TOPIC : APPLICATION OF PARTIAL

DERIVATIVESPREPARED BY:CHIRAG BANSAL (31)

GUIDED BY:BHAVIXA BHAGAT

INTRODUCTION:A partial differential equation is an equation involving a function of two or more variables and some of its partial derivatives. Therefore a partial differential equation contains one dependent variable and more than one independent variable. Here z will be taken as the dependent variable and x and y the independent variable so that .

We will use the following standard notations to denote the partial derivatives.

yxfz , .

,, qyzp

xz

t

yzs

yxzr

xz

2

22

2

2

,,

APPLICATIONS OF PARTIAL DERIVATIVES:

Tangent Plane and Normal LineLinearization and Linear

ApproximationsLagrange’s MultipliersLocal Maximum & Local Minimum

Tangent Plane and Normal Line Equation of the Tangent plane and Normal line can be made with the

help of partial derivation.

An equation of the tangent plane to the surface z = f(x, y) at the point P(x0, y0, z0) is:

fx(x0, y0, z0)(x – x0) + fy(x0, y0, z0)(y – y0) + fz (x0, y0, z0) ( z– z0 )= 0

And the equation of the normal to the surface at P(x0, y0, z0) which is a line through P is:

x - x0/ (fx)p = y - y0 /(fy)p = z - z0 /(fz )p

Example: Find the equations of the tangent plane and normal to the surface z=x²+y² at the point (1,-1,2).

At (1,-1,2), ∂f/∂x=-2, ∂f/∂y=2, ∂f/∂z=1Therefore equation of the tangent plane at (1,-1,2) is :(x-1)(-2)+(y+1)(2)+(z-2)(1)=0Or -2x+2+2y+2+z-2=0Or 2x-2y-z=2And,Equations of the normal are:x-1/-2 = y+1/2 = z-2/1

Linearization and Linear Approximations

As for one variable, we use the tangent line at (a, f(a)) as an approximation to the curve y =f(x) when x is near to a.

• Equation of this tangent line is

y = f(a) + f’(a)(x - a)

So the linear function whose graph is this tangent line, that is, L(x) = f(a) + f’(a)(x – a)

Similarly, Now The linearization of a function f(x , y) (Two variables) at a point (x0,y0) where f is differentiable is given by the function:

L(x , y)=f(x0,y0)+ fx(x0,y0)(x –x0)+fy(x0,y0)(y-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).

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.

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) L(x ,y)= 4x- y -2.

Lagrange’s Multipliers Let w = f(x,y,z), function of 3 independent variables (x,y,z) which is

collected by the condition:Φ (x,y,z) = 0

Then, L(x,y,z) = f(x,y,z) + (λ). Φ(x,y,z)

Where L = Lagrangian, L(x,y,z) = Lagrange’s Function, λ – Lagrange’sMultiplier, Φ (x,y,z) = Constraint Function

NOTE: In this method, we’ve to find the extreme values of f(x,y,z) subjected to given constraint.

EXAMPLE: Find the minimum of f(x,y) = 5x2 + 6y2 – xy subject to the

constraint x+2y = 24.

F(x,y, ) = 5x2 + 6y2 - xy + (x + 2y - 24)

Fx = F = 10x - y + ; Fxx = 10xFy = F = 12y - x + 2 ; Fyy = 12yF = F = x + 2y - 24 ; Fxy = -1

11

The critical point,10x - y + = 012y - x + 2= 0x + 2y - 24= 0

The solution of the system is x = 6, y = 9, = -51

D*=(10)(12)-(-1)2=119>0

Fxx = 10>0

We find that f(x,y) has a local minimum at (6,9).

f(x,y) = 5(6)2 + 6(9)2- 6(9) f(x,y) = 720

Local Maximum & Local Minimum

A function f has a local maximum (or relative maximum) at c if f(c) ≥ f(x) when x is near c. This means that f(c) ≥ f(x) for all x in some open interval containing c.

Similarly, f has a local minimum at c if f(c) ≤ f(x) when x is near c.

This means that f(c) ≤ f(x) for all x in some open interval containing c.

Working rule to find maximum and minimum values of a function f(x,y): Find ∂f/∂x and ∂f/∂y A necessary condition for maximum or minimum value

is ∂f/∂x=0 , ∂f/∂y=0 Solve simultaneous equations ∂f/∂x=0 , ∂f/∂y=0 Let (a₁,b₁) , (a₂,b₂) … be the solutions of these

equations. Now Take ∂²f/∂x²=r , ∂²f/∂x ∂y=s , ∂²f/∂y²=t. A sufficient condition for maximum or minimum value

is rt-s²>0.

If r>0 or t>0 at one or more points then those are the points of minima.

If r<0 or t<0 at one or more points then those points are the points of maxima.

If rt-s²<0 ,then there are no maximum or minimum at these points. Such points are called saddle points.

If rt-s²=0 or r=0, nothing can be said about the maxima or minima.

After getting maxima or minima i.e. values of x and y, put the values in the function f(x,y) and the result obtained will be maximum or minimum values of the function respectively.

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 to make ∂f/∂x=0 , ∂f/∂y=0.So by equating ∂f/∂x and ∂f/∂y equal to 0 gives, x=y=3 Now finding the values of r,s and t,∂²f/∂x²=r =54/x³ ; ∂²f/∂x ∂y=s=1 ; ∂²f/∂y²=t=27/y³At point (3,3)) => r(3,3)=3 ; s(3,3)=1 ; t(3,3)=3rt-s²=9-1=8 >o , Since r,t are both >0, we get minima at x=y=3 , so now putting

(3,3) into the function gives –Minimum value = 27