quiz3sol

quiz3sol - ~ b t~ c t~a × ~ b t~ c ~ c(b ~ r t =< e t...

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Math 251, Spring 2008, Quiz 3 Solutions (a) Find the derivative of ~ r ( t ) = t~a × ( ~ b + t~ c ) with respect to t . ( ~a etc neednt be constant vectors). (In essence, you have to use product rule a couple of times here. You may represent derivative of ~a by ~a 0 ) (b) Find the parametric equation of the tangent line to the given curve at the speciﬁed point x = e t , y = e - t sin t, z = 1 1 + t 2 , t = 0 Solution: (( t~a ) × ( ~ b + t~ c )) 0 = (( t~a ) 0 × ( ~ b + t~ c )) + (( t~a ) × ( ~ b + t~ c ) 0 ) = (( t~a 0 + ~a ) × (
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Unformatted text preview: ~ b + t~ c )) + (( t~a ) × ( ~ b + t~ c + ~ c )) (b) ~ r ( t ) = < e t , e-t cos t-e-t sin t,-2 t 1+ t 2 > and hence direction vector of the line would be < 1 , 1 , > which is ~ r ( t ) evaluated at t = 0 and a point on the line would just be ~ r (0) = (1 , , 1), and hence, equation of the line would be x = 1 + t, y = t, z = 1. 1...
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