Revision Notes For Core Page 3
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1
2
3
4Implicit Equations
These are where the x’s and y’s occur more than once so cannot be rearrange to make
x or y the subject.
dy
To find
of an curve defined implicitly, differentiate with respect to x but the y terms
dx
dy
dy
differentiate from f(y) to f’(y)
using the chain rule. Then rearrange to make
the
dx
dx
subject and this is the gradient formula.
2
2
Example If the equation on a curve is x
– 2xy + 4y
= 12, differentiating with respect to
dy
dy
x gives you 2x – 2y –2x
+ 8y
=0 (the +2xy is differentiated using the product rule)
dx
dx
−
+
−
+
dy
dy
2
x
2
y
x
y
(-2x + 8y) = -2x + 2y and so
=
=
(this is the gradient formula)
−
+
−
+
dx
dx
2
x
8
y
x
4
y
For the coordinates of a point where the gradient is 0, then –x + y =0 so rearrange and
substitute in to the equation of the curve to find the point that solves them both.
Vectors
4
Describes a translation either as a column vector
or a letter e.g p or multiples of
5
1
0
basic unit vectors e.g. 4 i + 5 j where i =
and j =
0
1
To get vectors of directions you aren’t given, use the commutative rule which means
draw the network and get from A to B using a different route not direct. To show that a
point B is on the line AC, show that the vector AB is a multiple of the vector AC.
The vector equation of a line is r = a + t p ( a is the position vector of a point on the line
and p is the displacement vector of the line (the direction the line is going in or the
movement in the x, y and z direction to get from point A to point B)
Find p by subtracting the vectors of 2 points on the line. a is the vector of the first point.
Example
If points A and B have coordinates (-5,3,) and (-2,9,) then the vector equation of the line
−
−
−
2
5
3
5
3
between them is p =
–
=
so the equation is r =
+ t
9
3
6
3
6
Same process if in 3D .
t is a scalar quantity that tells you how far along the line to move.
You can write this equation using basic unit vectors or column vectors.
To find if and where 2 lines intersect, make one equal to the other and solve for the x,y
and z values (or the i,j and k values) to find the size of t.
Do for all 3 coordinates to check that they do cross in all directions with the same value
for t. If they do not cross in all directions, the lines are skew.
Scalar Product of Vectors
a.b = |a| |b| cos
where |a| and |b| is the magnitude (length) of the displacement vectors
and found using Pythagoras and
is the angle between the vectors.
a.b means find the product of the x, y and z values and sum them.
l
u
.
.
= lu + mv + nw so combining gives |a| |b| cos
= lu + mv + nw
m
v
n
w
If 2 vectors are perpendicular, then p.q = 0
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