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91Ó°ÊÓ

Let r1(f)and r2(t) be differentiable vector functions with three components each. Prove that

ddtr1(t)-r2(t)=r1'(t)-r2(t)+r1(t)-r2'(t)∣(This is Theorem 11.11 (c).)

Short Answer

Expert verified

Ans:r1(t)r2(t)=x1(t),y1(t),z1(t)·x2(t),y2(t),z2(t)=x1(t)x2(t)+y1(t)y2(t)+z1(t)z2(t)=r1('t)·r2(t)+r1(t)r2'

Step by step solution

01

Step 1. Given information: 

r1(t)and r2(t)are differentiable vector functions with three components each .

02

Step 2. Proving :

Then

r1(t)r2(t)=x1(t),y1(t),z1(t)·x2(t),y2(t),z2(t)=x1(t)x2(t)+y1(t)y2(t)+z1(t)z2(t)ddtr1(t)·r2(t)=ddtx1(t)x2(t)+y1(t)y2(t)+z1(t)z2(t)=ddtx1(t)x2(t)+ddty1(t)y2(t)+ddtz1(t)z2(t)=x1('t)x2(t)+x1(t)x2'(t)+y1('t)y2(t)+y1(t)y2('t)+z1'(t)z2(t)+z1(t)z2'(t)=x1('t)x2(t)+y1'(t)y2(t)+z1'(t)z2(t)+x1(t)x2'(t)+y1(t)y2'(t)+z1(t)z2'(t)=x1('t),y1('t),z'(t)x2(t),y2(t),z2(t)+x1(t),y1(t),z1(t)x2'(t),y2'(t),z2'(t)=r1('t)·r2(t)+r1(t)r2'

Thus ddtr1(t)·r2(t)=r1('t)·r2(t)+r1(t)·r2('t)

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