James F. A Level Maths tutor

James F.

Currently unavailable: for regular students

Degree: Economics (Bachelors) - Nottingham University

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About me

Hi there everyone!

My name is James and I'm a tutor for A-Level Maths. I have loved maths ever since I was young, but realise how daunting it can be for some, and would love to try to transfer some of my own methods and reasonings to help simplify the work and make it more accessible.

I am extremely friendly, commited and patient. Whatever topics or questions you are struggling with you can feel free to send me and I will happily look at them ready to explain, and believe I would be the perfect tutor to help to build your confidence in the subject.

I have a wealth of experience in part-time jobs and voluntary roles meaning I am good at communicating effectively and building relationships with all types of people. Having completed A-Levels in the past year myself, my understanding of the demands, the stress, the syllabus and the most efficient revision technqiues for success is fresh and expansive.

Feel free to contact me for any information or to set up a meeting to see that I am right for you!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr

Qualifications

QualificationLevelGrade
MathematicsA-LevelA*
BiologyA-LevelA*
English LiteratureA-LevelB
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CRB/DBS Standard

No

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No

Currently unavailable: for regular students

Questions James has answered

Find the cartesian equation of a curve?

A curve has parametric equations: x = 2 + t2                           y = 4t Find the cartesian equation of this curve. A cartesian equation of a curve is simply finding the single equation of this curve in a standard form where xs and ys are the only variables.  To find this equation, you ...

A curve has parametric equations:

x = 2 + t2                           y = 4t

Find the cartesian equation of this curve.

A cartesian equation of a curve is simply finding the single equation of this curve in a standard form where xs and ys are the only variables. 

To find this equation, you need to solve the parametric equations simultaneously:

If y = 4t, then divide both sides by 4 to find (1/4)y = t.

This newly found value of t can be substituted into the equation for x:

x = 2 + (1/4(y))2 - expand the bracket (square both 1/4 and y) to derive x = 2 + 1/16 y2

Technically, this final equation is already in cartesian form as it only includes variables x and y, however to further rearrange the equation to find the standard 'y =' form:

x = 2 + 1/16 y2 (minus 2 from both sides)

x - 2 = 1/16 y2 (multiply each side by 16)

16x - 32 = y2    (and finally take square roots of both sides)

y = SQRT(16x-32)

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10 months ago

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Find the coordinates of the minimum point of the curve y = 3x^(2) + 9x + 10

(Note: in this answer I will use the notation y' to represent the differential of y, and by the same reasoning y'' denotes the differential of y', or the second derivative of y.) A curve's maximum or minimum will be found at the stationary point of the curve (for a continuous function), where...

(Note: in this answer I will use the notation y' to represent the differential of y, and by the same reasoning y'' denotes the differential of y', or the second derivative of y.)

A curve's maximum or minimum will be found at the stationary point of the curve (for a continuous function), where the gradient of the curve is flat and equal to zero.

We can find the gradient of a curve by differentiation, where the first derivative is found by the rule of 'bring the power down and minus 1 from the power.' Formally, this can be written as if y = Axn, then the differential of y with respect to x is y' = nAxn-1. Consider the curve y = 3x2 + 9x + 10 to see how this works in practice, first including the equation's hidden powers for more clarity:

y = 3x+ 9x+ 10x0

y' = (2)(3)x2-1 + (1)(9)x1-1 + (0)(10)x0-1

= y' = 6x + 9, as x= 1, and the last term disappears as it is multiplied by zero, constants (terms with no 'x' included) will always drop out once differentiated.

As stated earlier, the maximum or minimum is found where the curve is stationary, and the gradient is equal to zero, and so we set our newly derived gradient equal to zero, and rearrange the equation to find the value of x at this point:

6x + 9 = 0 (minus 9 from both sides)

6x = -9 (divide both sides by 6)

x = -9/6 = -3/2, and so the curve is stationary at the point where x is = to -3/2. What is the value of y at this point? Simply substitute this newly found x value into the original equation of the curve:

y = 3(-3/2)+ 9(-3/2) + 10 = (27/4) - (27/2) + 10 = 13/4. So there is a stationary point at (-3/2, 13/4).

Finally, in order to ensure that this point is a minimum point, we need to perform the second derivative test. This test states that we must differentiate y again (differentiate y'), if:

y'' > 0, then the curve is convex and minimised at that point.

y'' < 0, then the curve is concave and maximised at that point.

Therefore, lets differentiate y' = 6x + 9 again:

y'' = 6 which is > 0 and so the curve is in fact minimised, thus we can conclude that the minimum point of the curve is found at the point (-3/2, 13/4)

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11 months ago

451 views
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