CHAPTER 3 DIFFERENTIATION VECTORS OF Example 3.6 C1  Show that satisfies the differential equation cos 2t sin 2t r   C2 d r d r 2  2 2 dt C1 C2 where and are constants.  5r  0 d r   dt cos 2t sin 2t 2C1 sin 2t 2C2 cos 2t    C1 C2 d r 2 dt 2 On substituting these derivatives; on the left, the claim follows PARTIAL DIFFERENTIATION VECTORS: OF A A=A(x1,y2,z3) If a vector is dependent on more than one scalar variable i.e. say, A then we differentiate with respect to each independent variable partially denoted i by 1, 2, 3  For , the partial derivative of is    A A1iˆ A2 ˆj A3kˆ A w.r.t. x1 etc   ˆ ˆ k A  Similarly, the second partial derivative is w.r.t. x1 A   2 A1 ˆ A2 ˆ A3 k 2 2 2 2 2 2 2  Example 3.7  2 at (1,0, -2). If 2xz A  x yziˆ  xz kˆ and ˆj 2 3 2 , 2ziˆ B   yˆj x kˆ 2  iˆ kˆ ˆj 2xz   We have A B x yz xz 2 3 2 2z y x 2 2x3 ˆ x ˆ x k    2xz   4xz z xyz yz y z 3 2 4 3 2 2 4
CHAPTER 3 DIFFERENTIATION OF VECTORS Example 3.6 Show that r  C1 cos 2t  C2 sin 2t  satisfies the differential equation d 2 r 2  2 d r dt  5r  0 where C1 and C2 are constants. d r dt  C1 cos 2t C2 sin 2t   2C1 sin 2t  2C2 cos 2t  d 2 r dt 2 On substituting these derivatives; on the left, the claim follows PARTIAL DIFFERENTIATION OF VECTORS: If a vector A is dependent on more than one scalar variable i.e. A=A(x1,y2,z3) say, then we differentiate A with respect to each independent variable partially denoted by i  1, 2, 3 For A  A1iˆ  A2 ˆj  A3kˆ , the partial derivative of A w.r.t. x1 is A  ˆ  ˆ  k etc Similarly, the second partial derivative w.r.t. x1 is 2 A 2  2 A1 ˆ  2 2 A2 ˆ  2 2 A3 k 2 Example 3.7 If A  x 2 yziˆ 2xz 3 ˆj  xz 2 kˆ and B  2ziˆ  yˆj x 2 kˆ , 2   at (1,0, -2). iˆ ˆj kˆ We have A  B  x 2 yz 2xz 3 xz 2 2z y x 2  2x3 z 3 xyz 2 ˆ  x 4 yz  2xz 3 ˆ  x 2 y 2 z  4xz 4 k