Well, I'll try with a more simple question then, I guess. All hypocycloid can be described as a smaller circle rotating inside a larger, with a point P on the rim of the circle creating the curve. Why is then the smaller circle radius (b) * the angle in the small circle (ϑ) the same as the big...
Could someone give me a hint how the formula
s(θ) = (8*(a-b)*b*(sin (aθ/4b))^2)/a
is derived. s is the arc length of the astroid's curve.
where θ is the angle of the circle with (a-b) radius.
Thanks in advance!
the formula can be find at http://mathworld.wolfram.com/Hypocycloid.html
Alright, I might have the answer to this now but I'm not quite sure. I have integrated the expression:
L = ∫ √ (a(2/3)/x(2/3)) dx =
= ∫ a(1/3)/x(1/3) dx
to
[ ∛a * (3/2) * x(2/3) ]
Known by the graph on my analog paper a is represented as the max value of x and y, both positive...
Thanks LCKurtz!
Now I get the formula:
√ (a^(2/3)/x^(2/3)) dx = The length of the curve between two x-values
I guess I should integrate this next. I'll do it tomorrow and if I encounter any problems I'll ask for further guidance.
Can't really get the hang of those TEX stuff
Homework Statement
Calculate the length of the graph/equation: x^(2/3) + y^(2/3) = a^(2/3)
The graph is formed as an s.c. asteroid, almost like a diamond. It seems to be some sort of modified unit circle.
Homework Equations
The length of the graph between x1 and x2 can be described as L=∫...