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Gradient Confusion

  1. Mar 8, 2012 #1
    Hi I am having trouble getting my head around the definition of a gradient. I know a gradient tells us the direction of steepest slope that one must follow to arrive at a maximum and I know it is defined as:

    [​IMG]

    However I haven't got a gutt feeling for it, I need these questions answering before I can accept it:

    Where is the definition derived from?

    Why does adding the partial derivatives tell us the direction of maximum gradient?....I know this sounds stupid but if a function has gradients of 4,5,6 (x,y,x) what does that exactly mean? and why does adding them up points in the direction of the greatest rate of increase ?

    Can someone explain this in simply laymens terms to me, preferably using an example...


    I thank you in advance guys and gals
     
  2. jcsd
  3. Mar 8, 2012 #2

    HallsofIvy

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    If you were draw a line through point [itex](x_0, y_0, z_0)[/itex] in the direction of vector [itex]v= <v_x, v_y, v_z>[/itex] you can write the line in parametric equations as [itex]x= v_xt+ x_0[/itex], [itex]y=v_yt+ y_0[/itex], [itex]z= v_zt+ z_0[/itex]. On that line we can write the function f(x,y,z) as f(x(t),y(t), z(t)). Applying the chain rule to that, we have [itex]df/dt= (\partial f/\partial x)(dx/dt)+ (\partial f/\partial y)(dy/dt)+ (\partial f/\partial z)(dz/dt)[/itex].

    We can write that as a dot product: [itex]\left<\partial f/\partial x, \partial f/\partial y, \partial f/\partial z\right>\cdot\left<dx/dt, dy/dt, dz/dt\right>[/itex].

    That is why it is useful to define the gradient [itex]\nabla f= \partial f/\partial x\vec{i}+ \partial f/\partial y \vec{j}+ \partial f/\partial z\vec{k}[/itex].
     
  4. Mar 10, 2012 #3
    Thanks for your reply.....you lost me at the chain rule. I know how to perform the chain rule, but I don't know how you managed to apply the chain rule on:

    [itex]x= v_xt+ x_0[/itex], [itex]y=v_yt+ y_0[/itex], [itex]z= v_zt+ z_0[/itex]

    to arrive at:

    [itex]df/dt= (\partial f/\partial x)(dx/dt)+ (\partial f/\partial y)(dy/dt)+ (\partial f/\partial z)(dz/dt)[/itex]

    This is the part I am getting stuck on, can you break it down for me...


    Thanks alot
     
  5. Mar 12, 2012 #4
    I have a question on the application of the gradient vector:

    say A textbook question told me to "Find the directional derivative of f(x,y,z)=xy+z^2 at (1,1,1) in the direction towards (5,-3,3)."

    I have attempted this problem 3 times yet my answer is not the answer in the back of the answer guide: 2/3

    I found the gardient vector:<y,x,2z >; @ the point (1,1,1) => (1,1,2)
    I found the unit vector in the given direction: <5/√43,-3/√43,3/√43>
    and the dot product of the gardient and unit vector: 5/√43 - 3/√43 + 6/√43= 8/√43


    8/√43 is not the correct answer in the back of the book... I know typos exist but I suspect I've done something wrong or misunderstood the question. any help?
     
  6. Mar 12, 2012 #5

    arildno

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    Don't worry. I don't get it, either. your textbook is WRONG, no matter what it thinks about.
     
  7. Mar 12, 2012 #6
    Thanks for confirming...
     
  8. Mar 12, 2012 #7
    haha you completely hi-jacked my thread....:cry:
     
  9. Mar 12, 2012 #8
  10. Mar 12, 2012 #9

    HallsofIvy

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    I'm not sure how to "break it down" any more- that is the "chain rule" for functions of more than one variable: if f is a function of x, y, and z, and x, y, and z are themselves functions of the variable t, then we could replace each by its expression as a function of t so that f is itself a function of t and then
    [tex]\frac{df}{dt}= \frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt}+\frac{\partial f}{\partial z}\frac{dz}{dt}[/tex]
    and variations on that:
    If x, y, and z are functions of u, v, and w, say, then
    [tex]\frac{\partial f}{\partial u}= \frac{\partial f}{\partial x}\frac{\partial x}{\partial u}+\frac{\partial f}{\partial y}\frac{\partial y}{\partial u}+\frac{\partial f}{\partial z}\frac{\partial z}{\partial t}[/tex]
    etc.

    Or if f is a function of the single variable x and x is a function of u, v, and w,
    [tex]\frac{\partial f}{\partial u}= \frac{df}{dx}\frac{\partial x}{\partial u}[/tex]
    etc.
     
  11. Mar 12, 2012 #10

    mathwonk

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    If you hve a simple function like f(x,y,z) = x+2y+3z, then does it make sense that the direction in which it increases fastest is perpendicuklar to the directions in which it is constant?

    If so then the direction of fastest increase say at (0,0,0) is perpendicular to the plane x+2y+3z = 0.

    Do you know why this direction is along the vector (1,2,3)?


    In general any smooth function f(x,y,z) is constant along a surface f(x,y,z) = c,

    whose tangent plane is parallel to ∂f∂x x + ∂f/∂y y + ∂f/∂z z = 0.

    then the direction of greatest increase is prpendicular to that surface and hence to that plane

    and hence is parallel to the gradient (∂f/∂x, ∂f/∂y, ∂f/∂z).
     
  12. Mar 13, 2012 #11

    lavinia

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    For intuition about the gradient, I like to think about ideas of potential energy in classical physics.

    The work done by a force field on a particle - e.g. gravity on a point mass or and electric field on an electron - is the integral of the inner product of the field with the velocity vector of the particle along the curve. This is just the law work = Force x distance on a curved path.

    In many situations, this work represents a change in the potential energy of the particle e.g. a change in gravitational potential or electrostatic potential. The force field is just the gradient of the potential.

    Along a surface where the potential is constant, the force field does no work on the moving particle since the potential does not change. This means that the inner product of the force field with the velocity vector of a curve is zero if the curve lies on the constant potential surface. In other words the gradient of the potential is perpendicular to the surface. This perpendicular direction must be the direction of maximum change in potential energy since any other direction will have a component tangent to the surface that will have no effect on the change in potential.
     
    Last edited: Mar 13, 2012
  13. Mar 13, 2012 #12

    mathwonk

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    good illustration. e.g. you do no work against gravity unless you move something perpendicular to the surface of the earth, i.e. up or down.
     
  14. Mar 13, 2012 #13


    thanks for that. I understand that the chain rule but I don't understand the why you start adding the gradients in each direction, why not just leave the comma between them in place......?
     
  15. Mar 15, 2012 #14
    OkI have identified where I am getting confused, I do not know how to derive the Multivariable Chain Rule. I am going to try and figure out how this works.
     
  16. Mar 15, 2012 #15

    HallsofIvy

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    You both may be misunderstanding the phrase "in the direction towards (5, -3, 3)". I interpret that as the vector from (1, 1, 1) to (5, -3, 3) which is <4, -4, 2> which has length [itex]\sqrt{16+ 16+ 4}= 6[/itex]. The unit vector in that direction is <2/3, -2/3, 1/3>. The directional derivative is <1, 1, 2>.<2/3, -2/3, 1/3>= 2/3.
     
  17. Apr 2, 2012 #16
    Ok I have gone away and done some work. So far I understand the explanation up to applying the chain rule and deriving :

    [itex]df/dt= (\partial f/\partial x)(dx/dt)+ (\partial f/\partial y)(dy/dt)+ (\partial f/\partial z)(dz/dt)[/itex].

    After that can you explain, in a little more detail, how to get from that ^ to this :

    [itex]\nabla f= \partial f/\partial x\vec{i}+ \partial f/\partial y \vec{j}+ \partial f/\partial z\vec{k}[/itex].[/QUOTE]

    Things I am confused about:

    Why have you applied the dot product?
    Is it correct to just cancel out all the dx's, dy's and dz's, in the chain rule representation and just replace them with a unit vector?


    Thanks a bunch

    jon
     
  18. Apr 3, 2012 #17

    arildno

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    There is no worthwhile answer to your "why" other than that it is VALID to represent the sum you understand in terms of the dot product. (Multiplying out the dot product gives you the sum back!)
    Do you see that?
     
  19. Apr 3, 2012 #18
    just think of it as a generalization of the derivative operator to higher dimensions
     
  20. Apr 3, 2012 #19
    Actually yes I understand that. What about when you cancel the dx's, dy's and dz,s in the chain rule representation. Is it valid and necessary to add the unit vectors, i, j and k in to represent the direction of each derivative of df. Is adding the unit vectors just a way of representing the grad in vector form?????
     
  21. Apr 3, 2012 #20

    arildno

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    Well, if you did NOT have the vectors, you would not have a "dot product".
    Note that giving this product in vectorial form immediately performs the necessary cancelletion og terms of the type (df/dx)*(dy/dt) that does not occur in your original sum,( but would appear if you tried a a scalar produc like (df/dx+df/dy)*(dx/dt+dy/dt)), due to 0 in dot product value between orthogonal vectors.
    Having the vectors of unit length as well prevents multiplicative factors in each term other than 1.
     
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