{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2220pracprelim1sol

# 2220pracprelim1sol - Math 222 Prelim I Solutions Spring...

This preview shows pages 1–2. Sign up to view the full content.

Math 222 Prelim I Solutions Spring 2008 1. Let S be the level surface f ( x, y, z ) = x 2 + 3 y 2 + z 2 + xz = 6. (5 pts) (a) Find the tangent plane to S at the point (1 , 1 , 1). Solution . We have f = (2 x + z, 6 y, 2 z + x ), and at (1 , 1 , 1) this equals (3 , 6 , 3). The tangent plane is therefore 3( x 1) + 6( y 1) + 3( z 1) = 0. This can be simplified to x + 2 y + z = 4 if you want. (5 pts) (b) Find all points on S where the tangent plane is horizontal (parallel to the xy -plane). Solution . The tangent plane is horizontal where the normal vector f is vertical, i.e., of the form (0 , 0 , z ). Thus we have the conditions ∂f ∂x = 2 x + z = 0 and ∂f ∂y = 6 y = 0, so y = 0 and z = 2 x . Plugging y = 0 and z = 2 x into x 2 + 3 y 2 + z 2 + xz = 6 gives x 2 + 4 x 2 2 x 2 = 6, which has the solutions x = ± 2. Since y = 0 and z = 2 x we get the two points ± ( 2 , 0 , 2 2) where the tangent plane is horizontal. (5 pts) (c) Show that the points on S where the tangent plane is vertical (parallel to a plane containing the z -axis) form an ellipse in some sloping plane. Solution . Vertical tangent planes occur where f is horizontal, of the form ( x, y, 0) so we get the condition ∂f ∂z = 0. This gives the plane x + 2 z = 0, or x = 2 z . Plugging x = 2 z into the equation x 2 + 3 y 2 + z 2 + xz = 6 gives 4 z 2 + 3 y 2 + z 2 2 z 2 = 3 y 2 + 3 z 2 = 6, or y 2 + z 2 = 2. This is a cylinder (with axis the x -axis), and intersecting this cylinder with the plane x = 2 z gives an ellipse in this plane. 2. Let f ( x, y ) = x 2 + y 2 2 x . (5 pts) (a) Show that the level curves of f are circles passing through the origin. Solution . A level curve is f ( x, y ) = x 2 + y 2 2 x = c . This equation can be rewritten as x 2 + y 2 = 2 cx or x 2 2 cx + y 2 = 0. Completing the square, this can be written as ( x c ) 2 + y 2 = c 2

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern
{"found_math": false, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0}