Math 114 – Rimmer Section 16.2 Vector Fields 16.1/16.2 Vector Fields and Line Integrals Velocity vector fields showing the wind speed and direction Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals 1 Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals F ( xy,) = − yx , 2 Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals F( xy,) = y ,sin x F()xy,= ln( 1 + y2) ,ln( 1 + y 2 ) Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals A) F( xy,) = − yx , II B) F( xy,) = 1,sin y IV C) F( xy,) = x − 2, x + 1 I 1 D)() F xy,= y , x III 3 Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals y− x z F ( xyz,,) = yzx ,, F ( xyz, ,) = y , − 2, x F ()x,, y z = ,, z z 4 Math 114 – Rimmer Physics applications 16.1/16.2 Vector Fields and Line Integrals Velocity Field Dynamics Fluid dynamics Aerodynamics Thermodynamics Force Field Gravitational Field Electric Field Magnetic Field 4 Math 114 – Rimmer f x, y 16.1/16.2 Vector Fields ( ) and Line Integrals ∇fxy( ,) = fxyx( ,,,) f y ( xy ) Gradient Vector Field fxy( , ) = xy2 − y 3 ∇fxy( ,) = 2, xyx2 − 3 y 2 contour map with of f with the gradient field ∇f As we saw in 15.6, the gradient vectors are perpendicular to the level curves Gradient vectors are long where the level curves are close and short where they are far apart Directional derivative = ∇ f maximum value tightly packed level curves⇒ steep surface ⇒ longer gradient vectors Math 114 – Rimmer 16.1/16.2 Vector Fields A vector field F is called conservative if it is the and Line Integrals gradient of some scalar function. If there exists a function f such that F = ∇ f , then F is conservative. f is called the potential function for F . fxy( , ) = xy2 − y 3 F =2xy , x2 − 3 y 2 In 17.3 we learn how to tell whether a vector field is conservative and how to find f w hen it is. 5 Math 114 – Rimmer Section 16.1 Line Integrals 16.1/16.2 Vector Fields and Line Integrals Parametric Curve x= ft()(), y = gt f′ and g ′ continuous Consists of a finite Starts and ends at the same Starts and ends for t in [] a, b number of smooth curves point and doesn't cross itself at the same pt. Orientation of the curve Math 114 – Rimmer Geometric Interpretation of a Line Integral 16.1/16.2 Vector Fields and Line Integrals Definite Integral Line Integral 6 b Math 114 – Rimmer =′ + ′ 16.1/16.2 Vector Fields ∫ P( xydx,) + Q( xydy , ) ∫Ut()()()() f tdt Vtg tdt and Line Integrals C a Parametrize the path Substitute everything x= ft()(), ygtatb = ≤≤ Pxy()()(),= Pftgt() , = Ut() dx= f′( tdt) , dy = g ′ ( tdt) Qxy()()(),= Qftgt() , = Vt() Work done by a vector field F as its poin t of application moves along C from A to B . Calc II C→ line W=F ⋅ d Now C→ curve r the parametrization of C as a vector W=∫ F ⋅ d r C Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals Evaluate ∫ ydx+ xdy C on C : line segments from ()()0,0 to 1,0 and from ()()1,0 to 1,1 C2 ∫ydx+= xdy ∫ ydx ++ xdy ∫ ydx + xdy C C1 C 2 C1 x= t y = 0 x=1 y = t C: 0≤ t ≤ 1 C: 0≤ t ≤ 1 1 dx= dt dy = 0 2 dx=0 dy = dt 1 1 ∫ydx+ xdy = ∫ ()0 + 0 dt = 0 ∫ydx+ xdy = ∫ ()0 + 1 dt = 1 C1 0 C2 0 ∫ ydx+ xdy = 1 C 7 Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals Evaluate ∫( x2+ y 2 ) dx − 2 xydy on the given closed curve C . C y= x C2 y= x 2 C1 ∫( x22+ y) dx −2 xydy =+ ∫( x 22 y) dx −++ 2 xydy ∫ ( x 22 y) dx − 2 xydy C C1 C 2 Math 114 – Rimmer 16.1/16.2 Vector Fields and Line Integrals Evaluate ∫( x2+ y 2 ) dx − 2 xydy on the given closed curve C . C y= x 2 xt= yt = C2 C: 01≤ t ≤ 1 dx= dt dy = 2 tdt y= x 2 1 1 C ∫()()()x22+ ydx −2 xydy =+− ∫ t 244 t 4 tdt =− ∫ t 24 3 tdt 1 C1 0 0 1 t33 t 5 1359− − 4 =− =−= = 3 5 3 5 15 15 xt= y = t 0 C2 :1 t starts at 1 and ends at 0 dx= dt dy = dt 2 t 1 0 1 t3 1 ∫()()x2+ y 2 dx −2 xydy = ∫ t 2 +− t t dt =−∫t2 dt =− =− 0 3 3 C2 1 0 −4 1 −4 − 5 − 9 −3 ∫ ()x2+ y 2 dx −2 xydy =− = = = C 15 3 15 15 5 8 2 Math 114 – Rimmer Find the work done by the force F( x, y) = 2 xy i + 4 y j acti ng 16.1/16.2 Vector Fields and Line Integrals along the piecewise smooth curve consist ing of line segments from ()()()()−2,2 to 0,0 and from 0,0 to 2,3 . C2 C1 Work=∫F ⋅= dr ∫ F ⋅+ dr ∫ F ⋅ dr C C1 C 2 Find the work done by the force F(x, y) = 2 xy i + 4 y 2 j acti ng Math 114 – Rimmer 16.1/16.2 Vector Fields along the piecewise smooth curve consist ing of line segments and Line Integrals from ()()()()−2,2 to 0,0 and from 0,0 to 2,3 . F ⋅dr ∫ F ⋅dr ∫ C2 C1 3 xt= y = t xt= y = − t 2 C2 : 0≤ t ≤ 2 C1 : −2 ≤ t ≤ 0 3 dx= dt dy = − dt dx= dt dy = dt 2 3 3 rtt=i − j ⇒ dr=1, − 1 dt rt=i + t j ⇒ dr= 1, dt 2 2 2 2 2 2 F on C1 :− 2 t ,4 t F on C2 : 3 t ,9 t 0 2 2 2 27 33 ∫F ⋅dr = ∫ () −2 t − 4 tdt ∫F ⋅=dr ∫3 t2 + tdt 2 = ∫ tdt 2 C −2 2 2 1 C2 C 2 0 0 0 112 11 =−6tdt2 =− 2 t 3 =− 208() −−() =− 16 =t3 =()8 −= 0 44 ∫ −2 0 −2 2 2 Work=∫ F ⋅ dr =−16 + 44 = 28 C 9.