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

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Unformatted text preview: MA1104 Multivariable Calculus Lecture 18 Dr KU Cheng Yeaw Thursday March 26, 2009 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. 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 . 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 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 = (- 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 = . 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 = . We can restate the preceding theorem for (continuous) conservative vector field: Suppose F is continuous. If F is conservative then curl F = ....
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Slide_18 - MA1104 Multivariable Calculus Lecture 18 Dr KU...

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