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Mathematics Core 4

The final module of Maths. Pass this exam and you'll have an A-level. Isn't that enough incentive to revise?

Partial Fractions

You'll remember adding and subtracting algebraic fractions from Core 3. For example:

The reverse of this process is called splitting into partial fractions. It can only occur if the original fraction is proper, meaning you'll have to divide if they present something improper.

For example - split the following into partial fractions:

The first step is to let the fraction equal two fractions, using the two parts of the denometer. We don't know what's on the top of the fractions, so we call them A and B:

Now, we merge them together again:

If we just consider the numerators, we can say:

13 - x = A(x + 2) + B(3 - x)

We can remove either A or B by making the brackets equal 0.

When x = -2

13 - -2 = B(3 - -2)

15 = 5B

B = 3

When x = 3

13 - 3 = A(3 + 2)

10 = 5A

A = 2

Finally, we write down the final fractions and we're finished:

It's exactly the same method whether the denometer contains two, three or more brackets.

For fractions that contain repeated factors, the method is slightly different. Example: Split the following into partial fractions:

We need to use A B and C, but for the repeated root, we can't use (x - 1) twice. Instead, we use (x - 1) and (x - 1)2:

Merge these together, but don't include both (x - 1) and (x - 1)2. We only need to include (x - 1)2.

Now from the numerator, we can say:

1 = A(x - 1)2 + B(x + 2)(x - 1) + C(x + 2)

Stick some numbers in to solve:

When x = 1

1 = C(1 + 2)

3C = 1

C =

When x = -2

1 = A(-2 - 1)2

1 = 9A

A =

When x = 0

1 = A - 2B + 2C

1 = - 2B +

B = -

Finally, writing it out as partial fractions:

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Parametric Equations

We've been using y=f(x) for a long time. It's called a cartesian equation.

There is another type called parametric equations, where x and y are given in terms of another letter (a parameter).

For example:

x = 1 - t
y = t2 - 4

We can sketch the graph by putting in values for t, and working out the corresponding x and y values:

t
-3
-2
-1
0
1
2
3
x
4
3
2
1
0
-1
-2
y
5
0
-2
-4
-3
0
5

If we plot these coordinates, we get:

If we sub one t equation into another, we can get a cartesian equation:

x = 1 - t ... t = 1 - x
y = t2 - 4

y = (1 - x)2 - 4

y = x2 - 2x - 3

As we see from our graph, it's a quadratic.

However, you can't just find the cartesian equation first...

Look at the following:

x = 2t2
y = 1 - t2

When we find a cartesian equation, we get the straight line y= 1 - .

However, because x = 2t2, x must be positive, and so the domain of the graph is x ≥ 0.

Parametric equations are easy enough to manipulate.

If the graph crosses the x-axis, just put y to 0, and solve to find a value for t. Then use this to get the coordinates of x and y.

Question: A curve is defined by:

x = b(2t - 3)
y = b(1 - t2)

Where "b" is a constant. The curve passes through the point (0, -5). Find the value of "b".

So, we just put x = 0 and y = -5:

0 = b(2t - 3)
- 5 = b(1 - t2)

From the first equation, we can see that 2t - 3 = 0, and so t = . (b can't also be 0, because of the y coordinate).

If we put t = into the second equation, we get:

- 5 = b(1 - )
-5 = - b
b = 4

If two lines intercept, we just substitute on into the other, as you would with cartesian, and then solve.

Trig

If we get given a curve given by parametric equations which involves trig and we're asked to find the cartesian equation, we have to connect the two parts with a trig identity.

Example:

x = a cos x
y = a sin x

First, we rearrange them to get the trig alone on one side of the equations:

= cos x
= sin x

What trig identity involves sin and cos? Easy: sin2x + cos2x = 1. So, subbing in sin x and cox x, we get:

()2 + ()2 = 1

Expand the brackets and rearrange a bit to get:

x2 + y2 = a2

A circle with radius a!

Area under a curve

You might recall from Core 2 that area under a curve was:

Area = y dx

In parametrics, this principal is identical, but the formula is naturally slightly different to cope with the parameters:

y dt

It makes sense if you think about it. If you do x dt, the two dts cancel and leave dx, like in the original formula.

Example:

A curve is defined by the parametric equations:

x = t - 1
y = 4 - t2

Find the area between the curve and the x-axis when the curve is bordered by x=1 and x=-3.

First of all, we need t1 and t2. We find these by sticking out two x values in:

1 = t - 1
t = 2
-3 = t - 1
t = -2

Next we need to differentiate x to get .

= 1

Now stick everything into the formula y dt, remembering that t1 is the most negative value of x:

4 - t2 dt

Next, we integrate:

[4t - ]

(4x2 - ) - (4x-2 - )

8 - + 8 -

=

That's all there is to parametrics!

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Binomial Expansion

More stuff from Core 2, revisited. Remember the formula for binomial expansion? No? Well, it doesn't matter because it's in the formula booklet -- just as long as you know how to use it!

(1 + x)n = 1 + nx + + ... +

When we had positive integers, the series came to an end on the final power.

What they didn't tell us back then is that binomial expansion can also be used on negative integers or fractions. The only difference is that the series is infinite, only converging at certain values of x.

To converge, each term must get smaller, and so x must be between -1 and 1.

Example: Find the first 4 terms in the expansion of (1 + x) and state the values of x for which the expansion is valid.

First off, we need to write it properly: (1 + x)1/2.

Now we use the binomial formula with n=1/2.

1 + x + .x + .-.x2 + .-.-.x3

= 1 + x + x - x2 + x3 + ...

This is valid when |x| < 1.

Easy enough, hm? The only thing to be careful about is the plus or minus signs.

If we have an expansion that doesn't involve 1, such as (2 - x)1/2, we need to factorise it to get to the form, remembering that the factor needs raising to the power too:

21/2(1 - )1/2

Use the binomial expansion to find the first 3 terms:

2 (1 + . + ..()2)
2 (1 + + )

= 2 + +

It's valid when || < 1 or |x| < 2

A question might also ask you to use an expansion to work out something else. For example:

Expand (1 - 3x)1/2 in ascending powers of x, up to and including x3. Using this, with x=0.01, find an approximate answer for 97 to 5 d.p.

Firstly, we do the expansion:

1 + .-3x + ..-.(-3x)2 + .-.-.(-3x)3

1 - x - x2 - x3

So we can say:

(1 - 3x)1/2 1 - x - x2 - x3

Now, we consider the LHS, and put in x=0.01:

(1 - 0.03)1/2 = 0.971/2 = ()1/2 =

Notice how the bottom of the fraction has been square rooted from 100 to 10, so the square root only affects the 97.

Next we put x=0.01 into the RHS:

= 0.9848858125

From this, we can say:

= 0.9848858125

Times by ten:

97 = 9.84886 (5 d.p.)

Questions might throw partial fractions at you first, and then ask you to use binomial expansion, but there shouldn't be anything that hard. Right?

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Differentiation

Hah, we did hardly any in Core 3, 2 and 1! It's back for a fourth round...

Remember the parametrics from earlier? We need to know how to differentiate them. It's quite simple, really, just a case of:

= x

(If you think about it, the two 'dt's would cancel, leaving ).

Example:

If x=cos2t and y=sin2t, find the equation of the tangent to the curve where x=.

For a tangent, of course, we need to find the gradient, so:

x=cos2t

= -2sin2t

We need the reciprocal of this:

=

y=sin2t

= 2cost

 

Times these together:

= = -cot2t

So, when t=

x = cos

x =

y = sin

y =

= -cot = -1

Shoving all these into the equation y - y1 = m(x - x1) we get:

y - = -1(x - )

y = - x + 2

Implicit Differentiating

We've been differentiating explicitly for ages, such as y= 4x2 - 4x. Implicit differentiating is where you're given something like:

x2 + xy - 2y2 = 1

To differentiate, we state which letter we'll diff with respect to:

Diff w.r.t. x:

For any power of x, we differentiate normally. For any power of y, we differentiate and then times it by . From the above example, we get:

2x + y + x - 4y = 0

Next, we rearrange to get on one side of the equation:

x - 4y = -2x - y

(x - 4y) = -2x - y

=

Product Rule for the xy:

u=x v=y
= 1

= 1.

=

 

That's really all there is to it!

Differentiating ax

Make sure you learn this proof:

Let y = ax, a>0

Take logs to base e:

ln y = ln ax
ln y = x ln a

Diff w.r.t. x:

= ln a
= y ln a

If we think back, we remember that y=ax so:

= axln a

Example:

y = 2x
= 2xln 2

Rate of Change

This means differentiating a variable with respect to time. For example, the rate of change of Area, A, is .

The radius, r, of a circular hole is increasing at a rate of 0.5cms-1. Find the rate at which the area is increasing when r = 5cm.

The first stage is to extract the information that we know "rate of radius increase":

= 0.5 cms-1

Now we think about what we need: "rate of area increase". This would be:

So, what can we times by to get ? They work just like fractions -- the tops and bottoms will cancel if they're the same. Therefore:

= x

Now we think, how can we get ? Well, obviously, we need a formula with both of them, and of course, that's the area of a circle: πr2. So, we differentiate with respect to r:

A = πr2

= 2πr

We're wanting the area when r = 5, so stick this into the equation:

= 0.05 x 10π

= 5π cm2s-1

Sometimes a question will say something like:

The rate of increase is proportional to the radius.

This means we can say:

r

= kr

If the rate of decrease is proportional, we stick a negative in:

-r

= -kr

And if it's inversely proportional, we have the reciprocal of r:

=

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Vectors

They come to haunt us from Mechanics...

Remember, a unit vector has a length of 1 and is found from:

= unit vector

= magnitude of vector a.

Adding non-parallel vectors gives us a resultant vector...

Find the values of α and ß in the following:

(α - 1)a + (ß + 2)b = -ßa + (α - 1)b

Using vector a, we can say:

α - 1 = -ß

Using vector b, we can say:

ß + 2 = α - 1

Solving these two simultaneous equations will get us alpha and beta:

ß + 2 = -ß

ß = -1, α = 2

The direction of a vector can be written in the i and j form. For example: a = is the same as a = 5i + 1j.

To find the unit vector for direction, we divide by the magnitude. For 5i + 1j, magnitude = (52 + 12) = 26 = 25.

Therefore:

3D

We've been working in 2D for a long time... 3D is the next step. The third dimension coordinate is the k component:

Consider points A(x1,y1,z1) and B(x2,y2,z2)

The midpoint of these two points would be:

The length of AB would be:

Finding the unit vectors, etc. are all the same, but with the additional k.

Scalar (or dot) product of 2 vectors

If we have two vectors as shown to the right, we can find the acute angle between them using the scalar product formula. The two vectors must go from a point, or both go to a point in order for angle to be acute.

a.b = |a||b|cos θ = a1b1 + a2b2 + a3b3

If a and b are perpendicular, the angle between them is 90o. cos90o = 0 and so a.b must be 0.

Example: Find the angle between the two vectors:
a = 2i + 4j - 4k and b = 3i - 12j - 4k.

a1b1 + a2b2 + a3b3 = |a||b|cosθ

In the formula, the 1 refers to the i components, the 2 to the js and the 3 to the ks. So:

(2 x 3) + (4 x -12) + (-4 x -4) = (22 + 42 + (-4)2)(32 + (-12)2 + (-4)2)cosθ

6 - 48 + 16 = 36169 cosθ

-26 = 78cosθ

- = cosθ

θ = 109.5

Vector equation of a line:

r = a + λb

r is the position vector of a general point.

a is a point on the line.

λ is a scalar number.

b is the direction of the line.

Example: Find a vector equation of the line through A(3, 1, -4) and B(2,4,2).

First, we need the direction. This is given by B - A:

2i + 4j + 2k - 3i - j + 4k

= -i + 3j + 6k

Now we have a fixed point (A or B) and the direction, so we can say:

r = 2i + 4j + 2k + μ( -i + 3j + 6k)

And that's the answer!

Find the equation of the line parallel to AB which passes through the point C(2, -1, 4).

If it's parallel, the direction is equal. Therefore, we have our point C and direction, so the equation of the line is:

r = 2i + 4j + 2k + λ( -i + 3j + 6k)

Determine whether the point A(-6,-5, 3) lies on the line with equation:
r
= 2i - 3j + k + μ(4i + j - k)

Using i:

-6 = 2 + μ4
μ = -2

Using j:

-5 = -3 + μ
μ = -2

Using k:

3 = 1 - μ
μ = -2

Because μ = -2 for each equation, we know that the point lies on the line.

The principal is similar if two lines intercept -- for this to occur, at some point, they must be equal.

r1 = (2 + 2s)i + (s - 1)j + (9 - s)k
r2 =
(2 + 3t)i + tj + (2 + 2t)k

Using i:

2 + 2s = 2 + 3t

Using j:

s - 1 = t

Using k:

9 - s = 2 + 2t

If we solve the equations for i and j simultaneously, we can find t and s:

2 + 2s = 2 + 3(s - 1)
2s = 3s - 3
s = 3, t = 2

If these work in k, we know the lines intercept:

9 - s = 2 + 2t

LHS: 6
RHS: 6

Therefore, these two lines intercept. To find the point where they intercept, we just put t=2 and s=3 into one of the equations and find the coordinates:

(2 + 2s)i + (s - 1)j + (9 - s)k
8i + 2j + 6k

The scalar product formula can also be used with lines. The only difference is that we use the directions of the two lines:

Find the angle between the following lines:

r1 = 2i + j + k + s(i + 2j - 2k)

r2 = -i - 4j - 3k + t(3i - 12j - 4k)

So, we use (i + 2j - 2k) and (3i - 12j - 4k) in a1b1 + a2b2 + a3b3 = |a||b|cosθ:

(1 x 3) + (2 x -12) + (-2 x -4) = (12 + 22 + (-2)2)(32 + (-12)2 + (-4)2)cosθ

3 - 24 + 8 = 9169cosθ

cosθ = -

θ = 109.5o

The hardest question they might throw at you is something like:

Find the coordinate of the foot of the perpendicular from the point (-5,2,3) to the line r = 3i - 5j + k + t(i + 4j + 3k).

The first question that might strike you: what's the "foot of the perpendicular"? This is the point where a line crosses another perpendicularly.

We know that there is a line perpendicular to r which intercepts r and also crosses (-5,2,3).

We can find the direction from P(-5,2,3) to the the point A on r by using A - P. How is that helpful? The unknown coordinate lies on r and so has the formula 3i - 5j + k + t(i + 4j + 3k). So:

3i - 5j + k + t(i + 4j + 3k) - (-5i + 2j + 3k)

= 8i - 7j - 2k + t(i + 4j + 3k) = (8 + t)i + (-7 + 4t)j + (-2 + 3t)k

This direction is perpendicular to r, so a.b = 0. From this, we can work out t:

a1b1 + a2b2 + a3b3 = 0

Stick in the direction (8 + t)i + (-7 + 4t)j + (-2 + 3t)k and the direction of r (i + 4j + 3k):

((8 + t)x1) + (4(-7 + 4t)) + (3(-2 + 3t)) = 0

8 + t - 28 + 16t - 6 + 9t = 0

26t = 26

t = 1

Finally, stick t=1 into the equation for r to work out the coordinate:

r = 3i - 5j + k + t(i + 4j + 3k)

Coordinate: (4,-1,4)

That's all the vector stuff done! Take a deep breath before plunging into integration...

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Integration

Remember all the new ways of differentiating we learned in Core 3? Well, we need to learn how to undo all those now...

Basic integration has the equation:

xn dx =

and it works for all values of n, except for -1.

From the reverse of differentiating, we get a load more:

f'(x)ef(x) dx = ef(x) + c

dx = ln|f(x)| + c

f'(x)[f(x)]n dx =

f'(x)cos f(x) = sin f(x) + c

f'(x)sin f(x) = - cos f(x) + c

+ all the other trig ones, tan, cot, sec, etc. formed from cos and sin... check them out in the formula booklet!

If we are given something trig that can't be integrated, we may be able to use a trig identity to change it. Don't forget them!

What happens if you don't have the integral placed before something? The magic touch -- times and divide a number in.

For example:

cos 3x dx

In order to integrate, we need 3cos 3x, and so we times that part by 3. In order to make it equate to the original, however, we need to divide everything by 3, or times by :

3 cos 3x dx

Now we can integrate: sin 3x + c

Integrate:

We need the differential on the top, which would be -4, so we times by -4 inside and times by - outside:

-

Now we integrate using the above ln form:

= - ln(7-4x) + c

Integrate:

(3x - 9)4

We need the differential of the bracket, 3, so we times by 3 inside and outside:

3(3x - 9)4 dx

Now we integrate using the standard rule:

= + c

Nothing too hard there, is there? The only thing to remember is that sin and cos as the opposite to differentiating... sin -> -cos, cos -> -sin. And don't forget + c!

sin2x

This can't be integrated, but we can use a trig formula on it. You must learn this!

Remember: cos2x = 1 - 2sin2x

and therefore: 2sin2x = 1 - cos2x

Times it by 2 inside and divide by outside and then sub in:

2sin2x dx

1 dx - cos2x dx

For the second integration, we need 2 inside and another outside, so:

1 dx - 2cos2x dx

Finally integrate each part:

- sin2x + c

Make sure you know it for cos2x too - using cos2x = 2cos2x + 1.

Integration by substitution

Things like x(1 - x)5 can't be integrated normally, but we can if we use a substitution. The exam question will usually tell you what to use as the substitution, but make sure you understand the method in case you have to think up the substitution yourself...

x(1 - x)5 dx

From the RHS, we sub in the values accordingly, in order to eliminate all reference to x:

(1 - u)u5 -du

- -u5 + u6 du

Now, we can integrate:

-

Reverting back to x, we have:

- + c

Using: u = 1 - x

= -1

-du = dx

x = 1 - u

Remember "definite integrals," where you're given numbers to put in and work out at the end, rather than leaving +c. For integration by substitution, there is another step, and that is remembering to convert the numbers:

Sticking those in:

Now think about sin2θ. What trig identity connects it to cos? Obviously: siny: sin2θ + cos2θ = 1.

Rearranged: 1 - sin2θ = cos2θ, and so:

Square root the bottom:

And canceling, we're left with:

1 dθ

Using x = 2sinθ

= 2cos&theta

dx = 2cosθdθ

x
θ
2

0
0

 

Integrate:

[θ]

=

Integration by Parts

Remember the product rule from differentiation? = v + u

Integrating gives us a similar rule:

uv -vdx = udx, which is given in the formula booklet.

How does it work?

x cox x dx

This cannot be done normally, so we'll use parts, to the RHS.

With those, we do uv -vdx, so:

x sin x - sin x dx

Now, if we finally integrate the sin x, we get:

x sin x + cos x + c

u = x = cos x
(differentiate) (integrate)
= 1 v = sin x

Remember the pattern - differentiate u, integrate .

Be careful if the question includes ln or e. If it includes ln, make sure you differentiate ln (rearranging if necessary) because it can't be integrated. Things get interesting with e...

excos x dx

= exsin x - exsin x dx

= exsin x - [ - excos x - -excos x dx]

= exsin x + excos x - excos x dx

If you look now, you'll see we're back to the beginning with - excos x dx. So, if we add it, to move it to the other side, we get:

2excos x dx = = exsin x + excos x

Divide by two to get the answer:

excos x dx =

 

u = ex = cos x
= ex v = sin x

u = ex = sin x
= ex v = - cos x

 

Trapezium Rule

Remember this from Core 2? It's not changed...

Use the Trapezium rule to find an approx value for cot x dx to 4 d.p with 2 strips.

Remember:

h= and area=h{(y0 + yn) + 2(y1 + y2 + ... + yn-1)}

So, we have:

h== 0.4

Then draw up a table and calculate:

x
0.6
1
1.4
y
1.461696
0.642909
0.172477
 
y0
y1
y2

Now, stick these into the formula:

.0.4.{(1.461696 + 0.172477) + 2(0.642909)}

=0.5843 (4 d.p.)

If we integrate cot x, we get:

[ln|sin x|] = 0.5569 (4 d.p.)

The question might ask you to find the percentage error.

This is found by:

error real value x 100

= (0.5843 - 0.5569) 0.5569 x 100 = 4.4%

Area and volume

We already know that the area under a curve if found by y dx and in parametrics, y dx.

The volume occurs when we rotate an area under the curve by 360o.

Volume = π y2 dx

(parametrically = π y2 dx)

Example: Find the volume generated when the area bounded by the x-axis and the curve y=x3 - 2x2 is rotated completely about the x-axis.

First, we sketch the graph, to see what out boundaries are:

From this we can see we need to integrate the following:

π (x3 + x2)2 dx

If we expand the bracket, we get: x6 - 4x5 + 4x4:

πx6 - 4x5 + 4x4 dx

Now integrate:

π[ - + ]

= 1π

Differential Equations

To find the general solution of these, we need to integrate. For example:

2y = cos x

First, we need to get it into an integrating form by splitting up the by multiplying by dx:

2y dy = cos x dx

Then we integrate both sides and add c:

y2 = sin x + c

Given that y=1 when x=2, find y in terms of x for:

2 =

This is slightly different, because there are x and y on the RHS as well. We need to move e2y across as well as dx:

2e2y dy = ex dx

Now integrate:

e2y = ex + c

If we stick in x and y now, we can find c:

e2 = e2 + c

c = 0

Therefore, we can say:

e2y = ex

Manipulate the logs to get y alone:

2y = x

y =

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That is pretty much everything you need for Core 4! Good luck!