雜記

vector a x b : vector a sweeps in direction of vector b creates the area
0=(ai+bj)x(ai+bj)=ab (i x j + j x i)
i x j = - j x i


vector(a,b) moved as vector (c,d) painted green area
green area=blue-yellow=orange-pink=ad-bc


red area = green area



(a,b) & (-b,a) make ad-bc becomes a^2+b^2

(fgh)'= f'gh + fg'h + fgh'
(fg)''=f''g + 2f'g' + fg''
(fg)'''=f'''g + 3f''g' + 3f'g'' + fg'''

Z(x(t),y(t))  ; dZ = Zx dx + Zy dy  ; dZx = Zxx dx + Zxy dy :  dZy= Zxy dx + Zyy dy
則 ddZ = Zxx (dx)^2 + Zyy (dy)^2 + 2 Zxy dxdy + Zx ddx + Zy ddy (Zxx 表Z對x偏微分2次)


sin(x)=cos(x-pi/2)
cos(x)=sin(x+pi/2)

( f(t)e^(kt) ) '=k( f(t)e^(kt) )+ f'(t)e^(kt)
>  f(t)e^(kt) dt= d( f(t)e^(kt)/k )-  f'(t)e^(kt) /k dt>  f'(t)e^(kt) dt= d( f'(t)e^(kt)/k )-  f''(t)e^(kt) /k dt
f(t)e^(kt) dt
= d( f(t)e^(kt)/k-f'(t)e^(kt)/k^2 ) +  f''(t)e^(kt) /k^2 dt
= d [(f(t)/k-f'(t)/k^2+f''(t)/k^3+.......+fn-1(t)/k^n) e^kt] + (-1)^n fn(t)e^(kt) /k^n dt
當k=-1 => f(t)e^(-t) dt = d [ -(f(t)+f'(t)+f''(t)+............+fn-1(t)) e^(-t) ] + fn(t)e^(-t) dt

G(t)=f(t)e^(k1t)
> G'(t)--kG(t) = (k1-k) f(t)e^(k1t) + f'(t)e^(k1t)

( ln(1-e^(-x)) )'=  1/(e^x-1)

1/[(x-a)(x-b)]=[1/(x-a)-1/(x-b)]/(a-b)=1/(x-a)/(a-b)+1/(x-b)/(b-a)
1/[(x-a)(x-b)] dx = d ln(x-a)/(a-b)+ d ln(x-b)/(b-a) = d {ln [(x-a)/(x-b)]}/(a-b)

1/[(x-a)(x-b)(x-c)]=[1/(x-a)-1/(x-b)]/(a-b)/(x-c)=[1/(x-a)/(x-c)-1/(x-b)/(x-c)]/(a-b)
= {[1/(x-a)-1/(x-c)]/(a-c) - [1/(x-b)-1/(x-c)]/(b-c)}/(a-b)
=1/(x-a)/(a-c)/(a-b)+1/(x-b)/(b-c)/(b-a)+1/(x-c)/(c-a)/(c-b)

1/[(x-a)(x-b)(x-c)] dx = d ln {(x-a)^(1/(a-c)/(a-b)) (x-b)^(1/(b-c)/(b-a)) (x-c)^(1/(c-a)/(c-b))}

曲率半徑for U(x,y)=k
(UxUy)^2/ (Ux^2 +Uy^2)^(3/2) [Uxx/Ux^2 + Uyy/Uy^2 -2 Uxy/(UxUy)]
曲率半徑為正->凹向原點, 曲率半徑為負->凸向原點,