FL2009 Prelim1 solutions

# FL2009 Prelim1 solutions - Math 3210 first prelim Wednesday...

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

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.

Unformatted text preview: Math 3210 first prelim Wednesday 30 September 2009 10:10-11:00 pm Solutions 1. (a) d α = dx 1 dx 2 + dx 3 dx 4 . (b) α d α = x 1 dx 2 dx 3 dx 4 + x 3 dx 1 dx 2 dx 4 + dx 1 dx 2 dx 5 + dx 3 dx 4 dx 5 , ( d α ) 2 = 2 dx 1 dx 2 dx 3 dx 4 , α ( d α ) 2 = 2 dx 1 dx 2 dx 3 dx 4 dx 5 . (c) * α =- x 1 dx 1 dx 3 dx 4 dx 5- x 3 dx 1 dx 2 dx 3 dx 5 + dx 1 dx 2 dx 3 dx 4 α ( * α ) = ( x 2 1 + x 2 3 + 1 ) dx 1 dx 2 dx 3 dx 4 dx 5 . 2. (a) d dx Z x ln ( x + y ) dy = ln ( x + x ) + Z x ∂ ∂ x ln ( x + y ) dy = ln 2 x + Z x 1 x + y dy = ln 2 x + ln ( x + y ) fl fl x y = = ln 2 x + ln 2 x- ln x = 2 ln 2 x- ln x = 2 ( ln 2 + ln x )- ln x = 2 ln 2 + ln x . (b) c ( t ) = p c ( t ) · c ( t ) is constant, so c ( t ) · c ( t ) = c 1 ( t ) 2 + ··· + c n ( t ) 2 is constant, so d dt ( c ( t ) · c ( t )) = 2 c 1 ( t ) c 1 ( t ) + ··· + 2 c n ( t ) c n ( t ) = 0, i.e. c ( t ) · c ( t ) = 0. 3. (a) * ( dx 1 dx 3 dx 4 ) = dx 2 , * ( dx 1 dx 3 dx 4 dx 5 ) =- dx 2 ....
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

FL2009 Prelim1 solutions - Math 3210 first prelim Wednesday...

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

View Full Document
Ask a homework question - tutors are online