• Feb 6th 2011, 04:12 PM
Hello,
I'm trying to get a better feel for the geometric interpretation of the gradient of a function. Assuming you have a function in the 3 dimensional Cartesian coordinate system. I'm having a problem seeing these three geometric facts of the gradient.

1.) The gradient is a vector at each point along the surface that points in the direction of the maximum rate of change.

2.) The gradient is orthogonal to the level surface of the function

3.) The gradient vector is orthogonal to the surface.

I'm confused about how these three geometric properties can be all be true. If the gradient vector points in the direction of the maximum rate of change doesn't that imply that it is tangent to the plane? How can the gradient vector also then be orthogonal to the surface. Any help or links to good visuals would be greatly appreciated.

Thanks,
• Feb 6th 2011, 04:34 PM
TheEmptySet
Quote:

Hello,
I'm trying to get a better feel for the geometric interpretation of the gradient of a function. Assuming you have a function in the 3 dimensional Cartesian coordinate system. I'm having a problem seeing these three geometric facts of the gradient.

1.) The gradient is a vector at each point along the surface that points in the direction of the maximum rate of change.

2.) The gradient is orthogonal to the level surface of the function

3.) The gradient vector is orthogonal to the surface.

I'm confused about how these three geometric properties can be all be true. If the gradient vector points in the direction of the maximum rate of change doesn't that imply that it is tangent to the plane? How can the gradient vector also then be orthogonal to the surface. Any help or links to good visuals would be greatly appreciated.

Thanks,

The directional derivative in the direction $\vec{u}$ where $||\vec{u}||=1$ (it is a unit vector) then

$\nabla f(x,y,z) \cdot \vec{u}$ is the directional derivative.

First recall the geometric definition of the dot product $\vec{a} \cdot \vec{b}=||a||||b|\cos(\theta)|$

First note that this function takes on it maximum value when $\cos(\theta)=1 \implies \theta=0$

Also since $\vec{u}$ is a unit vector we get

$\nabla f \cdot \vec{u}=||\nabla f||\cos(\theta)$

Using the first observation this has a maximum when $\vec{u}$ is parallel to $\nabla f$ so the gradient points in the direction of max rate of change.

For 2) since the function is constant on a level surface the directional derivative must be equal to zero. This would give

$\nabla f \cdot \vec{u}=||\nabla f||\cos(\theta) \implies \theta =\frac{\pi}{2}$
• Feb 7th 2011, 11:02 AM
HallsofIvy
Quote:

Hello,
I'm trying to get a better feel for the geometric interpretation of the gradient of a function. Assuming you have a function in the 3 dimensional Cartesian coordinate system. I'm having a problem seeing these three geometric facts of the gradient.

1.) The gradient is a vector at each point along the surface that points in the direction of the maximum rate of change.

Along what surface?

Quote:

2.) The gradient is orthogonal to the level surface of the function

3.) The gradient vector is orthogonal to the surface.
orthogonal to what surface?

Quote:

I'm confused about how these three geometric properties can be all be true. If the gradient vector points in the direction of the maximum rate of change doesn't that imply that it is tangent to the plane? How can the gradient vector also then be orthogonal to the surface. Any help or links to good visuals would be greatly appreciated.

Thanks,
(2), where you refer to the "level surface of the function" is the only one that makes sense because it is the only one in which you specify which surface you are talking about. And I think that is where your confusion is. If your function is f(x,y,z), then the gradient, $\nabla f$ is a vector pointing in the direction of fastest increase- but that is a vector in space, not "along" any surface because f, by itself, does not define a surface. The rate of change of f in the direction of a unit vector $\vec{u}$ is $\nabla f\cdot\vec{u}$. Assuming that $\nabla f$ is not itself 0 (that the function is not a constant function) that will be 0 if and $\vec{u}$ is perpendicular to $\nabla f$. That is, if you move perpendicular to $\nabla f$, the function is constant along such a surface- that surface is a "level surface" of the function.