Exam 3 - Version 057 – Exam03 – Gilbert – (56380) 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 057 – Exam03 – Gilbert – (56380) 1 This print-out should have 12 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points From the contour map of f shown below de- cide whether f x and f y are positive, negative, or zero at P . 2 2 4 4 6 6 P x y 1. f x = 0 , f y > 2. f x < , f y < 3. f x = 0 , f y < 4. f x > , f y < 5. f x < , f y > correct 6. f x > , f y > Explanation: When we walk in the x-direction from P we are walking downhill, so f x < 0. On the other hand, when we walk in the y-direction from P we are walking uphill, so f y > 0. Consequently, at P f x < , f y > . keywords: contour map, contours, partial derivative, slope, 002 10.0 points Which equation has the surface x y z as its graph in the first octant? 1. x 3 + y 5 + z 4 = 1 2. x 5 + y 3 + z 4 = 1 correct 3. x 5 + y 4 + z 3 = 1 4. x 3 + y 4 + z 5 = 1 5. x 4 + y 5 + z 3 = 1 6. x 4 + y 3 + z 5 = 1 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 057 – Exam03 – Gilbert – (56380) 2 003 10.0 points Which one of the following equations has graph 1. x + z 2 − 4 = 0 2. z + y 2 − 4 = 0 3. y − z 2 + 4 = 0 4. x − y 2 + 4 = 0 5. z − x 2 + 4 = 0 correct Explanation: The graph is a parabolic cylinder that has constant value on any line parallel to the y- axis, so it will be the graph of an equation containing no y-term. This already eliminates the equations x − y 2 + 4 = 0 , y + x 2 − 4 = 0 , y − z 2 + 4 = 0 , z + y 2 − 4 = 0 . On the other hand, the intersection of the graph with the xz-plane, i.e. the y = 0 plane, is a parabola opening upwards on the z-axis as shown in x z Consequently, the graph is that of the equa- tion z − x 2 + 4 = 0 ....
View Full Document

This note was uploaded on 09/27/2010 for the course M 408 D taught by Professor Textbookanswers during the Spring '07 term at University of Texas.

Page1 / 7

Exam 3 - Version 057 – Exam03 – Gilbert – (56380) 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