Given by Prof. Alan Zehnder on 9/8/95 at 10:10 in Kaufmann Auditorium
Surfaces, Gradients and Contours
In Lecture 3 we finished up the topic of line
integrals. In this lecture we move on to the subject of the Gradient
function, and how it relates to surfaces in space.
There are two ways to define a surface in 3-dimensional space.
- You can define a function f(x,y,z) and require that it always have the
same constant value; f(x,y,z)=C.
For Example, a sphere would have f(x,y,z)=x^2 + y^2 + z^2, and C=R^2,
where R is the radius of the sphere.
- You can also set z to be a function of x and y; z=g(x,y). For the same
sphere as above, this would give you:
z=+sqrt(R^2-x^2 - y^2) and z=-sqrt(R^2-x^2 - y^2)
The gradient of a function is a Vector defined as:grad f =
Since the gradient can be taken of any function, a physical interpretation
is specific to what the function represents and how it represents it.
- If u is a unit vector, the derivative of a function f(x,y,z) in
the u direction is f (dot)
u. The derivative of f in the u direction is defined to be
the limit below:
and u is a unit vector.
- The direction of f(x,y,z) is the
direction of "steepest ascent" of the surface.
This can be seen by looking back at the first usage of the gradient. If
you attempt to maximize f (dot)
u, you can see that this is the case when f is in the same direction as u. Thus, the direction
of most severe change in f is the direction of f.
- Given z=g(x,y) representing a surface (hill),
g will point in the direction in the (x,y) plane you would walk
to go up the steepest part of the hill, and ||g|| would be the slope of the hill.
Level Curves are curves in the x-y plane that trace out g(x,y)=C, where C
is a constant. That is, curves which yield a constant height on the hill
created by z=g(x,y).
g points perpendicular to the level curves.
This can be proven by thinking about the first use of gradients again.
g (dot) u is the
derivitive in the u direction. g (dot) T, a tangent to a level curve, is the change
of g when one moves in the T direction. Since T is along a
line of g(x,y) = A constant, there is no change. So g (dot) T = 0. Thus, g must be perpendicular to T, and
therefore to the level curve.
- Let f(x,y,z) = C define a surface.
f is perpindicular to the surface.
The proof of this is also fairly simple.
Take C, an arbitrary path along the surface, and T, the unit tangent to
that path. The derivative of f along C is 0, since f(x,y,z) = a constant on
the surface, and thus does not change as you move along the surface. So,
f (dot) T = 0.
Since C is an arbitrary path, this implies that f is perpendicular to the surface.
Thus the normal to a surface defined by f(x,y,z) = C is f / ||f||
Given the paraboloid z= g(x,y) = x^2 + y^2 - C or f(x,y,z) = x^2 - y^2 - z
Find f, g, observe the relationship between them, and find the
derivitive of f in the x direction.
g = 2xi + 2yj =
2(xi + yj)
f = 2xi + 2yj - k
Note that g is the projection of f onto the x-y plane.
To find the derivative of f in the x (i) direction:
f (dot) i = 2x