C4 is an odd unit, largely comprised of the bits left over when the other 3 units were complete
It is, however, vital to understand it if one is planning to do well in FP4, as vectors and parametric equations feature heavily.
It comprises of:
-Binomial Theorum
-Differentiation
-Exponentials and Logs
-Integration
-Parametric Equations and Curves
-Trigonometry
-Vectors
Binomial Theorum
Differentiation
Exponentials and Logarithms
Integration
Parametric Equations and Curves
Trigonometry
Vectors
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Vectors are a difficult yet pivotal topic of C4
A solid understanding of C4 vectors will help a lot in FP4 where it is the main topic that carries over.
C4 vectors include:
-Vectors in 2- and 3-D
-Magnitude of a vector
-Geometrical interpretations of Operations
-Position Vectors
-Distance Between Two Points
-Vector Equations of lines
-Scalar Product
Vectors also play a pivotal role in M2 where they turn up in most questions in both 2- and 3-D forms.
Forms of Vectors
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Vectors are written in two main forms.
-Column form (which cannot be shown here)
-or i, k, j form
A vector which is 2 units long in the x-direction, 1 in the y-directions and 4 in the z-direction can be written as
v=(2i+j+4z)
The column form would have 2 on top, 1 in the middle and 4 at the bottom.
Magnitude of a Vector
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If v is a vector then |v| is its magnitude.
|v| is a scalar value.
If v=(2i+j+4k)
then |v|=Sqrt((2)^2)+1+(4)^2))=Sqrt(21)
This can be standardised by writing the vector without numerical values.
If v=(ai+bj+ck)
then
|v|=sqrt(a^2+b^2+c^2)
Scalar Product
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The scalar product, also known as the dot product, gives a rough idea of the relationship between two vectors.
The scalar product is named as such because it always gives a scalar.
If we have vector a and b and we want to get the scalar product it is written as a.b
There are two ways to find a.b
The first is using their magnitudes and the cosine of the angle between them.
a.b=|a||b|cosx where x is the angle between them (this should be done as theta but I lack the facilities).
The second form is far simpler if you have the two vectors.
If a=(xi+yj+zk)
and b=(di+ej+fk)
then a.b=xd+ye+zf
If both vectors are known but theta is not, the scalar product can be used to find it.
First find a.b using the second method.
Then rearrange the first to get cosx=(a.b)/(|a||b|)
Plug in the scalar product and the product of the magnitudes, use inverse cosine and bam, you have theta.
The scalar product can also test to see if two vectors are perpendicular.
As cos90 (cos(pi/2)) is zero, any perpendicular vectors will have a dot product of 0.
This can also be used to prove they are not perpendicular as any non-zero dot product means the vectors are not perpendicular.
Position Vectors
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Position vectors can be thought of as coordinates relative to a point. This point is typically the origin.
For example the position vector (i+2j+5k) has coordinates (1, 2, 5) if it is relative to the origin.
If, however, it was relative to the point (1, 0, 1) then it would have coordinates (2, 2, 6).
Position vectors of points can be subtracted to find direction vectors.
If we have point A with coordinates (1, 3, 2) and point B with coordinates (2, 2, 6) then AB (with an arrow above it when written) is B-A or (1, -1, 4). This will be very important when we reach the vector equation of a line.
Magnitude of AB would give the distance between these two points as a scalar value.
Vector Equation of a Line
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Until C4 all lines were written in Cartesian form. In C4 both vector equations and parametric versions are introduced.
Vector line equations can be both 2-D and 3-D, but are most commonly used for the later.
A vector equation of a line will look something like the one below
r=a+tb
where a is a point on the line, and b the direction vector. t is a parameter and a scalar.
The equation for r can be thought of as giving every point on the line relative to a known point.
When the parameter t is used it fixes a point on the line and gives the position vector of this line relative to the origin, NOT THE KNOWN POINT.
If a=(ci+dj+ek) and b=(fi+gi+hk) then r can be written in terms of its coordinates in a parametric form where:
x=c+ft
y=d+gt
z=e+ht
Often you may be asked to find a point on the line given one coordinate. In this case the simultaneous equations must be solved for t or the other coordinate. If it is to be solved for t then find t and plug it into the parametric equations to get the coordinates. Otherwise check that t is consistent for all your final answers.
Often the consistency of t will be used to test if a point is on the line. If t is consistent then the point given lies on the line. If t is different for one or more equations then the point does not lie on the line.
Geometrical Properties
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These are also a bit tacked on the end, but they are still important.
If a vector can be multiplied by a scalar to be the same as another vector, then the two are parallel.