quiz3sol - ~ b + t~ c )) + (( t~a ) ( ~ b + t~ c + ~ c ))...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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 ) × (
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

Ask a homework question - tutors are online