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The Partial Derivative

  1. Jun 3, 2005 #1
    A couple of quickies on the interpertation of the partial derivative I want to clear up with myself.

    If we have a parametric function:

    r(u,v)= x(u,v)i + y(u,v)j+z(u,v)k

    then the partial derivative W.R.T u or v is regarded as the tangent vector, and we can think of it as the speed, or velocity vector at the point (u,v) along the curve.

    Now for a function defined by,

    xi + yj + z=f(x,y)k

    When we take the partial derivative with respect to x or y, can we still regard it as the speed or velocty vector at the point (x,y)?

    I dont see any reason not to call it that, they are both legitamite vectors tangent at the point (x,y) or (u,v).
     
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  3. Jun 3, 2005 #2

    OlderDan

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    What does this mean?

    [tex]x\widehat i + y\widehat j + z = f(x,y)\widehat k[/tex]

    I have no trouble conceiving of r(u,v) as a position vector represented in an orthogonal basis [itex](\widehat i, \widehat j, \widehat k)[/itex] and the partial derivatives being "velocity" relative to a single changing parameter (a.k.a. u = "time" with constant v), but if you take the partial wrt x at constant y,z of your equation you get

    [tex] \widehat i = \frac{\partial f(x,y)}{\partial x}\widehat k [/tex]

    which is a fundamental violation of everything we know about orthogonal vectors. Did you mean

    [tex]z(x, y) = -x\widehat i - y\widehat j + f(x,y)\widehat k[/tex]

    with z being some vector function of x,y?
     
  4. Jun 3, 2005 #3
    Sorry for being unclear, I meant that the the z component was a function of x and y, basically, f(x,y)k . I dont see why you put a negative sign infront of x and y though.
     
  5. Jun 4, 2005 #4
    Any help would be appreciated.
     
  6. Jun 4, 2005 #5
    maybe you could restate the second equation that you'd like help with? I agree with OlderDan with regards to everything else.

    If you clarify the last part of your post, perhaps we could help more.
     
  7. Jun 5, 2005 #6
    Ok, lets try this again from the top;

    A couple of quickies on the interpertation of the partial derivative I want to clear up with myself.

    If we have a parametric function:

    r(u,v)= x(u,v)i + y(u,v)j+z(u,v)k

    then the partial derivative W.R.T u or v is regarded as the tangent vector, and we can think of it as the speed, or velocity vector at the point (u,v) along the curve.

    Now for a surface defined by,

    x i + yj + f(x,y)k

    When we take the partial derivative with respect to x or y, can we still regard it as the speed or velocty vector at the point (x,y)?

    The tangent vector would be,

    1i + 0j + f_x(x,y) k <---(Can this not be considered the speed as well?)

    I dont see any reason not to call it that, they are both legitamite vectors tangent at the point (x,y) or (u,v).
     
  8. Jun 5, 2005 #7
    Your surface: x i + y j + f(x,y) k exists in R^3, the way you have written it is equivalent to writing it as

    f(x,y)k = -x i - y j, or omitting the k unit vector (as it is assumed)

    f(x,y) = - x i - y j. Note x and y are both independant variables. The partials are

    f_x(x,y) = -1. f_y(x,y) = -1


    That is my understanding of your question. For z(x,y) to be an independant variable, you would need a function f(x,y,z) in order to represent it's partials.
     
  9. Jun 5, 2005 #8

    arildno

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    Correct.
    Let us parametrize a surface S in the following manner:
    [tex]\vec{S}(x,y)=x\vec{i}+y\vec{j}+f(x,y)\vec{k}[/tex]
    then the tangent vectors are given by:
    [tex]\frac{\partial\vec{S}}{\partial{x}}=\vec{i}+\frac{\partial{f}}{\partial{x}}\vec{k}[/tex]
    [tex]\frac{\partial\vec{S}}{\partial{y}}=\vec{j}+\frac{\partial{f}}{\partial{y}}\vec{k}[/tex]
     
    Last edited: Jun 5, 2005
  10. Jun 5, 2005 #9
    Im sorry, but I have to disagree. A vector is not equal to its k component that you can move around and set equal to like that whozum. Its not an equation, its a vector, with meerely functions as components.

    To arildno, can we regard it as the "speed" like we do for parametric surfaces? I.e. is the "speed" W.R.T x always 1 in the i direction, and f_x(x,y) in the z dierction? Likewisee the speed W.R.T y, always 1 in the y dirction, 0 in the x direction, and f_y(x,y) in the z direction?

    Or is there something fundamentally wrong to call it the "speed"
     
    Last edited: Jun 5, 2005
  11. Jun 5, 2005 #10

    arildno

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    Why call it speed??

    Why muddle the issue in that manner; the concept of tangent vectors is good enough, isn't it?
     
  12. Jun 5, 2005 #11
    Because I want to know if calling something "speed" is reserved for a special condition on the derivative, or can we call this situation speed as well. I want to have a rigerious definition of when and were speed can be used to describe a derivative.
     
  13. Jun 5, 2005 #12
    Im sorry if I seem anal about it, but I want to be very precise in when and where I can call it the speed. The book calles the partial of a parametric function the speed, but it does not call a function in terms of x,y the speed, which is why I ask about the rigidity (is that a word?) of the use of the term speed.
     
  14. Jun 6, 2005 #13

    OlderDan

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    I have read the conversation that follow this post of yours, but I want to go back here to respond to this, and make some additional observations. If you have not discovered the answer yourself, the negative signs are there because I moved those terms to the other side of the equation to solve it for what you had called z. The equation only made sense if z is a vector function of x and y.

    By changing your statement to say

    x i + yj + f(x,y)k

    you have changed from an equation to an expression. That expression can be interpreted as defining a surface in three dimensional space. arildno reasonably turned your expression into an equation with S as a vector equivalent to your expression. I assume S was chosen because your expression represents a set of termination points of vectors from the origin to the surface defined by the function f(x,y). By making this equality as a reasonable way to represent your expression, a sensible way of representing the partial derivatives with respect to x and y is achieved.

    But notice what has been done here. Your "new" problem has been mapped into your "old" problem with S in place of r, x in place of u, y in place of v, and f(x,y) in place of z(u,v).

    [tex]\vec{r}(u,v)=x(u,v)\widehat{i}+y(u,v)\widehat{j}+z(u,v)\widehat{k}[/tex]

    [tex]\vec{S}(x,y)=x(x,y)\widehat{i}+y(x,y)\widehat{j}+f(x,y)\widehat{k}[/tex]

    This is nothing more than a special case of your more general starting equation, with the x and y coordinates being functions of only one of the two parameters. What holds in the general case will hold in any specific case. If you want to think of the parameter x as "time" then the special case becomes

    [tex]\vec{S}(t,y)=t\widehat{i}+y\widehat{j}+f(t,y)\widehat{k}[/tex]

    and

    [tex]\frac{\partial\vec{S}}{\partial{t}}=\widehat{i}+\frac{ \partial{f}}{\partial{t}}\widehat{k}[/tex]

    and sure enough, this is a "velocity" relative to the time parameter with the y parameter held constant. You could do the same thing wrt y instead of x. In this special case, what you have shown is that "velocity" wrt to one of the two parameters is always perpendicular to either the x axis or the y axis

    As arildno has noted, there is really nothing to be gained by thinking in terms of velocity, but it does fit with your general construct. When the parameters are the spatial variables x and y, a change dx with y held constant is equivalent to moving a distance dx along the surface in the x direction, and a distance [tex]\frac{ \partial{f(x,y)}}{\partial{x}}dx[/tex] in the z direction. The partial derivative is the rate of change of the vector S with respect to a change in the spatial variable x. If you want to think of x as "time" than it is also a velocity.
     
  15. Jun 6, 2005 #14
    Thanks! I was considering that but I wanted to wait for one of your guys anwsers. That helps me alot.
     
  16. Jun 6, 2005 #15
    Ok, another question on interpretations of partial derivatives. It shows in the text that one way of thinking about the partial derivative W.R.T x or y is the slope of the tangent vector.

    But as you two have pointed out, this vector is:

    [tex] \frac{\partial \vec{S}}{\partial{x}}=\widehat{i}+\frac{ \partial{f}}{\partial{x}}\widehat{k}[/tex]


    So If we look at this vector, it has only components in the i and k direction. So in this SPECIAL case, we can see that the slope is the rise in z over the run in x, or f_X(x,y)/ 1, so we see that the partial of z is the slope of the tangent line.

    But this is true only because the y component is zero. If the y component were not zero, it would be meaningless to mention the word slope. Is this reasoning correct?
     
  17. Jun 6, 2005 #16

    OlderDan

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    The situation you are now looking at is that of a vector pointing to a location on the surface. That vector will always have the form

    [tex]\vec{S}(x,y)=x\widehat{i}+y\widehat{j}+f(x,y)\widehat{k}[/tex]

    One of the most important things about partial derivatives is keeping track of what is being held constant. In your case, taking a partial with respect to x means that y is being held constant, and taking the partial with respect to y means that x is being held constant. If you combine the two you can write the differential

    [tex]d\vec{S}(x,y) = \frac{\partial{\vec{S}(x,y)}}{\partial{x}}dx + \frac{\partial{\vec{S}(x,y)}}{\partial{y}}dy = \left[\widehat{i} + \frac{ \partial{f}}{\partial{x}}\widehat{k}\right]dx + \left[\widehat{j} + \frac{ \partial{f}}{\partial{y}}\widehat{k}\right]dy = dx\widehat{i} + dy\widehat{j}+\left[\frac{\partial{f}}{\partial{x}}dx + \frac{\partial{f}}{\partial{y}}dy\right]\widehat{k} [/tex]

    What does this mean? Imagine yourself at the point on the surface (x, y, z=f(x,y)). Now you want to move to another point on the surface. The path you take is quite arbitrary, as long as you wind up where you want to go. You can walk along the surface, or you can walk in the x direction some distance, then walk in the y direction some distance, then look "up" and "down" (z direction) to see if you need to climb up to or down to get back to the surface. The differential vector [tex]d\vec{S}[/tex] tells you the components of the vector pointing from [tex]\vec{S}[/tex] to [tex]\vec{S} + d\vec{S}[/tex], both of which terminate on the surface. The distance you have to move in the z direction is the sum of the changes in the z coordinate of the surface resulting from the displacements dx and dy. If your path to the new location only requires you to move in the x direction, then in the z direction it is because dy has been chosen to be zero. If your path to the new location only requires you to move in the y direction, then in the z direction it is because dx has been chosen to be zero. It is simply a matter of the (x,y) coordinates of your destination relative to your starting point.

    And notice what happens if you divide the differential displacement by the time dt it takes you to move. You get the velocity of the motion

    [tex]\frac{d\vec{S}(x,y)}{dt} = \vec{v} = \frac{dx}{dt}\widehat{i} + \frac{dy}{dt}\widehat{j}+\left[\frac{\partial{f}}{\partial{x}}\frac{dx}{dt} + \frac{\partial{f}}{\partial{y}}\frac{dy}{dt}\right]\widehat{k} [/tex]
     
  18. Jun 6, 2005 #17
    Wow that is a neat trick. Could you explain how you arrived at your 3rd equation. Where did you get ds= ds/dx dx + ds/dy dy from?
     
  19. Jun 14, 2005 #18
    Hey Dan, a quickie. It seems that in this line you are using the tangent plane approximation to write:

    [tex]d\vec{S}(x,y) = \frac{\partial{\vec{S}(x,y)}}{\partial{x}}dx + \frac{\partial{\vec{S}(x,y)}}{\partial{y}}dy = \left[\widehat{i} + \frac{ \partial{f}}{\partial{x}}\widehat{k}\right]dx + \left[\widehat{j} + \frac{ \partial{f}}{\partial{y}}\widehat{k}\right]dy = dx\widehat{i} + dy\widehat{j}+\left[\frac{\partial{f}}{\partial{x}}dx + \frac{\partial{f}}{\partial{y}}dy\right]\widehat{k} [/tex]

    but you wrote it in vector form instead of scalar form. I have not seen that done before, could you please expand upon how that is correct?

    It makes sense to me, a more direct way would be to use the direct linearization for the scalar function. And thats obviously the z component. And thats associated with a displacement of dx in the i direction and dy in the j direction, so you cant intuit how you came to the formula. Its your notation that I have not seen before, i.e chanigng a scalar formula into a vector formula like that.
     
    Last edited: Jun 15, 2005
  20. Jun 16, 2005 #19
    Hello, are you there dan. lol :-)
     
  21. Jun 20, 2005 #20

    OlderDan

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    This comes from fundamental properties of differentials of multivariate functions. By definition of a function, an arbitrary function in 3-space, f(x, y, z), must have a unique value at any coordinate point (x, y, z). The change in f corresponding to changes in x, y and z is

    [tex] \Delta f(x, y, z) = f(x + \Delta x, y + \Delta y, z + \Delta z) - f(x, y, z) [/tex]

    If you hold y and z constant, this becomes

    [tex] \Delta f(x, y, z) = f(x + \Delta x, y, z) - f(x, y, z) [/tex]

    If you divide by [tex] \Delta x [/tex] and take the limit as [tex] \Delta x [/tex] approaches zero you have

    [tex] \frac{df(x, y, z)}{dx} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{f(x + \Delta x, y, z) - f(x, y, z)}{\Delta x} = \frac{\partial f}{\partial x}[/tex]

    [tex] df(x, y, z) = \frac{\partial f}{\partial x} dx [/tex]

    Changing only y and changing only z yields similar equations in those variables. The justification for adding these results to get the total differential is that the change in the function when moving a differential distance from one point in space to another must be path independent. So if I move a small distance dx, followed by a small distance dy, followed by a small distance dz, or make those same changes in position in any other order, the sum of the changes in f must always be the same, and that must be the same as making all three changes at the same time. Putting it together you have

    [tex] df(x, y, z) = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy + \frac{\partial f}{\partial z} dz [/tex]

    where in each term the partial derivative implies that the other two variables are being held constant. Always remember that a partial derivative implies something is being held constant, and it is very important to know what that is.

    That is not a very formal derivation, but the most rigorous derivation will lead to the same result. In your problem you have a surface defined by the function of two variables z = f(x, y). By application of the above to a function of two variables you have

    [tex] dz = df(x, y) = \frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy [/tex]

    Notice that in this case the partial derivatives imply only one other variable is being held constant, not two, because z is constrained to have only one specific value at every coordinate point (x, y) rather than being allowed to vary independently. dz is thus uniquely determined by the function f(x, y) and the changes dx and dy.

    The vector introduced by arildno

    [tex]\vec{S}(x,y)=x\vec{i}+y\vec{j}+f(x,y)\vec{k}[/tex]

    is just another way to describe the surface defined by f(x, y). Think of it as a sum of two vectors, one from the origin to the point (x, y) in the x-y plane and then a vertical vector from the x-y plane to the surface. To move to a new position on the surface, you must move to a new coordinate (x + dx, y + dy) and then move a distance dz = df(x, y) to get back to the surface. You can draw a differential vector from the initial point to the final point. That vector is the change in the vector to the surface from one point to the next. If z could change arbitrarily, moving from one point to another in 3-space would be

    [tex]d\vec{S} = dx \widehat{i} + dy \widehat{j} + dz \widehat{k} [/tex]

    but z cannot change arbitrarily because the S vector must terminate on the surface. Since z is a function of x and y, the vector S is a function of x and y and this becomes

    [tex]d\vec{S}(x,y) = dx \widehat{i} + dy \widehat{j} + df(x, y) \widehat{k} [/tex]

    [tex]d\vec{S}(x,y) = dx \widehat{i} + dy \widehat{j} + \left[\frac{\partial f}{\partial x} dx + \frac{\partial f}{\partial y} dy \right] \widehat{k} [/tex]

    If x and y are functions of time, or some other parameter t, then by the chain rule you have

    [tex]\frac{d\vec{S}(x,y)}{dt} = \vec{v} = \frac{dx}{dt}\widehat{i} + \frac{dy}{dt}\widehat{j}+\left[\frac{\partial{f}}{\partial{x}}\frac{dx}{dt} + \frac{\partial{f}}{\partial{y}}\frac{dy}{dt}\right ]\widehat{k} [/tex]

    The identical result is obtained by taking the derivative wrt to t of the defining equation for the S vector in terms of partial derivatives wrt x and y as done in the previous post.
     
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