{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# Slide_18 - MA1104 Multivariable Calculus Lecture 18 Dr KU...

This preview shows pages 1–15. Sign up to view the full content.

MA1104 Multivariable Calculus Lecture 18 Dr KU Cheng Yeaw Thursday March 26, 2009

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

View Full Document
Recap Last Lecture ... we study Green’s Theorem and see how to apply it to compute line integrals. Green’s Theorem can be regarded as the counterpart of the Fundamental Theorem of Calculus for double integral
Overview In this lecture ... we define two operations that can be performed on vector fields. They play a basic role in the applications of vector calculus to fluid flow, electricity, and magnetism. Each operation resembles differentiation. However, one produces a vector field whereas the other produces a scalar field. We use these operations to formulate Green’s Theorem in vector forms. we use vector functions to describe more general surfaces, called parametric surfaces.

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

View Full Document
Curl Suppose: F = P i + Q j + R k is a vector field on R 3 The partial derivatives of P , Q , and R all exist Then, the curl of F is the vector field on defined by: Curl curl F = ∂R ∂y - ∂Q ∂z i + ∂P ∂z - ∂R ∂x j + ∂Q ∂x - ∂P ∂y k .
Whoosh ... what is that? do you expect me to remember this! Surprisingly (or unfortunately) this seemingly ugly expression turns out to have some physical significance. So we’d better find a way to memorize it. We introduce the vector differential operator O (‘del’) as Operator del O = i ∂x + j ∂y + k ∂z .

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

View Full Document
It has meaning when it operates on a scalar function to produce the gradient of f : O f = i ∂f ∂x + j ∂f ∂y + k ∂f ∂z .
If we think of O as a vector with components ∂x , ∂y , ∂z , we can also consider the formal cross product of O with the vector field F as follows: O × F = i j k ∂x ∂y ∂z P Q R = ∂R ∂y - ∂Q ∂z i + ∂P ∂z - ∂R ∂x j + ∂Q ∂x - ∂P ∂y k = curl F

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

View Full Document
Hey!, that is not too bad. Well ... if you insist ... The easiest way to remember curl is by means of the symbolic expression curl F = O × F .
Example 1. If F ( x, y, z ) = xz i + xyz j - y 2 k , find curl F . Solution. curl F = O × F = i j k ∂x ∂y ∂z xz xyz - y 2 = ∂y ( - y 2 ) - ∂z ( xyz ) i - ∂x ( - y 2 ) - ∂z ( xz ) j ∂x ( xyz ) - ∂y ( xz ) k

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

View Full Document
= ( - 2 y - xy ) i - (0 - x ) j + ( yz - 0) k = - y (2 + x ) i + x j + yz k .
Recall that the gradient of a function f ( x, y, z ) is a vector field on R 3 , so we can compute its curl. If f is a function of three variables that has continuous second order partial derivatives, then curl O f = 0 .

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

View Full Document
Proof. We have, by Clairaut’s Theorem, curl O f = O × O f = i j j ∂x ∂y ∂z ∂f ∂x ∂f ∂y ∂f ∂z = 2 f ∂y∂z - 2 f ∂z∂y i + 2 f ∂z∂x - 2 f ∂x∂z j + 2 f ∂x∂y - 2 f ∂y∂x k = 0 .
We can restate the preceding theorem for (continuous) conservative vector field: Suppose F is continuous. If F is conservative then curl F = 0 . This gives us a way to verify a vector field is not conservative.

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

View Full Document
Example 2. Show that the vector field F ( x, y, z ) = xz i + xyz j - y 2 k is not conservative.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 77

Slide_18 - MA1104 Multivariable Calculus Lecture 18 Dr KU...

This preview shows document pages 1 - 15. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online