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Curl and Divergence.

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Presentation on theme: "Curl and Divergence."β€” Presentation transcript:

1 Curl and Divergence

2 Curl Let 𝐅= 𝑃,𝑄,𝑅 be a vector field on 𝐑 3 and assume the partial derivatives of 𝑃, 𝑄 and 𝑅 all exist. Let 𝛻= πœ• πœ•π‘₯ , πœ• πœ•π‘¦ , πœ• πœ•π‘§ (the β€œdel” operator) The curl of the vector field 𝐅 is Example 1: Let Find curl F. P Q R

3 Curl Example 2: Let F π‘₯,𝑦,𝑧 = π‘₯,𝑦,𝑧 . Evaluate curl 𝐅.
A field with 𝛻×𝐅=𝟎 is called curl free or irrotational. Theorem 1: A conservative vector field with continuous derivatives is irrotational. Proof: Suppose 𝐅 is a conservative vector field and let 𝑓 be its potential function: 𝐅=𝛻𝑓= 𝑓 π‘₯ , 𝑓 𝑦 , 𝑓 𝑧 Clairaut’s Theorem

4 Curl The converse of Theorem 1 is also true if we assume the domain of the vector field is simply connected (in particular, the whole 𝑹 3 ) Theorem 2: If 𝐅 is a vector field on all of 𝑹 3 with continuous partial derivatives and 𝛻×𝐅=𝟎, then 𝐅 is a conservative vector field. REMARKS: Theorem 2 provides a quick and easy way to check whether a 3D vector field is conservative. The 2D vector field 𝐅 π‘₯,𝑦 = 𝑃 π‘₯,𝑦 ,𝑄(π‘₯,𝑦) can be written as a 3D vector field 𝐅 π‘₯,𝑦 = 𝑃 π‘₯,𝑦 ,𝑄 π‘₯,𝑦 ,0 . Then 𝛻×𝐅= 0,0, 𝑄 π‘₯ βˆ’ 𝑃 𝑦 . Therefore, in 2D, 𝛻×𝐅=0 is equivalent to 𝑄 π‘₯ = 𝑃 𝑦 The Fundamental Theorem of Line Integrals holds also in 3D for conservative vector fields.

5 Curl – Example 3 Show that 𝐅= 𝑦,π‘₯+𝑧,𝑦+6𝑧 is conservative and find a potential function. 𝐅 is irrotational, therefore there exists a potential function 𝑓 such that 𝐅=𝛻𝑓 𝑓 π‘₯ =𝑦 𝑓 𝑦 =π‘₯+𝑧 𝑓 𝑧 =𝑦+6𝑧 Integrate w.r.t π‘₯ 𝑓 π‘₯,𝑦,𝑧 =π‘₯𝑦+𝑔(𝑦,𝑧) differentiate w.r.t 𝑦 𝑓 𝑦 =π‘₯+ 𝑔 𝑦 (𝑦,𝑧) =π‘₯+𝑧 𝑔 𝑦 𝑦,𝑧 =𝑧 g 𝑦,𝑧 =𝑧𝑦+β„Ž(𝑧) 𝑓 π‘₯,𝑦,𝑧 =π‘₯𝑦+𝑧𝑦+β„Ž(𝑧) differentiate w.r.t 𝑧 𝑓 𝑧 =𝑦+β„Žβ€²(𝑧) =𝑦+6𝑧 β„Ž β€² 𝑧 =6𝑧 β„Ž 𝑧 =3 𝑧 2 +𝐾 𝑓 π‘₯,𝑦,𝑧 =π‘₯𝑦+𝑧𝑦+3 𝑧 2 +𝐾

6 Curl - Example 3 continued
Evaluate the line integral of 𝐅 along the curve comprised of the line segment from (0,0,0) to (1,1,2) and the line segment from (1,1,2) to (3,1,4) Initial point: (0,0,0) Terminal point: (3,1,4) Potential Function: 𝑓 π‘₯,𝑦,𝑧 =π‘₯𝑦+𝑧𝑦+3 𝑧 2 Fundamental Theorem for Line Integrals:

7 Curl Interpreting the curl: If we think of the vector field as a velocity vector field of a fluid in motion, the curl measures rotation. At a given point, the curl is a vector parallel to the axis of rotation of flow lines near the point, with direction determined by the Right Hand Rule. Example 4: Determine whether the curl of each vector field at the origin is the zero vector or points in the same direction as Β± i, Β± j, Β± k. (a) Based on the direction of the rotation and the Right Hand Rule, the curl will point in the βˆ’π€= 0,0,βˆ’1 direction.

8 Curl – Example 4 continued
(b) The vector field is clearly irrotational, thus the curl is the zero vector: 0,0,0 . (c) Based on the direction of the rotation and the Right Hand Rule, the curl points in the direction of +𝐣= 0,1,0

9 Divergence Let be a differentiable vector field on a region in 𝐑 3 .
Let 𝛻= πœ• πœ•π‘₯ , πœ• πœ•π‘¦ , πœ• πœ•π‘§ (the β€œdel” operator) The divergence of the vector field 𝐅 is (a scalar) Example 5: Let 𝐅= 𝑒 π‘₯𝑦𝑧 ,𝑦𝑧 sin π‘₯ , π‘₯ 2 + 𝑦 2 + 𝑧 Find div F. Example 6: Let 𝐅= π‘₯,𝑦,𝑧 . Find div F.

10 Divergence Interpreting the Divergence: The divergence at a given point measures the net flow out of a small box around the point, that is, it measures what is produced (source) or consumed (sink) at a given point in space. Example 7: Determine whether the divergence of each vector field at the indicated point P is positive, negative or zero (a) Draw a box around the point P. We can clearly see from the magnitude of the vectors that the net flow out of the box is negative (more β€œstuff” coming in than out). Thus the divergence is negative and we say the point P is a sink.

11 Divergence – Example 7 continued
Draw a box around the point. We can see that the net flow out of the box is zero (same amount of β€œstuff” is coming in than coming out). (b) Thus the divergence is zero and we say that the vector field is incompressible at the point. Draw a box around the point. The net flow is positive (less β€œstuff” coming in than out). (c) Thus the divergence is positive and we say that the point is a source.


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