calc3_gradientstest

calc3_gradientstest - 18.02, Spring 2008 Practice Test 2...

Info iconThis preview shows pages 1–2. 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: 18.02, Spring 2008 Practice Test 2 Solutions 1. Find the equation of the tangent plane to the surface x 3 y + z 2 = 3 at the point (- 1 , 1 , 2). Solution. The surface is defined by the equation f ( x, y, z ) = x 3 y + z 2 = 3. The normal to the tangent plane is given by f = 3 x 2 y i + x 3 j + 2 z k , and f (- 1 , 1 , 2) = (3 ,- 1 , 4). The equation of the plane is thus 3( x + 1)- ( y- 1) + 4( z- 2) = 0 . 2. Let w = f ( u, v ), where u = xy and v = x/y . Using the chain rule, express w x and w y in terms of x , y , f u , and f v . Solution. We have w x = f u u x + f v v x = yf u + 1 y f v and w y = f u u y + f v v y = xf u- x y 2 f v . 3. Let f ( x, y ) = x 2 y 2- x . (a) Find f at (2 , 1). Solution. We have f = (2 xy 2- 1 , 2 x 2 y ) so f (2 , 1) = (3 , 8). (b) Write the equation for the tangent plane to the graph of f at (2 , 1 , 2)....
View Full Document

This note was uploaded on 04/26/2008 for the course MATH 263 taught by Professor Lucas during the Spring '08 term at Loyola Chicago.

Page1 / 3

calc3_gradientstest - 18.02, Spring 2008 Practice Test 2...

This preview shows document pages 1 - 2. Sign up to view the full document.

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