Mar1-12 - Math 260, Spring 2012 Jerry L. Kazdan Class...

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: Math 260, Spring 2012 Jerry L. Kazdan Class Outline: March 1, 2012 Functions of Several Variables Reading: Marsden and Tromba, Vector Calculus , Chapter 2, Chapter 3.1–3.4 and Sections 8.1-8.2 in the notes http://www.math.upenn.edu/ ∼ kazdan/260S12/notes/math21/math21-2012-2up.pdf Exam. 2 will be on Tuesday, March 13 during class. As usual, it will be closed book, no calculators or cell phones etc., but you may use one 3 × 5 card with notes on both sides. The Exam will cover the mater covered through class on Tuesday, Feb. 28, with emphasis on the material since Exam. 1. 1. The Chain Rule a) Simplest Case Say you have a function f ( X ) := f ( x, y, z ) which might givde the temperature at a point X = ( x, y, z ) in R 3 . If we have a curve X ( t ) = ( x ( t ) , y ( t ) , z ( t )), then h ( t ) := f ( X ( t )) gives the temperature at points of the curve. We want to compute dh/dt . So we need the chain rule . For h ( t ) = f ( x ( t ) , y ( t ) , z ( t )) it states dh ( t ) dt = ∂f ∂x dx dt + ∂f ∂y dy dt + ∂f ∂z dz dt = ∇ f ( X ( t )) · X ′ ( t ) . Example f ( X ) := x 2 y + e 2 yz and the curve is X ( t ) = (cos t, sin t, 2 − t ). If h ( t ) := f ( X ( t )) then ∇ f ( X ) = (2 xy, x 2 + 2 ze 2 yz , 2 ye 2 yz ) and X ′ ( t ) = ( − sin t, cos t, − 1) . Thus, for instance at t = 0, X (0) = (1 , , 2), X ′ (0) = (0 , 1 , − 1) and ∇ f ( X (0)) = (0 , 5 , 0) so h ′ (0) = 5. Application Say we have a surface in R 3 on which f ( X ) is constant, f ( x, y, z ) = c . This is called a level surface of f (if f ( X ) specifies the temperature, this surface is often called an isotherm ). Let X ( t ) be a smooth curve on this surface, so f ( X ( t )) = c . Since c is a constant, if we take the derivative of this we find Theorem On any level surface, ∇ f ( X ) · X ′ ( t ) = 0 so the vector ∇ f ( X ) is orthogonal to the surface . Proof Since X ( t ) is a curve on the surface, its tangent vector, X ′ ( t ) is tan- gent to the surface. But ∇ f ( X ) · X ′ ( t ) = 0 means that ∇ f ( X ) is orthogonal to all these tangent vectors and hence to the surface....
View Full Document

This note was uploaded on 03/06/2012 for the course MATH 260 taught by Professor Staff during the Spring '12 term at UPenn.

Page1 / 5

Mar1-12 - Math 260, Spring 2012 Jerry L. Kazdan Class...

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