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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
Textbook Problem

State the Second Derivatives Test.

To determine

To state: The Second Derivative Test.

Explanation

Statement:

Second Derivative Test:

“Suppose the second partial derivatives of f are continuous on a disk with center (a,b) , and suppose that fx(a,b)=0 and fy(a,b)=0 (that is (a,b) is a critical point of f).

Let D=D(a,b)=fxx(a,b)fyy(a,b)[fxy(a,b)]2

(a) If D>0 and fxx(a,b)>0 , then f(a,b) is a local minimum.

(b) If D>0 and fxx(a,b)<0 , then f(a,b) is a local maximum.

(c) If D<0 , then f(a,b) is not a local maximum or minimum and it is called a saddle point”.

Example:

Consider the function is, f(x,y)=sinxsiny .

Calculation:

Take the partial derivative in the given function with respect to x and obtain fx .

δfδx=δδx(sinxsiny)=sinyδδx(sinx)=sinycosx

Thus, δfδx=sinycosx . (1)

Take the partial derivative in the given function with respect to y and obtain fy .

δfδy=δδy(sinxsiny)=sinxδδx(siny)=sinxcosy

Thus, δfδy=sinxcosy . (2)

It is given that the values of x and y lies between, π<(x,y)<π .

Set the above partial derivatives to 0 and find the values of x and y.

From the equation (1),

sinycosx=0siny=0(or)cosx=0y=sin1(0)(or)x=cos1(0)y=0,x=π2,π2

From the equation (2),

sinxcosy=0sinx=0(or)cosy=0x=sin1(0)(or)y=cos1(0)x=0,y=π2,π2

Thus, the critical points are, (0,0),(π2,π2),(π2,π2),(π2,π2) and (π2,π2) .

Obtain the second derivatives as follows.

Take the partial derivative of the equation (1) with respect to x and obtain fxx .

δ2fδx2=δδx(sinycosx)=sinyδδx(cosx)=siny(sinx)=sinysinx

Hence, δ2fδx2=sinysinx .

Take the partial derivative of the equation (2) with respect to y and obtain fyy .

δ2fδy2=δδy(sinxcosy)=sinxδδy(cosy)=sinx(siny)=sinxsiny

Hence, δ2fδy2=sinxsiny .

Take the partial derivative of the equation (1) with respect to y and obtain fxy .

δ2fδxδy=δδy(sinycosx)=cosxδδy(siny)=cosx(cosy)=cosxcosy

Hence, δ2fδxδy=cosxcosy .

Use second derivative test and obtain the value of D for the critical point (0,0) .

D(0,0)=fxx(0,0)fyy(0,0)[fxy(0,0)]2=[sin(0)sin(0)][sin(0)sin(0)][cos(0)cos(0)]2=(0)(0)[(1)(1)]2=1

Since, D(0,0)<0 and hence there exists a saddle point (0, 0)

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Chapter 14 Solutions

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