Wesley M.

Wesley M.

£40 /hr

Mathematics (Masters) - Bristol University

37 completed lessons

About me

I am a current Mathematics student at The University of Bristol. Maths can be a love-hate sort of subject for many people, but with my absolute love and enthusiasm for both maths itself and the idea of inspiring others, I hope to be able to change that for the better. I'm a friendly and outgoing person, who aims to inspire prospective pupils to not only enjoy their tutor sessions, but to look forward to them, giving them a positive outlook on maths. I believe that this is the way to achieve the best things from the subject. As i am currently just a first year in my course, I have fresh experience of what it's like to be at the point in school life that my pupils are at. Teaching and leading children as young as 5 right through to 18 comes as second nature, through my experiences in summer schemes, or from my elected school roles as Deputy Head Boy, First Year Mentor, and Captain of Debating Society. I look forward to meeting you soon!

I am a current Mathematics student at The University of Bristol. Maths can be a love-hate sort of subject for many people, but with my absolute love and enthusiasm for both maths itself and the idea of inspiring others, I hope to be able to change that for the better. I'm a friendly and outgoing person, who aims to inspire prospective pupils to not only enjoy their tutor sessions, but to look forward to them, giving them a positive outlook on maths. I believe that this is the way to achieve the best things from the subject. As i am currently just a first year in my course, I have fresh experience of what it's like to be at the point in school life that my pupils are at. Teaching and leading children as young as 5 right through to 18 comes as second nature, through my experiences in summer schemes, or from my elected school roles as Deputy Head Boy, First Year Mentor, and Captain of Debating Society. I look forward to meeting you soon!

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About my sessions

I think structure for tutorials is quite student-specific, and so I will try and figure out what methods work best for each individual, and tailor the tutorials towards that. I believe in a teaching style based on positive reinforcement, and trying to build an eagerness within the student to ask questions, as I think that this helps with creating a broad understanding of the topics at hand.

I think structure for tutorials is quite student-specific, and so I will try and figure out what methods work best for each individual, and tailor the tutorials towards that. I believe in a teaching style based on positive reinforcement, and trying to build an eagerness within the student to ask questions, as I think that this helps with creating a broad understanding of the topics at hand.

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Personally interviewed by MyTutor

We only take tutor applications from candidates who are studying at the UK’s leading universities. Candidates who fulfil our grade criteria then pass to the interview stage, where a member of the MyTutor team will personally assess them for subject knowledge, communication skills and general tutoring approach. About 1 in 7 becomes a tutor on our site.

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Sarah Parent from Morden

15 May, 2018

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Sarah Parent from Morden

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Sarah Parent from Morden

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Sarah Parent from Morden

28 Feb, 2017

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Qualifications

SubjectQualificationGrade
MathematicsA-level (A2)A*
Further MathematicsA-level (A2)A*
ChemistryA-level (A2)A
PhysicsA-level (A2)A

General Availability

MonTueWedThuFriSatSun
Pre 12pm
12 - 5pm
After 5pm

Pre 12pm

12 - 5pm

After 5pm
Mon
Tue
Wed
Thu
Fri
Sat
Sun

Subjects offered

SubjectQualificationPrice
MathsA Level£40 /hr
ChemistryGCSE£40 /hr
MathsGCSE£40 /hr
PhysicsGCSE£40 /hr
Maths13 Plus£40 /hr

Questions Wesley has answered

A curve is defined by the parametric equations x = 3 - 4t, and y = 1 + 2/t. Find dy/dx in terms of t.

At first glance, this looks quite tricky, as usually when we are asked to find dy/dx, we have one equation, but here we have 2.

So in this case, we need to use the statement that dy/dx = (dy/dt) * (dt/dx)

Then, we just need to find dy/dt and dy/dx.

dy/dt = -2/t^2

dx/dt = -4, and therefore dt/dx = -1/4

So, (dy/dt)*(dt/dx) = (-2/t^2)*(-1/4)

= 1/2t^2.

At first glance, this looks quite tricky, as usually when we are asked to find dy/dx, we have one equation, but here we have 2.

So in this case, we need to use the statement that dy/dx = (dy/dt) * (dt/dx)

Then, we just need to find dy/dt and dy/dx.

dy/dt = -2/t^2

dx/dt = -4, and therefore dt/dx = -1/4

So, (dy/dt)*(dt/dx) = (-2/t^2)*(-1/4)

= 1/2t^2.

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2 years ago

1762 views

What is the integral of (6x^2 + 2/x^2 + 5) with respect to x?

When we think of integration, we should be thinking of the same method every time; adding one to the power, and then dividing by that new power.

The integral of the sum of each of these terms is equal to the sum of the integrals of each individual term, and therefore we can take them on one-by-one.

When integrating 6x^2, we use the method i mentioned above, and add one to the power (which is 2 here) and divide by it. So this becomes (6x^3)/3, which equals 2x^3.

Now, for the second term, we know that 2/x^2 is the same as 2x^-2. This makes it so much easier to deal with. So with the same method as before, this becomes (2x^-1)/(-1), remembering that when we add 1 to -2, we get -1 and not -3. This then equals -2x^-1, or -2/x.

With the last term, 5, there is no power of x. So writing 5 on it's own is technically the same as writing 5x^0, as anything to the power of 0 is 1, and 5*1=5. This makes it easier to visualise what we're trying to do when we integrate this term. So, with the same method as always, this becomes (5x^1)/1, which is just 5x.

Now a fairly common mistake when it comes to indefinite integrals is made at this last part. There could always have been a constant value 'C' that was differentiated to give 0 in the term that we're integrating, so we have to add it back on when we've integrated all the other terms, to make sure we cover this possibility.

So, adding all the terms back together, the answer is 2x^3 - 2x^-1 + 5x + C.

When we think of integration, we should be thinking of the same method every time; adding one to the power, and then dividing by that new power.

The integral of the sum of each of these terms is equal to the sum of the integrals of each individual term, and therefore we can take them on one-by-one.

When integrating 6x^2, we use the method i mentioned above, and add one to the power (which is 2 here) and divide by it. So this becomes (6x^3)/3, which equals 2x^3.

Now, for the second term, we know that 2/x^2 is the same as 2x^-2. This makes it so much easier to deal with. So with the same method as before, this becomes (2x^-1)/(-1), remembering that when we add 1 to -2, we get -1 and not -3. This then equals -2x^-1, or -2/x.

With the last term, 5, there is no power of x. So writing 5 on it's own is technically the same as writing 5x^0, as anything to the power of 0 is 1, and 5*1=5. This makes it easier to visualise what we're trying to do when we integrate this term. So, with the same method as always, this becomes (5x^1)/1, which is just 5x.

Now a fairly common mistake when it comes to indefinite integrals is made at this last part. There could always have been a constant value 'C' that was differentiated to give 0 in the term that we're integrating, so we have to add it back on when we've integrated all the other terms, to make sure we cover this possibility.

So, adding all the terms back together, the answer is 2x^3 - 2x^-1 + 5x + C.

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2 years ago

682 views

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