This document provides information on several multivariable calculus topics:
1) Finding maxima and minima of functions of two variables using partial derivatives and the second derivative test.
2) Finding the tangent plane and normal line to a surface.
3) Taylor series expansions for functions of two variables.
4) Standard expansions for common functions like e^x, cosh(x), and tanh(x) using Maclaurin series.
5) Linearizing functions around a point using the tangent plane approximation.
6) Lagrange's method of undetermined multipliers for finding extrema with constraints.
1. Index
MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE.
TANGENT PLANE AND NORMAL LINE TO A
SURFACE
TAYLOR’s EXPANSION FOR FUNCTION OF TWO
VARIABLES.
LAGRANGE’s METHOD OF UNDETERMINED
MULTIPLES.
2. MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE
The Function f(x,y) is maximum at (x,y) if for all
small positive or negative values of h and k; we have
f(x+h , y+k) – f(x,y) < 0
Similarly f(x,y) is minimum at (x,y) if for all small
positive or negative values of h and k, we have
f(x+h , y+k) – f(x,y) > 0
3. Thus ,from the defination of maximum of f(x,y) at
(x,y) we note that f(x+h , y+k) – f(x,y) preserves the
same sign for a maximum it is negative and for a
minimum it is positive
Working rule to find maximum and minimum values
of a function f(x,y)
(1) find ∂f/∂x and ∂f/∂y
(2) a necessary condition for maximum or minimum
value is ∂f/∂x=0 , ∂f/∂y=0
MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE
4. solve simultaneous equations ∂f/∂x=0 , ∂f/∂y=0
Let (a₁,b₁) , (a₂,b₂) … be the solutions of these
equations.
Find ∂²f/∂x²=r ,
∂²f/∂x ∂y=s ,
∂²f/∂y²=t
MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE
5. (3) a sufficient condition for maximum or minimum
value is rt-s²>0.
(4 a ) if r>0 or t>0 at one or more points then those are
the points of minima.
(4 b) if r<0 or t<0 at one or more points then those
points are the points of maxima.
(5) if rt-s²<0 ,then there are no maximum or minimum
at these points. Such points are called saddle points.
MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE
6. (6) if rt-s²=0 nothing can be said about the maxima
or minima .it requires further investigation.
(7) if r=0 nothing can be said about the maximum or
minima . It requires further investigation.
MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE
7. Example
discuss the maxima and minima of
xy+27(1/x + 1/y)
∂f/∂x=y-(27/x²) ,
∂f/∂y=x-(27/y²)
For max. or min ,values we have ∂f/∂x=0 , ∂f/∂y=0.
y-(27/x²)=0…(1)
x-(27/y²)=0…(2)
Giving x=y=3
MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE
8. ∂²f/∂x²=r =54/x³
∂²f/∂x ∂y=s=1 ,
∂²f/∂y²=t=27/y³
r(3,3)=3
s(3,3)=1
t(3,3)=3
rt-s²=9-1=8>o , since r,t are both >0
We get minimum value at x=y=3 which is 27.
MAXIMA AND MINIMA OF FUNCTION OF
TWO VARIABLE
9. TANGENT PLANE AND NORMAL LINE
LET THE EQUATION OF THE SURFACE BE
f(x,y,z)=0
The equation of the tangent plane at P(x₁,y₁,z₁) to
the surface is
(x-x₁)(∂f/∂x)p + (y-y₁)(∂f/∂y)p +(z-z₁)(∂f/∂z)p=0
And the equations of the normal to the surface at
P(x₁,y₁,z₁) which is a line through P are:
10. x-x₁/ (∂f/∂x)p = y-y₁/(∂f/∂y)p = z-z₁/(∂f/∂z)p
Example
find the equations of the tangent plane and normal
to the surface z=x²+y² at the point (1,-1,2).
∂f/∂x=-2x
∂f/∂y=-2y
∂f/∂z=1
TANGENT PLANE AND NORMAL LINE
11. At (1,-1,2),
∂f/∂x=-2
∂f/∂y=2
∂f/∂z=1
Therefore equation of the tangent plane at (1,-1,2) is
(x-1)(-2)+(y+1)(2)+(z-2)(1)=0
Or -2x+2+2y+2+z-2=0
Or 2x-2y-z=2
Equations of the normal are
x-1/-2 = y+1/2 = z-2/1
TANGENT PLANE AND NORMAL LINE
12. TAYLOR’S & MACLAURIN’S SERIES
(Taylor’s series):-If f(x) is an infinitely
differentiable function of x which can be expanded as
a convergent power series in (x-a), then
f(x)= f(a)+ (x-a)f’(a)/1! +(x-a)2 f’’(x)/2!+ (x-a)3
f’’’(x)/3!+……….+(x-a)n fn(a)/n!+……….
Where a is constant.
13. By putting x-a=h; that is, x=a+h in equation,
we get
f(a+h)= f(a)+ h f’(a)/1!+ h2 f’’(a)/2!+
h3f’’’(a)/3!+….+hn fn(a)/n!+……
By putting a=0 in above equation, we have
f(x)=f(0)+ x f’(0)/1! +x2f’’(0)/2! +……+
xn fn(0)/n!+……
TAYLOR’S & MACLAURIN’S SERIES
14. Statement of Maclaurin’s Series:-If f(x) is an
infinitely differentiable function of x which can be
expanded as a convergent power series in x ,then
f(x)=f(0)+ x f’(0)/1! +x2f’’(0)/2! +……+ xn
fn(0)/n!+……
TAYLOR’S & MACLAURIN’S SERIES
15. Expansions of some standard function
The following are some expansion of standard
treated as standard expansions,obtained with the
help of maclaurin’s series. These standard
expansions are useful in obtaining the expansion of
oher functions.
16. 1.Expansion of e x , e –x ,cosh(x) ,sinh(x)
Let f(x)=e x, then f’(x)=f’’(x)=f’’’(x)=….=ex
Also, f’(0)= f’’(0)= f’’’(0)=….=1
By substituting the values of f(0),
f’(0), f’’’(0),…. In Maclaurin’s series
,We get
e x =1+ x+ x2/2!+ x3 /3!+…….
Expansions of some standard function
17. Replacing x by –x, yields
e –x= 1- x+x2/2! – x3/3!+……..
Adding e x and e –x , we obtain
Cosh(x)=( e x +e –x)/2
=1+ x2/2! + x4 /4!+……
Again, by subtracting e x and e –x ,we obtain
Sinh(x)=(e x- e –x)/2
=x+ x3/3!+ x5/5!+……
Note The expansions of e –x, cosh(x), sinh(x) can also
be obtained by using Maclaurin’s series.
Expansions of some standard function
18. Example:-
1. Expansion of tan x
Let y= f(x)= tanx,
Y1=sec2x = 1+tan2x=1+y2,
Y2=2yy1,
Y3=2y1
2+2yy2,
…… ………
……. .……..
Expansions of some standard function
20. By putting these values of y(0),y1(0),y2(0),….. in
Maclaurin’s series,
Y(x)=y(0)+ xy1(0)+ x2y2(0)/2!+ x3y3(0)/3!+…….
we get,
tan x=x+x3/3+ x52/15+…….
Expansions of some standard function
21. LINEARIZATION
(Linearization):-Linearization means to replace
given function of two variables. This can be achieved
by tangent plane approximation.
The linearization of a function f(x , y) at a point
(x0,y0) where f is differentiable is the function
L(x , y)=f(x0,y0)+ fx(x0,y0)(x –x0)+fy(x0,y0)(y-
y0)
22. Which is the equation of the tangent plane to the
graph of a function f of two variables at the point
(x0,y0, f(x0,y0).
The approximation
f(x , y)= L(x ,y )
is called the Standard Linear approximation or
Linear approximation or the tangent plane
approximation of f at (x0,y0).
LINEARIZATION
23. EXAMPLE:-
Find the linearization of f(x ,y) =X2 –xy+ (1/2) y2+
3 at the point (3, 2).
Let us first evaluate f, f x and f y at (3 ,2).
f(3 ,2)=(3)2 -(3)(2)+(1/2)(2)2+3=8.
F x =2x- y=> (f x)(3, 2)= 2(3)-2 =4.
F y=-x+ y=> (f y) (3, 2) =-3 +2=-1.
LINEARIZATION
24. Using differentiable function, the required
linearization is
L(x ,y)=f(3 ,2)+ f x (3 ,2)(x -3) + f y (3,2) (y-2)
=8+ (4) (x -3)+ (-1)(y-2)
= 4x- y -2.
LINEARIZATION
25. To find extreme values of a function we consider a function
of three variables with one restriction.
Lagrange’s Method of Undetermined
Multipliers