.

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. I request that all tutees send me an email at least 18 hours prior to our tutorials on the given topic to cover, in order that I can tailor information to the needs of individual exam boards/syllabuses.

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Ratings & Reviews

4.9from 71 customer reviews
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Marion (Parent)

October 11 2017

is good

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Marion (Parent)

August 1 2017

is good

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Marion (Parent)

May 4 2017

is good and clear

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Marion (Parent)

March 6 2017

he is good and helpful

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Qualifications

SubjectQualificationGrade
PhysicsA-level (A2)A*
MathematicsA-level (A2)A*
ChemistryA-level (A2)A
Theoretical PhysicsDegree (Bachelors)FIRST CLASS

General Availability

Before 12pm12pm - 5pmAfter 5pm
mondays
tuesdays
wednesdays
thursdays
fridays
saturdays
sundays

Subjects offered

SubjectQualificationPrices
Further Mathematics A Level£36 /hr
MathsA Level£36 /hr
PhysicsA Level£36 /hr
ChemistryGCSE£36 /hr
Further Mathematics GCSE£36 /hr
MathsGCSE£36 /hr
PhysicsGCSE£36 /hr
ScienceGCSE£36 /hr

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 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.

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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3 years ago

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