<p>If you're taking A-Level Maths - well done to you. It's one of the trickier A-levels but also one of the most well respected, and it will set you up for all sorts of career paths. That said, it can be a surprising jump up from the Maths you've done before. And because the course progresses so quickly, it's easy to feel like you're getting left behind.</p>
<p><strong>With one-to-one online support from a Maths tutor, you can spend time focusing on the topics you want to work on.</strong> Whether it's differentiation that's got you in a pickle, or logarithms that are giving you trouble, our tutors are on hand to help. So by the time it comes round to your exams, you'll feel confident, well prepared, and able to achieve the results you want.</p>

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Why limit yourself to someone who lives nearby, when you can choose from tutors across the UK?

By removing time spent travelling, you make tuition more convenient, flexible and affordable

We've combined live video with a shared whiteboard, so you can work through problems together

All your Online Lessons are recorded. Make the most out of your live session, then play it back after

Usually we cover both subject knowledge and exam technique, although that can change depending on each individual student. Then we go through diagrams, and they ask questions, and we go from there.

Lots of students say that the classes are too big in school, or that they don't have time to ask teachers after Online Lessons. In my Online Lessons, we take time to explore things in a little in a bit more detail.

I always look up the board my students are taking so the Online Lessons are really relevant. Then we go through past papers or set texts, whatever the student finds helpful.

I use the shared whiteboard. We make diagrams together and label them, and often the student prints it off because they know it's right and they completely understand it.

After tutoring one girl went and told all her friends the new explanation I gave her. And she was so excited about what she wrote in the exam she emailed me immediately afterwards.

There was one girl who had her exam on Monday. She wanted tuition on Friday, Saturday and Sunday beforehand. It was very intense, but she said the exam went well.

cos(x+a)=cosxcosa-sinxsina
so cosa=5
sina=3 which means tana=3/5
a=0.54 RADIANS
R=sqrt(5^{2}+3^{2}). =sqrt(34)
so sqrt(34)cos(x+0.54)

cos(x+a)=cosxcosa - sinxsina
therefore: cos(a)=5 and sin(a)=3
so tan(a)=3/5
into calcultor: a=0.54
to find R=square root(5^{2}+3^{2})
=root(34)
so 5cosx - 3sinx = root(34)cos(x+0.54)

d/dx (xy) = x dy/dx + y
d/dx (y^2) = 2y dy/dx [This is from the chain rule]
So, d/dx (2x^2 + xy + y^2 = 14)
=> 4x + x dy/dx + y + 2y dy/dx = 0
set dy/dx = 0 as stationary point has gradient 0
Obtains 4x+y=0
y=-4x
Sub this back into our original equation
14x^2 = 14
x^2 = 1
This is only satisfied by +1 and -1
When x=1 y=-4, when x=-1 y=4
So stationary points are (1,-4) and (-1,4)

When dealing with complex numbers and trigonometric functions, always turn to DeMoivre's Theorem that states [cos(θ)+isin(θ)]^{n} = [cos(nθ)+*i*sin(nθ)].
If we set n=2, the we see a combination of cos(2θ) and sin(2θ) on the right hand side. From here, we can expand the left hand side, just like we would with a normal quadratic expression, giving us: cos^{2}(θ) + 2cos(θ)(*i*sin(θ)) + (*i*sin(θ))^{2}. This can then be simplified to cos^{2}(θ) - sin^{2}(θ) + 2cos(θ)(*i*sin(θ)) as i^{2 }= -1 by definition.
Combining the right hand side and the left hand side gives:
cos^{2}(θ) - sin^{2}(θ) + 2cos(θ)(*i*sin(θ)) = cos(2θ)+*i*sin(2θ)
We can then equate real and imaginary parts of the equations to give:
cos^{2}(θ) - sin^{2}(θ) = cos(2θ) and
2cos(θ)(*i*sin(θ)) = *i*sin(2θ), and therefore 2cos(θ)sin(θ) = sin(2θ).

To solve this question we use the product rule, where we differentiate one variable whilst keeping the other constant, and vice versa, adding the two results together to get our answer.
A helpful formula is (dy/dx)=u(dv/dx)+v(du/dx), which is easy to remember as it is very catchy!
To start, we need to choose u and v. You can make the solution easier to get to if you choose suitable u's and v's (Although you will still arrive at the same answer regardless), so here the best option personally would be making u=y, and v=2x^2. As the formula suggests, we need to find (dv/dx) and (du/dx), so we start off by differentiating our v with respect to x. This obtains (dv/dx)=4x. Next, we find (du/dx), but this is slightly more difficult as we have to differentiate y with respect to x. The way I like to think about this is if you were differentiating something simple such as y=x. You would write (dy/dx)=1 without hesistating, but without knowing it you've differentiated y with respect to x here, and this is the same in this case of u=y, where differentiating obtains (du/dx)=(dy/dx). Now we have all the parts to our equation, and now we just sub them in.
After subbing u, v, (du/dx) and (dv/dx) into the equation, we arrive at our answer of:
(dy/dx)=4xy+2(x^2)(dy/dx)
I find after a lot of practice, you will learn not only how to choose your u and v as effectively as possible, but also how to do it in your head!

First we recall the general formula for the volume of revolution:

V = π ∫ [f(x)]² dx

Substituting our function, this gives:

V = π ∫₀¹ (3^x)² x dx

We could write (3^x)² as 3^(2x), or we could notice that this is equal to 9^x. This avoids unnecessarily over-complicating our integration which could lead to mistakes. So we have:

V = π ∫₀¹ (9^x) x dx

Given that we have two different functions of x under the integral, we must use the integration by parts method. Deciding which element to differentiate and which to integrate is made easier when we realise that:

∫ (9^x) dx = (9^x)/ln(9) = (9^x)/2ln(3)

To prove this, let y = a^x and take the natural logarithm of both sides:

ln(y) = ln(a^x)

Recall that one of our rules for logarithms is that ln(a^b) = bln(a):

ln(y) = xln(a)

Differentiate both sides with respect to x:

y’/y = ln(a)

dy/dx = yln(a) = (a^x)ln(a)

Now multiply by dx, divide by ln(a) and integrate both sides:

∫ (1/ln(a)) dy = ∫ a^x dx

Therefore ∫ a^x dx = y/ln(a) = (a^x)/ln(a)

As such if we differentiate x and integrate (9^x), we should be left with an integral we can immediately solve given that d/dx[x] = 1.

u = x, v’ = 9^x

u’ = 1, v = (9^x)/2ln(3)

Recall that the IBP formula is:

I = vu - ∫ vu’ dx

V = π [[ (9^x)x/2ln(3) ]₀¹ - (1/2ln(3)) ∫₀¹ 9^x dx ]

Note that we have simply taken a constant out of the integral to simplify it. When evaluating the first term, we only need to consider when x =1 as the second term goes to 0 due to the x in the numerator.

V = π [ 9/2ln(3) – (1/2ln(3)) [ (9^x)/2ln(3) ]₀¹]

V = π [ 9/2ln(3) – (1/2ln(3)) [ 9/2ln(3) – 1/2ln(3) ]]

Notice that we can factorise out 1/2ln(3) and there is a common denominator in the far-right bracket allowing us to sum the numerators, leaving 8/2ln(3). Cancel the 8 with the 2 and we’re left with:

V = π/2ln(3) [ 9 - 4/ln(3) ]

V = π ∫ [f(x)]² dx

Substituting our function, this gives:

V = π ∫₀¹ (3^x)² x dx

We could write (3^x)² as 3^(2x), or we could notice that this is equal to 9^x. This avoids unnecessarily over-complicating our integration which could lead to mistakes. So we have:

V = π ∫₀¹ (9^x) x dx

Given that we have two different functions of x under the integral, we must use the integration by parts method. Deciding which element to differentiate and which to integrate is made easier when we realise that:

∫ (9^x) dx = (9^x)/ln(9) = (9^x)/2ln(3)

To prove this, let y = a^x and take the natural logarithm of both sides:

ln(y) = ln(a^x)

Recall that one of our rules for logarithms is that ln(a^b) = bln(a):

ln(y) = xln(a)

Differentiate both sides with respect to x:

y’/y = ln(a)

dy/dx = yln(a) = (a^x)ln(a)

Now multiply by dx, divide by ln(a) and integrate both sides:

∫ (1/ln(a)) dy = ∫ a^x dx

Therefore ∫ a^x dx = y/ln(a) = (a^x)/ln(a)

As such if we differentiate x and integrate (9^x), we should be left with an integral we can immediately solve given that d/dx[x] = 1.

u = x, v’ = 9^x

u’ = 1, v = (9^x)/2ln(3)

Recall that the IBP formula is:

I = vu - ∫ vu’ dx

V = π [[ (9^x)x/2ln(3) ]₀¹ - (1/2ln(3)) ∫₀¹ 9^x dx ]

Note that we have simply taken a constant out of the integral to simplify it. When evaluating the first term, we only need to consider when x =1 as the second term goes to 0 due to the x in the numerator.

V = π [ 9/2ln(3) – (1/2ln(3)) [ (9^x)/2ln(3) ]₀¹]

V = π [ 9/2ln(3) – (1/2ln(3)) [ 9/2ln(3) – 1/2ln(3) ]]

Notice that we can factorise out 1/2ln(3) and there is a common denominator in the far-right bracket allowing us to sum the numerators, leaving 8/2ln(3). Cancel the 8 with the 2 and we’re left with:

V = π/2ln(3) [ 9 - 4/ln(3) ]

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Time: | 2018-02-20T13:06:53Z |