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Exam02-solutions-1

# Exam02-solutions-1 - Version 126 – Exam02 – gilbert...

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Unformatted text preview: Version 126 – Exam02 – gilbert – (55485) 1 This print-out should have 14 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Determine f xy when f ( x, y ) = 1 2 y tan − 1 ( xy ) . 1. f xy = − x (1 + x 2 y 2 ) 2 2. f xy = − xy 2(1 + x 2 y 2 ) 3. f xy = x (1 + x 2 y 2 ) 2 4. f xy = y (1 + x 2 y 2 ) 2 correct 5. f xy = − y 2(1 + x 2 y 2 ) 6. f xy = xy 2(1 + x 2 y 2 ) Explanation: Since we can choose whether to differentiate with respect to x or y first, for simplicity we will choose to differentiate first with respect to x because then the algebra is simpler. Indeed, by the Chain Rule, f x = 1 2 y ∂ ∂x parenleftBig tan − 1 ( xy ) parenrightBig = 1 2 y 2 parenleftBig 1 1 + ( xy ) 2 parenrightBig = y 2 2(1 + x 2 y 2 ) . Thus by the Quotient Rule f xy = 1 2 parenleftBig 2 y (1 + x 2 y 2 ) − y 2 (2 x 2 y ) (1 + x 2 y 2 ) 2 parenrightBig . Consequently, f xy = y (1 + x 2 y 2 ) 2 . keywords: partial differentiation, mixed par- tial derivative, Chain Rule, inverse tangent, PartialDiffMV, PartialDiffMVExam, 002 10.0 points Which equation has the surface x y z as its graph in the first octant? 1. x 5 + y 4 + z 3 = 1 2. x 4 + y 5 + z 3 = 1 3. x 3 + y 5 + z 4 = 1 4. x 3 + y 4 + z 5 = 1 5. x 4 + y 3 + z 5 = 1 6. x 5 + y 3 + z 4 = 1 correct Explanation: As the surface is a plane, it must be the graph of a linear function which can be writ- ten in intercept form as x a + y b + z c = 1 . But by inspection we see that the x-intercept is x = 5, the y-intercept is y = 3 and the z- intercept is z = 4. Consequently, the surface is the graph in the first octant of the equation x 5 + y 3 + z 4 = 1 . Version 126 – Exam02 – gilbert – (55485) 2 003 10.0 points Find parametric equations for the line pass- ing through the point P (2 , − 1 , 4) and perpen- dicular to the plane 3 x + 3 y − 4 z = 4 . 1. x = − 2 + 3 t, y = 1 + 3 t, z = − 4 − 4 t 2. x = 3 + 2 t, y = 3 + t, z = 4 − 4 t 3. x = 3 − 2 t, y = − 3 + t, z = − 4 + 4 t 4. x = 3 + 2 t, y = 3 − t, z = − 4 + 4 t 5. x = 2 + 3 t, y = − 1 + 3 t, z = 4 − 4 t correct 6. x = 2 − 3 t, y = 1 − 3 t, z = 4 − 4 t Explanation: A line passing through a point P ( a, b, c ) and having direction vector v is given para- metrically by r ( t ) = a + t v , a = ( a, b, c ) . Now for the given line, its direction vector will be parallel to the normal to the plane 3 x + 3 y − 4 z = 4 ....
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Exam02-solutions-1 - Version 126 – Exam02 – gilbert...

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