This problem is making me think, deeply about continuity and differentiability

[SammyS]

Guys please come back...

Okay looking through my book again, I went back to the definition of derivative.

f_y (0,0) = \lim_{h\to0} \frac{f(0,h+0) - f(0,0)}{h} = \lim_{h\to0} \frac{h - 0}{h} = \lim_{h\to0} 1 = 1

Alright, clearly there is some voodoo going on here.

How come I have to go back to the definition of the limit to get this?

No voodoo here.

The question was raised by BruceW as to whether this limit exists. Is the limit the same as h→0⁺ as it is for h→0^- ?

f(x,y) = \sqrt[3]{x^2 + y^3}

Therefore, \displaystyle f(0,y) = \sqrt[3]{ y^3}=y\,. This works fine for odd roots, not for even roots.

Thus for this function, f(0,h) = h, whether h is positive or negative.

I realize that this doesn't answer the larger question you asked, but I though I'd clear up this little detail. Yes, holding x fixed at a value of zero gives derivative w.r.t. y, which when evaluated at y=0 is equal to 1.

I hope to answer your question about 'voodoo' soon - unless I'm beaten to it.

 

[flyingpig]

So this is path-independent...

Just imagine (and believe), there are things even PF doesn't know.

 

[SammyS]

So this is path-independent...

Just imagine (and believe), there are things even PF doesn't know.

Not path independent.

Merely that the limit defining the derivative (w.r.t. y, evaluated at y=0) of f(0,y) exists .

 

[vela]

But why don't you get the same answer if you take the derivative first and then plug in the value?

When you writeF_y(x,y) = \frac{y^2}{(x^2+y^3)^{2/3}}you're assuming that x^2+y^3\ne 0. That's why you can't just naively differentiate and plug in x=y=0 to find \partial F/\partial y at (0,0). Instead, you have to do what they did in the solution and what SammyS explained back in post #4.

I've attached plots using F(x,y) below. The first one is a top view where you're looking down at the surface from a point on the z-axis. The white stripe corresponds to the curve x²+y³=0, which is where the F(x,y) is not differentiable. The second plot is a side view from a point on the x-axis, and the front edge is in the yz-plane. You can see the slope of the edge is constant. This is because ∂F/∂y(0,y) equals 1 for all y. The last plot is of the intersection of the surface with the xz-plane. You can see there's a cusp at the origin, so ∂F/∂x doesn't exist at the origin.

Are the definitions from Single Variables of Continuty and Differentiably not apply to multi-variables?

In the multivariate case, the derivative of a function depends on the path. The rate of change of the function may be different if you're moving along the x-axis or the y-axis or the curve y=x², etc. If you're thinking that the derivative should exist only when you get the same value independent of the path, that's wrong. You're generalizing what you learned about two-sided limits in the one-dimensional case to the multi-dimensional case incorrectly. I'm not sure that was your question, but that's what I think you were asking.

So there's a question of what it means for a function to be differentiable in the multivariate case. Look it up and tell us your understanding of this concept.

 

[flyingpig]

You can see the slope of the edge is constant. This is because ∂F/∂y(0,y) equals 1 for all y

Do you mean this part?

[PLAIN]http://img571.imageshack.us/img571/5297/xview.png

I'm not sure that was your question, but that's what I think you were asking.

Nope that was my question! I'll do my best to look up the definitions, but my theory is that there are no side limits.

 

[SammyS]

Do you mean this part?

The area you've circled is the label for values of x. The values of z are along the vertical edge at the left. The values of y are along the bottom.

Nope that was my question! I'll do my best to look up the definitions, but my theory is that there are no side limits.

What's a side limit?

 

[vela]

Do you mean this part?
[PLAIN]http://img571.imageshack.us/img571/5297/xview.png[/QUOTE]
No, in that plot, the y-axis runs horizontally, and the z-axis is vertical. The front edge of the surface is in the plane x=0. If you look at the labels, you can see that y goes from -1 to 1 while z also goes from -1 to 1.

 

[flyingpig]

No, in that plot, the y-axis runs horizontally, and the z-axis is vertical. The front edge of the surface is in the plane x=0. If you look at the labels, you can see that y goes from -1 to 1 while z also goes from -1 to 1.

But if y is the horizontal, then isn't it slanted?

 

[vela]

I take it you're referring to the surface. Yes, it is slanted because ∂F/∂y(0,y)=1. If it were horizontal, you'd have ∂F/∂y(0,y)=0.

 

[SammyS]

Here is a graph which is similar to vela's middle graph, the one you (flyingpig) and vela have posted recently as an image.

What we have here are a set of graphs with y on the horizontal axis and z on the vertical axis. The graphs are of a "family" of functions: z = f_n(y), with n=0,..., 5. (In this case, the subscript, n, is an index; n does not indicate differentiation.)

Each f_n is defined as: \displaystyle f_n(y)=\sqrt[3]{(n/5)^2+y^3}\,.

So, what we have here is that \displaystyle f_n(y)=F((n/5),y)\,, i.e., \displaystyle f_n(y)=F(x,y)\,, where x = 1/n, with x held constant for each value of n.

Hopefully, you can see that \displaystyle f_n'(y=0)=0\,, for n≠0. Also, f₀(y) = y , so f₀'(y) = 1, for all y.

Note: I spent some time yesterday getting WolframAlpha to graph the cube root as a real valued function. (It treats the cube root of a negative number as a complex valued function.) By the time I got the above graph, I noticed that vela had posted a more complete reply than I would have posted.

Perhaps this will help you understand what's going on with this interesting function & its partial derivatives.

 

[rude man]

dF = ∂F/∂x dx + ∂F/∂y dy
= 1/3 (x^2 + y^3)^(-2/3) (2x)dx + 1/3 (x^2 + y^3)^(-2/3) (3y^2)dy

dF/dy = 1/3 (x^2 + y^3)^(-2/3) (3y^2) since dx/dy = 0 if x,y independent variables.

So dF/dy at (0,0) = lim y^2/(X^2 + y^2)^(2/3) = lim y^(2/3) = 0

where limits are to x = y = 0.

 

[SammyS]

dF = ∂F/∂x dx + ∂F/∂y dy
= 1/3 (x^2 + y^3)^(-2/3) (2x)dx + 1/3 (x^2 + y^3)^(-2/3) (3y^2)dy

dF/dy = 1/3 (x^2 + y^3)^(-2/3) (3y^2) since dx/dy = 0 if x,y independent variables.

So dF/dy at (0,0) = lim y^2/(X^2 + y^2)^(2/3) = lim y^(2/3) = 0

where limits are to x = y = 0.

That's an interesting way to get ∂F/∂y .

I see that you fixed a typo. One remains. The exponent on y is 3, not 2, on your next to last line, i.e. .. = lim y^2/(X^2 + y^3)^(2/3) = ...

More importantly, (I was hopping that vela would have responded by now --LOL) that final limit is not correct.

\displaystyle \lim_{(x,y)\to(0,0)}\frac{y^2}{(x^2+y^3))^{2/3}}\ \text{Does Not Exist}\, as vela has adequately demonstrated.


.

 

[flyingpig]

No he didn't, I did it first!

 

[vela]

I was hopping that vela would have responded by now --LOL

I was waiting for you to post.

 

[flyingpig]

So there's a question of what it means for a function to be differentiable in the multivariate case. Look it up and tell us your understanding of this concept.

Sorry for the very very late response, but you guys have absolutely no idea how hard it was to dig this up. I'll post what I found

[PLAIN]http://img7.imageshack.us/img7/3756/unledzpe.png

Here is what we have

E(x,y) = (x^2 + y^3)^{\frac{1}{3}} - (mx + ny)

\lim_{(x,y) \to (0,0)} \frac{ (x^2 + y^3)^{\frac{1}{3}} - (mx + ny)}{x^2 + y^2}

Basically if that goes to 0, then we showed that f is differentiable at (0,0)

There is absolutely no continuity condition here. WHICH IS SOMETHING I DID NOT EXPECT.

The limit is unfortunately beyond me, but my attempt wasI called H = \frac{ (x^2 + y^3)^{\frac{1}{3}} - (mx + ny)}{x^2 + y^2}

So basically I took x = 0, then H = \frac{1-n}{y}. I tried doing y = 0, but that was even worse...

 

[deluks917]

A function is continuous if it is differentible. The important point is that the existence of partial derivatives does not imply a function has a "derivative". A function is differentible if you can "linearize" it, this is a good way to think about it.