Essential formulae 665
Maximum and minimum values:
If y = f (x) then
dy
dx
= 0 for stationary points.
Let a solution of
dy
dx
= 0 be x = a; if the value of
d 2 y
dx 2 when x = a is: positive, the point is a minimum,
negative, the point is a maximum.
Velocity and acceleration:
If distance x = f (t ), then
velocity v = f (t ) or
dx
dt
and
acceleration a = f (t ) or
d 2 x
dt 2
Tangents and normals:
Equation of tangent to curve y = f (x) at the point
(x 1 , y 1 ) is:
y − y 1 = m(x − x 1 )
where m = gradient of curve at (x 1 , y 1 ).
Equation of normal to curve y = f (x) at the point
(x 1 , y 1 ) is:
y − y 1 = −
1
m
(x − x 1 )
Partial differentiation:
Total differential
If z = f (u, v, ...), then the total differential,
dz =
∂z
∂u
du +
∂z
∂v
dv + ... .
Rate of change
If z = f (u, v, ...) and
du
dt
,
dv
dt
, … denote the rate of
change of u, v, … respectively, then the rate of change
of z,
dz
dt
=
∂z
∂u
·
du
dt
+
∂z
∂v
·
dv
dt
+ ...
Small changes
If z = f (u, v, ...) and δx, δy, … denote small changes
in x, y, … respectively, then the corresponding change,
δz ≈
∂z
∂x
δx +
∂z
∂ y
δy + ... .
To determine maxima, minima and saddle points for
functions of two variables: Given z = f (x, y),
(i) determine
∂z
∂x
and
∂z
∂ y
(ii) for stationary points,
∂z
∂x
= 0 and
∂z
∂ y
= 0,
(iii) solve the simultaneous equations
∂z
∂x
= 0 and
∂z
∂ y
= 0 for x and y, which gives the co-ordinates
of the stationary points,
(iv) determine
∂ 2 z
∂x 2 ,
∂ 2 z
∂ y 2 and
∂ 2 z
∂x∂ y
(v) for each of the co-ordinates of the stationary points, substitute values of x and y into
∂ 2 z
∂x 2 ,
∂ 2 z
∂ y 2 and
∂ 2 z
∂x∂ y
and evaluate each,
(vi) evaluate
∂ 2 z
∂x∂ y
2
for each stationary point,
(vii) substitute the values of
∂ 2 z
∂x 2 ,
∂ 2 z
∂ y 2 and
∂ 2 z
∂x∂ y
into
the equation =
∂ 2 z
∂x∂ y
2
−
∂ 2 z
∂x 2
∂ 2 z
∂ y 2
and evaluate,
(viii) (a) if >0 then the stationary point is a saddle
point
(b) if <0 and
∂ 2 z
∂x 2 < 0, then the stationary
point is a maximum point, and
(c) if <0 and
∂ 2 z
∂x 2 > 0, then the stationary
point is a minimum point