Exam02-solutions-1 - Version 126 Exam02 gilbert (55485) 1...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

Page1 / 8

Exam02-solutions-1 - Version 126 Exam02 gilbert (55485) 1...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online