Exam 2 - S09-1

Exam 2 - S09-1 - SOLUTIONS TO EXAM 2 – MATH 10260 SPRING...

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Unformatted text preview: SOLUTIONS TO EXAM 2 – MATH 10260, SPRING 2009 1. ( z- 3) = 2( x- 1) + 5( y + 2) = 2 x + 5 y + 11 . 2. p ((2- 1) 2 + (1- 1) 2 + (1- 2) 2 = √ 2 . 3. We have ∂f ∂x = y- e x , ∂f ∂y = x. ∂f ∂y = 0 ⇒ x = 0 , ∂f ∂x = 0 ⇒ y = e x = 1 . Hence (0 , 1) is the only critical point of the function. 4. The x-slope is- 3 , the y-slope is- 2 and the plane contains the point (0 , , 1) . Hence its equation is z- 1 =- 3 x- 2 y + 1 , i.e. , z =- 3 x- 2 y + 1 . 5. We have ∂f ∂x = 2 x x 2 + y 2 , ∂ 2 f ∂x 2 = 2( x 2 + y 2 )- (2 x ) · (2 x ) ( x 2 + y 2 ) 2 = 2( y 2- x 2 ) ( x 2 + y 2 ) 2 . 6. The number of vans is approximatively f (25 , 170) + ∂f ∂x (25 , 170) · (0 . 2) + ∂f ∂y (25 , 170) · (- 10) = 900 + (- 180) · (0 . 2) + (- 12) · (- 10) = 900- 36 + 120 = 984 . 7. y = 1 is an equilibrium solution of the differential equation y = g ( y ) , where g ( y ) = y (1- y ) , because g (1) = 0 and g (1) = 1- 2 < ....
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Exam 2 - S09-1 - SOLUTIONS TO EXAM 2 – MATH 10260 SPRING...

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