so I'm confused on this:

If the following is true, find fx(0, 0).

f(x,y) = (x^3+y^3)^(1/3)

If I differentiate this with respect to x, then wouldn't I have
(3x^2)/((x^3+y^3)^(2/3))? Taking the point (0,0) would then result in a
0/0, however, the answer is neither 0 nor NONE. (yes this is for webassign...)

I have also determined (or so I believe), that the graph at point (0,0)
does NOT have a limit and is therefore NOT continuous. What am I missing
here?

Thanks in advance... I've already e-mailed the prof. last week, no response. I don't see him till tomorrow evening, and I can't figure this out and it's killing me...

9/29/2008 10:43:43 PM

the answer is 1

when I worked it out I ended up with 1+ 0/0 so i dunno if that's still 1 or not.

[Edited on September 30, 2008 at 1:05 AM. Reason : .]

9/30/2008 1:03:36 AM

I get 0/0 as well.

^ How did you get 1 + 0/0? And BTW, that's not = 1. That's undefined.

9/30/2008 2:17:18 AM

you should use the limit definition of the partial derivative to find f_x (0,0).

f_x (x,y) = lim_(h->0) [f(x+h,y)-f(x,y)]/h

9/30/2008 6:40:55 AM

so then the answer is NONE

9/30/2008 8:46:13 AM

I got (3x^2)/(3(x^3+y^3)^(2/3))

that became

(x^2)/((x^3+y^3)^(2/3))

and that became

(x^2)/(x^2+y^2) (pretty sure this step is wrong)

1+(x^2)/(y^2)

that's as close as I could come. But it's still wrong because you can't distribute the exponent like that. So I guess it's none and another webassign glitch.

9/30/2008 10:44:55 AM

^wrong

9/30/2008 11:41:02 AM

Quote :

"(x^2)/((x^3+y^3)^(2/3)) and that became (x^2)/(x^2+y^2)"

holy shit dude, that's a classical grade 7 (and up till college) algbera mistake right there

(a + b)^2 != a^2 + b^2

9/30/2008 12:24:00 PM

Quote :

"f(x,y) = (x^3+y^3)^(1/3) If I differentiate this with respect to x, then wouldn't I have (3x^2)/((x^3+y^3)^(2/3))? Taking the point (0,0) would then result in a 0/0, however, the answer is neither 0 nor NONE. (yes this is for webassign...)"

Actually I found an error. If we differentiate the equation with respect to x, you get (1/3)(3x^2)/((x^3+y^3)^(2/3)) which is actually just (x^2)/((x^3+y^3)^(2/3)). Too bad this doesn't help me I'm going to talk to the teacher tonight and hopefully get this corrected ( I am certain that the answer is NONE) Thanks guys

9/30/2008 1:50:12 PM

^ that's what I got initially and it didn't help me so I tried some voodoo math.

^^ it was 1 in the morning and I was on a study break for a test today, give me a fucking break, I pointed out my own mistake.

[Edited on September 30, 2008 at 4:36 PM. Reason : .]

9/30/2008 4:35:59 PM

zomg. For shits and giggles, I decided to submit "1" as my answer. It took it and said it was right... WTF? Even my calc III teacher said it's not defined and therefore there is no answer.. this is strange...

9/30/2008 9:07:26 PM

way to go jerk. you broke math

9/30/2008 9:30:38 PM

Duck tape... anyone?

[Edited on September 30, 2008 at 9:42 PM. Reason : ]

9/30/2008 9:42:04 PM

what now bitches.

lol I got used to just doing random stuff to cancel things out to find an answer for webassign.

[Edited on September 30, 2008 at 9:46 PM. Reason : .]

9/30/2008 9:45:37 PM

I would like to think that the answer is not 1 because you said 1+0/0. I just believe the answer is 0/0 and when you divide it out, it's um, 1... you know... the same way x/x =1 or 8/8 =1

9/30/2008 10:12:15 PM

it's not 1... I did it wrong the first time, but if something absolutely wouldn't work in webassign i'd do something to try to cancel something out, despite whether or not the algebra is right. I'm sure it's a typo in the equation.

your professor is right.

9/30/2008 10:18:24 PM

The answer is 1.

You can show this by using the limit definition.

f_x (x,y) = lim_(h->0) [f(x+h,y)-f(x,y)]/h

So, f_x (0,0) = lim_(h->0) [f(h,0)-f(0,0)]/h = lim_(h->0) [ (h^3)^(1/3) - 0]/h

=lim(h->0) [h]/h = lim(h->0) 1 = 1

Thus, f_x (0,0) = 1.

I have taught Calc 3 at NC State before and I am a professor at WVU now. I hope your Calc 3 professor would know how to do such a problem.

[Edited on October 1, 2008 at 5:46 AM. Reason : x]

10/1/2008 5:44:47 AM

that totally makes sense. You, sir, are the man. Thank you.

Gosh, I didn't realize I had to go back to "basics"...

[Edited on October 1, 2008 at 8:50 AM. Reason : .]

10/1/2008 8:50:17 AM

f(x,y) = (x^3+y^3)^(1/3)

gives,

f_x (x,y) = (1/3)(x^3+y^3)^(1/3-1)*[3x^2]

which simplifies to,

f_x (x,y) = x^2 / [ (x^3 + y^3 )^(2/3) ]

then bring the x^2 inside the 2/3 power term it becomes x^3 by laws of exponents,

f_x (x,y) = { x^3 / [ x^3 + y^3] }^(2/3)

then simply rewrite by dividing top and bottom of fraction by x^3,

f_x (x,y) = { 1 / [ 1 + y^3/x^3] }^(2/3)

curious, there ought to be a way to see this without limits...

10/1/2008 9:01:47 PM