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

[flyingpig]

Homework Statement

differentiability is a tough word to spell.

F(x,y) = (x^2 + y^3)^{\frac{1}{3}}

Find F_y (0,0)

The Attempt at a Solution

http://www.wolframalpha.com/input/?i=D[%28x^2+%2By^3%29^%281%2F3%29%2Cy]

But I get 0/0

I found the answer to be

F_y (0,0) = \frac{\mathrm{d} }{\mathrm{d} y} F(0,y) = \frac{\mathrm{d} }{\mathrm{d} y} y = 1

I am guessing they set x = 0 because they aer looking at a particular case of a partial derivative by keeping x constant?

Oh how you can't take the derivative and then plug in the number and not get the same answer? I am thinking this has to do with limits so I did the following

\lim_{(x,y)\to (0,0))} \frac{y^2}{(x^2 + y^3)^{2/3}}

\lim_{(x,0)\to (0,0))} \frac{0}{(x^2 + 0)^{2/3}} = 0\lim_{(0,y)\to (0,0))} \frac{y^2}{(0 + y^3)^{2/3}} = 1

So the limit does not exist? I mean I got a 1 in the limit where I kept x constant??

Also since the limit at (0,0) does not exist, doesn't that mean it is not continuous and hence no differentiable?? Or is this rule only applies to single variable calculus?

Sammy, Mark, HallsofIvy, other smart people whom I have rudely forgotten about your name, please help.

 

[Hariraumurthy]

since the value of the limit is dependent on the x,y path, this means the limit does not exist.

 

[flyingpig]

since the value of the limit is dependent on the x,y path, this means the limit does not exist.

Okay...you aren't answering my question.

 

[SammyS]

Remember. what \displaystyle \frac{\partial f(x,y)}{\partial y} means. Hold x fixed at some value, say x₀, then take the derivative (w.r.t y.) of f(x₀, y).
So, it then makes no sense to take the limit as x → 0.

In other words: \displaystyle \lim_{y\to 0} \frac{y^2}{((x_0)^2 + y^3)^{2/3}}= \frac{0}{((x_0)^2 + 0)^{2/3}}\ \text{ if }\ x_0\ne0. This, of course is zero.

If x₀ = 0, then you simply get \displaystyle \frac{\partial f(0,y)}{\partial y}=\frac{y^2}{((0)^2 + y^3)^{2/3}}=1

 

[flyingpig]

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

 

[Ray Vickson]

Because the derivative is not continuous at (0,0): if it was, you would get the same value either way.

RGV

 

[flyingpig]

If it is not continuous at that point, how can it even be differentiable in the first place?

 

[Ray Vickson]

The function is continuous but it's derivative is not. Besides, in the >= 2 variable case there ARE examples where a function is not continuous at a point, but it has all the partial derivatives \partial{f}/\partial{x_j},\: j=1,\ldots,n at that point.

RGV

 

[flyingpig]

I asked this once before, but no one raelly answered.

In terms of "units" (say F is a position function?)

What do the partials and normal derivatives mean?

 

[SammyS]

What do you mean by "position function"?

 

[flyingpig]

Hmm maybe that was a mistake. Like in single variables have s(t)...? Doesn't seem to work here since partials and normal derivatives mean the same thing.

 

[BruceW]

Are you talking about a general scalar function of position? (i.e. temperature) T(x,y,z)
dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z} dz
So the derivative of T with respect to x (for example) is:
\frac{dT}{dx} = \frac{\partial T}{\partial x} + \frac{\partial T}{\partial y} \frac{dy}{dx} + \frac{\partial T}{\partial z} \frac{dz}{dx}
So this is what the derivative means when there are several variables.
You also asked about units. \frac{dT}{dx} will have units of temperature per length. And so will the partial derivative of T with x. For \frac{dT}{dt} we have units of temperature per unit of time. (and same for partial derivative).
So to find units, its just units of the quantity on top divided by units of quantity on the bottom.

 

[flyingpig]

Argh not quite what I want.

Like say I have

F(x,y) = x^2 y + 2y = C

Obvisouly the derivatives and partials are different

 

[SammyS]

...
Like say I have

F(x,y) = x^2 y + 2y = C

Obviously the derivatives and partials are different

What derivatives & partial derivatives are you referring to?

 

[flyingpig]

Let's say partial with respect to x vs normal derivative with respect to x. Clearly different.

 

[HallsofIvy]

If you have, say, F(x,y)= x^2y+ y= C, then you are defining y "implicitely" as a function of x: y= C/(x^2+ 1). We can write z= F(x,y)= x^2y+ y and think of this as defining a surface in an xyz- coordinate system. Setting z= F(x,y)= C, a constant, we are looking at a "level curve" (a "contour line" on a geodetic map).

If we were, say, walking along such a contour line we would be walking along a curve of constant elevation. By the chain rule,
\frac{dz}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}
For this particular function, F(x,y)= x^2y+ y that gives
\frac{dz}{dt}= 2xy\frac{dx}{dt}+ (x^2+ 1)\frac{dy}{dt}
and, because z= C,. a constant, that derivative must be 0:
\frac{dz}{dt}= 2xy\frac{dx}{dt}+ (x^2+ 1)\frac{dy}{dt}= 0

If we think of that as the dot product of two vectors, 2xy\vec{i}+ (x^2+ 1)\vec{j} and a "velocity vector" (dx/dt)\vec{i}+ (dy/dt)\vec{j}, tangent to the path, then that dot product is 0. That says that the vector
grad(F)= \nabla F= \frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}
is perpendicular to the level curve itself. That is, \nabla F(x,y) is always perpendicular to the curve defined by F(x,y)= constant.

Of course, this can be extended to three (or more) dimensions: if F(x,y,z) is a differentiable function of the three variables, then F(x,y,z)= constant defnes a surface and \nabla F is perpendicular to that surface. Since a plane is determined by a single point and a normal to the plane, the fact that \nabla F is normal to the surface helps find the tangent plane at any point.

(By the way, a function of more than one variable can have partial derivatives at a given point and not be "differentiable" there so it is important to know that your function is differentiable before using its gradient. Strictly speaking the gradient of function F(x,y,z) is only defined by
\frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}+ \frac{\partial F}{\partial z}\vec{k}
if F is differentiable.)

 

[flyingpig]

Okay, but the partial wasn't continuous at that point, yet a derivative exist.

 

[flyingpig]

I am back from my cold.

Would it be more better to ask what if E(x,y) = x^2 * y + y electric field (yay gauss's law again...)

What would dE/dx vs partial derivative with respect to x mean?

 

[BruceW]

The partial derivative with respect to x is:
\frac{ \partial E }{\partial x} \mid_{y \ constant} = 2xy
and the full derivative with respect to x is:
\frac{dE}{dx} = \frac{ \partial E }{\partial x} \mid_{y \ constant} \ + \ \frac{ \partial y}{ \partial x} \mid_{x \ constant} \ \frac{dy}{dx}
But we don't know the dependence of y on x, so it can't be evaluated. However, if y doesn't depend on x at all, then the second term disappears, so the partial derivative with respect to x equals the full derivative with respect to x.

 

[flyingpig]

I am asking for a physical meaning, computationally I know there is a difference.

 

[BruceW]

If you think of the function z(x,y) As the height of the terrain, then the partial derivative with respect to x is the change in z when y is kept constant. (i.e. you just go in the x direction, and see how the height changes).
But the full differential with respect to x depends on what path you're taking along the terrain. So you won't be going in just the x direction and seeing how the height changes, you will be going along some path which is specified by y(x). For example, if we have y=x, then the path will be north-east, and we are seeing how the height changes when we move north-east.

 

[flyingpig]

Hmm I think this question is getting derailed...

 

[HallsofIvy]

Well, yes, because you keep changing the question!

 

[flyingpig]

Okay forget the question about partial vs normal derivative.

 

[flyingpig]

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?

 

[BruceW]

The partial differential with respect to y at (0,0) does not exist, since taking h+0 or h-0 gives different answers.
According to wikipedia: If a function is differentiable at a point, then all of its partial derivatives must exist at that point.
So I guess the function is not differentiable at (0,0).

Edit: I really don't know what I'm talking about, I do physics, not maths.
Maybe the partial derivative does exist, but its not continuous at (0,0) so the function is not continuously differentiable at (0,0)?

Another Edit: Wikipedia says that y = \sqrt[3]{x} is not differentiable at x=0, so it would make sense to me that your function is not differentiable at (0,0)

 

[flyingpig]

You can have side limits in multi-variable??

 

[BruceW]

Yes, as far as I understand it, you must get the same answer when coming in from any direction. So instead of just left or right sided, you have an infinite number of different directions which all must agree.

 

[flyingpig]

Let me try this again.

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

 

[Syrus]

consider the unit step function. Obviously it is not continuous nor differentiable at x = 0. HOWEVER, the limit from the left (approaching x = 0) appears to be y = 0, while the limit approaching x = 0 from the right appears to be y = 1. Since these two limit values do not match, clearly the limit does not exist.

I believe your post above inquires as to the many-valued analog of such an argument?

 

[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.