# sol3 - Solution Assignment 3 Math237 Fall 2006 Set 1 A4...

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Unformatted text preview: Solution Assignment 3. Math237 Fall 2006 Set 1 A4 Since 3 * 1- 2 * 2 =- 1 6 = 0, the denominator term (3 x- 2 y ) is non zero in a sufficiently small neighborhood of (1 , 2 ,- 2). Note also that x = 1 , y = 2 ⇒ z = xy 3 x- 2 y =- 2 as a check. We have : ∂z ∂x = (3 x- 2 y ) y- 3 xy (3 x- 2 y ) 2 , ∂z ∂y = (3 x- 2 y ) x + 2 xy (3 x- 2 y ) 2 . Consequently, at (1 , 2 ,- 2), ∂z ∂x =- 8 and ∂z ∂y = 3. the equation of the tangent plane is therefore z =- 2 + ( ∂z ∂x ) a ( x- 1) + ∂z ∂y a ( y- 2), where the subscript ’a’ indicates evaluation at a = (1 , 2 ,- 2). Thus the tangent plane at (1 , 2 ,- 2) has for equation z =- 2- 8( x- 1) + 3( y- 2) =- 8 x + 3 y. Clearly, for x = 0 and y = 0 we get z = 0 for the tangent plane equation. Consequently, (0 , , 0) lies on the tangent plane. Set 1 A5(i) a) L a ( x ) = f ( a ) + ∇ f ( a ) • ( x- a ) see page 41 course notes....
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sol3 - Solution Assignment 3 Math237 Fall 2006 Set 1 A4...

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