Rafe L.

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Degree: Mathematics (Masters) - Warwick University

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I am a maths student at the University of Warwick. From a young age I have always l been passionate about maths and I thoroughly enjoy teaching and helping others.

I have volunteered as a football coach and as a maths tutor at school so I have a lot of experience teaching and mentoring.

The Sessions:

During our sessions, it will be up to you as to what we cover. With maths, understanding the basic concepts behind everything is key and so we will cover this before tackling exam questions.

Different people learn in different ways and I will use a variety of ways of explaining a concept untill we find what suites you best.

Maths, believe it or not, can be fun! I'll try to make all our sessions fun and enjoyable as well as productive and beneficial.

What happens next?

If you have any questions, send me a direct message or book a 'Meet the Tutor Session'! I would appreciate it if you told me what area you are struggling with and your exam board.

I look forward to meeting you!

#### Subjects offered

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

#### Qualifications

MathematicsA-LevelA*
Further MathematicsA-LevelA*
PhysicsA-LevelA*
ChemistryA-LevelA
 CRB/DBS Standard No CRB/DBS Enhanced No

### How to find the equation of a tangent to a curve at a specific point.

The first thing to remember is that the tangent to the curve at a given point has the same gradient as the curve at that point. Let's consider the general case y=f(x) at a point c=(x1,y1). Step 1:We have to differentiate to find dy/dx. Step 2: Calculate dy/dx when x=x1. This is the gradient a...

The first thing to remember is that the tangent to the curve at a given point has the same gradient as the curve at that point.

Let's consider the general case y=f(x) at a point c=(x1,y1).

Step 1:We have to differentiate to find dy/dx.

Step 2: Calculate dy/dx when x=x1. This is the gradient at the point c.

Step 3: Use the equation of a general straight line with gradient m at a point c 'y-y1=m(x-x1)', where m is the value of the gradient calculated in Step 2.

Here is an example. Find the equation of the tangent to the curve y=x^3-4x at the point (1,-3).

In this example x1=1 and y1=-3

Step 1: dy/dx=3x^2-4

Step 2: When x=1, dy/dx =-1, so m=-1

Step 3: Use the equation y-y1=m(x-x1) to obtain y--3=-1(x-1)

We can then rearrange this to the nicer form of y=-x-2.

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2 years ago

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### Integrating cos^2(x)+5sin^2(x)

Firstly, note that cos^2(x)+5sin^2(x)= cos^2(x) +sin^2(x) +4sin^2(x). By trignoemtric identies, cos^2(x)+sin^2(x)=1 and so we can just integrate 1+4sin^2(x) since this is equal to cos^2(x)+5sin^2(x). Again, by trignometric identities, 4sin^2(x)=4(1/2-1/2 cos(2x))=2-2cos(2x), and so 1+4sin^2(x...

Firstly, note that cos^2(x)+5sin^2(x)= cos^2(x) +sin^2(x) +4sin^2(x).

By trignoemtric identies, cos^2(x)+sin^2(x)=1 and so we can just integrate 1+4sin^2(x) since this is equal to cos^2(x)+5sin^2(x).

Again, by trignometric identities, 4sin^2(x)=4(1/2-1/2 cos(2x))=2-2cos(2x),

and so 1+4sin^2(x)=3-2cos(2x).

We can now integrate this much more easily...

3 integrates to 3x and -2cos(2x) integrates to -sin(2x).

Hence the integral, remembering the constant of integration, is...

3x -sin(2x) +c

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2 years ago

537 views
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