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Question Number 25783 by lizan 123 last updated on 14/Dec/17
Ifrisaunitvectorthenshowthat∣r×drdt∣=∣drdt∣
Answered by ajfour last updated on 15/Dec/17
r¯=cosθi^+sinθj^dr¯dt=(dθdt)(−sinθi^+cosθj^)⇒∣dr¯dt∣=absolutevalueof(dθdt).....(a)whiler¯×dr¯dt=(cosθi^+sinθj^)×(dθdt)(−sinθi^+cosθj^)=(dθdt)(cos2θ+sin2θ)k^=(dθdt)k^⇒∣r¯×dr¯dt∣=absolutevalueof(dθdt)=∣dr¯dt∣[see(a)].
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