Essential formulae 665
Maximum and minimum values:
Ify=f(x)then
dy
dx= 0 for stationary points.Let a solution of
dy
dx=0bex=a;ifthevalueof
d^2 y
dx^2whenx=ais:positive, the point is aminimum,
negative, the point is amaximum.Velocity and acceleration:
If distancex=f(t),then
velocity v=f′(t)ordx
dt
andacceleration a=f′′(t)or
d^2 x
dt^2Tangents 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 )wherem=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
Ifz=f(u,v,...), then the total differential,
dz=∂z
∂udu+∂z
∂vdv+....Rate of change
Ifz=f(u,v,...)and
du
dt,dv
dt,... denote the rate ofchange ofu,v,... respectively, then the rate of change
ofz,
dz
dt=∂z
∂u·du
dt+∂z
∂v·dv
dt+...Small changes
Ifz=f(u,v,...)andδx,δy,... denote small changes
inx,y,... respectively, then the corresponding change,δz≈∂z
∂xδx+∂z
∂yδy+....To determine maxima,minima and saddle points for
functions of two variables:Givenz=f(x,y),(i) determine∂z
∂xand∂z
∂y(ii) for stationary points,∂z
∂x=0and∂z
∂y=0,(iii) solve the simultaneous equations∂z
∂x=0and
∂z
∂y=0forxandy, which gives the co-ordinatesof the stationary points,(iv) determine∂^2 z
∂x^2,∂^2 z
∂y^2and∂^2 z
∂x∂y(v) for each of the co-ordinates of the station-
ary points, substitute values ofxand yinto
∂^2 z
∂x^2,∂^2 z
∂y^2and∂^2 z
∂x∂yand evaluate each,(vi) evaluate(
∂^2 z
∂x∂y) 2
for each stationary point,(vii) substitutethevalues of∂^2 z
∂x^2,∂^2 z
∂y^2and∂^2 z
∂x∂yintothe 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 asaddle
point(b) if< 0 and∂^2 z
∂x^2< 0 , then the stationary
point is amaximum point,and(c) if< 0 and∂^2 z
∂x^2> 0 , then the stationary
point is aminimum point