Directional Derivatives

A similar result involves the derivative in the direction of any vector. Note that the partial derivative with respect to x is the derivative in the x (or i or [1,0,0]) direction. We can take the derivative in the direction of any vector u. All we ask is that u be a unit vector, that is, it has length 1. You saw the theory in Calc II, here is the result:
Du(f)=gradient(f).u (Du(f) stands for the directional derivative of f in the direction of u.)

To find the derivative of x^2 y + y^2 z + z^2 x at the point (1,0,1) in the direction of [0,3,-1], do the following:


  Clear[u,v,f]
  f[x_,y_,z_]:=x^2 y + y^2 z + z^2 x;
  v={0,3,-1}
  u=v/norm[v]


  grad3[f][x,y,z].u


  grad3[f][x,y,z].u /.{x->1, y->0,z->1}

To find the derivative of f(x,y,z) above at (1,-3,2) in the direction of the curve r(t)=t^2 i +3t j + (1-t^3) k, do the following:

The direction of a curve at a point is given by its derivative at that point! To be at the point (1,-3,2) we need t=-1 in r(t).


  r[t_]:={t^2,3 t,1-t^3}


  dir=D[r[t],t]/.t->-1
  u=dir/norm[dir]


  grad3[f][x,y,z].u


  grad3[f][x,y,z].u /.{x->1, y->-3 ,z->2}

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