proving gradient of a function is always perpendicular to the contour lines Can someone give an explanation of how such a proof would go, given a function
example:   $y = f(x)$

Gradient Vector and level curves
The gradient vector is orthogonal to the tangent vector of a level curve.
Proof.  Intuitively as the greatest value of the directional derivative is in the direction of the gradient vector it seems logical that the steepest slope would be across rather than parallel to a contour line.
The following is a mathematical proof to confirm this.
$y=f_x$
The slope of the tangent is           $f_x^1$
The equivalent level curve is
$$z=0=f_x-y$$
$$∇f_xy=<├ ∂/(∂x,)-1>$$
The slope of this vector is the "$j$" component over the "$I$" component
$(-1)/(∂/∂x)$
This is at right angles to $f_x^1$
eg    $y=2x^2$
slope$= 4x$
level curve is 
$$z=0=2x^2-y$$
$$∇f_xy=<├ 4x-1>$$
slope$=(-1)/4x$
