Report

VC.04-VC.08 Redux Vector Fields Acting on a Curve VC.04: Find a surface with the given gradient (Ex 1 From VC.04 Notes) Example 1: Given f(x, y) x 1, y 1 , find a possible equation for the surface f(x, y). x2 y 2 f(x, y) x y? 2 2 VC.04: Gradient Fields Example 2: Plot the gradient field for f(x, y) 2 f e x y2 xy e x2 y2 : 1 2x(x y),1 2y(x y) How many points stand out to you in this vector field? This vector field has a source and a sink. What do the source and sink correspond to on the surface? Source Local Minimum Sink Local Maximum VC.04: Gradient Fields Example 2: Finally, compare the gradient field and f(x, y ) xy e x2 y2 : VC.04: Breaking Field Vectors into Two Components Let Field(x, y) (m(x, y),n(x, y) be a vector field acting on the curve (x(t),y(t)). Forward/Backward Push: Left/Right Push: The push of the field The push of the field vector in the direction of the tangent vector to the curve: vector in the direction of the normal vector to the curve: Field(x(t),y(t)) (x'(t),y'(t)) (x'(t),y'(t)) (x'(t),y'(t)) (x'(t),y'(t)) Field(x(t),y(t)) (y'(t), x'(t)) (y'(t), x'(t)) (y'(t), x'(t)) (y'(t), x'(t)) "The flow of the vector field "The flow of the vector field ALONG the curve." ACROSS the curve." Let the vector field (3x2 3x)(y 3),(3y 2 3y)(x 3) act on the ellipse (x(t), y(t)) ( 4,2) (4cos(t),2sin(t)). Use the push of the field vectors in the direction of the tangent vectors to the curve to determine whether the net flow ALONG the curve is clockwise or counterclockwise: Field(x(t),y(t)) (x'(t),y'(t)) Plot (x'(t),y'(t)) (x'(t),y'(t)) (x'(t),y'(t)) The net flow of the vector field ALONG the curve is clockwise. Let the vector field (3x2 3x)(y 3),(3y 2 3y)(x 3) act on the ellipse (x(t), y(t)) ( 4,2) (4cos(t),2sin(t)). Use the push of the field vectors in the direction of the normal vectors to the curve to determine whether the net flow ACROSS the curve is "inside to outside" or "outside to inside." Field(x(t),y(t)) (y'(t), x'(t)) Plot (y'(t), x'(t)) (y'(t), x'(t)) (y'(t), x'(t)) The net flow of the vector field ACROSS the curve is "inside to outside." VC.04: Flow Along and Flow Across "The flow of the vector field "The flow of the vector field ALONG the curve." ACROSS the curve." Field(x(t),y(t)) (x'(t),y'(t)) Plot (x'(t),y'(t)) (x'(t),y'(t)) (x'(t),y'(t)) Counterclockwise: Field(x(t),y(t)) (y'(t), x'(t)) Plot (y'(t), x'(t)) (y'(t), x'(t)) (y'(t), x'(t)) Inside to outside: Clockwise: Outside to Inside: *Could be 0! *Could be 0! For this chapter, our only measurement tool is to plot and then eyeball the results... VC.05: Measuring the Flow of a Vector Field ALONG a Curve If C is a closed curve with a counterclockwise parameterization: If C is not closed: Integral : Integral : b Field(x(t), y(t)) (x '(t), y '(t))dt a b m(x(t), y(t))x '(t) n(x(t), y(t))y '(t) dt a Ñ C m(x, y)dx n(x, y)dy These integrals are used to compute net flow of the vector field along the closed curve (clockwise or counterwise). They can be modified to measure flow of the vector field across the closed curve (inside to outside or outside to inside) with relative ease. b Field(x(t), y(t)) (x '(t), y '(t))dt a b m(x(t), y(t))x '(t) n(x(t), y(t))y '(t) dt a m(x, y)dx n(x, y)dy C Without a closed curve, the integral is measured whether the net flow along the open curve is in the direction of parameterization or against it, and whether the net flow across the open curve is from “above to below” or “below to above.” These integrals are usually called line integrals or path integrals. VC.05: Measuring the Flow of a Vector Field ALONG a Curve Let C be a closed curve with a COUNTERCLOCKWISE parameterization : Ñ C m(x, y)dx n(x, y)dy 0 describes a net flow of the vector field along the curve in the counterclockwise direction. Ñ C m(x, y)dx n(x, y)dy 0 describes a net flow of the vector field along the curve in the clockwise direction. Ñ C m(x, y)dx n(x, y)dy can equal 0. VC.05: Measuring the Flow of a Vector Field Along Another Closed Curve 2 x 1 2 Let Field(x, y) (3y, x ) be a vector field acting on the ellipse y 1. Compute Ñ C m(x, y)dx n(x, y)dy : 2 2 Component of Field Vectors in the Direction of the Field and Curve : Vectors on Curve : Tangent Vectors VC.05: Measuring the Flow of a Vector Field Along Another Closed Curve Let Field(x, y) (3y, x 2 ) be a vector field acting on the 2 x 1 2 ellipse m(x, y)dx n(x, y)dy : y 1. Compute Ñ C 2 Counterclockwise Field(x, y) m(x, y),n(x, y) (3y, x ) E(t) (2cos(t), sin(t)) (1, 0) 2 b Ñ C m(x, y)dx n(x, y)dy m(x(t), y(t))x'(t) n(x(t), y(t))y'(t) dt a 2 2 3 2 cos(t) 4 cos (t) 4cos (t) 6 si n (t)dt 0 10 Negative! The net flow of the vector field along the curve is against the direction of the parameterization (clockwise). VC.05: Measuring the Flow of a Vector Field ACROSS a Curve If C is a closed curve with a counterclockwise parameterization: Integral : If C is not closed: Integral : b b Field(x(t), y(t)) (y '(t), x '(t))dt a n(x(t), y(t))x '(t) m(x(t), y(t))y '(t) dt Ñ n(x, y)dx m(x, y)dy a a b b Field(x(t), y(t)) (y '(t), x '(t))dt C With a closed curve, the integral measures whether the net flow ACROSS the closed curve is from “inside to outside” or from “outside to inside.” n(x(t), y(t))x '(t) m(x(t), y(t))y '(t) dt a n(x, y)dx m(x, y)dy C Without a closed curve, the integral measures whether the net flow ACROSS the open curve is from above to below the curve or from below to above. VC.05: Measuring the Flow of a Vector Field Across Another Closed Curve Let Field(x, y) ( x cos(y), y sin(x)) be a vector field acting on the circle x2 y2 1. Compute Ñ n(x, y)dx m(x, y)dy : C Counterclockwise m(x, y),n(x, y) (x cos(y), y sin(x)) c(t) (cos(t), sin(t)) Ñ C n(x, y)dx m(x, y)dy n(x(t), y(t))x'(t) m(x(t), y(t))y'(t) dt b a 2 Negative! The net flow of the vector field across the curve is from outside to inside. VC.05: The Gradient Test A vector field, Field(x,y) m(x, y),n(x, y , is a gradient field if and only if: m(x, y) n(x, y) y x (The Gradient Test) Proof of "if" Part of Theorem: If Field(x,y) is a gradient field, then Field(x,y)= fx , fy . So fxy fyx . VC.05: The Flow of a Gradient Field Along a Closed Curve Let Field(x,y)= m(x, y),n(x, y) be a gradient field, and let C be a simple closed curve with a parameterization (x(t),y(t)) for a t b. b 1) Field(x(t), y(t)) (x'(t), y'(t))dt 0 a 2) Ñ m(x, y)dx n(x, y)dy 0 C 3) The flow of a gradient field along a simple closed curve is 0. Why is this intuitively true? How do we know a closed curve can't be a trajectory of a gradient field? Is the flow of a gradient field ACROSS a closed curve 0? VC.05: Path Independence: The Flow of a Gradient Field Along an Open Curve Let Field(x,y)= m(x, y),n(x, y) be a gradient field, and let C1 and C2 be different curves that share the same starting and ending point: m(x, y)dx n(x, y)dy m(x, y)dx n(x, y)dy C1 C2 A gradient field is said to be path independent. The flow of the vector field along any two curves connecting two points is the same... VC.06: The Gauss-Green Formula Let R be a region in the xy-plane whose boundary is parameterized by (x(t),y(t)) for tlow t thigh . Then the following formula holds : thigh tlow n m m x(t), y( t) x'(t) n x(t), y( t) y'(t) dt dA x y R With the proper interpretation, we can use this formula to help us compute flow along/across measurements!! The basic interpretation of Gauss-Green is that it is a correspondence between a line integral of a closed curve and a double integral of the interior region of the closed curve. So for this to work correctly, we just need to make sure the vector field has no singularities in the interior region! VC.06: Measuring the Flow of a Vector Field ALONG a Closed Curve Let C be a closed curve with a counterclockwise parameterization. Then the net flow of the vector field ALONG the closed curve is measured by: b Field(x(t), y(t)) (x '(t), y'(t))dt a b m(x(t), y(t))x '(t) n(x(t), y(t))y'(t) dt a Ñ C m(x, y)dx n(x, y)dy Let region R be the interior of C. If the vector field has no singularities in R, then we can use Gauss-Green: n m dx dy x y R rotField dx dy R Let rotField(x, y) n m . x y VC.06: Summary: The Flow of A Vector Field ALONG a Closed Curve: Let C be a closed curve parameterized counterclockwise. Let Field(x,y) be a vector field with no singularities on the interior region R of C. Then: Ñ C m(x, y)dx n(x, y)dy rotField dx dy R This measures the net flow of the vector field ALONG the closed curve. We define the rotation of the vector field as: n m rotField(x, y) D[n[x, y], x] D[m[x, y], y] x y VC.06: Measuring the Flow of a Vector Field ACROSS a Closed Curve Let C be a closed curve with a counterclockwise parameterization. Then the net flow of the vector field ACROSS the closed curve is measured by: b Field(x(t), y(t)) (y '(t), x '(t))dt a b n(x(t), y(t))x '(t) m(x(t), y(t))y '(t) dt a Ñ C n(x, y)dx m(x, y)dy Let region R be the interior of C. If the vector field has no singularities in R, then we can use Gauss-Green: m n dx dy x y R divField dx dy R Let divField(x, y) m n . x y VC.06: Summary: The Flow of A Vector Field ACROSS a Closed Curve: Let C be a closed curve parameterized counterclockwise. Let Field(x,y) be a vector field with no singularities on the interior region R of C. Then: Ñ C n(x, y)dx m(x, y)dy divField dx dy R This measures the net flow of the vector field ACROSS the closed curve. We define the divergence of the vector field as: m n divField(x, y) D[m[x, y], x] D[n[x, y], y] x y VC.06: The Divergence Locates Sources and Sinks Let C be a closed curve with a counterclockwise parameterization with no singularities on the interior of the curve. Then: If divField(x,y)>0 for all points in C, then all these points are sources and the net flow of the vector field across C is from inside to outside. If divField(x,y)<0 for all points in C, then all of these points are sinks and the net flow of the vector field across C is from outside to inside. If divField(x, y) 0 for all points in C, then the net flow of the vector field across C is 0. VC.06: The Rotation Helps You Find Clockwise/Counterclockwise Swirl Let C be a closed curve with a counterclockwise parameterization with no singularities on the interior of the curve. Then: If rotField(x,y)>0 for all points in C, then all these points add counterclockwise swirl and the net flow of the vector field along C is from counterclockwise. If rotField(x,y)<0 for all points in C, then all of these points add clockwise swirl and the net flow of the vector field across C is clockwise. If rotField(x, y) 0 for all points in C, then these points have no swirl, and the net flow of the vector field along C is 0 (irrotational). VC.06: Avoiding Computation Altogether Let Field(x, y) 7x 2, y 6 and let C be a closed curve given by 3 C(t) (x(t), y(t)) sin (t), cos(t) sin(t) for t . 4 4 Is the flow of the vector field across the curve from inside to outside or 2 outside to inside? m n divField(x, y) 71 8 x y Ñ C n(x, y)dx m(x, y)dy divField dx dy 8 dx dy R R Since divField(x,y) is ALWAYS positive for all (x,y) and there are no singularities for any (x,y), this integral is positive for any closed curve. That is, for ANY closed curve, the net flow of the vector field across the curve is from inside to outside. VC.06: Find the Net Flow of a Vector Field ACROSS Closed Curve Let Field(x, y) x 2 2xy, y 2 x and let C be the rectangle bounded by x 2, x 5, y 1,and y 4. Measure the flow of the vector field across the curve. m n divField(x, y) 2x 4y x y Ñ C n(x, y)dx m(x, y)dy divField dx dy R 4 5 2x 4y dx dy 1 2 105 Negative. The net flow of the vector field across our closed curve is from outside to inside. VC.06: A Flow Along Measurement With a Singularity y x Let Field(x, y) 2 , and let C be the curve described by 2 2 2 x y x y 1 1 C(t) sin2 (t) cos(t) ,cos(t)+sin(t)+ for t 2. Compute the 2 2 2 flow of the vector field along the curve. n m rotField(x, y) 0 x y Since rotField(x,y)=0, your field is a gradient away from singularities. The only swirl can come from singularities! There is a singularity at (0,0). We can replace our curve with any curve that encapsulates the singularity: C2 (t) cos(t), sin(t) for 0 t 2 Note: Had there been no singularities in the curve, how would we know that the net flow of the vector field ALONG the curve would be 0? VC.06: A Flow Along Measurement With a Singularity y x Field(x, y) 2 , 2 and C(t) cos(t), sin(t) for 0 t 2 2 2 x y x y Because of the singularity, we can't use rotField(x, y) dx dy. Instead, we will need to compute Ñ m(x, y)dx n(x, y)dy the old-fashioned way: C 2 Field(x(t), y(t)) (x'(t), y'(t)) dt 0 2 sin(t) cos(t) , 0 cos2 (t) sin2 (t) cos2 (t) sin2 (t) ( sin(t), cos(t)) dt 2 2 2 sin (t) cos (t) dt 0 2 1 dt 0 2 So the net flow of the vector field along the curve is counterclockwise! VC.06: A Flow Along With Multiple Singularities? No Problem! y y 1 x x Let Field(x, y) 2 2 , 2 2 and let C be the 2 2 2 2 x (y 1) x y x (y 1) x y curve pictured below. Compute the flow of the vector field along the curve. rotField(x, y) n m 0 x y Since rotField(x,y)=0, your field is a gradient away from singularities. The only swirl can come from singularities! There are singularities at (0,0) and (0,1). We can encapsulate the singularities with two little circles and sum our results! C1 (t) 0.5 cos(t), sin(t) for 0 t 2 C2 (t) 0.5 cos(t), sin(t) (0,1) for 0 t 2 VC.06: Flow Along When rotField(x,y)=0 n m 0. Here are some conclusions about the net flow x y of the vector field along various closed curves: Let rotField(x, y) If C doesn't contain any singularities, then Ñ m(x, y)dx n(x, y)dy 0. C If C contains a singularity, then Ñ m(x, y)dx n(x, y)dy Ñ m(x, y)dx n(x, y)dy C C1 for any substitute curve C1 containing the same singularity (and no new extras). If C contains n singularities, then Ñ m(x, y)dx n(x, y)dy Ñ m(x, y)dx n(x, y)dy ... Ñ m(x, y)dx n(x, y)dy C C1 Cn for little circles, C1 ,..., Cn , encapsulating each of these singularities. VC.06: Flow Along When divField(x,y)=0 m n 0. Here are some conclusions about the net flow x y of the vector field across various closed curves: Let divField(x, y) If C doesn't contain any singularities, then Ñ n(x, y)dx m(x, y)dy 0. C If C contains a singularity, then Ñ n(x, y)dx m(x, y)dy Ñ n(x, y)dx m(x, y)dy C C1 for any substitute curve C1 containing the same singularity (and no new extras). If C contains n singularities, then Ñ n(x, y)dx m(x, y)dy Ñ n(x, y)dx m(x, y)dy ... Ñ n(x, y)dx m(x, y)dy C C1 Cn for little circles, C1 ,..., Cn , encapsulating each of these singularities. VC.07: The Area Conversion Factor: If we let r andt tend to zero, we can use this to get T1 T2 t t dA r T1 T1 r T1 t r drdt T2 T2 r is called the Jacobian matrix, and T2 t T1 You can let A xy (r, t) r T1 T1 T2 t t r T1 r is called the Jacobian determinant. T2 T2 r . Think of the Jacobian determinant as an T2 t t Area Conversion Factor that lets us compute an xy-space integral in rt-space: Hence, f(x, y)dA f(x(r,t), y(r,t)) A xy (r,t) dr dt Rxy Rrt VC.07: The Area Conversion Factor: f(x, y)dA f(x(r,t), y(r,t)) A Rxy Rrt xy (r,t) dr dt Let T(r, t) be a transformation from rt-space to xy-space. That is, T(r,t) T1 (r, t), T2 (r, t) (x(r, t), y(r, t)). T1 T2 t t Then A xy (r, t) r T1 r . T2 Note: Since we derived A xy (r, t) as the magnitude of a cross product, we need it to be positive. This is why we put absolute value bars into the formula above. VC.07: : Fixing Example 2 Remember we wanted 1 dA for a circle of radius 4 centered at (0,0). R T(r, t) (x(r, t), y(r, t)) (r cos(t),r sin(t)) x y A xy (r, t) r r x y t t cos(t) sin(t) r sin(t) r cos(t) 1 dA R r 1 r dr dt 0 0 2 4 r dt 2 0 0 2 2 r cos2 (t) r sin2 (t) r cos2 (t) sin2 (t) 2 4 8 dt 0 16 Phew! We did it! VC.07: Mathematica-Aided Change of Variables (Parallelogram Region) Use Mathematica to compute e y dA for R given by the parallelogram: R These lines are given by y x 1, y x 4 , y 1 13 1 x , and y x 6. 4 4 4 We can rewrite them as 1 y x , 4 y x , 1 13 1 xy ,and x y 6. 4 4 4 1 13 Then we can let u y x and v x y for 4 u 1 and v 6! 4 4 But to compute our integral, we need the map from uv-space to xy-space... That is, we have u(x, y) and v(x, y ), but we need x(u, v ) and y(u, v). Let Mathematica do the work: Solve[{u u[ x , y ], v v[ x , y ]},{ x , y }] 4 1 x (u v ) and y (u 4v ) 5 5 VC.07: Mathematica-Aided Change of Variables (Parallelogram Region) Use Mathematica to compute e y dA for R given by the parallelogram: R 4 1 x (u v ) and y (u 4v ) 5 5 x A xy (u, v) u x v y 4 u 5 y 4 v 5 4 So we use 5 1 4 5 5 4 5 VC.07: Mathematica-Aided Change of Variables (Parallelogram Region) Use Mathematica to compute e y dA for R given by the parallelogram: R 4 4 1 , x (u v), y (u 4v), 5 5 5 13 for 4 u 1 and v6: 4 A xy (u, v) e R y dA 6 1 e 13/ 4 4 (u 4v )/5 4 du dv 5 417.1 (from Mathematica) VC.08: 3D Integrals: f(x, y, z) dx dy dz f(x, y, z) V R xyz xyz Ruvw x u x Vxyz (u, v, w) v x w y u y v y w (u, v, w) du dv dw z u z v z w In my opinion, a good way to think about f(x, y, z) dx dy dz is R xyz as a calculation of the mass of R xyz where f(x, y, z) is the density of the solid at any given point (x, y, z). VC.08: A 3D-Change of Variables with an Integrand Compute 9y dx dy dz where R R xyz xyz is the parallelepiped that is between the planes z 3x and z 3x 2, y x and y x 4, and y 2x and y 2x 3. wv x 3 All from Example 4: u z 3x 0 u 2 2v w y v yx 0 v 4 1 3 Vxyz (u, v, w) w y 2x 0 w 3 z u w v 3 3 4 2 9y dx dy dz 9y(u, v, w) V 0 0 0 3 4 2 R xyz (u, v, w) du dv dw 2v w 1 9 0 0 0 3 3 du dv dw 3 4 2 xyz 2v w du dv dw 0 0 0 132