Solve Geometry Q w/ L and c: Find R in Terms of L and c

AI Thread Summary
The discussion revolves around finding the radius R in terms of L and c, with c mistakenly identified as the arc-length of quarter circles instead of an ellipse. The user initially attempts to relate R, L, and c using trigonometric identities, deriving expressions for a and R. However, the realization that the arcs belong to an ellipse complicates the problem, as ellipses have two radii, requiring a more precise formulation of the problem. The conversation emphasizes the need to clarify known variables and the significance of R in the context of an ellipse. Further analysis is needed to accurately solve for R given the new understanding of the geometric shape involved.
madgorillaz15
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Homework Statement


In the attached drawing, find R in terms of L and c. Also, at the bottom of the picture I wrote something wrong. I said c, which equals 0.5, is the arc-length of each semi-circle, but I really meant to say each quarter circle. My bad. I'm not given a number for L so that can just remain a variable.

Can anyone see the solution? I'd appreciate any help.

Homework Equations

The Attempt at a Solution


I think I need to take advantage of c somehow, but I wasn't able to figure it out. Obviously, R=(a^2 + L^2)^(1/2), so I just need to figure out a in terms of L and c (or something else)?
 

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From your figure a=R⋅sin(c) and L=R⋅cos(c). Then...
 
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First,

a / R = sin (c) => a = R sin (c) --- (i)

Next,

R^2 = a^2 + L^2 => a = (R^2-L^2)^1/2 ---(ii)

Equating (i) and (ii) gives: R sin (c) = (R^2-L^2)^1/2 => R = L / cos(c)
 
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Svein said:
From your figure a=R⋅sin(c) and L=R⋅cos(c). Then...

Can I ask, are you treating c like an angle? Also, after speaking to a friend, I was mistaken in assuming that those arcs belonged to quarter circles. In fact, they belong to an ellipse (so the entire big arc you see is actually half an ellipse). How does that change your answer?
Thanks though.
 
titasB said:
First,

a / R = sin (c) => a = R sin (c) --- (i)

Next,

R^2 = a^2 + L^2 => a = (R^2-L^2)^1/2 ---(ii)

Equating (i) and (ii) gives: R sin (c) = (R^2-L^2)^1/2 => R = L / cos(c)

Hi titasB,
I apologize, but as in my other reply, does this answer change if the big arc is actually half an ellipse, and not have a circle as I first assumed?
 
madgorillaz15 said:
Can I ask, are you treating c like an angle?
Yes. If not, we need to find some other way. If it is part of a circle (as you mentioned in your original post), then the angle is c/R. Then you get a slightly more complicated expression. From the note on your figure, c=0.5, so the angle is 0.5/R. Inserting this, you get L=R⋅cos(0.5/R) - which is fine for finding L, but complicated for finding R. On the other hand, tg(0.5/R) = a/L - which gives 0.5/R = arctg(a/L).
 
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Svein said:
Yes. If not, we need to find some other way. If it is part of a circle (as you mentioned in your original post), then the angle is c/R. Then you get a slightly more complicated expression. From the note on your figure, c=0.5, so the angle is 0.5/R. Inserting this, you get L=R⋅cos(0.5/R) - which is fine for finding L, but complicated for finding R. On the other hand, tg(0.5/R) = a/L - which gives 0.5/R = arctg(a/L).

Yeah, I just checked and we aren't allowed to treat it like an angle. Furthermore, I was mistaken in saying that c was a part of a circle-it's actually an ellipse. How does that change the answer?
 
madgorillaz15 said:
I was mistaken in saying that c was a part of a circle-it's actually an ellipse. How does that change the answer?
Well, you will have to state your problem again, this time more precisely. What are the known parts and what are you supposed to determine? What do you know about the ellipse? What is the significance of R (an ellipse has two radii)?
 

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