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SuicideFuel Math thread problem (official)

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wei#3959

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Pass a hard entrance exam, become a STEM graduate in a good Uni, make >150k working at a big tech company
@Ahnfeltia

oh whoops for some reason i thought this was the "Fun facts about incels" thread
 
Ahnfeltia

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@Ahnfeltia

oh whoops for some reason i thought this was the "Fun facts about incels" thread
You don't say. Your post was pretty apt tho. I just wanted to make a shitty vaguely-math-related quip.
 
manletogre

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I might learn math from Duolingo
 
Ahnfeltia

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I might learn math from Duolingo
I thought you were joking but after a quick search Duolingo does actually have a math-oriented whatchamacallit.
 
uzuki7777

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I should be right at home here with a masters in math but I don't know how to solve the penultimate problem, and I was always weak in geometry. I'm really good at calculus, though. Taking a derivative or inverse just seems to make it worse.
 
CountBleck

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I should be right at home here with a masters in math but I don't know how to solve the penultimate problem, and I was always weak in geometry. I'm really good at calculus, though. Taking a derivative or inverse just seems to make it worse.
you have a masters in math. you should be able to solve it. i could definitely solve it, i just need to read a wikipedia article on ellipses so i can understand wtf the terminology means
 
uzuki7777

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Yeah I can certainly cheat and google it lol
 
Ahnfeltia

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I should be right at home here with a masters in math but I don't know how to solve the penultimate problem, and I was always weak in geometry. I'm really good at calculus, though. Taking a derivative or inverse just seems to make it worse.
I tried to solve it by introducing Cartesian coordinates, but it don't seem like the math is working out.
you have a masters in math. you should be able to solve it. i could definitely solve it, i just need to read a wikipedia article on ellipses so i can understand wtf the terminology means
How did you do it?
 
CountBleck

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I tried to solve it by introducing Cartesian coordinates, but it don't seem like the math is working out.

How did you do it?
i haven't actually given it a try yet. but bashing it out with cartesian coordinates doesn't seem to be the move. i'll spend today working on the problem
 
trying to ascend

trying to ascend

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Physics problem:
79c2fd1daef8424bbb4cae4c9e40529a


A sphere of mass m plugs a circular hole of radius r at the bottom of a container filled with water of density ρ. Lowering the water level slowly, at a given moment the sphere detaches itself from the bottom of the container. Find the height h of the water, in terms of the constants given, level for this to happen, knowing that the top of the sphere, at a height a from the bottom of the container, remains always covered with water
 
trying to ascend

trying to ascend

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An easy one: Let the curve be determined by the locus of the centers of the circles in R² , which are tangent to the line x = 2 and pass through the point (6,4) . Therefore, the tangent line to this curve through the point (6,8) has which equation?
 
Ahnfeltia

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An easy one: Let the curve be determined by the locus of the centers of the circles in R² , which are tangent to the line x = 2 and pass through the point (6,4) . Therefore, the tangent line to this curve through the point (6,8) has which equation?
The curve is the parabola given by (x - 2)^2 = (x - 6)^2 + (y - 4)^2 -- i.e., 8x = y^2 - 8y + 48. Now, dx/dy = y/4 - 1, so the desired tangent has equation x = (8/4 - 1)*(y - 8) + 6 = y - 2.
 
trying to ascend

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The curve is the parabola given by (x - 2)^2 = (x - 6)^2 + (y - 4)^2 -- i.e., 8x = y^2 - 8y + 48. Now, dx/dy = y/4 - 1, so the desired tangent has equation x = (8/4 - 1)*(y - 8) + 6 = y - 2.
Correct :feelsokman:
 
trying to ascend

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A ninja, with his super sharp sword, splits a 1 meter long staff into three parts with 2 random blows. What is the probability that these parts can form a triangle?
 
Ahnfeltia

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A ninja, with his super sharp sword, splits a 1 meter long staff into three parts with 2 random blows. What is the probability that these parts can form a triangle?
Very flavorful problem. By virtue of the triangle inequality, we want to find the probability that that longest of the three pieces doesn't exceed the combined length of the other two. Let X & Z denote the positions of the two random cuts (in metres) and let m & M be their minimum and maximum resp. I'll assume X & Z are independent. The probability we seek can be expressed as 1 - P(m > 1/2) - P(M < 1/2) - P(M - m > 1/2). Now,
  • P(m > 1/2) = P(X > 1/2)*P(Z > 1/2) = (1/2)*(1/2) = 1/4
  • P(M < 1/2) = P(X < 1/2)*P(Z < 1/2) = (1/2)*(1/2) = 1/4
  • P(M - m > 1/2) = P(|X - Z| > 1/2) = (1/2)^3 + (1/2)^3 = 1/4 by geometric considerations (utilizing that (X,Z) is uniformly distributed on the unit square)
All in all, the desideratum is therefore 1/4.
 
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Ahnfeltia

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i haven't actually given it a try yet. but bashing it out with cartesian coordinates doesn't seem to be the move. i'll spend today working on the problem
Did you end up solving it?
 
trying to ascend

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Very flavorful problem. By virtue of the triangle inequality, we want to find the probability that that longest of the three pieces doesn't exceed the combined length of the other two. Let X & Z denote the positions of the two random cuts (in metres) and let m & M be their minimum and maximum resp. I'll assume X & Z are independent. The probability we seek can be expressed as 1 - P(m > 1/2) - P(M < 1/2) - P(M - m > 1/2). Now,
  • P(m > 1/2) = P(X > 1/2)*P(Z > 1/2) = (1/2)*(1/2) = 1/4
  • P(M < 1/2) = P(X < 1/2)*P(Z < 1/2) = (1/2)*(1/2) = 1/4
  • P(M - m > 1/2) = P(|X - Z| > 1/2) = (1/2)^3 + (1/2)^3 = 1/4 by geometric considerations (utilizing that (X,Z) is uniformly distributed on the unit square)
All in all, the desideratum is therefore 1/4.
Correct :feelsokman:
 
Fallenleaves

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Very flavorful problem. By virtue of the triangle inequality, we want to find the probability that that longest of the three pieces doesn't exceed the combined length of the other two. Let X & Z denote the positions of the two random cuts (in metres) and let m & M be their minimum and maximum resp. I'll assume X & Z are independent. The probability we seek can be expressed as 1 - P(m > 1/2) - P(M < 1/2) - P(M - m > 1/2). Now,
  • P(m > 1/2) = P(X > 1/2)*P(Z > 1/2) = (1/2)*(1/2) = 1/4
  • P(M < 1/2) = P(X < 1/2)*P(Z < 1/2) = (1/2)*(1/2) = 1/4
  • P(M - m > 1/2) = P(|X - Z| > 1/2) = (1/2)^3 + (1/2)^3 = 1/4 by geometric considerations (utilizing that (X,Z) is uniformly distributed on the unit square)
All in all, the desideratum is therefore 1/4.
Your iq is mogger level
 
CountBleck

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Did you end up solving it?
not really, i suspect a change of coordinates would help, but i'm too lazy to learn the math behind cylindrical, polar coordinates?
or too lazy to learn the math behind conic sections.

i just wrapped up a full courseload of math courses and i'm dead and i'm trying to love my 3 weeks off hehhee

EDIT: no fuck this imma finish the problem. its my duty because its driving me crazy that i dont know how to prove
 
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Marathon

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how are some of you fuckers so quick?

My nigga @Caesercel you answer basically everything. How are you here with that Brain? I hope you weren’t my giga expert maths teacher whom I hated and was also inkwell.

What are you? Asian or Indian?

And yes you too greycel @Ahnfeltia
@trying to ascend @Fallenleaves

What do you guys do irl? Just curious
 
Fallenleaves

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@Ahnfeltia @trying to ascend Give me a basic integral question except improper ones (I've just started learning improper integrals)
 
Marathon

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Also, I expected you to be here @PPEcel :feelshehe:
 
trying to ascend

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@Ahnfeltia @trying to ascend Give me a basic integral question except improper ones (I've just started learning improper integrals)
(1 + x² + sqrt(1 - x²))/sqrt((1 - x4)(1 + x²))

It's meant to be x to the power of 4
 
CountBleck

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@Ahnfeltia @trying to ascend Give me a basic integral question except improper ones (I've just started learning improper integrals)
here
i wasted 3 minutes overleafing this. if you want solutions just ask for em, but give em a good try first. they aren't easy
Screenshot 117
 
CountBleck

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damn i give up on the ellipse problem, too much alphabet soup
 
trying to ascend

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Ok I failed

Integral of 1/sqrt (1-x^4) + integral of sqrt(1-x^2)/sqrt(1-x^4)*(1+x^2)
Simplify and you will end up with 1 + x² + sqrt(1 - x²)/(1 + x²)(sqrt(1 - x²).

If you call the term 1 + x² as a, and the other as b, you will notice that it's the same as 1/(1 + x²) + 1/sqrt(1 - x²), since it's (a + b)/ab.

Then you can just apply two triognometric integrals and you will end up with arctgx + arcsinx
 
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trying to ascend

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How does sqrt (1-x4) simplify into sqrt(1-x^2), provided all the other terms in the original function and simplified one remain the same
1 - x4 = (1 + x²)(1 - x²).

Since there is another (1 + x²), we can take it out of the root and just (1 - x²) will remain there
 
Fallenleaves

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1 - x4 = (1 + x²)(1 - x²).

Since there is another (1 + x²), we can take it out of the root and just (1 - x²) will remain there
I miswrote the question, damn it
@trying to ascend @Ahnfeltia send another one
 
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how are some of you fuckers so quick?
It's genetics; they have specifically chosen to further apply it in physics or mathematics, however. :society:
 
sigurankako

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I almost failed math in hs once
 
Ahnfeltia

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I miswrote the question, damn it
@trying to ascend @Ahnfeltia send another one
Integrate sin(x)*e^(-x) & cos(x)*e^(-x) from x = 0 to pi. Notice something interesting?
 
Ahnfeltia

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here
i wasted 3 minutes overleafing this. if you want solutions just ask for em, but give em a good try first. they aren't easy
View attachment 689619
(b) is just IBP
(c) is trivial after substituting x = sec(q)
(a) is trivial after substituting x = 1/y

Did I get the ideas right? I'm too lazy to actually calculate them
 
Ahnfeltia

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Consider the ellipse below, where DD' is a chord passing through its center, MM' is a locus chord, and the major axis of the ellipse is 2a.

Prove that: DD' ²= MM' . 2a


2e256071e436487e776e40241e7cc2cf
Got a hint?
 
CountBleck

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(b) is just IBP
(c) is trivial after substituting x = sec(q)
(a) is trivial after substituting x = 1/y

Did I get the ideas right? I'm too lazy to actually calculate them
yeah you did it the cleanest way possible; intended solution.
yea at one point you just know the method but don't bother to compute the integral because we aren't first year math students lol.
 

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