曲率求解。。。

Question

第五题

Answer 1

r= a.e^(nθ)
dr/dθ = an.e^(nθ)
dr/dθ | (a,0) = an
r=√(x^2+y^2)
dr/dx = (x + y.dy/dx )/√(x^2+y^2)
θ =arctan(y/x)
dθ/dx
= [ -y/x^2 + (1/x).dy/dx ]/[ 1+(y/x)^2]
------------
dr/dθ =(dr/dx) / (dθ/dx)
an.e^(nθ) = [(x + y.dy/dx )/√(x^2+y^2)] / { [ -y/x^2 + (1/x).dy/dx ]/[ 1+(y/x)^2] }
an = [(a + 0 )/√(a^2+0^2)] / { [ 0 + (1/a).dy/dx|(a,0) ]/[ 1+0] }
=1/ { (1/a) dy/dx|(a,0)}
dy/dx|(a,0) = 1/n本回答被网友采纳