Example 4.6 CHAPTER 4 GRADIENT, DIVERGENCE CURL AND f r  Prove that  r  r 1 2 Example 4.7 x . Find if      In r In y z 12 2 2 2 1 2 1    ˆ ˆ ˆ 2 1 r  2x   y z r 2 2 2 2 Example 4.8 Show that is a vector perpendicular to the surface where c is a  (x,  y, z) c constant. P(x,y,z) Let be the position vector to any point on the surface. Then r  xiˆ  yˆj  zkˆ P. lies in the tangent plane to the surface at d r  dxiˆ  dyˆj  dzkˆ But 0 dx     d dy dz or  i  ˆ i.e. so that is perpendicular to and therefore to the surface. 0 d r  d r  Example 4.9 x 2x 4x Let . Show that can be expressed as 2 4ziˆ 3y 2zkˆ       V y z y V ˆj the gradient of a scalar function. V   ˆ Assume 
CHAPTER 4 Example 4.6 GRADIENT, DIVERGENCE AND CURL Prove that r   f  r r 1 2 Example 4.7 Find if   In r  12 In x 2  y 2  z 2 . 1 2  1 ˆ  ˆ  ˆ 2 1 r 2x 2  y 2  z 2  r 2 Example 4.8 Show that is a vector perpendicular to the surface  (x, y, z)  c where c is a constant. Let r  xiˆ  yˆj  zkˆ be the position vector to any point P(x,y,z) on the surface. Then d r  dxiˆ  dyˆj  dzkˆ lies in the tangent plane to the surface at P. But d  dx  dy  dz  0 or i  ˆ  i.e.  d r  0 so that is perpendicular to d r and therefore to the surface. Example 4.9 Let V  x  2 y  4ziˆ  2x 3y z ˆj  4x y  2zkˆ . Show that V can be expressed as the gradient of a scalar function. Assume V   ˆ 