M252Quiz5R - Quiz 5 R1 REVIEW: CH. 18 Here, we deal with...

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Quiz 5 – R1 REVIEW: CH. 18 Here, we deal with Cartesian coordinates. (Schey and Marsden discuss other systems.) VECTOR FIELDS (18.1) We say a scalar function f is “nice” if its 1 st -order partial derivatives are continuous (and, therefore, f itself is) “where we care.” A vector field F is “nice” if its components are nice. F A is the vector associated with point A . GRAD, CURL, AND DIV (18.1) The del (or nabla) operator = x , y ,... grad f = f = f x , f y (See Sections 16.6 and 16.7 for more on gradients.) curlF = ∇ × F = i j k x y z M N P where F x , y , z ( ) = M x , y , z ( ) , N x , y , z ( ) , P x , y , z ( ) is a vector field in 3 . A is a vector whose … … direction indicates axis of rotation of field near P (use right-hand rule); … length indicates strength of rotational effect. These properties are justified by Stokes’s Theorem in Section 18.7.
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Quiz 5 – R2 div F = F = x , y ,... M , N = M x + N y + ... where F x , y ( ) = M x , y ( ) , N x , y ( ) is a vector field in n . div F A is a scalar: If this is negative, then there is a sink at A . If this is positive, then there is a source at A . (Think: Alphabetical order coincides with numerical order. Also, it measures the tendency of a fluid to diverge from A .) If this is 0, then A has neither. These properties are justified by the Divergence (or Gauss’s) Theorem in Section 18.6. Warning: Unlike grad and curl , div actually yields a scalar function, not a vector- valued function. Interesting Identities div curlF ( ) = 0 , where F is in 3 curl grad f ( ) = 0 , where f is a function of x , y , and z Technical Note: We assume that f and the components of F have continuous 2 nd -order partial derivatives.
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Quiz 5 – R3 LINE / PATH INTEGRALS (18.2) We will integrate along a piecewise-smooth (“ps”) path C . A ps path has no breaks, corners, cusps, or backtracks. Arrowheads indicate orientation along the path. Here, we will focus on formulas used for the 2D xy -plane. The formulas below are naturally extended to the 3D xyz case. Let’s say C is parameterized by: r t ( ) = x t ( ) , y t ( ) We may then rewrite problems in terms of t alone instead of both x and y . Technical Note: The smoothness condition requires that the tangent VVF r t ( ) be non- 0 along the path (except possibly at endpoints) and continuous along the path. For ps curves, replace “path” with “pieces of the path.”
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Quiz 5 – R4 ds , the differential of arc length, can be expressed in many ways: ds = dx ( ) 2 + dy ( ) 2 from Pythagorean Theorem ( ) = dx dt 2 + dy dt 2 dt if t is increasing, and, therefore, dt > 0 ( ) Warning: For now, let’s say we are required to parameterize paths (or pieces of paths) in such a way that t is increasing consistently with the orientation. Otherwise, we replace dt with dt or dt in these formulas.
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M252Quiz5R - Quiz 5 R1 REVIEW: CH. 18 Here, we deal with...

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