ATAR Notes: Forum

VCE Stuff => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematics => Topic started by: /0 on May 21, 2009, 11:24:07 am

Title: how to prove it is a cusp!?
Post by: /0 on May 21, 2009, 11:24:07 am: SAY YOU HAVE THE GRAPHS

$Y=X^2$ FOR $X\geq 0$

AND $Y=0$ FOR $X \geq 0$

THE GRADIENT AT X = 0 IS THE SAME FOR BOTH GRAPHS, YET THAT POINT IS A CUSP................................

HOW DOES ONE GO ABOUT PROOFING THAT IT IS A CUSP???????????????????????????????????????????????????
Title: Re: how to prove it is a cusp!?
Post by: Mao on May 21, 2009, 11:26:27 am: if $\lim_{x\to a^+}f'(x) \neq \lim_{x\to a^-} f'(x)$

then f(x) a cusp at x=a
Title: Re: how to prove it is a cusp!?
Post by: /0 on May 21, 2009, 11:29:26 am: But the limits are the same aren't they?????????????????????????????????
Title: Re: how to prove it is a cusp!?
Post by: Mao on May 21, 2009, 11:36:54 am: for a smooth join they are the same (hence differentiable)

but for a sharp turn they are two different values approaching from left and right.
Title: Re: how to prove it is a cusp!?
Post by: /0 on May 21, 2009, 11:44:33 am: But for the equation I put up, the equations are differentiable at x = O, and they approach the same values from left and right. HOWEVER, there is a cusp at x = O , so how does that work??????
Title: Re: how to prove it is a cusp!?
Post by: Mao on May 21, 2009, 11:54:16 am: neither of the two graphs have cusps. where is the question from?
Title: Re: how to prove it is a cusp!?
Post by: /0 on May 22, 2009, 08:38:37 am: Soz what I mean is, if you graph the two graphs as I have described in the first post, then visually, it will seem like there is a cusp at $x=0$ .
i.e. you could think of the relation as

$\{(x,y): y=x^2, y=0\ and \ x \geq 0\}$
Title: Re: how to prove it is a cusp!?
Post by: evaporade on May 22, 2009, 09:19:11 am: you look at lim (invf)'(y) as y approaches 0+ and 0-
Title: Re: how to prove it is a cusp!?
Post by: kamil9876 on May 22, 2009, 11:37:17 am: By using the basics of Mao's argument, you see that:

$\lim_{x\to 0^+} f'(x)= \lim_{x\to 0^-} f'(x)$ must NOT occur.

However: $\lim_{x\to 0^-} f'(x)$ doesn't even exist. Since $f(-\delta)$ doesn't exist, where $\delta>0$

In fact f isn't even a function.

If you define a cusp as a non-differentiable point the above is true.

However I wasn't sure about a formal definition of a cusp so i did some research and found: http://mathworld.wolfram.com/Cusp.html
Which is contrary to the popular statement y=|x| has a cusp at the origin... Meh lately these quasi-geometrical terms seem like crap to me... although if this is your hw I can see that you should be worried :P

Anyways, according to source above:

let $f(x)=x^2$
let $g(x)=0$

$f(0)=g(0)$
$\lim_{x\to 0^+} f'(x) = \lim_{x\to 0^+} g'(x)$ , but $\lim_{x\to 0^-} f'(x) = \lim_{x\to 0^-} g'(x)$ is not satisfied(because both limits do not exist) hence the two curves meet with same tangents however they are 'branches'(link uses this term) as they do not extend beyond this intersection.

You decide which definition of cusp to believe. I agree with the math but I'm not the best linguist ;D