Learn the Arc Length of a Circle Formula with Plainmath's Expert Help

I want to write an algorithm that morphs or transforms an arc of a circle to a straight line joining both ends of the arc. I would like to understand the maths behind this such a transformation. How can this be done?

I have something starting at (50, 10) it then rotates counter clockwise by 30 degrees, around the point at (50, 0), essentially mapping out an arc of a circle. How do I find the point it now lies on?

Let the three points A,B,C be the vertices of a moving spherical triangle on the surface of a sphere. The triangle moves so that while the vertices A,B remain fixed, the angle BCA at the vertex C stays constant. What is the locus of the moving vertex C? Is there a special name for the curve traced out by C? If A,B,C were the vertices of a plane triangle, the corresponding locus would be the arc of a circle. I have made some calculations, which-if they do not contain errors-lead me to believe that, in the spherical case, the locus I am seeking is not the arc of a small circle (on the sphere). But, if so, I do not know what type of curve it is.

Consider the function $r=f(\mathrm{\theta <!--\; \theta \; -->})$ in polar coordinates. The length of an arc of a circle is just
$S=\mathrm{\theta <!--\; \theta \; -->}r$
Where $r$ is the radius of the circle and $\mathrm{\theta <!--\; \theta \; -->}$ is the angle that represents this arc. But since $r=f(\mathrm{\theta <!--\; \theta \; -->})$ and $\mathrm{\theta <!--\; \theta \; -->}$ Should approach zero so that we can get the exact value of the arc, So
$dS=f(\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}$
Integrating from ${\mathrm{\theta <!--\; \theta \; -->}}_1$ to ${\mathrm{\theta <!--\; \theta \; -->}}_2$, we get :
$S={\int <!--\; \int \; -->}_{{\mathrm{\theta <!--\; \theta \; -->}}_1}^{{\mathrm{\theta <!--\; \theta \; -->}}_2}f(\mathrm{\theta <!--\; \theta \; -->})d\mathrm{\theta <!--\; \theta \; -->}$
But the actual formula for the length of a curve in polar coordinates is
${\int <!--\; \int \; -->}_{{\mathrm{\theta <!--\; \theta \; -->}}_1}^{{\mathrm{\theta <!--\; \theta \; -->}}_2}\sqrt{f^2(\mathrm{\theta <!--\; \theta \; -->})+f^{\prime }(\mathrm{\theta <!--\; \theta \; -->}{)}^2}d\mathrm{\theta <!--\; \theta \; -->}.$
I know that my approach isn’t rigorous enough, but it’s is still reasonable, so why it is different from the actual formula?

I have an arc of a circle, and I have some other point in space (this might lie on the arc or it might not). I am looking for a formula that will compute the closest point on the arc to the other point. I also need to be able to get the distance from the start of the arc to this closest point.
what I know about the arc is: (1) the circle centre, (2) the angle of the beginning of the arc, (3) the arc angle, (4) the circle radius, (5) begin and end points of the arc on the circle circumference, (6) the direction the arc is going (i.e. Clockwise or Counter-clockwise)
what I know about the other point is: (1) its position

Point C moves along the top arc of a circle of radius 1 centered at the origin O(0, 0) from point A(-1, 0) to point B(1, 0) such that the angle BOC decreases at a constant rate 1 radian per minute. How does the area of the triangle ABC change at the moment when |AC|=1? Answer: it increases at 1/2 square units per minute. Could you give me a hint how to solve this task? I don't even know what to begin with.

A circle has diameter AD of length 400.
B and C are points on the same arc of AD such that |AB|=|BC|=60.
What is the length |CD|?

Let $AB$ be an arc of a circle. Tangents are drawn at $A$ and $B$ to meet at $C$. Let $M$ be the midpoint of arc $AB$. Tangent drawn at $M$ meet $AC$ and $BC$ at $D$, $E$ respectively. Evaluate
$\underset{AB\to <!--\; \to \; -->0}{lim}\frac{\mathrm{\Delta <!--\; \Delta \; -->}ABC}{\mathrm{\Delta <!--\; \Delta \; -->}DEC}$
I don't think evaluating the limit will be a problem. But how do I find the areas of the triangles and in what parameters?

Consider a circle with a circumference of n. On this circle, I define two arcs of length $k<n$, $A_1$ and $A_2$. The centres of the two arcs are uniformly distributed on the circle.
Let ${\mathrm{\Omega <!--\; \Omega \; -->}}_1=A_1\setminus <!--\; \setminus \; -->A_2$ and ${\mathrm{\Omega <!--\; \Omega \; -->}}_2=A_2\setminus <!--\; \setminus \; -->A_1$ such that the length of ${\mathrm{\Omega <!--\; \Omega \; -->}}_1\cup <!--\; \cup \; -->{\mathrm{\Omega <!--\; \Omega \; -->}}_2$ is $2k$ minus the overlapping part of the two arcs.
What is the expectation of the length of ${\mathrm{\Omega <!--\; \Omega \; -->}}_1\cup <!--\; \cup \; -->{\mathrm{\Omega <!--\; \Omega \; -->}}_2$?

$arg(\frac{z-<!--\; -\; -->a}{z-<!--\; -\; -->b})=c$
My understanding is as follows. The angle c between the lines za and zb is constant. za and zb meet at z and the angle between the lines perpendicular to za and zb is 2c, which is only the case if z lies on a circle passing through z, a and b with centre where the lines perpendicular to za and zb intersect. So the locus is an arc of a circle through a and b except a and b (because $arg(\frac{z-<!--\; -\; -->a}{z-<!--\; -\; -->b})$ is undefined there). Is this reasoning correct?

Is there another way the statement could this be expressed (e.g. in terms of moduli)? Also, what is an efficient way of finding the centre of the circle?

Let the endpoint of a differentiable, vector-valued function a$(t):(t_a..t_b)\to <!--\; \to \; -->\mathbb{R}$ move on a circle in an Euclidean plane. Let $t_0\in <!--\; \in \; -->(t_a..t_b)$. Let $s(t)$ be the length of the arc between $t_0$ and point a$(t)$. Let there be a positive real number $\mathrm{\delta <!--\; \delta \; -->}$ such that $\mathrm{\forall <!--\; \forall \; -->}t:0<|t-<!--\; -\; -->t_0|<\mathrm{\delta <!--\; \delta \; -->}:s(t)-<!--\; -\; -->s(t_0)\ne <!--\; \ne \; -->0$.
My (physics) book uses a proof which seems to imply that $s$($t$), where $t$ is time, is differentiable in respect to $t$. How to prove its differentiability?

My teacher said that the central angles of a circle are equal to the measure of the arc, but I don't understand on how this could possibly work.
Can someone please explain how this is possible?

So, I have an arc that's part of a circle that I must "travel across".

The circle has a radius of 15 - a circumference of 94.248 (rounding). The arc length in question is equal to 15.708.

Now, normally, you could sort of walk along the arc, turning with it. But, if I could only travel straight, I need to know how I would get from one part of the arc to 15.708 units later in the arc of the circle (only being able to travel straight, or make 90 degree turns).

How to prove that the minor arc of a great circle is the shortest path connecting two points on the surface of a sphere?
The key to solving the problems is that we must take all curves connecting the two points , not just great-circle or parametrized differentiable curve, into account!
A similar problem is the straight line distance(shortest distance between two points): we must take all curves connecting the two points into account!

MathJax(?): Can't find handler for document MathJax(?): Can't find handler for document I'm having trouble finding out the first step to this problem.
So far all I have is:

Find parametric equations for the arc of a circle of radius 6 from $P=(0,0)$ to $Q=(12,0)$.

If segment mn is a diameter of a circle f, and v is a fixed point on mn, for which point E on MN is the length FE minimized?

An ant starts on (0,3) on a circle with 3 radius and walks 2 units counterclockwise along the arc of a circle. Find the $x$ and $y$ coordinates of where the ant ends up.
I only know how to move in a straight line (up-down, left-right) to determine coordinates.
How would I begin to attempt this problem? I imagine I have to plug in a 2 into some sort of formula.

Verify the circumference of the circle formula for a circle of radius r using the arc length formula.
How do I complete this problem? Do I use the ellipse equation or the proper circle equation with (x, h) and (y,k)?

To construct this shape, draw a circle. Place the compass on a point on the circle and draw an arc of the same radius as the circle. Now place the compass at the intersection of the arc and the circle and draw an arc. Repeat this process until you have gone around the entire circle. Which of the following shapes is created by connecting all arc and circle intersections with a ruler?
A) Regular pentagon
B) Regular hexagon
C) Regular Octagon
D) Regular Dodecagon
Does anyone else get an octagon?

I found this in a textbook without a solution and I wasnt able to solve it myself.

Let ABCD be a tetrahedron with all faces acute. Let E be the mid point of the longer arc AB on a circle ABD. Let F be the mid point of the longer arc BC on a circle BCD. Let G be the mid point of the longer arc AC on a circle ACD.

Show that points D,E,F,G lie on a circle.

My approach to that was to try to show these point were co-planear. Since they all lie on one sphere (the one with inscribed tetrahedron ABCD) that would solve the problem. Needles to say I failed at that.