The derivative of a function of one variable gives the slope of the
tangent line to the graph. The partial derivatives and of a function of two variables
determine the * tangent plane* to
the graph.

The graph of is a surface in 3 dimensions. Suppose we're trying to find the equation of the tangent plane at .

To write down the equation of a plane, we need a point on the plane and a vector perpendicular to the plane. We have a point on the plane, namely .

To find a vector perpendicular to the plane, we find two vectors in the plane and take their cross product. To do this, look at a small piece of the surface near the point of tangency. A small piece will be nearly flat, and will look like the parallelogram depicted below:

The vectors and which are the sides of the parallelogram are tangent to the surface at the point of tangency. Consider . It runs in the x-direction. A small change in x produces a change in z --- the amount the vector "rises".

How much does z change due to a change in x? The rate of change of z with respect to x is , so the change in z produced by changing x by is just .

Now is a vector with x-component , no y-component, and z-component . Therefore,

A similar argument shows that

The cross product is

I need *any* vector perpendicular to the surface. Since
vectors which are multiples are parallel, I may use this vector as
the perpendicular vector to the surface:

This is often referred to as the * normal vector*
to the surface and denoted by .

The tangent plane at is

The normal line to the surface at is the line which passes through and is perpendicular to the tangent plane. The normal line is parallel to the normal vector . Therefore, the parametric equations of the normal line are

* Example.* Find the equation of the tangent
plane and the parametric equations of the normal line to at .

The normal vector to the surface is

Plugging in and gives .

The tangent plane is

The normal line is

* Example.* Use a tangent plane to approximate
.

The idea is to think of this as the result of plugging numbers into a function . What is f? Well, the form of the expression suggests that 1.99 corresponds to one of the variables and 1.01 to the other. It's natural to use the function

I want to approximate . The point is close to , so I'll use the tangent plane at to approximate f.

The normal vector is

Plug in , . This gives .

When and , . The point of tangency is .

The tangent plane is

Now set , . This gives . (The actual value is 1.989803.)

Here is an equivalent way to think of things that is similar to the "approximation by differentials" technique you may have seen in first-year calculus. The change in f produced by small changes in in x and in y is approximated by

Thus,

Here denotes the "nice" point ( in the last example) and denotes the "ugly" point ( in the last example).

If you redo the example using this differential approach, you'd have

Suppose a surface is given parametrically:

Consider the vectors

These vectors are tangent to the *curves in the surface*
determined by letting one of u or v vary and holding the other
constant. For example, if u varies and is constant, I get the curve

The velocity vector for this curve is .

Likewise, consider the curve obtained by setting u to a constant:

The velocity vector for this curve is .

The cross product of and is a normal vector to the surface:

* Example.* Find the equation of the tangent
plane and the parametric equations for the normal line to

First, the point of tangency is obtained by plugging and into x, y, and z. I get , , and . The point is .

Next,

When and ,

The normal vector is

The tangent plane is

The normal line is

Copyright 2018 by Bruce Ikenaga