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Gradient Intuition

  1. Jul 23, 2013 #1
    Hi there, I just started to learn about gradients. I can calculate them with ease; but I don't think I really understand them conceptually. I understand the usual example of the temperature scalar field where the temperature in a room is a function of your position T(x, y, z). But when it comes to generic functions, I dont know what im doing! And it drives me crazy! Any/all help is welcomed! :)
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  3. Jul 23, 2013 #2


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    The gradient of a function of several variables is similar to the derivative of a function of a single variable.

    The gradient represents the rate of change of a multi-variable function with respect to one of the variables.

    The steeper the gradient, the more rapidly the function is changing with a given change in a particular variable.
  4. Jul 23, 2013 #3


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    The gradient vector of function f(x, y, z) always points in the direction of fastest increase and its length is that maximum increase.
  5. Jul 23, 2013 #4
    The (vector) gradient of a scalar function enables you to determine the rate of change of the function in any arbitrary spatial direction. To do this, you just dot the gradient with a unit vector in the arbitrary direction. If you want to determine the change in the function between two neighboring spatial points joined by a differential position vector, you just dot the gradient with the differential position vector. Thus, the vector gradient of a function is a very useful, and frequently used, tool to quantify directional changes in a function.

  6. Jul 24, 2013 #5


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    Think of the gradient as an n-vector when there are n variables. For intuition, consider the function f(x,y) = x^2. Can you interpret the gradient vector (sorry, I should say gradient vectors) as pointing in the direction of quickest increase? What about g(x,y) = x^2 + y^2?
    Last edited: Jul 24, 2013
  7. Jul 24, 2013 #6
    [A caveat: All the responses seen so far are true as long as the function (say [itex]f[/itex]) whose gradient you're computing is well-behaved enough, e.g. if the function [itex](x,y,...,z)\mapsto \nabla f(x,y,...,z)[/itex] is continuous.]

    Spatial metaphors are a very good way to get an intuition for these formal objects. Your temperature metaphor is a good one, and can serve your intuition even in situations where [itex]f[/itex] isn't a temperature.

    My analogy of choice: You're standing on a hill, and at coordinates (i.e. latitude and longitude) [itex]\vec x[/itex], the height of the hill is [itex]f(\vec x)[/itex]. At any location [itex]\vec x[/itex], the gradient [itex]\nabla f(\vec x)[/itex] somehow describes the "slope in all directions" all at once. How does it encode this? Well, consider any unit vector [itex]\vec u[/itex], thought of as a direction in which one could walk. If you walk a small amount in direction [itex]\vec u[/itex] from location [itex]\vec x[/itex], then [itex]\vec u \cdot f(\vec x)[/itex] is the slope of the incline you're walking. Again, this intuition might be helpful even if points [itex]\vec x[/itex] in the domain of [itex]f[/itex] are 19-dimensional instead of 2-dimensional.
  8. Jul 24, 2013 #7
    Okay its starting to come together now! Thanks everyone. On another note, if its not to much trouble; Could someone explain to me the uses of curl and divergence? Obviously theyre useful in someway but I can't really wrap my head around what they mean. Thanks again everyone!
  9. Jul 25, 2013 #8
    Whoops, typo. The corrected version (with an added [itex]\nabla[/itex]) is below.

    If you walk a small amount in direction [itex]\vec u[/itex] from location [itex]\vec x[/itex], then [itex]\vec u \cdot \nabla f(\vec x)[/itex] is the slope of the incline you're walking.
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