This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: B U Department of Mathematics Math 102 Calculus II Spring 2006 Final This archive is a property of Bo˘gazi¸ ci University Mathematics Department. The purpose of this archive is to organise and centralise the distribution of the exam questions and their solutions. This archive is a non-profit service and it must remain so. Do not let anyone sell and do not buy this archive, or any portion of it. Reproduction or distribution of this archive, or any portion of it, without non-profit purpose may result in severe civil and criminal penalties. 1. What horizontal plane is tangent to the surface z = x 2- 4 xy- 2 y 2 + 12 x- 12 y- 1 and what is the point of the tangency? Solution: Begin by rewriting the surface equation as F ( x, y, z ) = x 2- 4 xy- 2 y 2 + 12 x- 12 y- 1- z. Obviously, the gradient of this function at the point ( x , y , z ) is a normal vector to the surface at ( x , y , z ), where ~ ∇ F ( x , y , z ) = (2 x- 4 y + 12)ˆ ı + (- 4 x- 4 y- 12)ˆ + (- 1) ˆ k For ~ ∇ F ( x , y , z ) to be the normal vector to the horizontal plane, the first two components of...
View Full Document
- Spring '08
- Vector Calculus, Cos, dydx, sin xy