CHAPTER 4 GRADIENT, DIVERGENCE CURL AND Example 4.1  2x2 The gradient of  yz3 is   4xyz 2x 6x    z yz kˆ ˆj 3iˆ 2 3 2 2 x,   Now given that  is continuously differentiable with respect to its y, z variables x,y,z, then if we let x=x(t), y=y(t), z=z(t) where t is a parameter, we have t  x(t),    y(t), z(t) t, On differentiating with respect to we obtain d dt dt dt dt     dy or dz  d  dx  i ˆ  ˆ where r  xiˆ  yˆj  zkˆ  Hence total differential d of is given by  d grad  d r If is a small element of an arc, then ds d d r  grad  ds ds dt dt dt d d , Now if we take then r  rˆt r r ds dr   d r d r d r dr ds dt dt Meaning that d d r ˆ  , the result becomes ˆ  grad  ˆ ds ds If we denote this unit vector by i.e.
CHAPTER 4 GRADIENT, DIVERGENCE AND CURL Example 4.1 The gradient of   2x2 yz3 is    4xyz 3iˆ  2x 2 z 3 ˆj  6x 2 yz 2 kˆ Now given that    x, y, z is continuously differentiable with respect to its variables x,y,z, then if we let x=x(t), y=y(t), z=z(t) where t is a parameter, we have  t    x(t), y(t), z(t) On differentiating with respect to t, we obtain d dt  dt  dt  dt or d  dx  dy  dz i  ˆ  ˆ where r  xiˆ  yˆj  zkˆ Hence total differential d of  is given by d  grad  d r If ds is a small element of an arc, then d ds  d r ds  grad Now if we take r  rˆt , then d r dt  d r ds dt  d r dr dt Meaning that d r ds d r dr dt dt If we denote this unit vector by ˆ i.e. d r ds  ˆ , the result becomes d ds  ˆ  grad