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A segment of a circle is the region enclosed by an arc and its chord (See figure below). If $r$ is the radius of the circle and $x$ the angle subtended at the center of the circle, find the value of $x$ (correct to $4$ decimal places) for which the area of the segment is one-fourth the area of the circle. Use the Newton iteration method with $x_0 = 2$. enter image description here

Start wearing purple
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YuYu Huang
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  • You want to subtract the area of the circular sector minus the area of the triangle, set this equal to 1/4 the area of the circle, and then use Newton's method to solve for x. – user84413 Aug 28 '13 at 18:05

3 Answers3

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Hint: To apply Newton's Method, we first need an equation for the area of the segment:

Area of the sector of the circle between the radii: $\frac12xr^2$
Area of the triangle inside the sector outside the segment: $\frac12r^2\sin(x)$
Area of the segment: $\frac12r^2(x-\sin(x))$
Area of one-fourth of the circle: $\frac14\pi r^2$

So the equation we need to solve is $x-\sin(x)=\frac12\pi$ .

Apply Newton's Method to this equation.

robjohn
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\begin{align} A_{seg}&=\frac{1}{2}xr^2-\frac{1}{2}r^2\sin(x)\qquad (1)\\ A_{circle}&=\pi r^2\\ A_{seg}&=0.25\times A_{circle}\\ \frac{1}{2}xr^2-\frac{1}{2}r^2\sin(x)&=0.25\times \pi r^2\\ \frac{x}{2}-\frac{1}{2}\sin(x)&=\frac{\pi}{4} \\ 2x-{2}\sin(x)&={\pi} \\ \end{align} $(1)-$http://mathbits.com/MathBits/TISection/Trig/AreaTrigTri.htm

The iterations are-

$$\begin{pmatrix}x &f(x) &f'(x)& x_{new}\\2 &-0.480093754 &1.416146837& 2.339014106\\2.339014106& 0.049067584& 1.694854658& 2.310063197\\2.310063197& .000304169& 1.673746337& 2.309881467\\2.309881467& 1.2203E-08& 1.673612035& 2.30988146\\2.30988146& 0 &1.673612029 &2.30988146\\2.30988146& 0& 1.673612029 &2.30988146\\\end{pmatrix}$$

Inquest
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Given $x$, we can find the area of the section as $$ A = \frac{1}{2}r^2(x-\sin x) $$ So, given $A$, the equation we'd like to solve (using Newton's method) is $$ \frac{1}{2}r^2(x-\sin x)-A=0 $$ Or, as I will rewrite for convenience: $$ x-\sin x-\frac{2A}{r^2}=0 $$ Now, if $A$ is one quarter the area of the circle, the above becomes $$ x-\sin x-\frac{2(\frac{\pi r^2}{4})}{r^2}=0\implies\\ y(x) = x-\sin x-\frac{\pi}{2}=0 $$ Newton's algorithm dictates that we update $x$ by $$ x_{i+1} = x_i-\frac{y(x_i)}{y'(x_i)} $$ $y$ is as given above, so that $y'$ is given by $$ y'(x) = 1-\cos x $$ So, our method can be described recursively as $$ x_{i+1} = x_i-\frac{x_i-\sin x_i-\frac{\pi}{2}}{1-\cos x_i} $$ With the initial estimate $x_0=2$.

Ben Grossmann
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