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

Could someone explain to me why the following holds?
$\frac{\text{Area of sector}}{\text{Area of circle}}=\frac{\mathrm{\theta <!--\; \theta \; -->}}{2\mathrm{\pi <!--\; \pi \; -->}}$
When deriving the area of a sector my book just quotes the above but doesn't explain why it holds. Could someone explain?

Also, a similar argument is used when deriving arc length, which I don't understand.

By the way, I know that $\mathrm{\theta <!--\; \theta \; -->}$ is the angle subtended by the arc and $2\mathrm{\pi <!--\; \pi \; -->}$ is the angle in a full circle.

How does the ratio of the areas make it equivalent to the ratio of their angles?

The velocity of light above a hot surface decreases with the height from that surface. The velocity is given by $v=v_0(\frac{1-<!--\; -\; -->y}{\mathrm{\alpha <!--\; \alpha \; -->}})$ where $y$ is the vertical distance above the starting position. Show that the path taken by light from (0,0) to $(2X,0)$ is the arc of a circle centeblack at $(X,\mathrm{\alpha <!--\; \alpha \; -->})$. Note that the origin is chosen to be well above the ground, so that $y<0$ is allowable.

Attempt at a solution: After working through from the velocity relation given in the question, we get the following ODE: ${\displaystyle \frac{1}{v_0(\frac{1+y}{\mathrm{\alpha <!--\; \alpha \; -->}})\sqrt{1+y^2}}}=-<!--\; -\; -->C$ where $C$ is a constant.

Find a harmonic function on $|z|<1$ with value 1 on an arc of $|z|=1$ and zero on the rest.
Of course, if allow the arc with value 1 to be the whole $|z|=1$ circle it would be trivial.
Without that, how to construct such a function, I'm a bit lost. I suppose if I can find a holomorphic function $f(z)$ whose real part full fill that it shall work, but playing around with the normal functions such as $e^x$, $\sin <!--\; \; -->x$ etc, I can't find one with such property.

A parametrization of a circle by arc length may be written as
$\mathrm{\gamma <!--\; \gamma \; -->}(t)=c+r\cos <!--\; \; -->(t/r)e_1+r\sin <!--\; \; -->(t/r)e_2.$
Suppose $\mathrm{\beta <!--\; \beta \; -->}$ is an unit speed curve such that its curvature $\mathrm{\kappa <!--\; \kappa \; -->}$ satisfies $\mathrm{\kappa <!--\; \kappa \; -->}(0)>0$.
How to show there is one, and only one, circle which approximates $\mathrm{\beta <!--\; \beta \; -->}$ in near $t=0$ in the sense
$\mathrm{\gamma <!--\; \gamma \; -->}(0)=\mathrm{\beta <!--\; \beta \; -->}(0),{\mathrm{\gamma <!--\; \gamma \; -->}}^{{}^{\prime }}(0)={\mathrm{\beta <!--\; \beta \; -->}}^{{}^{\prime }}(0)\text{and}{\mathrm{\gamma <!--\; \gamma \; -->}}^{{}^{″}}(0)={\mathrm{\beta <!--\; \beta \; -->}}^{{}^{″}}(0).$
I suppose we must use Taylor's formula, but I wasn't able to solve this problem.

Let segment gh be a chord of a circle $\mathrm{\omega <!--\; \omega \; -->}$ which is not a diameter, and let n be a fixed point on gh. For which point b on arc gh is the length n minimized?

Let $S^1$ be a circle of perimeter equal to 1 (not radius 1) .Let $f:S^1\to <!--\; \to \; -->S^1$ be an arbitrary homeomorphism. By uniform continuity, it always possible to find an $\mathrm{ϵ<!--\; ϵ\; -->}>0$ such that $d(x,y)<\mathrm{ϵ<!--\; ϵ\; -->}$ implies
$d(f(x),f(y))<\mathrm{ϵ<!--\; ϵ\; -->},$
where d denote the shortest distance between two points on the circle.
My question is: Is it always possible to find an $\mathrm{ϵ<!--\; ϵ\; -->}>0$ such that for any arc $I$ (or call it interval if you like) on $S^1$ such that $length(I)<\mathrm{ϵ<!--\; ϵ\; -->}$ implies that
$length(f(I))<\frac{1}{4}?$
In other words, do homeomorphisms always send very short arcs to short arcs, rather than long arcs?
Since a homeomorphism can be expanding or contracting, I am really puzzled by this. There must be some tricks I don't know.

I have a circle with an arc beginning at point (x,y). The radius is r, the arc angle(w/ respect to center) is $\mathrm{\theta <!--\; \theta \; -->}$. How do I calculate the end point of the arc (a,b) ?
I know that the arc-length=radius*(arc angle)
I can't seem to find an easy way to solve this, I think the way to go is with parametric equations but I'm not sure.

I'm not great at mathematics so I'm sure this is trivial to most. I have been searching around however and not been able to find how to figure out the incircle of a circle sector, or, in other words, the point inside a sector that is furthest away from the radii and the arc. The simpler solution the better

If argument of $\frac{z-<!--\; -\; -->z_1}{z-<!--\; -\; -->z_2}$ is $\frac{\mathrm{\pi <!--\; \pi \; -->}}{4}$, find the locus of $z$.
$z_1=2+3i$
$z_2=6+9i$
Approach: I tried to solve the equation using diagram, basically plotting the points on the Argand plane. What I got is a circle with center $7+4i$ and a radius of $\sqrt{26}$ units. The two complex numbers given lie on this circle, and form a chord. Any point lying on the major arc of this chord satisfies the condition.
How exactly would I represent this as a locus of the point? And is there any other method that I can use that does not involve a diagram?

What is the higher dimensional analogue for the arc of a circle?
I'd like to work with the set of all points lying within a certain distance of a given point on an n-sphere, and I'd like to describe these sets by the (generalised) solid angles which they subtend.

Assume we have an arc-length parameterized curve $\mathrm{\beta <!--\; \beta \; -->}:I\to <!--\; \to \; -->{\mathbb{E}}^2$ with I a random interval. I want to show that if $\mathrm{\beta <!--\; \beta \; -->}(s)\cdot <!--\; \cdot \; -->{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }(s)=0$ for arc-length parameter $s$, then $\mathrm{\beta <!--\; \beta \; -->}$ is (part of) a circle.

I was thinking about using the Frenet-formulas for proving this. If we want to show something is a circle, one needs to show that the curvature is constant. In the Frenet-frame $(T,N)$, the curvature is: $\mathrm{\kappa <!--\; \kappa \; -->}=T^{\prime }\cdot <!--\; \cdot \; -->N$. So to be constant, it needs to be:
${\mathrm{\kappa <!--\; \kappa \; -->}}^{\prime }=0=T^{″}\cdot <!--\; \cdot \; -->N+T^{\prime }\cdot <!--\; \cdot \; -->N^{\prime }$
I was thinking about using the formulas $T={\mathrm{\beta <!--\; \beta \; -->}}^{\prime }$, $T^{\prime }=\mathrm{\kappa <!--\; \kappa \; -->}N$ and $N^{\prime }=-<!--\; -\; -->\mathrm{\kappa <!--\; \kappa \; -->}T$. But together with the given condition: $\mathrm{\beta <!--\; \beta \; -->}(s)\cdot <!--\; \cdot \; -->{\mathrm{\beta <!--\; \beta \; -->}}^{\prime }(s)=0$, I got stuck.

Find the radius of a circle in which the central​ angle, $a$, intercepts an arc of the given length s. Round to the nearest hundredth as needed.
$a=144,s=102$
The​ length, $s$, of an arc intercepted by a central angle of radians on a circle of radius $r$ is given by the formula below.
$s=ar$
This formula is only valid if a is measured in radians, so you must use the following formula to convert from degrees to radians.
$d\cdot <!--\; \cdot \; -->\frac{\mathrm{\pi <!--\; \pi \; -->}}{180}$
What I am confused about is that in the example guide that came along with the question, it gets $\frac{4\mathrm{\pi <!--\; \pi \; -->}}{5}rad$ from the degree to radian conversion. How did they get to that answer?

The coordinates of an arc of a circle of length $\frac{2pi}{p}$ are an algebraic number, and when $p$ is a Fermat prime you can find it in terms of square roots.
Gauss said that the method applied to a lot more curves than the circle. Will you please tell if you know any worked examples of this (finding the algebraic points on other curves)?

Can anyone help me with this problem? I think I probably miss some theorems.
In the figure angle AED = 30 deg, and the minor arc AB,BC, and CD are all equal. The degree measure of the minor arc AD is
a) not enought information, b) 30 deg, c) 105 deg, d) 45 deg, e) none of these.

Does $arg\left({\displaystyle \frac{z-<!--\; -\; -->1}{z+1}}\right)={\displaystyle \frac{\mathrm{\pi <!--\; \pi \; -->}}{3}}$ represent a circle or arc of a circle?
By substituting x+iy=z I do get equation of a circle. But I feel it should be one particular arc of the circle (the locus) depending on the direction of rotation of (z−1) to (z+1) by Coni's method.
Can someone clarify this issue? Thanks.

Given the hyperbolic metric $d{s}^2=\frac{d{x}^2+d{y}^2}{x^2}$ on the half plane $x>0$, find the length of the arc of the circle $x^2+y^2=1$ from $(\cos <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->},\sin <!--\; \; -->\mathrm{\alpha <!--\; \alpha \; -->})$ to $(\cos <!--\; \; -->\mathrm{\beta <!--\; \beta \; -->},\sin <!--\; \; -->\mathrm{\beta <!--\; \beta \; -->})$

I found that $d{s}^2={\displaystyle \frac{d{\mathrm{\theta <!--\; \theta \; -->}}^2}{{\cos }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}}$ but when I try to plug in $\mathrm{\pi <!--\; \pi \; -->}/3,-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}/3$, which should give me the arc length of $2\mathrm{\pi <!--\; \pi \; -->}/3$,
I get $4\mathrm{\pi <!--\; \pi \; -->}/3=\sqrt{{\displaystyle \frac{{(\mathrm{\pi <!--\; \pi \; -->}/3-<!--\; -\; -->(-<!--\; -\; -->\mathrm{\pi <!--\; \pi \; -->}/3))}^2}{co{s}^2(\mathrm{\pi <!--\; \pi \; -->}/3)}}}$
I feel like I'm making a simple mistake but I cant place it

Show that the curve defined by
$\mathrm{\gamma <!--\; \gamma \; -->}(s):=(a\cos <!--\; \; -->\left(\frac{s}{a}\right),a\sin <!--\; \; -->\left(\frac{s}{a}\right))$
Is on a circle with the radius $a$ and the center (0,0),also show that $\mathrm{\gamma <!--\; \gamma \; -->}(s)$ has been parameterized by its arc length.

I know that the parametric equation of $a$ circle with radius a and center $(x_0,y_0)$ is :
$x=x_0+a\cos <!--\; \; -->\left(t\right)$
$y=y_0+a\sin <!--\; \; -->\left(t\right)$
If we denote the parametric equation of a circle with $\mathrm{\gamma <!--\; \gamma \; -->}(t)=(x_0+a\cos <!--\; \; -->\left(t\right),y_0+a\sin <!--\; \; -->\left(t\right))$,then we have:
$\frac{d\mathrm{\gamma <!--\; \gamma \; -->}(t)}{dt}=(-<!--\; -\; -->a\sin <!--\; \; -->\left(t\right),a\cos <!--\; \; -->\left(t\right))$
$\Vert \frac{d\mathrm{\gamma <!--\; \gamma \; -->}(t)}{dt}\Vert =\sqrt{a^2}=a$
So the arc length is :
$s={\int <!--\; \int \; -->}_0^t\Vert \frac{d\mathrm{\gamma <!--\; \gamma \; -->}(\tau )}{dt}\Vert d\tau =a{\int <!--\; \int \; -->}_0^{t}d\tau$
Which implies $s=at$.

And if we parameterize the circle by its arc length then:
$\mathrm{\gamma <!--\; \gamma \; -->}(s)=(a\cos <!--\; \; -->\left(\frac{s}{a}\right),a\sin <!--\; \; -->\left(\frac{s}{a}\right))$
Which is the given parametrization.

But I have not shown that the center is at the origin.

I'm struggling with the following question for long. I tried to apply isoperimetric inequality $4\mathrm{\pi <!--\; \pi \; -->}A\le <!--\; \le \; -->L^2$, but my attempt has been unsuccessful. Could anyone give me a hint?

Let $AB$ be a segment of straight line and let $l>$ length of $AB$. Show that the curve $C$ joining $A$ and $B$, with length $l$, and such that together with $AB$ bounds the largest possible area is an arc of a circle passing through $A$ and $B$.

I would like to calculate the arc length of a circle segment, i.e. I know the start coordinates $(x/y)$ of the circle segment, the end coordinates $(x/y)$ and the x and y distances from the starting point to the center point of the circle segment.
I know that I can calculate the circumference with 2 * radius * PI. Consequently, I would have to calculate the radius and the angle of the circle segment via Pythagorean theorem and sin and cos. My question: Is there a simple formula where I just have to put in start-coordinates, end-coordinates and the circle origin point coordinates?
Thanks.

I have to show that the solution of a differential equation is an arc of a great circle. The differential equation is as follows (in spherical coordinates):
$\frac{{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}{\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }}{(1+{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}({\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }{)}^2{)}^{\frac{1}{2}}}=C$
where $C$ is an arbitrary constant and ${\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }$ denotes the derivative of $\mathrm{\varphi <!--\; \varphi \; -->}$ with respect to $\mathrm{\theta <!--\; \theta \; -->}$.

My reasoning:
By setting $\mathrm{\varphi <!--\; \varphi \; -->}(0)=0$, any arc of a great circle will have no change in $\mathrm{\varphi <!--\; \varphi \; -->}$ with respect to $\mathrm{\theta <!--\; \theta \; -->}$, so with this initial condition the answer follows by proving that ${\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }=0$. My issue is that upon working this round I end up with
$({\mathrm{\varphi <!--\; \varphi \; -->}}^{\prime }{)}^2=\frac{C^2}{{\sin }^4<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}-<!--\; -\; -->C^2{\sin }^2<!--\; \; -->\mathrm{\theta <!--\; \theta \; -->}}$
From this I can see no way forward.
Where do i go from here/ what should I do instead?