PremiumJames Y. A Level Maths tutor, GCSE Maths tutor, A Level Further Mathe...

James Y.

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Degree: MPhys: Physics with North American Study (Masters) - Exeter University

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About me

I'm James, a physics student at the University of Exeter. I'm currently in the final year my four year undergraduate master's degree, having completed a year abroad at Iowa State University in the United States. While in America I tutored my roommate in chemistry and maths, so I have a little experience in teaching. 

Having received tutoring for A-levels, I can honestly say that it helps tremendously with understanding difficult or challenging subject matter (particularly physics), which served to make me more interested in the subjects.

During the sessions, you will decide what is covered- this was something my tutors did with me, and it means that I don't end up wasting your time by teaching you something you already know. Please also let me know your exam board as well as whichever topics you're having issues with. (Note: For Maths/Further Maths A-level I did C1-4, S1-3, M1-3 and FP1-2, as such I will not be able to tutor any Decision modules).

If you have any questions or concerns, feel free to send me a message or book a session, I'm looking forward to helping you!

Subjects offered

SubjectLevelMy prices
Further Mathematics A Level £20 /hr
Maths A Level £20 /hr
Physics A Level £20 /hr
Chemistry GCSE £18 /hr
Further Mathematics GCSE £18 /hr
Maths GCSE £18 /hr
Physics GCSE £18 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA
Further MathematicsA-LevelA
PhysicsA-LevelB
ChemistryA-LevelB
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Questions James has answered

The curve C has the equation y = 2x^2 -11x + 13. Find the equation of the tangent to C at the point P (2, -1).

The first step is to differentiate the equation of the curve in order to find the gradient of the tangent at the curve. Remember that when differentiating polynomials, we multiply the index of the variable x, by its coefficient, then subtract 1 from the index. In addition, remember that x0 = 1...

The first step is to differentiate the equation of the curve in order to find the gradient of the tangent at the curve. Remember that when differentiating polynomials, we multiply the index of the variable x, by its coefficient, then subtract 1 from the index. In addition, remember that x0 = 1.

In this case, dy/dx = 4x - 11.

Now if we plug in the x-coordinate of P (2) into dy/dx, we will get the gradient of the tangent to the curve at P.

dy/dx = 4(2) - 11

dy/dx = 8 - 11

dy/dx = -3.

Now we find the equation of the tangent using the formula for the equation of a straight line, and plugging in the coordinates of P:

y - y= m(x - x1)

y - (-1) = -3(x - 2)

y + 1 = -3x +6

3x + y - 5 = 0.

This is the equation of the tangent to the curve C at P.

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3 months ago

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