CHAPTER 4 GRADIENT, DIVERGENCE CURL AND Example 4.2  4xz (1,-2,-1) Find the directional derivative of  x yz  at in the direction 2 2 2iˆ 2kˆ . ˆj  4xz  yz  2 2 2xyz iˆ x 4z 8zxkˆ    x zˆj  y  2 2 2  8iˆ 10kˆ at ˆj Unit vector in the direction of 2iˆ 2kˆ is ˆj 2iˆ 2kˆ ˆj ˆ  3 Required directional derivative is 1 3 3    163 13 37 3  20 x, If  , this relationship represents a surface in space, depending on y, z  constant the value ascribed to the constant. If d r is the displacement in this surface, then 0  d  since is constant over the surface. Thus from  grad d r d  we have   0 such small displacements d r in the surface. d grad d r  But  cos  0 grad d r grad d r   2   grad is perpendicular to d r The unit vector in the direction of is called the unit normal vector at P. Nˆ grad Unit Normal Vector Nˆ 
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL Example 4.2 Find the directional derivative of   x 2 yz  4xz 2 at (1,-2,-1) in the direction 2iˆ ˆj 2kˆ .  2 yz  4xz 2   2xyz  4z 2 iˆ  x 2 zˆj  x 2 y  8zxkˆ  8iˆ ˆj 10kˆ at Unit vector in the direction of 2iˆ ˆj 2kˆ is ˆ  2iˆ ˆj 2kˆ 3 Required directional derivative is 1 3  163  13  20 3  37 3 If  x, y, z  constant , this relationship represents a surface in space, depending on the value ascribed to the constant. If d r is the displacement in this surface, then d  0 since  is constant over the surface. Thus from grad  d r  d we have d  grad  d r  0 such small displacements d r in the surface. But grad  d r  grad d r cos  0    2 grad is perpendicular to d r The unit vector Nˆ in the direction of grad is called the unit normal vector at P. Unit Normal Vector Nˆ 