{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

ps12sol

# ps12sol - 18.02 Fall 2008 Problem Set 12 Part B Solutions 1...

This preview shows pages 1–2. Sign up to view the full content.

18.02 Fall 2008 – Problem Set 12, Part B Solutions 1. a) x = cos t , y = sin t , z = t , so dx = - sin t dt , dy = cos t dt and dz = dt ; the range of integration is 0 t 2 π . Hence F · dr = 2 π 0 ( a sin t + b cos t sin 2 t ) ( - sin t ) dt + (2 cos 2 t sin t ) cos t dt + (cos t cos t - t 2 ) dt = 2 π 0 ( - a sin 2 t - b sin 3 t cos t + 2 cos 3 t sin t + cos 2 t - t 2 ) dt = - - b 1 4 sin 4 t 2 π 0 + - 1 2 cos 4 t 2 π 0 + π - t 3 3 2 π 0 = (1 - a ) π - 8 π 3 / 3 . b) ∇× F = ( R y - Q z ) ˆ ı +( P z - R x ) ˆ +( Q x - P y ) ˆ k = (0 - 0) ˆ ı +( a cos z - cos z ) ˆ +(4 xy - 2 bxy ) ˆ k = 0 , ( a - 1) cos z, 2(2 - b ) xy . Therefore F is conservative if and only if a = 1 and b = 2. c) By integrating f x = sin z + 2 xy 2 we obtain f = x sin z + x 2 y 2 + g ( y, z ). Comparing f y = 2 x 2 y + g y = x 2 y we obtain g y = 0, so in fact g = g ( z ). Finally f z = x cos z + g ( z ) = x cos z - z 2 gives g ( z ) = - z 2 , so g = - 1 3 z 3 + c . The potential function is therefore f ( x, y, z ) = x sin z + x 2 y 2 - 1 3 z 3 + c . (Other method: f ( x 1 , y 1 , z 1 ) = f (0 , 0 , 0) + C F · dr where C is any curve from the origin to ( x 1 , y 1 , z 1 ).) For the helix in part (a): F · dr = f (1 , 0 , 2 π ) - f (1 , 0 , 0) = - 1 3 (2 π ) 3 = - 8 π 3 / 3, in agreement with the answer of (a) for a = 1.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 2

ps12sol - 18.02 Fall 2008 Problem Set 12 Part B Solutions 1...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online