Rate of Change and Gradient

Recall the directional derivative of f in the direction of a unit vector u, Du(f)=gradient(f).u.
This gives the rate of change of f in the direction of the vector u.
Recall that a.b=||a|| ||b|| cos(theta) where theta is the angle between the vectors a and b. Using this with the directional derivative, we see that
Du(f)=gradient(f).u=||gradient(f)|| ||u|| cos(theta) which implies (since ||u||=1 and cos(theta) is between -1 and 1)
-||gradient(f)||<=Du(f)<=||gradient(f)||

Now if u is in the direction of gradient(f), then theta=0 and we have Du(f)=||gradient(f)||
If u points in the opposite direction of gradient(f), then theta=Pi and we have
Du(f)=-||gradient(f)||

What does this say?

A (differentiable) function increases most rapidly in the direction of the gradient (u=gradient(f)) and the rate of change in this direction is ||gradient(f)||.

A (differentiable) function decreases most rapidly in the opposite direction of the gradient (u=-gradient(f)) and the rate of change in this direction is -||gradient(f)||.

Example

Suppose the temperature at each point of a metal plate is given by
T(x,y)=Exp[x]Cos[y] + Exp[y]Cos[x].
In what direction does the temperature increase most rapidly at (0,0)? What is the rate of increase?

In[77]:=

  T[x_,y_]:=Exp[x]*Cos[y] + Exp[y]*Cos[x]

In[78]:=

  grad2[T][x,y]

Out[78]=

    x           y          y           x
  {E  Cos[y] - E  Sin[x], E  Cos[x] - E  Sin[y]}

In[79]:=

  grad2[T][x,y]/.{x->0,y->0}

Out[79]=

  {1, 1}

Answer:

Answer: ...

Example

Up to The Gradient