Yaniv P. GCSE Maths tutor, GCSE Physics tutor, GCSE Further Mathemati...

Yaniv P.

Currently unavailable: until 22/12/2015

Degree: Mathematics (Masters) - Bristol University

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About me

I am a 1st year Mathematics undergraduate at the University of Bristol

I am an experienced tutor of Maths at Primary, Secondary, and A-level, as well as Further Maths at A-level. I also have a long history of tutoring through school programs, in Physics, Maths, and Further maths across all levels, and am a common source of aid for my little sister in her studies.

In addition to this, my years as a black belt in Karate have had me teach lower belts of all ages and skills, and have equipped me with patience and the ability to guide my students.

Tutoring a student and seeing them succeed is an incredibly satisfying feeling - to know that you were able to truly help them is fantastic, and that is why I have chosen to work on this site.

In mathematics, particularly, I am able to bring around the context and links that various methods have with other fields of mathematics. I apply this to my tutoring - a true understanding of maths and physics can only be had by knowing how to use an idea, and how it joins with other fields or concepts. This understanding is the only way to succeed in the sciences, and the moment it is gained is always excellent.

A typical session will consist of getting to grips with the areas for improvement of a student, and proceeding from there. Guided study of course material, and frequent student participation are essential - mathematics and physics are not subjects that can be learned by wrote, and the teaching of both hinges on exercises. Depending on urgency or student understanding, I may incorperate the study of past papers, or papers based off of their course content, that I have readied. But above all, I will focus on basic technique and knowledge - this is often the point of failing for most students, where a single basic principle is not understood, and the rest topple around it like a house of cards. So, I will always fall back into this focus - how one idea is set around another.

Subjects offered

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

Qualifications

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

CRB/DBS Standard

No

CRB/DBS Enhanced

09/02/2015

Currently unavailable: until

22/12/2015

Questions Yaniv has answered

Finding the derivative of a polynomial.

Take any polynomial, eg/ y=x3+1/2x2-3x+9. Then dy/dx=3x,+x-3, in this case. This is because, when deriving in this sense, you take each term in x, multiply it by its index, and reduce that index by 1. In a general sense, for y=(n0)xn+(n1)xn-1+...+(nn-1)xn-(n-1)+(nn),             dy/dx=(n)(n0)x...

Take any polynomial, eg/ y=x3+1/2x2-3x+9. Then dy/dx=3x,+x-3, in this case. This is because, when deriving in this sense, you take each term in x, multiply it by its index, and reduce that index by 1.

In a general sense, for y=(n0)xn+(n1)xn-1+...+(nn-1)xn-(n-1)+(nn),             dy/dx=(n)(n0)xn-1+(n-1)(n1)xn-2+...+(n-(n-1))(nn-1). Multiply the x term by the power, reduce the power by one. This works for all powers, even non-integers.

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2 years ago

372 views

How do you invert a 2x2 matrix?

Take a matrix A=(acbd), where a,b,c,d are numbers. First find the determinant. This is a*d-b*c. Now, rearrange the matrix to become (d-c-ba). Divide this by the determinant, to have A-1=1/(ad-bc)(d-c-ba). This is the inverse of A.

Take a matrix A=(acbd), where a,b,c,d are numbers.

First find the determinant. This is a*d-b*c.

Now, rearrange the matrix to become (d-c-ba). Divide this by the determinant, to have A-1=1/(ad-bc)(d-c-ba). This is the inverse of A.

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2 years ago

357 views

How do you solve the integral of ln(x)

This will use the process of integration by parts. First, notice that ln(x)=ln(x)*1. So, the integral of ln(x) is the integral of ln(x)*1. The process of integration by parts is;  int(v*du/dx)dx=vu - int(dv/dx*u)dx. Set ln(x)=v, 1=du/dx, so int(ln(x)*1)dx = ln(x)*x - int(1/x*x)dx = x*ln(x)-i...

This will use the process of integration by parts.

First, notice that ln(x)=ln(x)*1.

So, the integral of ln(x) is the integral of ln(x)*1. The process of integration by parts is;  int(v*du/dx)dx=vu - int(dv/dx*u)dx.

Set ln(x)=v, 1=du/dx, so int(ln(x)*1)dx = ln(x)*x - int(1/x*x)dx = x*ln(x)-int(1)dx = x*ln(x)-x+constant.

And you're done!

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2 years ago

307 views
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