Heffernann SM P1Solutions 2006

Heffernann SM P1Solutions 2006 - SPECIALIST MATHS TRIAL...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
SPECIALIST MATHS TRIAL EXAMINATION 1 SOLUTIONS 2006 _____________________________________________________________________ Question 1 ( 29 ( 29 2 1 3 cos 3 2 3 sin a. - = - = t a t t v a. (1 mark) b. ma R = where R is the magnitude of the resultant force. ( 29 ( 29 1 3 cos 6 2 1 3 cos 3 2 - = - = t t R R is a maximum when ( 29 1 3 cos = t ; that is, when ( 29 t 3 cos equals its maximum value. So, the maximum value of R is given by Newtons 5 1 1 6 = - × = R (1 mark) b. ii. © THE HEFFERNAN GROUP 2006 Specialist Maths Trial Exam 1 solutions (1 mark) u _ u (1 mark)
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 Question 2 a. (1 mark) correct shape with points ( 29 ( 29 ( 29 ( 29 3 , 1 3 , 1 , 0 , 3 0 , 1 - - shown (1 mark) correct centre ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 y x y x dx dy x dx dy y dx dy y x y x y x 4 1 9 8 1 18 1 18 8 0 8 1 18 36 4 1 9 1 9 4 1 2 2 2 2 - - = - - = - - = = + - = + - = + - b. c. If 0 4 then 0 y y __________( A ) For 0 1 ), 1 , 1 ( < - - x x so ) ( __________ 0 ) 1 ( 9 B x - - (1 mark) Using ( A ) and ( B ), we have © THE HEFFERNAN GROUP 2006 Specialist Maths Trial Exam 1 solutions 1 9 4 ) 1 ( 2 2 = + - y x (1 mark) for differentiating x term and constant term (1 mark) for differentiating y term (1 mark) (implicit differentiation)
Background image of page 2
3 0 4 ) 1 ( 9 - - y x Therefore 0 dx dy since y x dx dy 4 ) 1 ( 9 - - = (1 mark) So ( 29 1 , 1 for 0 , 0 for - x dx dy y . © THE HEFFERNAN GROUP 2006 Specialist Maths Trial Exam 1 solutions
Background image of page 3

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

View Full DocumentRight Arrow Icon
4 Question 3 a. Let ( 29 x e y 2 arctan = ( 29 x x x x x x x e e e e dx du du dy dx dy e e dx du u du dy e u u 4 2 2 4 4 2 2 2 1 2 2 1 1 rule) (chain 1 1 2 1 1 where arctan + = × + = = + = = + = = = as required. (1 mark) b. From a. ( 29 ( 29 x x x e e e dx d 4 2 2 1 2 arctan + = ( 29 ( 29 + = dx e e dx e dx d x x x 4 2 2 1 2 arctan so, ( 29 ( 29 ( 29 [ ] ( 29 ( 29 ( 29 { } ( 29 ( 29 ( 29 { } ( 29 - = - = - = = + + = + 4 25 arctan 2 1 1 arctan arctan 2 1 arctan arctan 2 1 arctan 2 1 1 Now constant a is 1 2 arctan 2 5 log 0 5 log 2 5
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/22/2010 for the course MATH specialist taught by Professor Akhmad during the Spring '10 term at Mt. Mercy.

Page1 / 14

Heffernann SM P1Solutions 2006 - SPECIALIST MATHS TRIAL...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online