This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Version 126 Exam02 gilbert (55485) 1 This printout should have 14 questions. Multiplechoice 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 xintercept is x = 5, the yintercept 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 ....
View Full
Document
 Spring '11
 Gilbert
 Chain Rule, Derivative

Click to edit the document details