Daniel W. A Level Maths tutor, GCSE Maths tutor, GCSE Chemistry tutor

Daniel W.

£22 - £24 /hr

Currently unavailable: for new students

Studying: Chemistry and Maths (Bachelors) - Leeds University

5.0
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32 reviews| 34 completed tutorials

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About me

Me, Myself & I I am studying Chemistry and Maths at the University of Leeds and I am about to go into the second year of my degree. From quite an early age I realised Maths and Science are where my interests lie.  I volunteered as a Young Leader in Scouts and Beavers for 2 years so I have a lot of experience of working with children, some as young as 5. Having had to organise a large group of 5-7 year olds taught me the importance in pre planning so every session we have, I will be fully prepared to answer any questions you may have. Tutoring During the tutoring sessions you will have complete control over what we cover, whether its revision over a large topic or something very specific, preperation for a test or help with a difficult piece of homework, no matter the problem by the end of the hour we will have the problem solved and hopefully you will have a good enough understanding of the problem to then teach someone else.  Now what? If you have any questions regarding anything you have just read or how the whole tutoring process works you can contact me on here via a book the tutor session which is completely free and gives you a chance to decided if I will be able to help you out. Hope to hear from you,  DanielMe, Myself & I I am studying Chemistry and Maths at the University of Leeds and I am about to go into the second year of my degree. From quite an early age I realised Maths and Science are where my interests lie.  I volunteered as a Young Leader in Scouts and Beavers for 2 years so I have a lot of experience of working with children, some as young as 5. Having had to organise a large group of 5-7 year olds taught me the importance in pre planning so every session we have, I will be fully prepared to answer any questions you may have. Tutoring During the tutoring sessions you will have complete control over what we cover, whether its revision over a large topic or something very specific, preperation for a test or help with a difficult piece of homework, no matter the problem by the end of the hour we will have the problem solved and hopefully you will have a good enough understanding of the problem to then teach someone else.  Now what? If you have any questions regarding anything you have just read or how the whole tutoring process works you can contact me on here via a book the tutor session which is completely free and gives you a chance to decided if I will be able to help you out. Hope to hear from you,  Daniel

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Ratings & Reviews

5from 32 customer reviews
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Paul (Parent)

January 31 2017

Excellent guidance and support given.

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Paul (Parent)

November 17 2016

Very helpful

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Dylan (Student)

October 19 2016

Excellent tutor, he has a clear concise and informative way of communication in lesson and puts a lot of time into organising the session itself. I have had an easy time learning things from him in AS Mathematics and therefore caught up with the classwork.

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Kevin (Student)

June 22 2017

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Qualifications

SubjectQualificationGrade
ChemistryA-level (A2)A
MathsA-level (A2)A
BiologyA-level (A2)A

General Availability

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

Subjects offered

SubjectQualificationPrices
ChemistryA Level£24 /hr
MathsA Level£24 /hr
ChemistryGCSE£22 /hr
MathsGCSE£22 /hr

Questions Daniel has answered

Find the equation of the tangent to the curve y=x^3-4x^2+2 at the point (3,-7)

y=x3-4x2+2

A tangent to a curve at a specific point along the line will have excatly the same gradient as the curve at that point. For example if the point was (0,0) the tangent would just be a horizontal line along the x-axis. To work out the gradient we simply differentiate the curve, set this to equal 0 and solve with the given value of x. 

Here dy/dx = 3x2-8x

Setting this to 0 and solving gives us 3(3)2-8(3)=3 and so our gradient at the point (3,-7) is 3

We are now able to use the equation y-y(1)=m(x-x(1)) where m is the gradient and x(1) & y(1) are the coordinates we've been given. 

Rearranging we get y=3(x-3)-7

y=3x-16 and this is the equation of the tangent to the curve y=x3-4x2+2 at the point (3,-7)

y=x3-4x2+2

A tangent to a curve at a specific point along the line will have excatly the same gradient as the curve at that point. For example if the point was (0,0) the tangent would just be a horizontal line along the x-axis. To work out the gradient we simply differentiate the curve, set this to equal 0 and solve with the given value of x. 

Here dy/dx = 3x2-8x

Setting this to 0 and solving gives us 3(3)2-8(3)=3 and so our gradient at the point (3,-7) is 3

We are now able to use the equation y-y(1)=m(x-x(1)) where m is the gradient and x(1) & y(1) are the coordinates we've been given. 

Rearranging we get y=3(x-3)-7

y=3x-16 and this is the equation of the tangent to the curve y=x3-4x2+2 at the point (3,-7)

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1 year ago

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