Exam 2 -- Solution

Exam 2 -- Solution - MAC2313 Exam 2 1. Let f ( u,v ) = u...

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Unformatted text preview: MAC2313 Exam 2 1. Let f ( u,v ) = u sin(2 u- v ). Show that f is differentiable and then find the linearization of f at ( , 0). Solution: Taking the 1st partial derivatives: f u ( u,v ) = sin(2 u- v ) + 2 u cos(2 u- v ) f v ( u,v ) =- u cos(2 u- v ) . Since f u and f v are continuous for every point, f is differentiable at every point. Now, f u ( , 0) = sin(2 )- 2 cos(2 ) = 2 f v ( , 0) =- cos(2 ) =- f ( , 0) = sin(2 ) = 0 . The linearization of f at ( , 0) is: L ( x,y ) = 2 ( x- )- ( y- 0)- . 2. Find the arclength function s ( t ) with respect to t = 0 for ~ r ( t ) = * t cos( t ) ,t sin( t ) , 2 2 3 2 t 3 + . Solution: The arlength function s with respect to t = 0 for ~ r is defined: s ( t ) = Z t || ~ r ( u ) || du. Taking the derivative of ~ r : ~ r ( t ) = D cos( t )- t sin( t ) , sin( t ) + t cos( t ) , 2 t E . Taking the magnintude of ~ r ( t ): || ~ r ( t ) || = q (cos( t )- t sin( t )) 2 + (sin( t ) + t cos( t )) 2 + ( 2 t ) 2 = q cos 2 ( t )- 2 t 2 sin( t )cos( t ) + t 2 sin 2 ( t ) + sin 2 ( t ) + 2 t 2 sin( t )cos( t ) + t 2 cos 2 ( t ) + 2 t = p t 2 + 2 t + 1 = q ( t + 1) 2 = t + 1 . Therefore, s ( t ) = Z t ( u + 1) du = t 2 2 + t....
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Exam 2 -- Solution - MAC2313 Exam 2 1. Let f ( u,v ) = u...

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