Föreläsning 16

Idag

Parametrisering av kurvor

x[t_]=t^2-Log[t+1]

y[t_]=t*Log[t+1]-2t^(3/2)

ParametricPlot[{x[t],y[t]},{t,0,1}]

Kurvintegraler

Starta med  y=(Sqrt[x])

Integrate[Sqrt[1+x],{x,0,a}]

Från x=0 till x=a har vi båglängden -(2/3)+(2/3) (((1+a))^(3/2))

Solve[s==-(2/3)+(2/3) (((1+a))^(3/2)),a]

Vi kan sätta t = -1+((((4+12 s+9 (s^2)))^(1/3))/(2^(2/3)))

Vi får en parametrisering

x(s)=-1+((((4+12 s+9 (s^2)))^(1/3))/(2^(2/3)))
y(s)= (((-1+((((4+12 s+9 (s^2)))^(1/3))/(2^(2/3)))))^2)
där 0 ≤ s ≤ 2/3 ((Sqrt[8])-1)

Ganska krångligt.

Kopplingar till potential

Exempel

x[t_]=t*(t-1)

y[t_]=t^3-t^2

ParametricPlot[{x[t],y[t]},{t,0,1}]

Några vektorfält

VectorPlot[{x^2*y,x*(y-x)},{x,0,1},{y,0,1}]

VectorPlot[{-x/(x^2+y^2)^(3/2),-y/(x^2+y^2)^(3/2)},{x,0.5,2},{y,0.5,2}]

VectorPlot[{-x,-y},{x,-2,2},{y,-2,2}]

VectorPlot[{-y,x},{x,-2,2},{y,-2,2}]

Exempel 1

P[x_,y_]=x+y

Q[x_,y_]=x^2/y

fig1=VectorPlot[{P[x,y],Q[x,y]},{x,1,4},{y,2,4}]

fig2=ParametricPlot[{t^2,2*t},{t,1,2}]

Show[fig1,fig2]

Exempel 2

fig1=VectorPlot[{x*y,x-y},{x,0,1},{y,0,1}]

fig2=ParametricPlot[{t,t},{t,0,1}]

fig3=ParametricPlot[{t,t^2},{t,0,1}]

Show[fig1,fig2,fig3]

Exempel 3

VectorPlot[{Exp[x*y],Cos[x*y]},{x,0,1},{y,0,1}]