supplement to BOB p4-1 - Supplement to BOB (page 4) 9.5, #3...

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Supplement to BOB (page 4) 9.5, #3 BOB's answer is correct, and so is this one: Ò#  &>ß !ß %  &>Ó 9.5, #10 Ò$ >ß $ >ß %>Ó cos sin 9.5, #12 line through in the direction of Ð%ß!ß$Ñ Ò#ß)ß&Ó 9.5, #16 hyperbola (or actually one branch of a hyperbola) BCœ"Dœ! ## 9.5, #17 BOB's answer is almost correct. It should read as follows: hyperbola BC œ " D œ ! (Without the “ ,” the equation represents a hyperbolic cylinder, which is a Dœ! surface, rather than a hyperbola, which is a curve.) 9.5, #22 r r > œ # Ð>Ñ œ Ò"ß #>ß !Ó Ð#Ñ œ Ò"ß %ß !Ó ÄÄ ! ww u q Ä Ð#Ñ œ ß ß ! ÐAÑ œ Ò#  Aß %  %Aß !Ó Ä ’“ "% "( "( ÈÈ 9.5, #24 r r > œ # Ð>Ñ œ Ò$ >ß $ >ß %Ó Ð# Ñ œ Ò!ß $ß %Ó ! 11 sin cos u q Ä Ð# Ñ œ !ß ß ÐAÑ œ Ò$ß $Aß )  %AÓ Ä $% && 9.5, #33 BOB's answer is incomplete. Here are the missing parts. a Ä Ð>Ñœ Ò"ß#>ß!Óœ
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This note was uploaded on 01/28/2010 for the course ESE 317 taught by Professor Hastings during the Fall '08 term at Washington University in St. Louis.

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