#Calculating the arc length of an ellipse
41 messages · Page 1 of 1 (latest)
I know the semiminor axis (at least I think that's what it's called) should be like 1.5 based on the information given, but I'm not entirely sure what to do with it. I heard something about calculating "T" for an "elliptic integral" but I'm not entirely sure what those are or how to calculate them. I tried doing what I understood to be the steps in this webpost I found, but I keep getting the wrong answer.
https://math.stackexchange.com/questions/433094/how-to-determine-the-arc-length-of-ellipse
x(t) = acosθ
y(t) = bcosθ
Yeah, I saw that, but I'm not entirely sure how to calculate those or use them to calculate the arc length
adonhs
I have no idea what most of this means
a and b are the length and width of the ellipse
And I'm guessing theta is the angle between the start of the arc and the end of the arc, but I have no clue what the rest of it is or what it even wants me to do with it
You mean the other radius? Because it was really close to 1.5 last I checked
x' is the derivative of x(θ)
y' is the derivative of y(θ)
and additinally both are being squared
I don't know what to do with those values, though
And I'm not sure what that means
I figured it to be 2
It might be. I probably erroneously assumed a right triangle
It's a parameterization
I don't know what that means either
There is an alternative without it but you will need the eclipse equation
The x2/a2 + y2/b2 = 1 one?
adonhs
The bottom equation still makes no sense to me
Since y = f(x) you would have to solve the eclipse equation for y and then choose the + or - version depending on the bounds and differentiate it.
Yea if you want a proof or the "why" do some self research
I don't know what the equation is even asking me to do
Yeah. I don't know what that is
I mentioned that being a term I've seen, but had no understanding of
It's a symbol
I don't know what kind of operation it's asking for
Why dont you use an online calculator then?
Do you know differentiation?
I don't know what that is either
Yeah, I'm aware