Rearranging a formula (segment area)

In summary, Pavadrin has been given a problem to solve involving the minor segment of a circle. He knows that the formula to solve for a segment is: S = \frac{1}{2}r^2 (\theta - \sin \theta ), but is confused on what to do with the sin theta part. He suggests graphing theta and sin(theta) over the interval {0; pi} and reading the answer off the graph. Alternatively, he suggests using derivatives to find the angle corresponding to the minimum segment.
  • #1
pavadrin
156
0
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:

[tex]S = \frac{1}{2}r^2 (\theta - \sin \theta )[/tex]

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:

[tex] \theta - \sin \theta = \frac{{2S}}{{r^2 }}[/tex]

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
 
Physics news on Phys.org
  • #2
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.
 
  • #3
okay thanks for that, didn't even consider graphing it :)
 
  • #4
pavadrin said:
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 [tex]\frac{df(x)}{dx} = 0[/tex]. 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 [tex]\frac{df(x)}{dx} = 0[/tex] changes from - to + , you are in a minimum.

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

this should get you started

regards
marlon
 
  • #5
If you can assume that the angle is tiny, you may approximate the sine with a finite sum, say [tex]\sin(\theta)\approx\theta-\frac{\theta^{3}}{6}[/tex]
This will yield:
[tex]\frac{\theta^{3}}{6}\approx\frac{2S}{r^{2}}\to\theta\approx(\frac{12S}{r^{2}})^{\frac{1}{3}}[/tex]
 
  • #6
marlon said:
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 [tex]\frac{df(x)}{dx} = 0[/tex]. 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 [tex]\frac{df(x)}{dx} = 0[/tex] changes from - to + , you are in a minimum.

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

this should get you started

regards
marlon

marlon, what does this problem have to do with finding a minimum?
 
  • #7
HallsofIvy said:
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
 
  • #8
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
 
  • #9
Office_Shredder said:
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
 
  • #10
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".
 
  • #11
HallsofIvy said:
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
 

1. How do I rearrange a formula to solve for segment area?

To rearrange a formula to solve for segment area, you need to isolate the variable representing the segment area on one side of the equation. This can be done by using algebraic operations such as addition, subtraction, multiplication, and division to move other variables and constants to the other side of the equation.

2. What is the formula for calculating segment area?

The formula for calculating segment area depends on the shape of the segment. For a circle, the formula is A = (θ/360) x πr², where θ is the central angle and r is the radius. For a triangle, the formula is A = (1/2) x b x h, where b is the base and h is the height. For a sector, the formula is A = (θ/360) x πr², where θ is the central angle and r is the radius.

3. How do I know when to use the formula for segment area?

You can use the formula for segment area when you know the necessary measurements for the specific shape of the segment. For example, if you know the central angle and radius of a circle segment, you can use the formula A = (θ/360) x πr² to calculate the area. Make sure to use the correct formula for the shape of the segment you are working with.

4. What are some common mistakes to avoid when rearranging a formula for segment area?

Some common mistakes to avoid when rearranging a formula for segment area include forgetting to apply the correct algebraic operations, not isolating the variable for segment area, and using the wrong formula for the specific shape of the segment. It is important to double-check your work and make sure all steps are accurate.

5. Can I use the formula for segment area to solve real-world problems?

Yes, the formula for segment area can be used to solve real-world problems. For example, if you are an architect designing a circular building, you can use the formula A = (θ/360) x πr² to calculate the area of a specific segment of the building. This can help you determine the size and layout of different sections of the building.

Similar threads

  • Precalculus Mathematics Homework Help
Replies
14
Views
219
  • Precalculus Mathematics Homework Help
Replies
18
Views
1K
  • Precalculus Mathematics Homework Help
Replies
8
Views
983
  • Precalculus Mathematics Homework Help
Replies
7
Views
1K
  • Precalculus Mathematics Homework Help
Replies
3
Views
2K
  • Precalculus Mathematics Homework Help
Replies
2
Views
1K
Replies
1
Views
689
  • Precalculus Mathematics Homework Help
Replies
16
Views
4K
  • Precalculus Mathematics Homework Help
Replies
9
Views
2K
  • Calculus
Replies
29
Views
652
Back
Top