# Del Operator

1. Feb 21, 2010

Can anyone please explain the Del operator. I am much confusesd about it.
One formula says that it gives the normal vector to the tangent plane and other tell me that it is gradient of the surface. What it actually is??

2. Feb 21, 2010

My question is basically what gradient of the function gives??
It gives normal to the surface or tangent to the surface??

3. Feb 21, 2010

### arildno

Suppose you have a function f(x,y,z), and some point P $(x_{0},y_{0},z_{0})[/tex] Now, let us evaluate the rate of change of f at P, that we get by walking away from P, along some line, with direction vector [itex]\vec{v}$

That is, we form the auxiliary function:
$$h(t)=f(x_{0}+t*v_{x},y_{0}+t*v_{y},z_{0}+t*v_{z}), \vec{v}=v_{x}\vec{i}+v_{y}\vec{j}+v_{z}\vec{k}, ||\vec{v}||=1$$

Now, AT P, we have:
$$\frac{dh}{dt}\mid_{t=0}=\nabla{f}\mid_{t=0}\cdot\vec{v}$$

Note here that there is NO rate of change of f AT P if that dot product is 0.

Thus, such vectors $\vec{v}$ must represent TANGENT VECTORS to the LEVEL SURFACE of f at P, i.e, the surface upon which f has the same constant value that it has at the point P itself!

The MAXIMAL rate of change of f will occur if we walk along a line that is PARALLELL to the gradient of f, and that must be a line that is ORTHOGONAL to the level surfaces.

And hence, the gradient of f at P is along the vector normal from the level surface to which P belongs.

4. Feb 21, 2010

### slider142

The del operator in basic vector calculus in Rn symbolises the "operator vector" $$\left(\frac{d}{dx^1}, \frac{d}{dx^2}, \cdots, \frac{d}{dx^n}\right)$$
This is such that for a function f from Rn into R, we have:
$$\nabla f = \left(\frac{df}{dx^1}, \frac{df}{dx^2}, \cdots, \frac{df}{dx^n}\right)$$
This operation is referred to as the gradient of f, and the vector generated, when taken as being attached to the point at which the gradient is taken, is called the gradient vector of f at that point. The gradient vector of f at a point is always normal to level sets of f, which can be shown by a short argument. (A level set of a function is a set of points for which f is a particular constant).
In the special case that f is a map from R3 into R, a level set of f is given by the equation f(x, y, z) = k where k is a constant. This is the equation of a surface in R3 and thus the gradient vector of f at each point will be normal to the surface (a level set of f).

5. Feb 21, 2010

### HallsofIvy

If a surface is give by f(x,y,z)= constant (i.e. you can think of this as a "level surface" for f), then $\nabla f= grad f$ is a vector normal to the surface.

Now, if you know a point on a surface, $(x_0,y_0,z_0)$ and a normal vector, <A, B, C>, there then the tangent plane is given by $A(x-x_0)+ B(y-y_0)+ C(z-z_0)= 0$.

That is, grad f is a normal vector to the surface which can then be used to find the tangent plane.

But understand that a surface does not have a "tangent line" or "tangent vector". There are many lines and vectors tangent to a surface at any point. A surface has a single normal line and a single tangent plane at any point.

6. Feb 21, 2010

Sir,one last question Normal to the level curve is parallel to normal to the surface in 2 variables??

7. Feb 21, 2010

### arildno

Sure. And in 4 variables, 17 and whichever other number of variables you have.

The principle is identical.

8. Feb 21, 2010

### HallsofIvy

I think a better word is "analogous". "Parallel" had me thinking about the geometry of the situation!