Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have such family of sets: $X_{a,b} =\{ (x,y)\in \mathbb R^{+2}: ax^2<y\le \sqrt[3]{ab^2} \}$, $a,b \in \Bbb R$, $a,b>0$ and I have to find $\bigcup_{a}X_{a,b}$. Even though i had already done this in the past i forgot how. All I remember is that the result is a curve dependent from a. Could anyone give me at least a hint?

share|improve this question
    
A curve depending on $b$ not $a$, no? –  1015 Feb 9 '13 at 15:12
    
Oh yeah, you're right - on b –  Max Feb 9 '13 at 15:13
    
I think it is a two-dimensional object, rather than a curve. –  1015 Feb 9 '13 at 15:17
    
I am not really familiar with mathemathical terms in english. What i meant is that the 2D object is outlined by a curve. –  Max Feb 9 '13 at 15:20

1 Answer 1

up vote 1 down vote accepted

Consider the intersection of $\{y=ax^2\}$ and $\{y=\sqrt[3]{ab^2}$ in the first the upper-right quadrant. This is $$ \left(\frac{b^{1/3}}{a^{1/3}},\sqrt[3]{ab^2}\right). $$ Then convince yourself that your set is the open region delimited by the lines $\{x=0\}$ and $\{y=0\}$, and the curve $$\{\left(\frac{b^{1/3}}{a^{1/3}},\sqrt[3]{ab^2}\right)\;;\; a>0\}=\{(x,b/x)\;;\; x>0\}. .$$

Note: to prove the claim, Look for instance at the segment joining $(0,0)$ and the point above, whose interior points are all in your set, and then take the union of these over $a>0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.