Problem of the day thread #1

>>8548304
I probably have a shitty methodology but here goes.
It looks like the area is progressing linearly.

A(L) = a*L + b

I will assume that 0 < L < 2r
If L = 2R, we have Area(2R) = [math](\pi*r^2)/2[/math]
If L = 0, we have area(0) = 0

So b = 0
So a = [math]\frac{\pi^2}{4R}[/math]

A(l) = [math]\frac{\pi^2}{4R}*l[/math]

Don't you need the angle?

>>8548337

The angle is not a given.

>>8548333

Good attempt and the methodology wasn't bad at all. Ultimately wrong though.

>>8548347
It might be because I can't use tex.

What I wanted to give was :

So a = [math]\frac{\pi^2*r^2}{4r}[/math]
A(l) = [math]\frac{\pi^2*r}{4}*l[/math]

However the area is not progressing linearly, I tried it. So it might be some kind of polynome ?

Given that a process with dissapative processes would result in any gain in free energy being less than the work put into the system derive the Jarzynski Equality. You may assume the Fluctation Theorem to hold true with out any derivation

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I came up with

(arcsin(L/2R)*R^2)/2-L*sqrt(R^2-(L/2)^2)/4

[math]\frac{1}{2}\pi r^2 - (r^2 cos^{-1}(\frac{L}{2})+\frac{L}{2}\sqrt{r^2+\frac{1}{4}L^2})[/math] where inverse cosine gives angles in radians.

>>8548393
typo, this should be the formula:

[math]\frac{1}{2}\pi r^2 - (r^2 cos^{-1}(\frac{L}{2r})+\frac{L}{2}\sqrt{r^2+\frac{1}{4}L^2})[/math]

>>8548369
Why didn't you apply the exponent to L/2 ?

>>8548304
>babby-tier HS questions
back in my day, when /sci/ was less meme'd, we had weekly threads with Putnam questions

>>8548368
Some one do my fucking homework please

>>8548304
Parametrize half circle as [math] \frac{L}{2}(x) = f(x) = \sqrt{R^2 - x^2} [/math], thus [math] x_L =\sqrt{R^2 - \frac{L^2}{4}}[/math]. Integrate from x_L to to r to get [eqn]A = 2\int_{x_L}^R \sqrt{R^2 - x^2} dx = \frac{1}{2} \pi R^2 + \frac{L}{4} \sqrt{-L^2 + 4 R^2} + 4 R^2 \mathrm{arccot}\left( \frac{L}{\sqrt{-L^2 + 4 R^2}} \right)[/eqn]

>>8548342
The angle doesn't change gents

>>8548304
draw lines from the center of the circle to the endpoints of the chord. the angle these lines make with the cord is
T = arccos(L/(2*R))
the area of the resulting triangle is
A = L/2 * R * sin(T)
the unknown angle of the isoceles triangle is
S = pi - 2*T
the area of the smaller resulting sector is
B = S/(2*pi) * pi * R^2
and the area of the shade region is
ans = B - A
can we get a more difficult problem please, this is trivial

>>8548469
So the line always has the slope of the OP?

>>8548473

Tomorrow there will be a more difficult one.

Any suggestions should goto [email protected]

>>8548481
Thank you for getting this started man!!

>>8548304
A = asin(L/R)*R^2-L*sqrt(R^2-L^2)

>>8548481
pull them from "Problem Solving through Problems" by Larson or "Putnam or Beyond" by some guys at Princeton.

>>8548498
Btw this is fucking easy

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Here ya go, fags

>>8548507
>Here ya go, fags
jesus christ simplify your shit mate

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>>8548304
Hopefully you are not triggered by bad handwriting. I tried to write big letters so that it would still be readable.

To solve I used resources from: https://www.mathsisfun.com/algebra/trig-sine-law.html

and

https://www.mathsisfun.com/geometry/circle-sector-segment.html

As I cannot be bothered to memorize this stuff in the age of information and if you want to understand my solution but don't remember those formulas then you can check em there.

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>>8548549
wtf. When I pass from phone to computer even though it looks okay, when I upload it gets rotated.

I tried fixing it by loading it into paint and then saving it. I call this advanced technique "Paint transform".

>>8548539
There doesn't seem to be any simplifying to do here, m8

>>8548549

> Can't be bothered to memorize this stuff in the age of information
> Can't be bothered to memorize
> memorize

Not rederiving all formulas as needed.

pleb.

>>8548570
>Not rederiving all formulas as needed.

You want me to rederive elementary geometry formulas so that I can solve a problem on the internet? Fuck that. Maybe if there was money on the line.

If you know what you are looking for then googling it is way faster.

>>8548578
>being this bad at mathematics

>>8548474
Is this a serious question?

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>>8548304
A=2[math]\int_{\sqrt{1-\frac{L^{2}}{4}}}^{R}\sqrt{1-y^{2}}[/math]dy

>>8548996
oops. what I meant was 'area equals two times the integral...'

too lazy to solve that integral though.

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i tihnk i got it

>>8549035
oops i fucked up

Aa and Ab are (R^2phi)/2

>>8549074
>>8549035
>phi

You are supposed to find the area in terms of R and L, boi.

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>>8549035
>labeling the three radii three different letters

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>>8548996
error. interchange 1 with R.

>>8549090
lol fuck man give me a break im a 26 year old whos just finishing high school

>>8549090
its not three different letters, its two different letters btw, and they aren't all equal

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>>8549117
You are forgiven. Congrats on fighting the good fight.

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>>8548304

>>8549124
thanks bro. i just applied to university today. wish me luck

>>8549130
Good luck broski.

You ex armed forces?

>>8549117
>>8549130
Good luck on this journey, curing retardation is hard but it can happen.

Here's a very cute problem: Show that each function from the real numbers to the real numbers can be written uniquely in the form f = e + o, where e is an even function and o is an odd function.

(Hint: Try and assume f(x) = e(x) + o(x), and use the properties of even/odd functions to solve for e or o).

>>8548304
[latex]
int_{R-cos(sin^{-1}(L/R))}^{R}sqrt{R^2-x^2}dx
[\latex]

>>8549130
Godspeed

>>8549191
[latex]
2int_{R-cos(sin^{-1}(L/R))}^{R}sqrt{R^2-x^2}dx
[/latex]

Edit, also forgot the 2

>>8549193
forgot the slashes too

[latex]2\int_{R-\cos(\sin^{-1}(L/R))}^{R}\sqrt{R^2-x^2}dx[\latex]

>>8549197 I hate myself
[math]2\int_{R-\cos(\sin^{-1}(L/R))}^{R}\sqrt{R^2-x^2}dx[\math]

>>8549190
Hm. I know this is true but I can't show it.

Like I know f(x) = asinx + bcosx can give all real numbers, but so can just any even/odd function on it's own no? They both vary between -1 and 1 so there's no reason (that I can see on initial inspection) that asinx can't be any number.

note a and b are other variables not arcsin

>>8549191
>>8549193
>>8549197
>>8549199
Press the TeX button in the upper left before posting senpai.

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>>8549199
R u memeing?

>>8549191
>>8549193
>>8549197
>>8549199
There's a preview LaTeX button m8

>>8549199
[eqn]2\int_{R-\cos(\sin^{-1}(L/R))}^{R}\sqrt{R^2-x^2}dx[/eqn]

Word thanks, I was not meming but sorry for ruining the thread lol

>>8549201
That only gives functions of the form Asin(x+B), I'm asking for a formula given an arbitrary function f (it can be anything, it doesn't need to be a continuous function or anything like that).

You want the areas themselves?

Draw a line through the line creating the segment of L.

Produce the line of the diameter out so the radius is to the (radius + other line segment) in the same ratio as the meeting of the base(middle) of Segment L with the end of the produced line to the middle of segment to the edge of the circle.

Now make a cone with the same base as the segment L's base and height equal to the line produced, starting from the middle of the segment. That cone's area will be equal to segment L. You can do the same thing with the other side.

That's not really finding the area of the portion of the circle that's shaded, but whatever lol. Just a cool geometric trick. It's what Archimedes used to solve the famous AC: a given length :: a given area : CB

>>8549117

How did you end up in this situation? I'm genuinely interested in hearing your story.

>>8549294
This. I too am curious

>>8549163
>>8549294
>>8549299
I really don't feel like I have any good excuse, but I'll try to tell the story as objectively as possible.

>Be me
>9 years old
>good grades, happy kid, loving family
>suddenly anxiety attacks
>don't want to be at school ever anymore
>refuse to leave house
>dad homeschools me for grades 4 and 5
>learned a lot more than if I was in school, coincidentally
>unfortunately, completely lost all confidence in myself and became socially awkward to the max
>back to school in grade 6
>everyone thinks i'm weird, do badly in school because I guess I have adhd
>my parents still trying to be supportive, encourage me to do gifted program
>in gifted program from grade 7 to 10, doing even worse in school but probably learning more at least
>in grade 9 start smoking pot, innocent at first, do it with my friends for fun
>start to smoke more and more pot, skip a ton of classes in grade ten because I figured out how to get away with it, fail grade ten math
>summer school
>basically just keep doing this thing where I barely pass each year until grade 12 when I get expelled because they knew I was selling pot but they couldn't prove it so they expelled me for absenteeism
>alternative program to finish diploma, work full time, things are cool
>doing more drugs now, spending all of my money on it

cont'd

>>8549613
>anxiety still following me from when I was 9, sent to therapist, say useless shit, even try medication because I'm stupid
>lifelong distrust of therapists abilities to actually help anyone permanently instilled
>eventually around age 21 start having anxiety attacks again, but this time I'm having delusions and basically i have to stop doing drugs because I begin to believe I'm being killed by them
>definitely all in my head, but in order to avoid the episodes I stop doing everything, including smoking pot, and start eating healthy and exercising
>you might think this is good but really anything done out of fear is bad, i'm just running away from my problems, whatever
>get job at golf course
>realizing i like working a lot because it keeps my mind off of things
>that winter, work at burger place, its a lot of fun but one of our regular customers is an electrical engineer, encourages me to go back to school and try again
>around new years, meet the woman that i am still with now almost 3 years later i believe
>since then, working on and off at the golf course and other places and going to night school and day school for several months here and there
>trying to catch up with the science and math shit
>finally almost finished like 10 classes later
>feelsgood.jpg

i only started doing this shit because i wanted to be an electrician. now im doing it because i love it and i want to study physics.

so theres my story. never really told anyone in it's entirety. i usually just tell people i smoked too much pot because then i dont have to tell them that i have anxiety attacks because it makes me feel like a pussy.

>>8549613
>>8549626
Keep fighting the good fight anon. You can do it

>>8549226
>>8549201
>>8549190
That's classical. It describes the equality F = O + E, where F = the set of functions from IR to any set S, and O and E are the sets of odd and even functions from IR to S.

Let f in F. Let us show the wanted equivalence.

Suppose there exists o and e such that f = o + e. Let x in IR, then, one can write : f(x) = o(x) + e(x) (E1)
f(-x) = o(-x) + e(-x)
The latter becomes :
f(-x) = -o(x) + e(x) (E2)

(E1) + (E2) yields e(x) = 1/2(f(x)+f(-x))
(E1) - (E2) yields o(x) = 1/2(f(x)-f(-x))

Reciprocically, e an o defined above clearly are such that e + o = f.

QED.

Is this thread serious? I honestly can't tell. This whole thing reminds me of a conversation between Bill Hitchert and his friends.

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>>8548304
Here's a brainlet tier solution:
To find the area under the line with length [math]L[/math] which is drawn from point [math]P[/math] to point [math]Q[/math] on the circle we first find the area of the circular sector with angle [math]2\varphi[/math] as shown in the figure then subtract the area of the two triangles [math]PCS[/math] and [math]QCS[/math] which by the way are identical triangles

To find the area of circular sector we just use the formula [math]A=\frac{\alpha r^2}{2}[/math] where [math]r[/math] is the radius and [math]\alpha[/math] is the angle of the sector

We know the radius is [math]R[/math] and to find the angle of the sector we need to find [math]\varphi[/math] in the figure then multiply it by two, to find [math]\varphi[/math] we use basic trigonometry:
[math]\sin(\varphi)=\frac{\frac{L}{2}}{R}=\frac{L}{2R}[/math]
[math]\varphi=\sin^{-1}(\frac{L}{2R})[/math]
and here we found [math]\varphi[/math] now we multiply it by two and plug it into the formula and we get:
[math]A_1=\frac{2\sin^{-1}(\frac{L}{2R})R^2}{2}=\sin^{-1}(\frac{L}{2R})R^2[/math]
and that's the area of the sector

Now for the other part which is the area of the triangles since the two triangles are identical we only need to find the area of one of them then multiply it by two, We know the base is [math]\frac{L}{2}[/math] and the height is [math]X[/math], to find [math]X[/math] we use the same basic trigonometric methods:
[math]\theta + \varphi + \frac{\pi}{2} = \pi[/math]
[math]\theta=\frac{\pi}{2}-\varphi=\frac{\pi}{2}-\sin^{-1}(\frac{L}{2R})[/math]
[math]\sin(\theta)=\sin(\frac{\pi}{2}-\sin^{-1}(\frac{L}{2R}))=\cos(\sin^{-1}(\frac{L}{2R}))=\sqrt{1-({\frac{L}{2R}})^2}=\frac{X}{R}[/math]
[math]X=R\sqrt{1-({\frac{L}{2R}})^2}[/math]
Now the area of one triangle is:
[math]A_2=\frac{LR}{4}\sqrt{1-({\frac{L}{2R}})^2}[/math]
And finally the final area would be:
[math]A=A_1-2A_2=\sin^{-1}(\frac{L}{2R})R^2-\frac{LR}{2}\sqrt{1-({\frac{L}{2R}})^2}[/math]
And that's it.

>>8549852
Take a breath anon. You want to be precise? That's good. Let us be precise.

>PCS and QCS which by the way are identical triangles
You did not justify this.

>know the base is L2 and the height is X
You are overusing a word. Only isocele triangles have a side called the base (or basis).

>θ=π2−φ=π2−sin−1(L2R)
You haven't justified why phi = sin^-1(L/2R) in terms of where phi is in the trigonometric circle, and why sin is a bijection on the segment where phi belongs to.

>A, A1, A2
Those are not correctly defined.

>>8548304
Geometry tiers
Algebraic > Differential > (Cartesian) Analytic > Affine > Training google botnet > Eating worms > Eating Someone else's shit > Eating dogshit > Euclidian

>>8549109
x^2+y^2= R^2 ?

>>8548304
Dumb way:We could use the cosine theory and determine the angle of the sector. Angle*R^2 /2 would be the area of the sector after that we just subtract the area of the triangle and we get the answer.
Another dumb way:if we take the area differential l*dr=2sqrt(R^2-r^2)dr and we integrate it from R to R1=sqrt(R^2-L^2/4)

>>8550057
fuggg forgot the square. Sorry anons. Thanks.

>>8549934
Ok the proof is little messed up but i'll try to explain as much as possible

>PCS and QCS which by the way are identical triangles
by identical i meant congruent, not sure if there is a difference but anyway:
CP=CQ (both are lines from a center of a circle so they have the same length R(the radius))
CS is shared by both triangles
because CS is an angle bisector so it cuts QP in halves so QP = QS/2 = PS/2 so QS=SP
Because the two triangles have all the same side lengths they are congruent(SSS) Q.E.D.

>You haven't justified why phi = sin^-1(L/2R) in terms of where phi is in the trigonometric circle, and why sin is a bijection on the segment where phi belongs to.
Not sure what you exactly mean by this but the upper bound for phi is pi/2 because the maximum value for L is the radius and phi will approach pi/2 when L approaches R, if you're asking how is phi=sin^-1(L/2R) well:
sin(phi)=L/2R
sin^-1(sin(phi))=sin^-1(L/2R)
phi=sin^-1(L/2R)
Duh

>A, A1, A2
I run out of symbols so i had to make some and i'm not sure why i used these in the first place, these should represent the areas because i thought they will make things a little tidy but they didn't, A1 is the area of the sector, A2 is the area of one triangle, A is the area under L

>>8550217
Oh what the fugg i'm doing it's 10:00 pm i can't focus and i really need to sleep just ignore what i said about phi
i somehow mistook the domain with the range sorry for that, L/2R will never be greater than 1/2 because as i said the max for L is R

>>8550241
ignore the last box, it was riddled with autism

>>8548368
Non equilibrium thermo bro, I didn't know that anyone else actually knows about this stuff

>>8548304
took me 5 mins
/($fract{2asin($fract{l}{2r})}{360°}-$fract{l}{2}\sqrt[2]{r^2-$fract{l^2}{4}}/)

>>8550260
It's hard being Patrician in a world of plebs, isn't it?

>>8548473
shouldnt T be arccos(1-(1/2)*(L^2/R^2))?

>>8550722
we have R*cos(T) = L/2 which implies T = arccos(L/(2*R))
i don't know where you're getting your T from