Anonymous at Thu, 12 Sep 2024 12:32:15 UTC No. 16376853

>>16376839
He was also a monster in other ways

Anonymous at Thu, 12 Sep 2024 15:00:24 UTC No. 16377034

>>16376839
What device is used to push the ball into the cup?

Anonymous at Thu, 12 Sep 2024 15:04:25 UTC No. 16377038

>>16376853
You are a closet fag

Anonymous at Thu, 12 Sep 2024 15:25:02 UTC No. 16377076

>>16376853
fucking kek

Anonymous at Thu, 12 Sep 2024 15:29:33 UTC No. 16377088

>>16376839
Don't you just model for the radius of the cup as a function of the depth?

Anonymous at Thu, 12 Sep 2024 17:27:36 UTC No. 16377280

can someone else try to solve this it's been a while since i took calc 3.

Anonymous at Thu, 12 Sep 2024 18:04:02 UTC No. 16377326

literal easy task, just think it a bit with geometry first. I guess you have two cases to check. One where the ball does not enter half way and another where it does. You also need to check if a ball that enters whole meets the requirement even it's probably not that case. As the discontinuity is at where ball is exactly half way, that is fourth case to check.

Anonymous at Thu, 12 Sep 2024 18:05:08 UTC No. 16377327

>>16376839
I get r = .00315mm, anybody else get this?

Anonymous at Thu, 12 Sep 2024 18:10:05 UTC No. 16377333

>>16376853
>Rapes college students with math to make them depressed
>Rapes children with his penis to make them gay
What the fuck was his problem

Anonymous at Thu, 12 Sep 2024 18:11:56 UTC No. 16377336

>>16377327
Enough about your testicles radius

Anonymous at Thu, 12 Sep 2024 19:12:34 UTC No. 16377402

>>16376839
Man fuck him.
Not gonna do the problem, but as a tip, there are 4pi steradians for a sphere, which is part of the V = 4pi r^3/3 and SA = 4pi r^2. This comes out of the 2pi[1-cos(x)], where x is the polar angle (can't use theta cuz it'd be confusing for the problem).

So if you want a segment of a sphere, 2pi(1-cosx) is needed.

Anonymous at Thu, 12 Sep 2024 19:58:59 UTC No. 16377464

>>16377327
Are you retarded? Where are h and theta?

Anonymous at Thu, 12 Sep 2024 20:22:51 UTC No. 16377497

>>16377464
If had solved the problem, you would already know the answer.

Anonymous at Thu, 12 Sep 2024 20:23:22 UTC No. 16377498

>>16377464
check your butthole

Anonymous at Thu, 12 Sep 2024 21:07:07 UTC No. 16377560

[math]
r = \frac{h}{\csc \theta + 1}
[/math]

Anonymous at Thu, 12 Sep 2024 22:54:25 UTC No. 16377674

>>16376839
>James Stewart
Is it normal to not be able to answer all the questions in a section?

Anonymous at Fri, 13 Sep 2024 01:47:53 UTC No. 16377899

>>16377674
It's literally just using a compass to draw a circle of radius √2. That's fucking trivial dude. You can determine any irrational number of the form √(x2 + y2) where x,y are natural numbers. So yes you can determine √5, no you cannot determine √6 and so on.

Anonymous at Fri, 13 Sep 2024 01:51:00 UTC No. 16377906

>>16377899
Actually you can probably determine all the algebraic numbers this way using recursion. Once you know where √2 is you can use that as an x value. You can construct another number line for the y axis to find the vertical positions if √2 as well. You can also use fractions on the number line.

Anonymous at Fri, 13 Sep 2024 02:48:55 UTC No. 16377953

>>16376839
How would you determine this without knowing the angle?
The radius changes according to the angle think θ=180 or even 360 which would make the cone a line.

Post the solution from the textbook

Anonymous at Fri, 13 Sep 2024 03:44:03 UTC No. 16377990

>>16377953
the solution is a function of theta, dumbass

Anonymous at Fri, 13 Sep 2024 06:43:14 UTC No. 16378124

>>16377906
>all the algebraic numbers
How do you determine the cube root of 2?

Anonymous at Fri, 13 Sep 2024 11:41:42 UTC No. 16378372

>>16378124
Use a sphere.

Anonymous at Sat, 14 Sep 2024 02:32:44 UTC No. 16379478

>>16376839
Not a math major but I'm assuming cones probably have a formula that relates the angle to h like spheres have 4/3pi r^3. I'm assuming you're supposed to convolute them to see the relationship between the formulas.
But Idk how you'd come up with the largest r value without using software since you're supposed to filter the oversized r's resulting from large or small degrees. The question literally doesn't say anything that limits the angle.

>>16377990
Post it

Anonymous at Sat, 14 Sep 2024 05:05:17 UTC No. 16379609

>>16379478
nta, software, wtf?? This is just normal calculus.
There only like 3 cases to check at max (the first being simple af) and 2 cases at min (i dunno if you can combine the last two together). What makes the problem annoying is that you (prob) need to check the extra two cases. A basic problem only makes you check one.

At least a good amount of the work for the last two are similar. Just use >>16377402 to find the volumes

🗑️ Anonymous at Sat, 14 Sep 2024 06:34:13 UTC No. 16379657

>>16377326
Discontinuity is at [math]r = \frac{h\sin (x)}{\cos ^2(x)}[/math], retard.

Anonymous at Sat, 14 Sep 2024 06:56:51 UTC No. 16379666

>>16377326
Discontinuity is at [math]r = \frac{h\sin (\theta)}{\cos ^2(\theta)}[/math], retard.

Anonymous at Sat, 14 Sep 2024 07:23:15 UTC No. 16379672

>>16377088
I'd just give him the middle finger and buy enough paper cups and balls of different sizes and higher densities than water to figure it out the old fashioned way. Why kill yourself with math when you can just do and plug the math in in retrospect?

Anonymous at Sat, 14 Sep 2024 11:31:52 UTC No. 16379836

Any way to do this problem without solving an integral? I say this because once you have a formula for the volume, you need to derive it to find a maximum
Any way this derivative just cancels the integral?
Not even going to do any of them, it just struck me as one of these tricks you use to get fast results in gotcha problems

Anonymous at Sat, 14 Sep 2024 17:12:48 UTC No. 16380262

>both the ball and cone sizes can change
What the fuck is going on

Anonymous at Sat, 14 Sep 2024 17:16:52 UTC No. 16380266

>>16376839
h = ∞
∴ r = ∞

Anonymous at Sat, 14 Sep 2024 22:23:42 UTC No. 16380807

>>16377464
Wtf is theta

Anonymous at Sun, 15 Sep 2024 03:02:10 UTC No. 16381165

>>16376839
r = h*sin(theta)
t. intuition chad

Anonymous at Sun, 15 Sep 2024 03:55:05 UTC No. 16381242

>>16376839
lol sci can't even solve this shit. what a bunch of larpers

Anonymous at Sun, 15 Sep 2024 20:38:49 UTC No. 16382277

>>16381165
Not too bad, considering you spent no time on the problem.

Anonymous at Sun, 15 Sep 2024 21:02:04 UTC No. 16382290

>>16382277
did you fix the angle or let it vary as well. the problem is for a fixed theta

Anonymous at Mon, 16 Sep 2024 16:59:55 UTC No. 16383477

>>16382290
No, I've solved the problem numerically for varying h and theta(phi in my case, for whatever reason)

Anonymous at Mon, 16 Sep 2024 17:02:00 UTC No. 16383483

>>16383477
forgot to show phi

Anonymous at Mon, 16 Sep 2024 18:00:16 UTC No. 16383556

>>16379666
good for you. I didn't even want to work this through. I just give you the keys. You welcome noobs.

Anonymous at Mon, 16 Sep 2024 20:08:13 UTC No. 16383758

>>16383483
nice i solved it and got the same answer approx 0.41435. i made a mistake with center of sphere should be at z = R/sin(a) instead of z = Rsin(a) in my earlier post >>16377280

Anonymous at Mon, 16 Sep 2024 20:23:50 UTC No. 16383773

solution is one of the roots of this quadratic. git gud noobs

Anonymous at Mon, 16 Sep 2024 20:34:49 UTC No. 16383783

>>16383758
Damn I guess I'm retarded, i never figured out that height of the ball is just r/sin(theta), I did some vector fuckey.
But when I plug in your approximation, the sphere/circle high-point seems to be at same level as the cone rim. For cones with small theta this is about right, but for wider cones this is far from opt. Maybe I got confused somewhere...

Anonymous at Mon, 16 Sep 2024 21:23:44 UTC No. 16383819

>>16383783
i think i found the solution

Anonymous at Mon, 16 Sep 2024 21:40:41 UTC No. 16383836

>>16383773
Your roots are not real for pic related.
>>16383819
Something is still wrong, but its close.
I'll try to improve my solution tomorrow, since it obviously accumulates some error along my convoluted calculation

Anonymous at Tue, 17 Sep 2024 02:41:16 UTC No. 16384106

I haven't touched an integral in ages (and in fact I never passed calc lmao) but this doesn't seem too difficult.
looking at the way the volume of a ball is calculated (https://math.hmc.edu/calculus/hmc-mathematics-calculus-online-tutorials/single-variable-calculus/volume/) and using >>, I guess you could get it easily.