# B Gradient and direction

1. Oct 12, 2018 at 4:09 PM

### Mathematicsresear

<Moderator's note: Moved from a homework forum.>

1. The problem statement, all variables and given/known data

Is the gradient perpendicular to all surfaces or just level surfaces?
For instance, if I I have a function f(x,y)=z where z is the dependent variable then that is a surface, wouldn't that be a level surface to a function of x,y,z so shouldn't the gradient also be perpendicular to the surface and level surface?

2. Oct 12, 2018 at 4:13 PM

### Orodruin

Staff Emeritus
Yes. In particular, it would be the level surface $g(x,y,z) = 0$, where $g(x,y,z) = f(x,y) - z$.

The gradient of $g(x,y,z)$ would be perpendicular to the surface. It is unclear what you mean by

3. Oct 12, 2018 at 4:16 PM

### Mathematicsresear

I mean, would it also be perpendicular to the function g(x,y,z)?

4. Oct 12, 2018 at 4:21 PM

### Orodruin

Staff Emeritus
You cannot just say "gradient". You must specify the gradient of what function.

5. Oct 12, 2018 at 4:23 PM

### Charles Link

If you take $g(x,y,z)=$ constant , the gradient of $g(x,y,z)$ is perpendicular to the surface defined by $g(x,y,z)=$ constant. It works also for the case of the constant equal to zero, where $g(x,y,z)=f(x,y)-z$. $\\$ In particular, $\nabla g(x,y,z)$ evaluated at $(x_o, y_o, z_o)$ is perpendicular to the surface $g(x,y,z)=g(x_o,y_o, z_o)$ at $(x_o,y_o, z_o)$.

6. Oct 12, 2018 at 4:23 PM

### Mathematicsresear

Alright, I understand. The gradient if perpendicular to a surface, but it also is pointing in the direction of greatest increase, I'm not sure what the link between those to are. So is it both, pointing in the direction of greatest increase, and perpendicular to the surface?

7. Oct 12, 2018 at 7:37 PM

### Charles Link

The reason for greatest increase is because $dg=\nabla g \cdot d \vec{s}=|\nabla g| \cos{\theta} |d \vec{s}|$, where $d \vec{s}=dx \, \hat{i}=dy \, \hat{j} +dz \, \hat{k}$. (Use the definition of the gradient and work out the partial derivatives, etc. and compute $\nabla g \cdot d \vec{s}$). $\\$ The dot product picks up a $\cos{\theta}$ factor that is equal to 1 if $d \vec{s}$ is parallel to $\nabla g$. $\\$ We can also compute $\frac{dg}{ds}=|\nabla g| \cos{\theta}$.

Last edited: Oct 12, 2018 at 7:46 PM
8. Oct 12, 2018 at 10:35 PM

### RPinPA

If you have a function $z = f(x,y)$, then the gradient is a 2-vector in (x,y) space. It lies in the (x,y) plane. It is perpendicular to the contour lines of z, which are curves in (x,y) space. But it makes no sense to talk about that vector being perpendicular to the surface.

With 2 variables you can envision this very easily as a surface, for instance a hillside, on the surface of the earth. We'll pretend the earth is flat. Imagine you are standing on a hillside, and your (x,y) directions are East and North. Look around the hill from where you are standing. One direction goes most steeply up the hill, assuming you're not at a local maximum. That's the direction of the gradient of your hill surface. It is a compass direction, like "northeast".

Is "northeast" perpendicular to the hillside? Is that a meaningful question?

9. Oct 12, 2018 at 10:53 PM

### Charles Link

@RPinPA makes a good point that was previously omitted/overlooked: $\\$ $dz=(\frac{\partial{f}}{\partial{x}}) \, dx+(\frac{\partial{f}}{\partial{y}}) \, dy =\nabla^{(2)} f \cdot d \vec{r}=|\nabla^{(2)} f| \cos(\theta) | d \vec{r}|$, where $d \vec{r} =dx \, \hat{i}+ dy \, \hat{j}$. $\\$ Here $\nabla^{(2)}$ refers to a two-dimensional gradient. $\\$ The previous 3 dimensional gradient is perpendicular to the surface of the "hill" $f(x,y)-z=0$. Meanwhile, my post 7 is correct in regards to the function $g$, but doesn't correctly answer the question in regards to height $z$.

Last edited: Oct 12, 2018 at 11:03 PM
10. Oct 13, 2018 at 8:08 AM

### wrobel

Offtop for those who started to study tensor calculus

It is important to add that one should not be confused with two different objects
1) differential of a function which is a covector with components $(\frac{\partial f}{\partial x^i})$
and
2) gradient of a function $(\nabla f)^i=g^{ij} \frac{\partial f}{\partial x^j}$ which is a vector
These are two different types of tensors but their components coincide in Cartesian frame as long as it exists. Here $g^{ij}$ is the inverse Gramian matrix

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted