Higher Engineering Mathematics

Ess-For-H8152.tex 19/7/2006 18: 2 Page 711

ESSENTIAL FORMULAE 711

Tangents and normals

Equation of tangent to curvey=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 curvey=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

∂u

du+

∂z

∂v

dv+....

Rate of change

Ifz=f(u,v,..) and

du

dt

,

dv

dt

, ... denote the rate of

change 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 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 station-

ary points, substitute values ofxandyinto

∂^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 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