Use the arc length formula to find the arc length of the upper half of the circle with center at $(0,0)$ and radius $3$. Also, find the arc length of the curve in the first question by using elementary geometry.
[Math] Arc Length Formulas
analytic geometrycalculusdefinite integralsplane-curves
Related Solutions
Here is a numerical solution. The parameters are taken as $A$,$B$,$C$,$M$,$K$,$c$. The equation of the arc is given by
$$(y-C)^2+(x-B)^2=A^2$$
and for the line
$$y=M\,x+K$$
The break-point is taken as $c$. Therefore the objective function to be minimized becomes
$$F=\sum_1^n\bigg((y_i-F_{1,i})^2+(y_i-F_{2,i})^2\bigg)$$
$$F_{1,i}=-\sqrt{A^2-(x_i-B)^2}+C\qquad \text{if }x_i\le c\text{ and }0\text{ elsewhere}$$
$$F_{2,i}=M\,x_i+K\qquad \text{if }x_i>c\text{ and }0\text{ elsewhere}$$
The first constraint function assures that the arc and line meet at the break-point $c$
$$-\sqrt{A^2-(c-B)^2}+C-M\,c-K=0$$
The second constraint that the line is tangent to the arc at $c$
$$\frac{c-B}{\sqrt{A^2-(x-B)^2}}-M=0$$
And the last constraint is that the center of arc lies on a line given by $y=m_1x+k$
$$C-m_1\,B+k_1=0$$
To test the algorithm I generated 120 points and added error terms of $N(0,0.1)$ as given in below graph. The black dots are for circle with $A=3$, $B=1$, $C=4$; and the blue dots for line with $M=0.8944$, $K=-0.9193$. The break-point is given as $c=3$
In Matlab I used "fmincon" with "interior-point" algorithm. The initial guess is randomly generated as $[A=0.9045;B=0.8663;C=0.9225;M=0.5637;K=0.1863;c=0.6766]$. After 170 iterations the algorithm stopped convergent with objective function value of $1.16322$ and maximum constraint violation of $1.66\cdot 10^{-16}$. The parameters are found to be
$$A=2.97\qquad B=0.99\qquad C=3.96\qquad M=0.88\qquad K=-0.88\qquad c=2.95$$
As you can see the calculated parameters are very close to the original parameters even though random error is added to the data. The final grap shows the data (red) and the calculated fitting function (blue)
-------------------ADDITION OF CODE------------------------
Please note that $A\rightarrow x(1)$, $B\rightarrow x(2)$, $C\rightarrow x(3)$, $M\rightarrow x(4)$, $K\rightarrow x(5)$, $c\rightarrow x(6)$
First the function file for objective function (save it as a seperate file as objfun.m)
function f = objfun( x )
global n
global Y_er X
sum=0;
for i=1:2*n
if X(i)<=x(6)
f1=-sqrt(x(1)^2-(X(i)-x(2))^2)+x(3);
sum=sum+(Y_er(i)-f1)^2;
else
f2=x(4)*X(i)+x(5);
sum=sum+(Y_er(i)-f2)^2;
end
end
f=sum;
end
As the second constraint equations (save it as a seperate file as constfun.m)
function [ c ceq ] = constfun( x )
c=[];
f1=-sqrt(x(1)^2-(x(6)-x(2))^2)+x(3);
f2=x(4)*x(6)+x(5);
g1=x(6)-x(2);
g2=x(4)*(sqrt(x(1)^2-(x(6)-x(2))^2));
ceq(1)=f1-f2;
ceq(2)=g1-g2;
ceq(3)=x(3)-3*x(2)-1;
end
And the final one is main file (save it as m.file with any name in the same folder with objfun.m and constfun.m)
clc
clear
global Y_er X n
global beta
%------------------- GENERATION OF SAMPLE POINTS -------------------------
% Generate 60 uniformly distributed random points between 0 and 3
n=60;
a=0;
b=3;
X1=a+(b-a)*rand(n,1);
% Generate y=-sqrt(A-(x-B)^2) + C function for random data
A=3;
B=1;
C=4;
for i=1:n
Y1(i,1)=-sqrt(A^2-(X1(i)-B)^2)+C;
end
plot(X1,Y1,'k.');
% Generate errors with normal dist (mean 0 stdev 0.1)
Er=0.1*randn(n,1);
Y1_er=Y1+Er;
hold on;
plot(X1,Y1_er,'ro');
% Now generate 60 more samples between 3 and 6
clear a b
a=3;
b=6;
X2=a+(b-a)*rand(n,1);
% Generate y=m*x+k function for random data
m=(-B+a)/sqrt(A^2-(a-B)^2);
k=-sqrt(-(a-A-B)*(a+A-B))-a*m+C;
for i=1:n
Y2(i,1)=m*X2(i)+k;
end
plot(X2,Y2,'b.');
% Generate errors with normal dist (mean 0 stdev 0.1)
Er=0.1*randn(n,1);
Y2_er=Y2+Er;
hold on;
plot(X2,Y2_er,'ro');
%Construct the final data set
X=[X1;X2];
Y_er=[Y1_er;Y2_er];
figure(2)
plot(X,Y_er,'b.');
%--------------------------------OPTIMIZATION---------------------------
x0=rand(6,1);
options = optimset('Display','iter-detailed','Algorithm','interior-point','TolFun',1e-12,'TolX',1e-12);
x = fmincon(@objfun,x0,[],[],[],[],[],[],@constfun,options);
figure(3)
hold on
for i=1:2*n
if X(i)<=x(6)
r=-sqrt(x(1)^2-(X(i)-x(2))^2)+x(3);
else
r=x(4)*X(i)+x(5);
end
plot(X(i),r,'b.')
end
PS: If you have your own data set put x-data in variable X and y-data in Y_er. In addition put n= your number of data points. Hope it works for you.
PS2: Please modify your constraint "center of arc on a given line" in "constfun.m" in ceq(3).
The length $L$ of the cardioid is given by \begin{align*}L&=2\int_0^{\pi} \sqrt{(dr/dt)^2 +r^2} dt\\ &= 2\int_0^{\pi}\sqrt{(-\sin(t))^2 +(1+\cos(t))^2} dt \\&=2\int_0^{\pi}\sqrt{\sin^2(t) +1+2\cos(t)+\cos^2(t)} dt \\&=2\sqrt{2}\int_0^{\pi}\sqrt{1+\cos(t)} dt =4\int_0^{\pi}\cos(t/2) dt=8[\sin(t/2)]_0^{\pi}=8. \end{align*}
P.S. As regards the hint, note that $$2\sqrt{2}\int_0^{\pi}\sqrt{1+\cos(t)} dt = 2\sqrt{2}\int_0^{\pi}\frac{\sin(t)}{\sqrt{1-\cos(t)}} dt= 4\sqrt{2}\int_0^{\pi}\frac{d(1-\cos(t))}{2\sqrt{1-\cos(t)}}\\ =4\sqrt{2}[\sqrt{1-\cos(t)}]_0^{\pi}=8.$$
Best Answer
HINTS: The parametrisation of a circle of radius $r$ in $\mathbb{R}^2$ is given by
$$\gamma(t) = (r \cos (t), r \sin (t)).$$
Arclength is calculated by working out
$$\int_a^b | \gamma'(t)| dt,$$
where $a, b$ are the two ends of your interval and $|\gamma ' (t)|$ is the velocity of the cure.
EDIT: Ok, assuming you have made some kind of attempt, here is how to answer the question:
We are told that the radius of the circle is $3$, so using my first hint, we can write the circle as
$$\gamma(t) = (3 \cos (t), 3 \sin (t)).$$
Now, we want to calculate the arc length of the upper half. We know that a regular circle goes from $0$ round to $2 \pi$ and so if we want just the upper half, we take half of this and we get our interval to be from $0$ and $\pi$, i.e we have $a = 0$ and $b = \pi$.
Next, we want to calculate $| \gamma ' (t)|$:
$$\gamma(t) = (3 \cos (t), 3 \sin (t)),$$ $$\gamma'(t) = (-3 \sin (t), 3 \cos (t)),$$ $$|\gamma'(t)| = |(-3 \sin (t), 3 \cos (t)| = \sqrt{(-3 \sin (t))^2 + (3\cos (t))^2} = \sqrt{9\sin ^2(t) + 9 \cos ^2(t)} = \sqrt{9(\cos ^2(t) + \sin ^2(t))} = 3,$$
which means to find the arc length, we now have to solve the equation
$$\int_{0}^{\pi} 3 dt$$ $$= [3t]_{0}^{\pi}$$ $$= 3(\pi) - 3(0) = 3\pi.$$
Now, if you wanted to check to see if your answer is correct, you can use a different method, i.e like one Andre Nicolas used, and you will see that you get the same answer.