# Normal to a level surface

1. Mar 25, 2017

### Taylor_1989

1. The problem statement, all variables and given/known data
Hi guys I am a bit stuck on how I am suppose to show, that thse funcitons are normal to level surface. I am I suppose to do a derivation of some sort, alls my notes say in a tiny box, that grad f is always in the direction to the normal of the surface. Do I have to calculate a direction vector, I really am not too sure. I am I have taken the partial derivative but what do I do from there?

2. Relevant equations

3. The attempt at a solution

2. Mar 25, 2017

### Staff: Mentor

Please show us what you have tried.

I should mention that how your textbook describes level surfaces is at least confusing, and at worst, wrong. For a function of two variables f(x, y), the level curves are curves in the x-y plane for which f(x, y) = c, for some constant c. For example, if f(x, y) = x2 + y2 - 4, the level curves are all circles if k > -4, and a single point if k = -4. Note that the graph of z = f(x, y) is a surface in three dimensions. The level curves are two dimensional curves that lie in various planes that are parallel to the x-y plane.

For a function of three variables f(x, y, z), the level surfaces are surfaces in three dimensions for which f(x, y, z) = c, with c a constant. It's not possible to graph of a function of three variables, w = f(x, y, z), as this would require four dimensions. One could graph the level surfaces f(x, y, z) = c, as each level surface requires only three dimensions.

As far as the part about showing that the gradient is normal to the level curve/level surface, you previously should have learned something about how to determine whether two vectors are perpendicular.

3. Mar 25, 2017

### Taylor_1989

First thanks for the clarification. Two vectors will be perpendicular we the dot product is equal to 0. But What I do understand is what vectors I am comparing?
See I get the partial derivative will give me grad f but where is the other vector. I cant see how I am suppose to show that the it perpendicular with out another vector. I am very new to this maths a day to be exact. So I am thinking that the arbitary postion that i choose say $(x_0,y_0,z_0$ is a postion vector?

4. Mar 25, 2017

### LCKurtz

I think you are overthinking this problem. If you want a vector perpendicular to a level surface, the answer is the gradient vector. That's all you have to do. Now, if you look back at the proof of the theorem that says the gradient is always perpendicular to a level surface, you might see an argument that shows the gradient is perpendicular to all the tangent vectors to the surface at a given point. That tells you the "other" vectors to which the gradient is perpendicular. But you don't usually reprove the theorem every time you want to use it. So if you want a vector perpendicular to a surface at a point $(x_0,y_0,z_0)$ on the surface $f(x,y,z)=C$, the answer is $\nabla f(x_0,y_0,z_0)$. That's all you have to do.

5. Mar 25, 2017

### vela

Staff Emeritus

I think you're supposed to figure out what the normal to each surface is based on your answers to exercise 3.1 and show $\nabla f$ is proportional to it.