Assignment 3 Solutions

# Assignment 3 Solutions - Department of Mathematics MATHS...

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Unformatted text preview: Department of Mathematics MATHS 208 Assignment 3 Solutions 1. (6 marks) a) The equation x 3- xy + 4 y 2 + xyz- z 3 = 11 defines a function z = f ( x,y ) implicitly. Use implicit differentiation to find the value of ∂z ∂x when ( x,y,z ) = (1 , 2 , 2) . Solution: F = x 3- xy + 4 y 2 + xyz- z 3- 11 = 0 , F x = 3 x 2- y + yz, at (1 , 2 , 2) F x = 3- 2- 4 =- 3 , F z = xy- 3 z 2 , at (1 , 2 , 2) F z = 2- 12 =- 10 , ⇒ ∂z ∂x =- F x F z =- 3 10 . b) If f ( x,y,z ) = y 2 ze x + y- sin( xyz ) = 0, find ∂z ∂x and ∂y ∂z . Solution: F x = y 2 ze x + y- yz cos( xyz ) , F y = (2 y + y 2 ) ze x + y- xz cos( xyz ) , F z = y 2 e x + y- xy cos( xyz ) , ⇒ ∂z ∂x =- F x F z =- y 2 ze x + y- yz cos( xyz ) y 2 e x + y- xy cos( xyz ) ⇒ ∂y ∂z =- F z F y =- y 2 e x + y- xy cos( xyz ) (2 y + y 2 ) ze x + y- xz cos( xyz ) 2. (5 marks) Given w = x 2 y + y 2 z + z 2 x, verify that ∂w ∂x + ∂w ∂y + ∂w ∂z = ( x + y + z ) 2 . Solution: w = x 2 y + y 2 z + z 2 x. Then ∂w ∂x = 2 xy + z 2 . ∂w ∂y = x 2 + 2 yz. ∂w ∂z = y 2 + 2 xz. Therefore ∂w ∂x + ∂w ∂y + ∂w ∂z = x 2 + y 2 + z 2 + 2( xy + yz + xz ) = ( x + y + z ) 2 . 3. (5 marks) Given f ( x,y,z ) = 3 x 2 + xy- 2 y 2- yz + z 2 , find the rate of change of f ( x,y,z ) at the point (1 ,- 2 ,- 1) in the direction of the vector (2 ,- 2 ,- 1) ....
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Assignment 3 Solutions - Department of Mathematics MATHS...

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