Tutorial 4 Solns

Tutorial 4 Solns - Math 116 - Lab 4 - Solutions - Fall...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 116 - Lab 4 - Solutions - Fall 2011. 1. d dx ( x ) = d dx ( ( y- x ) 4 ) ⇒ 1 = 4( y- x ) 3 ( y- 1). Therefore y = 1 4( y- x ) 3 + 1 . 2. Implicit differentiation gives: d dx 3 e cos( x ) + 7 e sin( y ) = d dx (10) 3 e cos( x ) (- sin( x )) + 7 e sin( y ) (cos( y )) y = 0 which implies y = 3 e cos( x ) sin( x ) 7 e sin( y ) cos( y ) . Hence the slope of the tangent line at ( π 2 , ) is 3 7 and the equation of the tangent line is y = 3 7 x + arcsin ln 10- 3 e 7 . 3. Implicit differentiation gives, d dx ( y 3 ) = d dx ( ( y- x ) 2 ) 3 y 2 y = 2( y- x )( y- 1) y = 2( x- y ) 3 y 2 + 2( x- y ) . A vertical tangent line occurs when y is undefined so solving 3 y 2 +2( x- y ) = 0 for x gives x =- y (3 y- 2) 2 . Substituting into the original equation gives y = 0 or y = 4 9 . Hence there is a vertical tangent line at the points (0 , 0) and ( 4 27 , 4 9 ) . 4. a) We have, ln( y ) = ln x cot( x ) = cot( x )ln( x ) ....
View Full Document

This note was uploaded on 12/28/2011 for the course MATH 116 taught by Professor Robertandre during the Spring '11 term at Waterloo.

Page1 / 3

Tutorial 4 Solns - Math 116 - Lab 4 - Solutions - Fall...

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

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