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

  1. Jul 28, 2011 #1
    1. The problem statement, all variables and given/known data

    differentiability is a tough word to spell.

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

    Find [tex]F_y (0,0)[/tex]

    3. The attempt at a solution


    But I get 0/0

    I found the answer to be

    [tex]F_y (0,0) = \frac{\mathrm{d} }{\mathrm{d} y} F(0,y) = \frac{\mathrm{d} }{\mathrm{d} y} y = 1[/tex]

    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

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

    [tex]\lim_{(x,0)\to (0,0))} \frac{0}{(x^2 + 0)^{2/3}} = 0[/tex]

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

    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 continous 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.
    Last edited: Jul 28, 2011
  2. jcsd
  3. Jul 28, 2011 #2
    since the value of the limit is dependent on the x,y path, this means the limit does not exist.
  4. Jul 28, 2011 #3
    Okay...you aren't answering my question.
  5. Jul 28, 2011 #4


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    Remember. what [itex]\displaystyle \frac{\partial f(x,y)}{\partial y}[/itex] 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: [itex]\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 [/itex]. This, of course is zero.

    If x0 = 0, then you simply get [itex]\displaystyle \frac{\partial f(0,y)}{\partial y}=\frac{y^2}{((0)^2 + y^3)^{2/3}}=1[/itex]
  6. Jul 28, 2011 #5
    But why don't you get the same answer if you take the derivative first and then plug in the value?
  7. Jul 28, 2011 #6

    Ray Vickson

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

  8. Jul 28, 2011 #7
    If it is not continuous at that point, how can it even be differentiable in the first place?
  9. Jul 28, 2011 #8

    Ray Vickson

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    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 [itex] \partial{f}/\partial{x_j},\: j=1,\ldots,n[/itex] at that point.

    Last edited: Jul 28, 2011
  10. Jul 28, 2011 #9
    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?
  11. Jul 28, 2011 #10


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    What do you mean by "position function"?
  12. Jul 29, 2011 #11
    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.
  13. Jul 29, 2011 #12


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    Are you talking about a general scalar function of position? (i.e. temperature) T(x,y,z)
    [tex] dT = \frac{\partial T}{\partial x} dx + \frac{\partial T}{\partial y} dy + \frac{\partial T}{\partial z} dz [/tex]
    So the derivative of T with respect to x (for example) is:
    [tex] \frac{dT}{dx} = \frac{\partial T}{\partial x} + \frac{\partial T}{\partial y} \frac{dy}{dx} + \frac{\partial T}{\partial z} \frac{dz}{dx} [/tex]
    So this is what the derivative means when there are several variables.
    You also asked about units. [itex]\frac{dT}{dx}[/itex] will have units of temperature per length. And so will the partial derivative of T with x. For [itex]\frac{dT}{dt}[/itex] 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.
  14. Jul 30, 2011 #13
    Argh not quite what I want.

    Like say I have

    [tex]F(x,y) = x^2 y + 2y = C[/tex]

    Obvisouly the derivatives and partials are different
  15. Jul 31, 2011 #14


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    What derivatives & partial derivatives are you referring to?
  16. Aug 1, 2011 #15
    Let's say partial with respect to x vs normal derivative with respect to x. Clearly different.
  17. Aug 1, 2011 #16


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    If you have, say, [itex]F(x,y)= x^2y+ y= C[/itex], then you are defining y "implicitely" as a function of x: [itex]y= C/(x^2+ 1)[/itex]. We can write [itex]z= F(x,y)= x^2y+ y[/itex] and think of this as defining a surface in an xyz- coordinate system. Setting [itex]z= F(x,y)= C[/itex], 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,
    [tex]\frac{dz}{dt}= \frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}[/tex]
    For this particular function, [itex]F(x,y)= x^2y+ y[/itex] that gives
    [tex]\frac{dz}{dt}= 2xy\frac{dx}{dt}+ (x^2+ 1)\frac{dy}{dt}[/tex]
    and, because z= C,. a constant, that derivative must be 0:
    [tex]\frac{dz}{dt}= 2xy\frac{dx}{dt}+ (x^2+ 1)\frac{dy}{dt}= 0[/tex]

    If we think of that as the dot product of two vectors, [itex]2xy\vec{i}+ (x^2+ 1)\vec{j}[/itex] and a "velocity vector" [itex](dx/dt)\vec{i}+ (dy/dt)\vec{j}[/itex], tangent to the path, then that dot product is 0. That says that the vector
    [tex]grad(F)= \nabla F= \frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}[/tex]
    is perpendicular to the level curve itself. That is, [itex]\nabla F(x,y)[/itex] 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 [itex]\nabla F[/itex] is perpendicular to that surface. Since a plane is determined by a single point and a normal to the plane, the fact that [itex]\nabla F[/itex] 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
    [tex]\frac{\partial F}{\partial x}\vec{i}+ \frac{\partial F}{\partial y}\vec{j}+ \frac{\partial F}{\partial z}\vec{k}[/tex]
    if F is differentiable.)
    Last edited by a moderator: Aug 15, 2011
  18. Aug 2, 2011 #17
    Okay, but the partial wasn't continuous at that point, yet a derivative exist.
  19. Aug 14, 2011 #18
    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?
  20. Aug 14, 2011 #19


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    The partial derivative with respect to x is:
    [tex]\frac{ \partial E }{\partial x} \mid_{y \ constant} = 2xy [/tex]
    and the full derivative with respect to x is:
    [tex] \frac{dE}{dx} = \frac{ \partial E }{\partial x} \mid_{y \ constant} \ + \ \frac{ \partial y}{ \partial x} \mid_{x \ constant} \ \frac{dy}{dx} [/tex]
    But we don't know the dependance 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.
  21. Aug 14, 2011 #20
    I am asking for a physical meaning, computationally I know there is a difference.
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