/sci/ - Science & Math

they are unit circles, so use the equation of a circle in the Cartesian plane and find where the four circles intersect to form the region A. then use the equations for arc length and the areas of sectors with some trigonometry and you're there.

I forget, how do you find the area of a circle with a radius of 1?

>>4328440

Finding the points of intersection seems pretty tedious, are you sure there's no other way?

>>4328446
you dont have to find their positions in the x-y plane, i just thought that might be a good way to determine how far apart each intersection is so you can determine the arc length. its actually not that tedious, since all the circles have radius 1!

The circumferences intersect at (0.5, y) and (x, 0.5), so, radius being 1, the y and x of upper and rightmost intersection points are, respectively
<span class="math">
\begin{equation}
y=\sin[\arccos(0.5)]=\sin\frac{\pi}{3}=\frac{\sqrt3}{2}
\end{equation}
\begin{equation}
x=\cos[\arcsin(0.5)]=\cos\frac{\pi}{6}=\frac{\sqrt3}{2}
\end{equation}
[/spoiler]
Area of circular segment is that of the circular sector minus triangle made by those points and center of circumference
<span class="math">
\begin{equation}
A_{seg}=\frac{\theta}{2\pi}\cdot\pi r^{2}-\frac{r^{2}\sin(\theta)}{2}=\frac{\left(\frac{\pi}{3}-\frac{\pi}{6}\right)-\sin\left(\frac{\pi
}{3}-\frac{\pi}{6}\right)}{2}=\frac{\pi}{12}-\frac{1}{4}
\end{equation}
[/spoiler]
Now add the area of the triangle made by the intersection points and the centre of the square and multiply everything by 4
<span class="math">
\begin{equation}
A_{tri}=\frac{\left(\frac{\sqrt3-1}{2}\right)^{2}}{2}=\frac{1}{2}-\frac{\sqrt3}{4}
\end{equation}
\begin{equation}
A=4\left(\frac{\pi}{12}-\frac{1}{4}+\frac{1}{2}-\frac{\sqrt3}{4} \right )=\frac{\pi}{3}-\sqrt3+1
\end{equation}
[/spoiler]

>>4329058
>fucking aliens messed with my latex
The circumferences intersect at (0.5, y) and (x, 0.5), so, radius being 1, the y and x of upper and rightmost intersection points are, respectively
<span class="math">
\begin{equation}
y=\sin[\arccos(0.5)]=\sin\frac{\pi}{3}=\frac{\sqrt3}{2}
\end{equation}
[/spoiler]
<span class="math">
\begin{equation}
x=\cos[\arcsin(0.5)]=\cos\frac{\pi}{6}=\frac{\sqrt3}{2}
\end{equation}
[/spoiler]
Area of circular segment is that of the circular sector minus triangle made by those points and center of circumference
<span class="math">
\begin{equation}
A_{seg}=\frac{\theta}{2\pi}\cdot\pi r^{2}-\frac{r^{2}\sin(\theta)}{2}=\frac{\left(\frac{\pi}{3}-\frac{\pi}{6}\right)-\sin\left(\frac{\pi
}{3}-\frac{\pi}{6}\right)}{2}=\frac{\pi}{12}-\frac{1}{4}
\end{equation}
[/spoiler]
Now add the area of the triangle made by the intersection points and the centre of the square and multiply everything by 4
<span class="math">
\begin{equation}
A_{tri}=\frac{\left(\frac{\sqrt3-1}{2}\right)^{2}}{2}=\frac{1}{2}-\frac{\sqrt3}{4}
\end{equation}
[/spoiler]
<span class="math">
\begin{equation}
A=4\left(\frac{\pi}{12}-\frac{1}{4}+\frac{1}{2}-\frac{\sqrt3}{4} \right )=\frac{\pi}{3}-\sqrt3+1
\end{equation}
[/spoiler]

>>4329062
>>4329058


Engineer. Here, I'm a high schooler.
Let the area of the quarter circle be x.
Subtract 1 by x. Let this be y. Take 2y-x = 2z. One more region, t, is equal to y-2z.

The area A = 1-4z-4t

Express A in terms of x. Substitute x with area of quarter circle.

I prolly made a mistake. Can't proofread and I'm on a phone. Hope this gives you an idea and you solve it. If you want, I'LLC comeback on a comp and visualize this.

Fuck you, engineer.

>>4329077

Shit. I just failed hard.

you could split that area into a square and 4 segments, and no calculus was used that day.

What on earth is going on with the hint?! It's half wrong!! There's absolutely no square root of 3 involved. The answer is 1-pi/2.

>>4329135

I'll define the area touching each quarter circle in the corner as Region Y.


Area of figure is 1 unit squared.

Area of circle is pi. Divide that by four to get the total area of one quarter circle.

Subtract pi/4 from 1 to get the area of Region Y.

Multiply this value by 4 to get the area of all four Region Y's. Then, subtract this value from the value of the square's area, which is 1,to get the area of the A.


1 - (1 - pi/4)4 =
1 - (4 - pi)/4 =
(4 - 16 + 4pi)/16
(4pi - 12)/16

(pi - 3)/4

Well shit, that's not root3, is it?

>>4329349
lol what the fuck are you doing, your regions Y all intersect. >>4329135 he already told you the correct answer.

>>4328431
Cool, I remember solving this as sort of a "special assignment" during 11th(?) grade or so.. Also made a small program to estimate the right answer, to verify my solution.. Ah good times.. When your problems were really no problems at all..

tl:dr is there an answer yet? or should I take this endeavor of finding one?

>>4328431
>>4328431
The only way you can get Pi is through calculus.
Pi does not come from simple algebra, If you use any kind of formula with Pi in it, that means that FORMULA IS DERIVED USING CALCULUS.

\thread

>>4329422
hint is pi

>>4329430
As stated, question c) is impossible to solve.

By using Pi you have automatically used calculus.

>>4329422

>>4329135
>The answer is 1-pi/2.

oh, a negative area?

no, the correct answer was (π/3)-sqrt(3)+1

ITT
>>4329448
>>4329448
>>4329448
>>4329448
>>4329422
>>4329422
>>4329422
>>4329422

>>4329373


Ah, fuck me. Hang on...

1 - (1 - pi/4)2 =
1 - (4 - pi)/2 =
(2 - 8 + 2pi)=

(2pi - 6) ≈ 0.283


I'm not sure about his answer because it appears to give a negative value.

>>4329349

just wondering what Y is in your definition.

"I'll define the area touching each quarter circle in the corner as Region Y."

This doesn't make sense to me. What is a quarter circle in the pic?

>>4329531

>>4329481

so now you just took 2 regions Y. You're missing the big chunks of the remaining two regions Y.

How would one solve this with calculus?
Srs question.

I can see how to do it. I'm not goingto bother actually doing it though.

the intersects between the arcs form a right angled triangle. use this to find out two of the corners of the shaded region. then calculate te straight-line length between these corners. calculate the area of the square within hte shaded region.
then consider the isoceles triangle going between one side of teh square and the opposite corner. find this angle. use this angle to find area of segment. subtract isoceles trangle are from this to get the area of he curvy parts you add to the square.

There's probably a faster way but I don;t care.

>>4329656

Yeah, okay. Probably should've been paying attention there.

It's <span class="math">4*(\frac \pi {12} - \frac \sqrt 3 4 + frac 1 4) = \frac \pi 3 - \sqrt 3 + 1[/spoiler]
>>4329062 is correct.
It's pretty easy.

>>4329682
You could define 4 functions which represent the quarter circles, then integrate.

I actually just tried that, my result is
pi/3 - sqrt(3) + 1

Now without calculus...

>>4330275

I know this is off topic, but yeah, I kind of suck at calculus. Would really appreciate if you could show the whole process.

Make an equilateral triangle from the lower corner, to the top of the circle intersection, to the bottom right corner.

Then it should be easy since the angles you're using are just pi/3.
Gogogo

>>4329683
What this man pointed out.
There is a triangle with unit sides (as drawn).

>>4330403

The 4 function which represent the circles:

a(x) = sqrt(1-x²)
b(x) = sqrt(1-(x-1)²)
c(x) = -sqrt(1-x²)+1
d(x) = -sqrt(1-(x-1)²)+1

If you don't understand, just plot them and you will see. Now.

Integrate a(x)-d(x) over their intersection point to 0.5, and then add it with the integral of b(x)-c(x) over 0.5 to THEIR intersection point.

>>4330431
the integrals would be b(x)-d(x) + a(x)-c(x)

>>4328431
Linear system of equations.
>You should be able to solve this.

>>4330431
Could you elaborate on the integration part, i'm really shitty with calc.

>>4330431
Fuck you, I've been leaning Mathematica for the past hour trying to do this. I finally plotted the shit. How do I integrate it god damnit