APPLICATION OF PARTIAL DIFFERENTIATION
- 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
- 19. Y(0)=0, Y1(0)=1+y2(0) =1, Y2(0)=2y(0)y1(0)=2(0)(1)=1, Y3(0)=2y1 2(0)+ 2y(0)y2(0) =2(1)2+2(0)(0)=2, ……… ……….. ……… ……….. 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
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