What is the Maximum Rate of Change in Scalar Fields?

[CoolDude420]

Homework Statement

Hi,

I'm having some doubts about the gradient. In my lecture notes the gradient of a scalar field at a point is defined to point in the direction of maximum rate of change and have a magnitude corresponding to the magnitude of that maximum rate of change of the scalar field.

Now, if I google online, people say that the Gradient is the maximum rate of increase.

Which is it? Maximum rate of increase or Maximum rate of decrease or maximum rate of change?
Say you are at the top of a hill and can't go any higher then clearly there is no maximum rate of increase then shouldn't the gradient there be zero? But if its maximum rate of change then shouldn't it be pointing downward?

Homework Equations

The Attempt at a Solution

 

[Charles Link]

A good example might be a hemisphere. (Edit: You can write $x^2+y^2+z^2=R^2$, so that $z=\sqrt{R^2-(x^2+y^2)}$). In this case the function is the height $z=f(x,y)$, and you take a two dimensional gradient: $\nabla f(x,y)=(\frac{\partial{f(x,y)}}{\partial{x}}) \hat{i}+(\frac{\partial{f(x,y)}}{\partial{y}}) \hat{j}$. $\\$ In general, $df=\nabla f \cdot d \vec{s}$. $\\$ For this 2-D case $dz=df= \nabla f \cdot (\hat{i} dx+\hat{j} dy)$. The change $dz$ is maximized for a given change $|d \vec{s} |$ when $d \vec{s} =\hat{i} dx+\hat{j} dy$ points along $\nabla f$, (gradient here in two dimensions), so that the dot product $\cos(\theta)$ factor is 1. $\\$ And you do have it right=at the top of the hill, the two dimensional gradient is zero. Anywhere else on the hemisphere, you will find $\nabla f$ points radially inward for the direction of steepest ascent. And if you travel at right angles to this direction, (on the hemisphere staying at the same height}, you will find $\frac{dz}{|ds|}=0$, just as it should, since the dot product $\cos(\theta)$ factor gives zero for $\theta =\pi/2$. $\\$ The same kind of calculation applies to a function $w=f(x,y,z)$ in taking a 3 dimensional gradient. $\\$ Editing: And for the above example, you could even make your direction of travel be in 3 dimensions, so that the differential distance traveled is $d \vec{s}'=\hat{i} dx+\hat{j} dy+\hat{k} dz$, but that adds some additional complexity. It's easier to just consider the x and y motion for the direction of travel from a mathematical viewpoint.

 

[Mark44]

A good example might be a hemisphere. (You can write $z=R \sqrt{x^2+y^2}$).

For the record, this is the equation of the upper half of a right circular cone. For a hemisphere that is the upper half of the sphere centered at the origin and of radius R, the equation would be $z = \sqrt{R^2 - x^2 - y^2}$.

 

[Charles Link]

For the record, this is the equation of the upper half of a right circular cone. For a hemisphere that is the upper half of the sphere centered at the origin and of radius R, the equation would be $z = \sqrt{R^2 - x^2 - y^2}$.

Thank you @Mark44 I now made the necessary correction. :)

 

[scottdave]

I recall it first being explained as walking through some hilly terrain. The "steepest path" is the gradient.

It could be a field of other values, such as the temperature at every point in a room, for example.

You may want to read this: http://mathworld.wolfram.com/Gradient.html

Also on Wolfram, there are some interactive apps that may help.