Currently unavailable: for regular students
Degree: Computer Science (Bachelors) - Warwick University
I am a Computer Science student currently studying at Warwick University. I've always had a passion throughout my education for Computing, as the possiblities through programming are incredible.
Why should I choose you for my sessions?
I feel that I have a wide range of experience that I can bring to one-on-one sessions to help you with any topic you are struggling with. I have previous experience teaching Maths to GCSE students while at secondary school, whereby I went through past exam problems to isolate areas in which students were having trouble and helped them understand where they were going wrong.
As someone who learns through understanding the core concepts at work, I don't encourage memorisation, but instead a full understanding so that you can tackle a given problem in a range of situations, even if the question looks slightly different. I will use a wide variety of techniques to teach a concept (anologies, diagrams, etc), until you feel you have fully grasped it. We will then go through example questions to test if your new found knowledge has sunk in.
I also hope to show through the lessons how fun and interesting Computing can be, showing how building on simple ideas can lead you to solving interesting mathematical problems or creating cool computer programs that can complete amusing tasks.
If you have any questions regarding tutoring,or even if you are struggling with a question and just want some quick help, don't hesitate to contact me. Remember to include your exam board and include any topics that you are strugging with.
I hope to hear from you soon!
|Computing||A Level||£20 /hr|
|Maths||A Level||£20 /hr|
|Before 12pm||12pm - 5pm||After 5pm|
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When thinking about stationary points, it is important to remember that these points happen when the gradient of the curve is 0 (-> dy/dx = 0), or if you prefer, when the rate of change of a graph is 0.
So the first step is to differentiate the given equation:
If we take the first part of the equation, 6ln(x), we can split this into two seperate parts, ln(x) and 6, where these variables are multiplied together. Using the chain rule, where u = 6 and v = ln(x), we can calculate the differential by calculating v.du + u.dv, where dv is the differential of v, and du is the differential of u.
As the differential of ln(x) is 1/x, and the differential of 6 = 0, v.du is equal to ln(x).0, and u.dv is equal to 6.(1/x) which is equal to 6/x.
Next we move on to the differentiation of x2 , and as differentating involves multiplying by the current power, and then taking away 1 from the previous power, we get 2x.
Next, 8x differentiates to 8, as x1 multiplied by 1 and minus 1 is equal to x0 , and 8.1 = 8.
Finally, 3 differentiates to 0, as the derivative represents the rate of change, and a constant factor has a rate of change of 0 (as it is constant).
This leaves us with:
dy/dx = 6/x + 2x - 8
Getting rid of the fraction by multiplying by x, we get:
dy/dx = 6 + 2x2 - 8x (which rearranges to give)
-> dy/dx = 2x2 -8x + 6
By factorising the equation out, we get that x is either equal to 1 or 3:
= 2x2 - 8x + 6
= (2x - 6)(x-1)
-> 2x - 6 = 0 or x - 1 = 0
-> x = 3 or x = 1
Substitute in the values we have calculated into the original equation:
x = 1
y = 6ln(1) +12 - (8.1) + 3 (ln(1) = 0)
y = 1 - 8 + 3
y = -4
So when x = 1, y = -4
x = 3
y = 6ln(3) + 32 - (8.3) + 3
y = 6ln(3) + 9 - 24 + 3
y = 6ln(3) - 12
So when x = 3, y = 6ln(3) - 12
So the coordinates of the stationary points are:
(1, -4) and (3, 6ln(3) - 12)see more