微积分福音-梯度、方向导数与切平面

11.1f(x,y)∇f(x,y)∇f(x,y)=(∂f∂x,∂f∂y)∇f(x,y,z)∇f(x,y,z)=(∂f∂x,∂f∂y,∂f∂z)nf(x1,x2,...,xn)n∇f(x1,x2,...,xn)=(∂f∂x1,∂f∂x2,...,∂f∂xn)f(x,y)=x2y+sin(xy)(2,3)∇f(x,y)=(∂f∂x,∂f∂y)=(2xy+ycos(xy),x2+xcos(xy))∇f(2,3)=(12+3cos(6),4+2cos(6))11.2方向導數directionalderivativez=f(x,y)xy−(a,b)⇀uz=f(x,y)(a,b)⇀uD⇀uf(a,b)a,b⇀u(cos(θ),sin(θ))F(t)=f(a+tcos(θ),b+tsin(θ))(a,b)(cos(θ),sin(θ))tF(0)=f(a+0,b+0)=f(a,b)F(t)tt=01f(a,b)(cos(θ),sin(θ))D⇀uf(a,b)=F′(0)F′(0)ddtf(xz}|{a+tcos(θ),yz}|{b+tsin(θ))t=01(x,y)=(a,b)t=02=∂∂xf(a,b)dxdt+∂∂yf(a,b)dydt=∂∂xf(a,b)cos(θ)+∂∂yf(a,b)sin(θ)(∂∂xf(a,b),∂∂yf(a,b))·(cos(θ),sin(θ))∇f(a,b)·(cos(θ),sin(θ))⇀uf(x,y)(a,b)⇀uD⇀uf(a,b)=∇f(a,b)·⇀uf(x,y)=x2+y2⇀v=(1,1)D⇀vf(−3,2)∇f(−3,2)=(∂∂xf(−3,2),∂∂yf(−3,2))=2x(−3,2),2y(−3,2)=(−6,4)⇀v√2√2⇀u=1√2(1,1)D⇀vf(−3,2)=(−6,4)·1√2(1,1)=−2√2=−√23f(a,b)D⇀uf(a,b)=∇f(a,b)·⇀u∇f(a,b)·⇀u·cos(φ)φ∇f(a,b)(a,b)∇f(a,b)C⇀u1φC·1·cos(φ)φcos(φ)=1φ=0φ=0∇f(a,b)⇀u2f(a,b)∇f(a,b)∇f(a,b)f(x,y)(a,b)cos(φ)=1C·1·cos(φ)=C·1·1=CC∇f(a,b)∇f(a,b)∇f(a,b)24z=f(x,y)(a,b)∇f(a,b)1.f(x,y)(a,b)2.x2+y2≤1(0,0)170cos(y+xy)+9(x2+x+2y)cos(φ)=1cos(φ)=−1∇f(0,0)=170ysin(y+xy)+9(2x+1)(0,0),170(1+x)sin(y+xy)+9(2)(0,0)=(9,18)(−1,−2)33(9,18)(−9,−18)(−1,−2)(−4,−8)5100(x2+y2−2x−3y+2xy)(x,y)=(1,2)∇f(1,2)=(100(2x−2+2y)(1,2),100(2y−3+2x)(1,2))=(400,300)√4002+3002=500g(x,y,z)=C∇g(x,y,z)=(∂f∂x,∂f∂y,∂f∂z)4g(x,y,z)=Cdifferenialdg=dCCdC=0dggxdx+gydy+gzdz=046(gx,gy,gz)·(dx,dy,dz)=0∇g(x,y,z)·d⇀X=0d⇀X=(dx,dy,dz)g(x,y,z)=Cd⇀X∇g(x,y,z)0法向量z=f(x,y)(a,b)∇f(a,b)1.f(x,y)(a,b)2.g(x,y,z)=C(a,b,c)∇g(a,b,c)(a,b,c)z=x2+2y+7zx2+2y+7−z=0y=3x+4y3x−y+4=0(3,−1)x2+y2=4(2x,2y)7z=f(x,y)z=c1,c2,...,cnf(x,y)=c1,...,f(x,y)=cn∇f(a,b)∇f(a,b)(x,y)=(a,b)∇f(a,b)(x,y)=(a,b)∇f(a,b)z=f(x,y)(x,y)=(a,b)f(x,y)=ck(x,y)=(a,b)(x,y)(x,y)=(2,1)(x,y)=(2,1)(x,y)=(2,1)8x3+y3−92xy=0(2,3)∇(x3+y3−92xy)=(3x2−9y2,3y2−9x2)(2,3)(12−272,27−9)=(−32,18)(2,3)(2,3)5−32(x−2)+18(y−3)=0591.3x2+4y2=z2(3,2,5)x2+4y2−z2=0∇(x2+4y2−z2)=(2x,8y,−2z)(3,2,5)(6,16,−10)3(x−3)+8(y−2)−5(z−5)=010z=ln(x2+y2)(1,0,0)ln(x2+y2)−z=0(2xx2+y2,2yx2+y2,−1)(1,0,0)(2,0,−1)2(x−1)−z=011