Gradient Field Plots with Contours
Dr. K. G. TeBeest

> with( plots ) :

> f := x^2 + 2*y^2 ;

> gp := gradplot( f, x = –8.0 .. 8.0, y = –8.0 .. 8.0, grid = [11,11], color = f, arrows = THICK, scaling = constrained ) :

> cp := contourplot( f , x = –8.0 .. 8.0, y = –8.0 .. 8.0, thickness = 3, scaling = constrained, color = black ) :

> display( { gp, cp } );

Example 2:   Hypberbolic Paraboloid

> f := x^2 – y^2 ;

> gp := gradplot( f, x = –4.0 .. 4.0, y = –4.0 .. 4.0, grid = [11,11], color = f, arrows = thick, scaling = constrained ) :

> cp := contourplot( f , x = –4.0 .. 4.0, y = –4.0 .. 4.0, thickness = 3, scaling = constrained, color = black ) :

> display( { gp, cp } );

NOTES: Recall that function z = f ( x, y) can be interpreted as a surface in 3-D.

1. The gradient vector of f at point (a,b) is orthogonal to the contour of f through (a,b).
2. The gradient vector of f at point (a,b) points in the direction in which function f increases most rapidly.
   So at point (a,b), f is steepest in the direction of the gradient vector.
3. The length of the gradient vector of f at point (a,b) gives the rate of change of f in the direction of the gradient vector.
   So f is steeper where the gradient vectors are longer (and where the contours are closely packed); f is less steep where the gradient vector is shorter (and where the contours are further apart).
4. If we view function f as a surface in 3-D, then water would flow along a path parallel to the gradient vectors, but in the opposite direction of the gradient vectors. The water's path would also be perpendicular (orthogonal) to the contours.