The Gradient Vector is Perpendicular to the Level Curve

In[65]:=

  Clear[f,x,y]
  f[x_,y_] = 5 - (x^2  + x y  +  y^2)

Out[65]=

       2          2
  5 - x  - x y - y

Here is a plot of some level curves f[x,y] = c.

In[66]:=

  a = -1;
  b = 1;
  levelcurves = ContourPlot[Evaluate[f[x,y]],{x,a,b},{y,a,b},
             ContourSmoothing->Automatic,
                AxesLabel->{"x","y"}];

The lighter shading indicates larger values of f[x,y].

Here is the 3D plot of the surface z = f[x,y]:

In[67]:=

  surfaceplot = 
  ParametricPlot3D[{x,y,f[x,y]},{x,a,b},{y,a,b},
  AxesLabel->{"x","y","f[x,y]"}];
  

The level curves are plots at the height f[x,y]=c.
Here is f[x,y] = 4.5:

In[68]:=

  c = 4.5;
  Show[levelcurves,Contours->{c},ContourShading->False];
  

Here is the plot of the intersection of the surface and the plane z = 4.5:

In[69]:=

  c = 4.5;
  plane = ParametricPlot3D[{x,y,c},{x,a,b},{y,a,b},
  PlotPoints->{2,2},DisplayFunction->Identity];
  Show[surfaceplot,plane,DisplayFunction->$DisplayFunction];
  

Consider f(x,y) = 2 x^2 + x y + y^2.

In[70]:=

  Clear[x,y,f]
  f[x_,y_] = 2 x^2 + x y + y^2;
  grad2[f][x,y]

Out[70]=

  {4 x + y, x + 2 y}

Here is a part of the level curve f[x,y] = f[1.5,2].

In[71]:=

  
  levelcurve = ContourPlot[f[x,y],{x,0,3},{y,1,4},
  Contours->{f[1.5,2]},ContourShading->False,
  DisplayFunction->$DisplayFunction];

  General::spell1: 

     Possible spelling error: new symbol name "levelcurve"

       is similar to existing symbol "levelcurves".

These next two will plot the gradient vector.

In[72]:=

  Clear[x,y,t]
  x0=1.5;
  y0=2;
  {x1,y1}=grad2[f][x,y]/.{x->1.5,y->2};
  p0={x0,y0};
  p1={x1,y1};
  x[t_]:=x0+t(x1-x0)
  y[t_]:=y0+t(y1-y0)

In[73]:=

  gradient=ParametricPlot[{x[t],y[t]},{t,0,1}]

  General::spell1: 

     Possible spelling error: new symbol name "gradient"

       is similar to existing symbol "Gradient".

Out[74]=

  -Graphics-

This plots the level curve and the gradient at (1.5,2)

In[75]:=

  Show[levelcurve,gradient,PlotRange->All,
  AspectRatio->Automatic,AxesLabel->{"x","y"},
  DisplayFunction->$DisplayFunction];
  

Well...

In[76]:=

  The gradient vector is perpendicular to the level curve!

  General::spell1: 

     Possible spelling error: new symbol name "the"

       is similar to existing symbol "The".

  General::spell1: 

     Possible spelling error: new symbol name "level"

       is similar to existing symbol "Level".

Out[76]=

  is level perpendicular the The to vector curve! -Graphics-

Proof:
If f[x,y] = c is your level curve and you give it a parametrization r[t]=x[t]i+y[t]j, then g[t]=f[x[t],y[t]]=c.
But then g'[t] = 0
By the chain rule g'[t] = gradf[x[t],y[t]].{x'[t],y'[t]}.
So gradf[x[t],y[t]].{x'[t],y'[t]} = 0.
This means that the gradient is perpendicular to the tangent vector at that point.
By the way, we cannot visualize it here, but this works for any dimension.

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