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
Physics problem:
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
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
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.
Correct
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)
Last edited:
Did you end up solving it?
Correct
Your iq is mogger level
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
Last edited:
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
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
@Ahnfeltia @trying to ascend Give me a basic integral question except improper ones (I've just started learning improper integrals)
is a locus chord a chord that passes through one of the foci?
Also, I expected you to be here @PPEcel
Yes
(1 + x² + sqrt(1 - x²))/sqrt((1 - x4)(1 + x²))
It's meant to be x to the power of 4
Are we finding the integral of the function above?
Yes, with respect to x
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
damn i give up on the ellipse problem, too much alphabet soup
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
Last edited:
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
Thanks
I miswrote the question, damn it
@trying to ascend @Ahnfeltia send another one
It's genetics; they have specifically chosen to further apply it in physics or mathematics, however.
I almost failed math in hs once
Haven't learnt it yet
Integrate sin(x)*e^(-x) & cos(x)*e^(-x) from x = 0 to pi. Notice something interesting?
oh my how embarassing, i misread your request. i thought you were asking specifically for improper integrals. whoops.
1/2(e^(-pi)+1)
(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
correctamundo
Got a hint?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
The other integral shows up in the integral after integration by parts for both integrals, great stuffcorrectamundo
Indeed. Hence why I suggested it
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.
Start off by evaluating points D, M, M’, D’ of the elipse.
You can solve it with elementary math
I tend to do too much in my head. I could benefit from writing out more stuff. In this case however, I'm confident I could solve those integrals off the top of my head. Nice assortment of integrals btw.
I think that's exactly what I tried. If I'm not mistaken, the Cartesian coordinates (taking the center of the ellipse as the origin) of M & M' are pretty ugly (involving quadratic surds). Is that right?
ln|1+x^5|+ C
Let g(x) = (x - 3)/(x + 1). Note that g(g(x)) = (x + 3)/(1 - x) and g(g(g(x))) = x. We are given that f(g(x)) + f(g(g(x))) = x. Ergo
- f(g(g(y)) + f(y) = g(y)
- f(z) + f(g(z)) = g(g(z))
False. If you differentiate this, you'll get a factor of 5x^4 because of the chain rule.
It's impossible, IBP, trig sub, u-sub and IBPF can't be used
Why are there 20 bluepillers
Similar threads
Blackpill
[STUDIES] Your problem as an Incel is NOT ACTIVATING FUNDEMENTAL REWARD CIRCUITRY IN HER BRAIN
- veryrare
- Inceldom Discussion
DepressionTookMyIQ
Users who are viewing this thread
