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I'm sorry, but this is the wrong question. The 'right'() question is here:>

http://web2.0calc.com/questions/both-circles-have-the-same-radius-nbsp-r-1-nbsp-if-area-nbsp-a-nbsp-area-nbsp-b-nbsp-area-nbsp-c-what-is-the-distance-between-the

 

Image result for overlapping circles area

civonamzuk  May 21, 2015

Best Answer 

 #1
avatar+18715 
+13

Both circles have a radius  r=1.  Can you make these 3 regions,  A  B  and  C, to occupy the equal areas ?

$$\small{\text{$
\rm{central~angle}= \varphi_\rm{rad}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\rm{area}_B &=&
2\cdot\left[
\pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )}
\right] \\
\rm{area}_A &=& \rm{area}_B=\rm{area}_C\\\\
\rm{area}_{circle_1} = \rm{area}_{circle_2} &=& \pi r^2\\
\rm{area}_{circle_1} - \rm{area}_B &=& \rm{area}_B \\
\rm{area}_{circle_1} &=& 2\cdot \rm{area}_B \\
\pi r^2 &=& 2\cdot \rm{area}_B \\
\pi r^2 &=& 2\cdot 2\cdot\left[
\pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )} \right] \\
\pi r^2 &=& 2\cdot r^2 \cdot \varphi_{\rm{rad}}
- 2 \cdot r^2 \cdot
\sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : r^2\\
\pi &=& 2 \cdot \varphi_{\rm{rad}} - 2 \cdot \sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : 2\\
\frac{\pi}{2} &=& \varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } \\\\
\end{array}
$}}$$

$$\boxed{
\varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } = \frac{\pi}{2}
}$$

The Newton-Raphson method in one variable is implemented as follows:


Given a function ƒ defined over the reals x, and its derivative ƒ',

we begin with a first guess x0 for a root of the function f.

Provided the function satisfies all the assumptions made in the derivation of the formula,

a better approximation x1 is

$$\small{\text{
$
x_{1} = x_0 - \frac{f(x_0)}{f'(x_0)} \,$.
}}$$

$$\varphi_{\rm{rad}_1} =
\varphi_{\rm{rad}_0} - \dfrac{ \varphi_{\rm{rad}_0} -\sin{ ( \varphi_{\rm{rad}_0} ) } - \frac{\pi}{2}
}
{1-\cos{ (\varphi_{\rm{rad}_0}) } }$$

Iteration:

$$\small{\text{$
\begin{array}{rcl}
\varphi_{\rm{rad}_0} &=& 2\\
\varphi_{\rm{rad}_1} &=& 2.33901410590 \\
\varphi_{\rm{rad}_2} &=& 2.31006319657 \\
\varphi_{\rm{rad}_3} &=& 2.30988146730 \\
\varphi_{\rm{rad}_4} &=& 2.30988146001 \\
\cdots \\
\varphi_{\rm{rad}} = 2.30988146001 \\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\rm{area}_B &=&
2\cdot\left[
\pi r^2 \cdot \frac{\varphi_\rm{rad} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_\rm{rad} )}
\right] \qquad | \qquad r=1 \\
\rm{area}_B &=&
2\cdot\left[
\pi \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot \sin{(\varphi_{\rm{rad}} )}
\right] \\
\rm{area}_B &=& \varphi_{\rm{rad}} - \sin{(\varphi_{\rm{rad}} )} \\\\
\rm{area}_B = 2.30988146001 -\sin{ ( 2.30988146001 )} &=& \frac{\pi}{2}\\
\rm{area}_A = \rm{area}_C &=& \pi r^2 - \rm{area}_B
\qquad | \qquad r=1 \\
\rm{area}_A = \rm{area}_C &=& \pi - \frac{\pi}{2} = \frac{\pi}{2}
\end{array}
$}}$$

heureka  May 21, 2015
Sort: 

1+0 Answers

 #1
avatar+18715 
+13
Best Answer

Both circles have a radius  r=1.  Can you make these 3 regions,  A  B  and  C, to occupy the equal areas ?

$$\small{\text{$
\rm{central~angle}= \varphi_\rm{rad}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\rm{area}_B &=&
2\cdot\left[
\pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )}
\right] \\
\rm{area}_A &=& \rm{area}_B=\rm{area}_C\\\\
\rm{area}_{circle_1} = \rm{area}_{circle_2} &=& \pi r^2\\
\rm{area}_{circle_1} - \rm{area}_B &=& \rm{area}_B \\
\rm{area}_{circle_1} &=& 2\cdot \rm{area}_B \\
\pi r^2 &=& 2\cdot \rm{area}_B \\
\pi r^2 &=& 2\cdot 2\cdot\left[
\pi r^2 \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_{\rm{rad}} )} \right] \\
\pi r^2 &=& 2\cdot r^2 \cdot \varphi_{\rm{rad}}
- 2 \cdot r^2 \cdot
\sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : r^2\\
\pi &=& 2 \cdot \varphi_{\rm{rad}} - 2 \cdot \sin{ ( \varphi_{\rm{rad}} ) } \qquad | \qquad : 2\\
\frac{\pi}{2} &=& \varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } \\\\
\end{array}
$}}$$

$$\boxed{
\varphi_{\rm{rad}} - \sin{ ( \varphi_{\rm{rad}} ) } = \frac{\pi}{2}
}$$

The Newton-Raphson method in one variable is implemented as follows:


Given a function ƒ defined over the reals x, and its derivative ƒ',

we begin with a first guess x0 for a root of the function f.

Provided the function satisfies all the assumptions made in the derivation of the formula,

a better approximation x1 is

$$\small{\text{
$
x_{1} = x_0 - \frac{f(x_0)}{f'(x_0)} \,$.
}}$$

$$\varphi_{\rm{rad}_1} =
\varphi_{\rm{rad}_0} - \dfrac{ \varphi_{\rm{rad}_0} -\sin{ ( \varphi_{\rm{rad}_0} ) } - \frac{\pi}{2}
}
{1-\cos{ (\varphi_{\rm{rad}_0}) } }$$

Iteration:

$$\small{\text{$
\begin{array}{rcl}
\varphi_{\rm{rad}_0} &=& 2\\
\varphi_{\rm{rad}_1} &=& 2.33901410590 \\
\varphi_{\rm{rad}_2} &=& 2.31006319657 \\
\varphi_{\rm{rad}_3} &=& 2.30988146730 \\
\varphi_{\rm{rad}_4} &=& 2.30988146001 \\
\cdots \\
\varphi_{\rm{rad}} = 2.30988146001 \\
\end{array}
$}}$$

$$\small{\text{$
\begin{array}{rcl}
\rm{area}_B &=&
2\cdot\left[
\pi r^2 \cdot \frac{\varphi_\rm{rad} }{2\pi} - \frac{1}{2} \cdot r^2 \sin{(\varphi_\rm{rad} )}
\right] \qquad | \qquad r=1 \\
\rm{area}_B &=&
2\cdot\left[
\pi \cdot \frac{\varphi_{\rm{rad}} }{2\pi} - \frac{1}{2} \cdot \sin{(\varphi_{\rm{rad}} )}
\right] \\
\rm{area}_B &=& \varphi_{\rm{rad}} - \sin{(\varphi_{\rm{rad}} )} \\\\
\rm{area}_B = 2.30988146001 -\sin{ ( 2.30988146001 )} &=& \frac{\pi}{2}\\
\rm{area}_A = \rm{area}_C &=& \pi r^2 - \rm{area}_B
\qquad | \qquad r=1 \\
\rm{area}_A = \rm{area}_C &=& \pi - \frac{\pi}{2} = \frac{\pi}{2}
\end{array}
$}}$$

heureka  May 21, 2015

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