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Degree: Mathematics (Bachelors) - Warwick University
Hi I'm Ben, I studied mainly maths and sciences at school, and went on to study maths at Warwick University. I have had some previous tutoring experience whilst I was at school, but it is a lot harder to find tutoring jobs here!
As I am only in my first year at university, I was taking my A Levels this time last year, so the memory is still all too fresh in my mind. I understand the want to get the exams over and done with and start enjoying the long post A level summer, but at the same time these exams are important for achieving offers from university, or for applying next year. By this point I expect that you will have covered most of your subject content, so it will only be little things that you need help with.
I hope to see you in the classroom, feel free to book a meet the tutor session if you want to find out more!
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a) The first part of this question is asking us to find points, (x,y), such that x = -8. By taking the equation and substituting in x = -8, we should find values, y, such that the points given are (-8,y).
(-8)3 - 4y2 = 12(-8)y
-512 - 4y2= -96y
The coefficients of these terms are relatively large, so we can look to divide them through by some number to simplify the problem, luckily all the numbers are divisible by 4. They are also negative, so multiplying the whole equations by -1 will turn them positive, which makes it nicer to handle.
128 + y2 = 24y
y2 - 24y +128 = 0
Now here using the quadriatic formula: [-b ± √(b2 - 4ac)]/2
We get y = [24 ± √(242 - 4*1*128)]/2, = [24 ± √(576 - 512)]/2 = [24 ± √(64)]/2
= [24 ± 8]/2 = 12 ± 4.
So the two values of y for which x = -8 are 8 and 16, so the point (x,y) where x = -8 are
b) This part of the question requires us to differentiate implicitly the formula given. The question asks for the gradient, in other words it wants us to find a value of dy/dx for the points we worked out in the first section. We will differentiate the equation term by term.
Starting with 12xy, we will need to use the product rule for differentiation as there are two variables multiplied together. In case you had forgotten, the product rules works by effectively differentiating one of the variables while leaving the other alone, taking the result of this and adding it to the result we get had we differentiated the other term whilst leaving the first term alone.
Ie. d/dx(12xy) = 12y + 12x(dy/dx)
The first term 12y comes from differentiating x and leaving y alone, d/dx(x) = 1
The second 12x(dy/dx) comes from differentiating y and leaving x alone, d/dx(y) = dy/dx
The in both cases the 12 can be ignored as it is just a constant, d/dx(12xy) = 12*d/dx(xy)
Now moving to the 4y2 term. Here we can also use the product rule by considering this term as 4y*y. Giving:
d/dx(4y*y) = 4*d/dx(y*y) = 4*y*(dy/dx) + 4*(dy/dx)*y = 8y(dy/dx)
And then for the x3 term, it is easy to see using normal differentiation that d/dx(x3) = 3x2
This gives the whole equation as:
3x2 - 8y(dy/dx) = 12y + 12x(dy/dx)
Rearranging and factoring out the (dy/dx) gives:
3x2 - 12y = (dy/dx)(12x+ 8y)
Dividing by (12x+ 8y) gives us an expression for dy/dx:
(3x2 - 12y)/(12x+ 8y) = (dy/dx)
Now substituting in the (x,y) values we got from the first part (-8,8) (-8,16)
For (-8,8), (3(-8)2 - 12(8))/(12(-8)+ 8(8)) = (dy/dx)
(192 - 96)/(-96 + 64) = 96/-32 = -3 = dy/dx
For (-8,16), (3(-8)2 - 12(16))/(12(-8)+ 8(16)) = (dy/dx)
(192 - 192)/(-96 + 128) = 0/32 = 0 = (dy/dx)
So at (-8,8), dy/dx = 3 and at (-8,16), dy/dx = 0