Rearranging a formula (segment area)

[pavadrin]

Hey
I’ve been given a problem to solve involving the minor segment of a circle. I know that the formula to solve for a segment is:

S = \frac{1}{2}r^2 (\theta - \sin \theta )

However in this problem I’ve been given the segment area and the radius of the circle and been ask to find the minor angle. So far the best I have managed in rearranging the formula is:

\theta - \sin \theta = \frac{{2S}}{{r^2 }}

I am confused on what to do with the sin theta part, so if somebody out there is able to help it will be greatly appreciated. Or am I wasting my time and this is not possible?
Many thanks in advance
Pavadrin

 

[verty]

The only way I can fathom to solve this is to plot the graphs of theta and sin(theta) over the interval {0; pi} and read the answer off the graph.

 

[pavadrin]

okay thanks for that, didn't even consider graphing it :)

 

[marlon]

okay thanks for that, didn't even consider graphing it :)

You can use derivatives. If you have been given a function f(x). this function has a local minimum/maximum at those x values where \frac{df(x)}{dx} = 0. To be sure you are looking at a local minimum, you need to make a sign chart of the first derivative you just calculated. At those x-points when the sign of the \frac{df(x)}{dx} = 0 changes from - to + , you are in a minimum.

In your case you have a function S of both r and \theta. The \theta corresponding to the minimum segment can be found by taking the first derivative of S with respect to r (\theta is considered to be a constant) and solve that equation for \theta.

this should get you started

regards
marlon

 

[arildno]

If you can assume that the angle is tiny, you may approximate the sine with a finite sum, say \sin(\theta)\approx\theta-\frac{\theta^{3}}{6}
This will yield:
\frac{\theta^{3}}{6}\approx\frac{2S}{r^{2}}\to\theta\approx(\frac{12S}{r^{2}})^{\frac{1}{3}}

 

[HallsofIvy]

You can use derivatives. If you have been given a function f(x). this function has a local minimum/maximum at those x values where \frac{df(x)}{dx} = 0. To be sure you are looking at a local minimum, you need to make a sign chart of the first derivative you just calculated. At those x-points when the sign of the \frac{df(x)}{dx} = 0 changes from - to + , you are in a minimum.

In your case you have a function S of both r and \theta. The \theta corresponding to the minimum segment can be found by taking the first derivative of S with respect to r (\theta is considered to be a constant) and solve that equation for \theta.

this should get you started

regards
marlon

marlon, what does this problem have to do with finding a minimum?

 

[marlon]

marlon, what does this problem have to do with finding a minimum?

Well, isn't he supposed to find the angle corresponding to the minimum segment surface ?

marlon

 

[Office_Shredder]

No, he's trying to find what angle a segment intersects on a circle given the radius of the circle and the length of the segment

 

[marlon]

No, he's trying to find what angle a segment intersects on a circle given the radius of the circle and the length of the segment

Opps then i misread the question.

marlon

 

[HallsofIvy]

Perhaps you misread " the minor segment ". Two radii divide a circle into two arcs. Unless the radii are part of the same diameter, one arc is smaller than the other: the "minor arc" and, by extension, forms the "minor segment".

 

[marlon]

Perhaps you misread " the minor segment ". Two radii divide a circle into two arcs. Unless the radii are part of the same diameter, one arc is smaller than the other: the "minor arc" and, by extension, forms the "minor segment".

KABOOMMMM... Indeed you are right : i completely misread that question. Thanks for the clarification.

regards
marlon