   Chapter 14, Problem 4P

Chapter
Section
Textbook Problem

For what values of the number r is the function f ( x , y , z ) = { ( x + y + z ) r x 2 + y 2 + z 2 if  ( x , y , z ) ≠ ( 0 , 0 , 0 ) 0 if ( x , y , z ) = ( 0 , 0 , 0 ) continuous on R3?

To determine

To find: The value of r for which the function f(x,y)={(x+y+z)rx2+y2+z2if(x,y,z)00if(x,y,z)=0 is continuous on 3 .

Explanation

Definition used:

A function f of two variables is continuous at (a,b) if lim(x,y)(a,b)f(x,y)=f(a,b) .

Result used:

“The inequality (a+b+c)24a2+4b2+4c2 .”

Calculation:

The given function is, f(x,y)={(x+y+z)rx2+y2+z2if(x,y,z)00if(x,y,z)=0 .

Condition 1:

If the function is continuous on 3 then by the definition of continuous,

lim(x,y,z)(0,0,0)f(x,y,z)=f(0,0,0)=0

Thus, the value of r should satisfy the condition 1.

From the given result,

(x+y+z)24x2+4y2+4z2(x+y+z)24(x2+y2+z2)

Consider the rational fraction (x+y+z)rx2+y2+z2 ,

The domain of this rational (x+y+z)rx2+y2+z2 is {(x,y,z)|(x,y,z)(0,0,0)} .

Consider (x+y+z)r , then its abosute value is, |(x+y+z)r|=(|(x+y+z)2|)r2 .

Now by the result the above equation can be expressed as,

(|(x+y+z)2|)r2=[(x+y+z)2]r2[(x+y+z)2]r2[4(x2+y2+z2)]r2[(x+y+z)2]r2(4)r2(x2+y2+z2)r2[(x+y+z)2]r22r(x2+y2+z2)r2

Compute the value of |f(x,y,z)0| for (x,y,z)(0,0,0) as follows,

|f(x,y,z)0|=|(x+y+z)rx2+y2+z2|=|(x+y+z)r|x2+y2+z2

Substitute the respective values in the above equation,

|f(x,y,z)0|=|(x+y+z)r|x2+y2+z22r(x2+y2+z2)r2x2+y2+z22r(x2+y2+z2)r2(x2+y2+z2)12r(x2+y2+z2)r21

The value of |f(x,y,z)0|=2r(x2+y2+z2)r21

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