/sci/ - Science & Math

Holy heck! In which Semester do you learn that in a european Uni?

>>10988740

High school, anon

>>10988752
FUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUCK!!

>>10988755

If you're born poor and American, you're essentially destined to never make it unless you're an anomalous brainchad. Our public k-12 system produces good low skill wage slaves, not academics.

>>10988794
The gif is meant to imply that even though I've been hobbled from day 1 by a shitty education, sowing only acrimony against all things mathematical, I remain resolute that I will eventually understand Topology competently. Sorry if you missed that.

>>10988740
elementary school. that's why the call it elementary maths.

>>10988802

Totally possible, but forget about ever competing with well prepared Chinamen and Yurops. They're starting off 500 steps ahead of you, and while you can certainly still finish the race, you probably won't do so quicker than them.

>>10989032
I'm not interested in math to get laid, noticed or make a lot of money. What kind of retarded plan would that be?

>>10988794
Can you solve this problem?

>>10988723
What can you use? Calculus?

>>10988755
not true lmao

>>10989108
Geometry and Algebra

I accidentally calculated the minimum value because I can't read, but that should be around
1.587

>>10988802
>that anon who wants to learn about wacky ultradimensional shape math!

Face it dude, you're just as much interested in math for the wrong reasons as any other number of people. There are a lot of boring and difficult calculations describing those shapes and you probably have no real interest in topology.

>>10989119
It is, this is babby linear algebra.

>>10989181
Can we get steps ?

Intuitively , I'd assume the answer would come from using the vector halfway between the normal vectors of the planes, and taking antipodal positions on the sphere. Probably a harder way to do it is setting up equations from projections and such, but I won't bother with that. Anyway, I'm getting 128/7.

>>10989181
Setting P=Q would yield zero so I doubt you're correct. Nothing says they must be on opposite sides, only that they must be on the sphere.

I have a theory that math and physics teachers desperately try to make the subject a lot more complicated than it actually is by writing explanations and problems in the most autistic way possible.

I remember my partial differential equations book in University was like written in ancient runes. All the professor did was literally copy everything from the book on the blackboard.

Found an American professor who explained the things online and it all made sense. It’s easy as fuck but these people make it seem like some genius shit and I never understood why

>>10989181
answer is 24

>>10989291
Professors arent teachers. They are good at what they have a degree in, but that doesnt make them good at teaching.

>>10989340
look up the definition of a professor. a teacher isn't inherently a good teacher.
>>10989291
it's written with the expectation that your prerequisite classes taught you to be an autist already.

>>10989100

Zero. Set P = Q.

>>10989291
this. I honestly think math people invented these autism symbols and shit just to circle jerk and make themselves feel smart because they can describe shit people already know intuitively using the greek alphabet

>>10988723
I took college geometry...
I should be able to do this cleverly using some clever cross ratio and spherical geometry...
Instead I just turn it into obnoxious and autistic calculus problem,..
But in real world, when I become professional mathematician, I let interns handle it.

>>10989650
>autistic symbols
The alternative is lengthy paragraphs and 20 pages for 1 problem.
>shit people already know intuitively
That's arrogant and ignorant. I bet that I can find an example of 3rd grade level math that you don't know intuitively.

>>10989291
>the professor teaches by copying from the book verbatim
That's because the book is perfect. He's mocking you for wasting money on uni instead of autodidacting. Also he wants the class brainlets to come to his office hours so he knows how much to dumb down his tests by (otherwise too many students will fail and he'll get shit for it.)

>>10989843
incredibly based.

Pretty late to this thread but this is an obvious case of lagrange multipliers

>>10989908
>muh lasagna multipliers

>>10988723
The answer should be zero

>>10989292
The answer is 6 you absolute brainlet!

>>10989292
The answer is 6 you brainlet. When PQ = <0, 1, sqrt(3)>

>>10988723
can anyone find a solution to tis i,m going insane

>>10991332
plz , for the love of god

>>10991390
see
>>10989908
set up an equation using the definitions given and maximize it using Lagrange multipliers. It's similar to the finding the extrema of functions with single variable equations using derivatives, but have more terms and operators, as well as constraints
see the Khan Academy materials (https://www.khanacademy.org/math/multivariable-calculus/applications-of-multivariable-derivatives/constrained-optimization/a/lagrange-multipliers-single-constraint)) and videos on it

>>10989137
It doesnt even need an algebra

A lot of replies on this thread. But I see no right answer. /sci/ can't be this dumb can it?

>>10992874
Do it yourself if you care so much. It's not that hard. Just a calc 2/3 problem that's probably quite lengthy. Nobody here wants to go back to solving a massive integral for (yous) on a Japanese dating forum. We're all past that in our intellectual careers. It's the agony of doing something objectively below us.

>>10992951
>>10990266
>>10989612
>>10989181
brainlet

>>10992951
>>10991984
I want an elegant solution

>>10993512
USing hish school math , vector relations or something like that

>>10989291
True that uni professors focus a lot on proofs, theorem and formalization, instead of having a pedagogic approach

>>10988723
>>10989100
>>10989181
>>10989244
>>10989292
>>10989785
>>10989908
>>10990266
>>10991332
>>10991984
>>10992874
>>10992951
>>10993512
>>10993517
>The plane [math]y+\sqrt{3}z+8=0[/math]
The absolute state of /sci/.

>>10993554
Answear is 24

you brainlet

>>10993554
Elaborate you piece of shit

>>10988794
>>10988752
what are you, retarded? this is a 6th grade geometry problem.

>>10993554
What's your problem with the plane?

>>10993554
x variable is 'missing' >>10993649
this could be a line if it was just the plane yz, but is the space xyz, so the plane is a linear equation of the three variables. This equation is 0x+y+sqrt3z+8=0 and still is a plane. There is no problem with the scalar 0.

>>10988752
>>10993736
I know you guys are memeing but isn't this just an optimization problem with a couple of analytic geometry thrown in?

>/sci/ defeated by an entry level course problem
LMAO the absolute state

>>10988723
>>10988723
Pleeez for the love of god., Somebody solve this with steps , i am going insane

Is this a high school problem or does it require some basic undergrad math knowledge?
Just enrolled at uni and I'm not sure I could solve this

I'm not gonna let this thread die till someone solves this

This question was made by simple mathmatica conception

But having knowledge of mathmatica conception is not enough to slove this problem

because these problem's real purpose is secretly checking your IQ

This question hurts my brain not because the problem is hard but because the way it is physically written on the paper is disorienting. There's missing words, capitals in the middle of the sentences, margins changing in the middle of a line, inconsistent spacing, no periods, and the font changes. It's the written equivalent of listening to someone who had a stroke.

Draw a circle which goes around the equator of the sphere. That circle should go through two points in particular: the place where the sphere is closest to plane 1, and where it's closest to plane 2.

Intuitively, the two points must be on this circle. That's how you'll minimize the "shadow".
Of course there's probably ways to prove this, but we can just say that intuitively.

Now we can also intuitively say that this problem is symmetric. So find the location on the circle which is halfway between those two points which the circle is running through (the points closest to the two planes), P and Q must be an even angle away from this middle place. Thus, we can define the position of P and Q by a single factor: an angle theta, where Q is located at theta, and P is located at negative theta.

Ok so how are you gonna do this? Well basically you will get PQ in terms of theta, P1-Q1 in terms of theta, and P2-Q2 in terms of theta, then take the derivative with respect to theta and set to 0. Then solve for theta.

That's how you're gonna do this.
Of course there are more rigorous ways to prove everything I just said above, but that's what you're going to end up finding in the end.

>>10994664
this is basically "optimize this function, given this input constraint". If you know multivariable calculus, you can solve this easily.

First of all, let's draw two lines which are perpendicular to each plane respectively, and run through the center of the sphere. Then find the angle between these lines.

Remember how to get a perpendicular vector knowing a plane?
So for the first plane, the perpendicular vector will be in the [math]\hat{\textbf{y}}[/math] direction, and for the other plane the perpendicular vector will be in the [math]\hat{\textbf{y}} + \sqrt{3} \hat{\textbf{z}}[/math] direction.

Defining
[math]\hat{\textbf{n}}_1 = \langle 0, 1, 0 \rangle[/math]
[math]\hat{\textbf{n}}_2 = \langle 0, \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle[/math]

the angle between them will be: [math]cos^{-1} (\hat{\textbf{n}}_1 \cdot \hat{\textbf{n}}_2)[/math]
thus, the angle between them
[math]\alpha = \frac{\pi}{3}[/math]

Let's put our origin at the center of the sphere, and define two arbitrary vectors leading out from the center. The only condition these vectors have is that their total length = 2.
Thus:
[math]\vec{\textbf{v}}_1 = \langle a_1, b_1, c_1 \rangle[/math]
[math]\vec{\textbf{v}}_2 = \langle a_2, b_2, c_2 \rangle[/math]
where
[math]\left(a_1^2 + b_1^2 + c_1^2\right)^{\frac{1}{2}} = 2[/math]
[math]\left(a_2^2 + b_2^2 + c_2^2\right)^{\frac{1}{2}} = 2[/math]

The vector between the two is:
[math]\vec{\textbf{d}} = \langle a_2-a_1,b_2-b_1,c_2-c_1 \rangle[/math]
with a squared length:
[math] \left| \overrightarrow{PQ} \right|^2 = (a_2-a_1)^2 + (b_2-b_1)^2 + (b_2-b_1)^2[/math]

>>10995612
The projection of a vector onto a plane can be found by subtracting the component of that vector which is perpendicular to the plane.
to find this component, do the dot product with the normal vector in each case.
[math]\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1= b_2 - b_1[/math]
[math]\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2 = \frac{1}{2}(b_2 - b_1) + \frac{\sqrt{3}}{2}(c_2-c_1)[/math]

[math]\left| \overrightarrow{P_1 Q_1} \right|^2 = \left( \vec{\textbf{d}} - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)\hat{\textbf{n}}_1 \right) \cdot \left( \vec{\textbf{d}} - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)\hat{\textbf{n}}_1 \right) = d^2 - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2[/math]
[math]\left| \overrightarrow{P_2 Q_2} \right|^2 = \left( \vec{\textbf{d}} - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)\hat{\textbf{n}}_2 \right) \cdot \left( \vec{\textbf{d}} - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)\hat{\textbf{n}}_2 \right) = d^2 - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)^2[/math]

Thus
[math]2\left| \overrightarrow{P Q} \right|^2 - \left| \overrightarrow{P_1 Q_1} \right|^2 - \left| \overrightarrow{P_2 Q_2} \right|^2 = 2d^2 - (d^2 - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2) - (d^2 - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)^2) = (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2 + (\vec{\textbf{d}} \cdot \hat{textbf{n}}_1)^2[/math]

So we are really trying to maximize [math](b_2 - b_1)^2 +\left( \frac{1}{2}(b_2 - b_1) + \frac{\sqrt{3}}{2}(c_2-c_1)\right)^2 = \frac{5}{4}(b_2 - b_1)^2 + \frac{\sqrt{3}}{2}(b_2 - b_1)(c_2-c_1) +\frac{3}{4}(c_2-c_1)^2 [/math]

We can reason that we want to maximize [math]b_2, c_2[/math], and minimize [math]c_1, b_1[/math].

We know that
[math]c_2 = \sqrt{4 - a_2^2 - b_2^2}[/math], [math]c_1 = \sqrt{4 - a_1^2 - b_1^2}[/math]

>>10995616
So plugging this into our equation, we find:
[math]m = \frac{5}{4}(b_2 - b_1)^2 + \frac{\sqrt{3}}{2}(b_2 - b_1)(\sqrt{4 - a_2^2 - b_2^2}-\sqrt{4 - a_1^2 - b_1^2}) +\frac{3}{4}(\sqrt{4 - a_2^2 - b_2^2}-\sqrt{4 - a_1^2 - b_1^2})^2[/math]

Assert that we want to find
[math]\frac{\partial}{\partial b_2} m = 0, \frac{\partial}{\partial b_1} m = 0, \frac{\partial}{\partial a_2} m = 0, \frac{\partial}{\partial a_1} m = 0[/math]

[math] \frac{\partial}{\partial a_2} m = 0 = -\frac{3}{2}a_2\frac{(b_2-b_1)}{\sqrt{4 - a_2^2 - b_2^2}} - \frac{3}{2}a_2\frac{\sqrt{4 - a_2^2 - b_2^2}-\sqrt{4 - a_1^2 - b_1^2}}{\sqrt{4 - a_2^2 - b_2^2}}[/math]
[math]a_2 = 0[/math]
[math] \frac{\partial}{\partial a_1} m = 0 = \frac{3}{2}a_1\frac{(b_2-b_1)}{\sqrt{4 - a_1^2 - b_1^2}} + \frac{3}{2}a_1\frac{\sqrt{4 - a_2^2 - b_2^2}-\sqrt{4 - a_1^2 - b_1^2}}{\sqrt{4 - a_1^2 - b_1^2}}[/math]
[math]a_1 = 0[/math]
[math]\frac{\partial}{\partial b_2} m = 0 = \frac{5}{2}(b_2-b_1) + \frac{\sqrt{3}}{2}\left(\sqrt{4 - b_2^2}-\sqrt{4 - b_1^2}\right) -\frac{\sqrt{3}}{2}b_2\frac{(b_2-b_1)}{\sqrt{4-b_2^2}}-\frac{3}{2}b_2\frac{(\sqrt{4-b_2^2}-\sqrt{4-b_1^2})}{\sqrt{4-b_2^2}}[/math]
[math]\frac{\partial}{\partial b_1} m = 0 = \frac{5}{2}(b_2-b_1) - \frac{\sqrt{3}}{2}\left(\sqrt{4 - b_2^2}-\sqrt{4 - b_1^2}\right) +\frac{\sqrt{3}}{2}b_1\frac{(b_2-b_1)}{\sqrt{4-b_1^2}}-\frac{3}{2}b_1\frac{(\sqrt{4-b_2^2}-\sqrt{4-b_1^2})}{\sqrt{4-b_1^2}}[/math]

I've probably made an algebra or differentiation error somewhere here. So from here on out so I won't continue to solve any further. But basically you have two equations, two variables. Solve for [math]b_1[/math] and [math]b_2[/math], then go back and plug in to get [math]c_1[/math] and [math]c_2[/math].
From here it is trivial (though not "easy")

>>10995616
> do the dot product with the normal vector in each case.
d ⋅n^1=b2−b1
d ⋅n^2=12(b2−b1)+3√2(c2−c1)

You forgot to divide by the length of the normal vector

>>10996067
it's a unit vector, length 1
that's why there is the hat symbol

>>10995620
Good work anon ,but there has to be a simpler expression ?
Btw you did not use the angle between the planes

>>10996118
you could do what this anon was saying: >>10995190

So the way you would do this:

[math]m = (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2 + (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)^2[/math]
assume:
[math](\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1) + (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)[/math]

since it is symmetric, the dot product doesn't change it's angle no matter how far apart the two points are moved
[math]\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1 = d \cdot cos(\frac{pi}{6}) = d\frac{\sqrt{3}}{2}[/math]
therefore
[math]m = \frac{3}{2}d[/math]

so when you maximize "d", you maximize "m".
The sphere has a radius of 2, thus the largest you can make "d" is 4

so it should be
[math]m = 6[/math]

of course you're just assuming a symmetrical answer there. I'm sure you can reason it out though.

>>10996118
you could do what this anon was saying: >>10995190

So the way you would do this:
[math]m = (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2 + (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)^2[/math]
assume:
[math](\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1) = (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)[/math]

since it is symmetric, the dot product doesn't change it's angle no matter how far apart the two points are moved
[math]\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1 = d \cdot cos(\frac{pi}{6}) = d\frac{\sqrt{3}}{2}[/math]
therefore
[math]m = \frac{3}{2}d[/math]

so when you maximize "d", you maximize "m".
The sphere has a radius of 2, thus the largest you can make "d" is 4

so it should be
[math]m = 6[/math]

of course you're just assuming a symmetrical answer there. I'm sure you can reason it out though.

>>10996118
you could do what this anon was saying: >>10995190

So the way you would do this:
[math]m = (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2 + (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)^2[/math]
assume:
[math](\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1) = (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)[/math]

since it is symmetric, the dot product doesn't change it's angle no matter how far apart the two points are moved
[math]\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1 = d \cdot \cos(\frac{pi}{6}) = d\frac{\sqrt{3}}{2}[/math]
therefore
[math]m = \frac{3}{2}d^2[/math]

so when you maximize "d", you maximize "m".
The sphere has a radius of 2, thus the largest you can make "d" is 4

so it should be
[math]m = 24[/math]
of course you're just assuming a symmetrical answer there. I'm sure you can reason it out though.

One way you could "prove" the symmetric answer is correct is to do this:

[math]\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1 = d \cdot \cos(\theta)[/math]
[math]\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2 = d \cdot \cos(\frac{\pi}{3} - \theta)[/math]

so
[math]m = d^2\cos^2(\theta) + d^2\cos^2(\frac{\pi}{3} - \theta)[/math]
then say that [math]\frac{\partial}{\partial \theta} m = 0[/math]

[math]0 = 2d^2(-\cos(\theta)\sin(\) + \cos(\frac{\pi}{3} - \theta)\sin(\frac{\pi}{3} - \theta))[/math]
[math]\cos(\theta)\sin(\theta) = \cos(\frac{\pi}{3} - \theta)\sin(\frac{\pi}{3} - \theta)[/math]
and this is true when [math]\theta = \frac{\pi}{6}[/math]

There, that should be sufficient.

SUMMARY

1) Find the (normalized) normal vectors to the planes
[math]\hat{\textbf{n}}_1 = \langle 0, 1, 0 \rangle[/math]
[math]\hat{\textbf{n}}_2 = \langle 0, \frac{1}{2}, \frac{\sqrt{3}}{2} \rangle[/math]
angle between them:
[math]\alpha = cos^{-1} (\hat{\textbf{n}}_1 \cdot \hat{\textbf{n}}_2) = \frac{pi}{3}[/math]

2) Find the projections of the vector "[math\vec{\textbf{d}}][/math], between P and Q
[math]\overrightarrow{P_1 Q_1} = \vec{\textbf{d}} - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)\hat{\textbf{n}}_1[/math]
[math]\overrightarrow{P_2 Q_2} = \vec{\textbf{d}} - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_2)\hat{\textbf{n}}_2[/math]

The length of these will be:
[math]\left| \overrightarrow{P_1 Q_1} \right|^2 = d^2 - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2[/math]
[math]\left| \overrightarrow{P_1 Q_1} \right|^2 = d^2 - (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2[/math]

Therefore:
[math]2\left| \overrightarrow{P Q} \right|^2 - \left| \overrightarrow{P_1 Q_1} \right|^2 - \left| \overrightarrow{P_2 Q_2} \right|^2 = (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2 + (\vec{\textbf{d}} \cdot \hat{\textbf{n}}_1)^2[/math]

3) Establish a basis. Assume [math]\hat{\textbf{n}}_1, \hat{\textbf{n}}_2[/math] are in the x-y plane. Split [math]\vec{\textbf{d}}[/math] into two pieces, call one [math]\vec{\textbf{p}}[/math] in the x/y plane, and one [math]\vec{\textbf{r}}[math] in the z direction.

[math]\hat{\textbf{n}}_1 \cdot \vec{\textbf{r}} =0 ; \hat{\textbf{n}}_2 \cdot \vec{\textbf{r}} =0[/math]
therefore, calling the total "m"
[math] m = (\hat{\textbf{p}} \cdot \vec{\textbf{n}}_1)^2 + (\hat{\textbf{p}} \cdot \vec{\textbf{n}}_2)^2[/math]

Since [math]d^2 = p^2 + r^2[/math], p is maximized when r =0.

>>10996326
r should be in the x direction ?