Olaoluwa O. GCSE Further Mathematics  tutor, A Level Further Mathemat...

Olaoluwa O.

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Degree: Natural Sciences (Bachelors) - Durham University

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About me

About me

I am a 19 year old Mathematics & Philosophy student at Durham University.

I am particularly interested in the analytic side of pure mathematics as well as Latin language, two subjects which I enjoyed studying through to A level and through which I have gained useful insight into good exam technique.

I feel it is very important for people to have a thorough understanding of the mathematical theories, instead of simply knowing ad hoc methods of solving problems. This will help a student to tackle any question thrown at them.

As for Latin, I am an extremely enthusiastic linguist and translator, and achieved 100% in the two GCSE language papers as well as full marks in the AS level language paper (there was no such paper for A2). However, my experience in Latin also allows me to be helpful with problems of literary analysis.

My teaching style

As much as I enjoy these subjects, I am also particularly keen for teaching and understand the importance of patienceencouragement, and constructive criticism when teaching students of all ages.

I will often ask students to talk me through how they have come to their answer. In this way, I will be able to discern whether they truly understand the principles behind the method, but it also leads to more clarity in the student's own mind of what they are doing. Being able to explain out loud their methodology will prove vital not only on exam papers, but also in academic interviews for universities.

More than anything, I fully believe that both mathematics and Latin can be extremely fun because analysing the logical connections involved is as satisfying as anything.

I look forward to being of help!

Subjects offered

SubjectLevelMy prices
Maths A Level £20 /hr
Further Mathematics GCSE £18 /hr
Latin GCSE £18 /hr
Maths GCSE £18 /hr
Maths IB £20 /hr
Latin 13 Plus £18 /hr
Maths 13 Plus £18 /hr
Maths 11 Plus £18 /hr

Qualifications

QualificationLevelGrade
Further MathematicsA-LevelA
MathematicsA-LevelA*
LatinA-LevelA
MusicA-LevelB
Disclosure and Barring Service

CRB/DBS Standard

No

CRB/DBS Enhanced

No

Currently unavailable: for new students

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Ratings and reviews

5.5from 8 customer reviews

Kabale (Parent) December 4 2016

Great lesson as always!

Kabale (Parent) November 26 2016

Great lesson!

Kabale (Parent) November 6 2016

Thank you Ola, L is doing brilliantly and I am so pleased. You are a great tutor! Kabale

Lola (Student) October 29 2016

Great session!
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Questions Olaoluwa has answered

How do I know which method of integration to use?

First of all, it is very important to simply learn some common integrations by rote (e.g trig functions, exponentials, polynomials, 1/x). If you know these well, it will often be easy to spot which technique to use when integrating a seemingly difficult function. If you are in a situation whe...

First of all, it is very important to simply learn some common integrations by rote (e.g trig functions, exponentials, polynomials, 1/x). If you know these well, it will often be easy to spot which technique to use when integrating a seemingly difficult function.

If you are in a situation where you have to integrate a function comprised of two different types of function, such as f(x)=xe^x, integration by parts can be useful (i.e substitute u for one part of the function and v' for the other, such that the you now have to integrate uv' with solution uv - integral(vu'dx)). It is often a good idea when choosing what to use as and v to choose as u whatever simplifies the most when you differentiate. For example, in the above function I would make u=x and v'=e^x so that u'=1 and v=e^x. Then the integral of vu' is simply the integral of e^x which is quite simple.

Unless you are given case like this, integration by inspection (just by looking at it and using your known results) is always possible but can sometimes be hard to spot. If this is the case, it can be useful to use a clever substitution to help you. (Have a look at my solution to the integral of f(x)=x(1-x)^6 for an example of this

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8 months ago

220 views

How do you translate long, complex Latin sentences?

Every sentence in Latin can be broken up into smaller, simple parts. The first step is always to identify and perhaps even highlight all the words which agree with each other in case, number and gender. Often times, if nouns and adjectives come from different declensions or one or more are irr...

Every sentence in Latin can be broken up into smaller, simple parts. The first step is always to identify and perhaps even highlight all the words which agree with each other in case, number and gender. Often times, if nouns and adjectives come from different declensions or one or more are irregular, it can be difficult to spot what agrees with what and mistakes can be made. For this reason, it is imperative that you learn your grammar tables and vocab.

Next, look for your main verb. This will usually be at the end of the sentence and tends to be indicative (not subjunctive) - again, learning verb tables thoroughly is vital. Generally, if a verb is subjunctive, that is a good sign that it is part of a subordinate clause.

Once you have completed these two quick steps, you then have to piece together all of the parts of the sentence in a way which makes sense. Usually there will only be one correct solution so if you think there is more than one possibility, consider two options:

1) Do your two solutions have precisely the same meaning (be very rigorous in determining the answer)? If so, then either is fine. If not;

2) Revise your grouping and specification of words. The chances are that you have misidentified the case/number/gender of a word or group of words or you have mistranslated the tense/mood of a verb.

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8 months ago

252 views

How would you integrate a function like f(x)=x(1-x)^6?

For a question like this, it would be far too time-consuming to expand the bracket and then multiply through by x. It might then seem that integration by parts is the optimal solution, but this is actually not necessary. If you make the substitution u=1-x then you have du=-dx. Then the integra...

For a question like this, it would be far too time-consuming to expand the bracket and then multiply through by x. It might then seem that integration by parts is the optimal solution, but this is actually not necessary.

If you make the substitution u=1-x then you have du=-dx. Then the integral of x(1-x)^6 with respect to x is equivalent to the integral of -(1-u)u^6 with respect to u. By multiplying through, we have u^7-u^6 which is not difficult to integrate.

An important thing to remember in this is always to remember to find dx in terms of du - do not assume that dx=du because it rarely does. Finally, make sure you sub x back into your final answer.

If you are given a definite integral where you need a substitution, always remember to change your limits appropriately. There is then no need to sub into your integral at any point in the working

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8 months ago

236 views
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