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Chapter 12: Q 63. (page 932)

Prove Theorem 12.10. That is, show that see=s whenS is a subset of R2or R3.

Short Answer

Expert verified

The theorem is proved "If Sis the subset ofR2orR3,thensee=s

Step by step solution

01

Step 1. Given information.

We have given subsetsee=s

02

To find the complement of S

If S is the subset of S2 then the complement of S is

Se=x,y鈭�R2:x,y鈭�s

Now the complement of Se is :

see=x,y鈭�R2:x,y鈭�Se

According to the definition of complement of a set, it is clear that elements which are included in a set are not included in its complement. Similarly, the elements which are not included in the complement must be included in the original set.

Thus, if x,y鈭�sethen x,y鈭�s

Therefore,

see=x,y鈭�R2:x,y鈭�ssee=s

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Most popular questions from this chapter

Construct examples of the thing(s) described in

the following.

Try to find examples that are different than

any in the reading.

(a) A function z = f(x, y) for which 鈭噁(0, 0) = 0 but f is

not differentiable at (0, 0).

(b) A function z = f(x, y) for which 鈭噁(0, 0) = 0 for every

point in R2.

(c) A function z = f(x, y) and a unit vector u such that

Du f(0, 0) = 鈭噁(0, 0) 路 u.

Given a function of n variables, z=fx1,x2,鈥�,xn,and a constraint equation, gx1,x2,鈥�,xn=0,how many equations would we obtain if we tried to optimize f by the method of Lagrange multipliers?

Evaluate the following limits, or explain why the limit does not exist.

lim(x,y)鈫�(0,0)x3+y3x2+y2

Sketch the level curves f(x, y) = c of the following functions for c = 鈭�3, 鈭�2, 鈭�1, 0, 1, 2, and 3:

f(x,y)=y2鈭�x2

Evaluate the following limits, or explain why the limit does not exist.

lim(x,y,z)鈫�(0,0,0)x2+y2x2+y2+z2
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