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2:20 AM
It is currently 7:20 PDT. I will be back in around 30-40 minutes
 
2:48 AM
I'm here, and will be for a while.
 
3:25 AM
@bobble Okay, you still here?
Just one question (well two) today
 
yep, here
sorry was chatting with Sciborg in main room
 
Yeah I saw
:P
Okay, I'll try to make it as quick as possible
A couple minutes to type the equation out
 
I'm here too, if that helps, although I'm also working on something in another tab
 
I will attempt to help with math, but most of the calculus I did in college has completely fallen out of my brain :p
 
Graph the function of v(theta)=(9/pi^2)*(theta)^2*sqrt(4pi^2-theta^2)
Let me paste the picture
 
3:33 AM
Have you tried the plugging-in-points method?
 
I'm trying to figure how to determine the shape
 
theta = 0, pi, 2pi, etc
 
Theta of 0 is 0
 
uhh hang on let me get some paper for this one
 
hold on, is 4pi^2-pi^2=3pi^2?
 
3:36 AM
I think I know what they want you to do
 
Is this a case of plugging in 0 and 2pi for theta and solving for each, or am i remembering wrong
 
This equation can be rewritten (using x instead of theta) as:
a * x^2 * sqrt(b - x^2)
the actual values of a and b don't matter
 
There are two factors that act separately: a * x^2 and sqrt(b - x^2)
 
3:39 AM
When one increases, what happens to the other?
 
To confirm, this is not supposed to be polar coord or anything right?
 
The other decreases
@Ankoganit Noooo no no
 
ah alright
 
vietnam flashbacks to polar coordinate calculus
 
decreases by a proportional amount
ish
 
3:40 AM
hm?
How?
 
when you increase one by a value, the other decreases by a distinct but constant value, no matter how much you increase by
I'm not sure I'm explaining this well
 
Ah okay I see
 
Simplifying further: x^2 * sqrt(b -x^2)
ish
clearer now?
 
@bobble this looks sketchy
nvm edited
 
3:41 AM
i'm confused as well
 
I put the b back in
 
@bobble Kind of
I'm not sure where you're going with this
 
How do you maximize the area of a rectangle when you have a constant perimeter?
(not the same, but a similar principle)
 
hold on processing notes
i forgot
 
Can you guess? Intuit your way to the answer?
 
3:44 AM
sure
let me thin
 
If you make it very long and skinny, with same perimeter, does that increase or decrease area?
 
So what shape would have the maximum area, for same perimeter?
 
Where the value of the length and the width are as close to each other as possible
 
Can you see how this applies to x^2 * sqrt(b - x^2)?
 
3:47 AM
let me see
Oh if we make this into a rectangle
just say l=x^2, and w=sqrt(b-x^2)
 
Again, it's not exact because the "sides" don't increase/decrease by the same amount
 
yeah
It's just to organize my thoughts
something like
b-x^2=x^2?
 
We're trying to figure out the shape, not exact numbers
 
Hm
Oh, we're finding the maximum, aren't we
 
Going back to the rectangle analogy, if you change any side lengths away from square, what happens to the area?
 
3:51 AM
One increases, the other decreases
 
what happens to the area of the rectangle?
 
It decreases
 
So if you made a graph, what would it look like? (length of one side on x, area of rectangle on y)
 
As x increases, y decreases
 
How about the overall shape of the graph, from the side being negligible length to being all the length?
 
3:56 AM
uh hm
What do you mean negligible length to being all the lengths?
 
length = 0 to length = perimeter / 2
 
I think the easiest way to solve this one is honestly just to solve for theta = 0, pi and 2pi and then look at the graph shape.
 
basically the extremes when the "rectangle" is essentially a line
 
@Sciborg I'm trying to solve this intuitively :P
Does it look like a sine graph?
 
That's fair :p
I was gonna post my solution where I just did it by that method
 
3:58 AM
What do you mean by "sine graph"? Does it increase? Decrease? If so, where? For how long?
 
So as x increases, y increases until halfway through, before decreasing at a proportional rate to how it increased (looking like an upside-down x^2 graph?). This would make sense, because, you would want to essentially minimize the lengths urm hold on
How do I explain this
 
You're exactly correct
 
You've basically got it!
 
So what does your original graph look like?
 
4:02 AM
uh..
Wait, is it just a sine graph?
It can't be 4(theta), we know that
 
It's not a sine graph because it doesn't to wavy thingy
 
At zero, it'll be zero. The graph of theta can be negative, which is odd
 
it's never negative
no factors in the equation can be negative - see ^2
 
Oh, right right
@bobble I forgot PEMDAS
Oh, there's an asymptote at 2(pi)
 
4:06 AM
no.... isn't it just 0?
 
Wait, yes
Sorry I was thinking x/0 from last unit :P
It would peak at pi
 
And -pi
Since ^2
 
because the halfway point between zero and 2pi is pi
 
no
 
@bobble No to...
 
4:09 AM
It peaks at theta=5.13 according to desmos
 
because increasing x^2 decreases sqrt(b - x^2) by less
 
????
 
because of the "b" offset for that term
 
4:10 AM
oh ffs
 
So the maximum point is biased towards higher x^2
 
yeah peak would not be at pi
i was confused for a sec there
 
Do I plug in pi for theta?
 
correction: sqrt(b - x^2) decreases by the same numerically but not the same relatively
 
pi was just what i happened to solve for, but yeah the actual peak is at 5.13
grumble grumble
 
4:11 AM
It's 9sqrt(3pi^2)
 
If you just want the shape all you have to realize is the peak is biased towards higher numbers
^^ the shape, for reference
 
ahh, it was goofier than i drew it as lol
 
Note how the x-axis is stretched out
because the y numbers get honkin' big
 
Wait but aren't we sketching things by theta and pi?
 
For sake of graphing theta = x and f(theta) = y
Because you've got that v(theta) which is your y
And pi is a constant, not a variable
 
4:14 AM
x axis in terms of pi
thank you, desmos
 
because your range is 0 < theta < 2pi, so you only want right half of graph.
mentally replace theta with x
 
booming voice from the heavens Does it all make sense now?
 
I'm not sure how I would get the value without graphing
 
You can solve to find the maxima and minima, that was why i solved for roots because roots are where y = 0
 
4:17 AM
urmmmm without calculous
 
Well, you could take the derivative and set it equal to 0 :) But that's calc
 
I was about to suggest that yeah lol. But it's calc
You can find maximum values without calc tho
 
yeah. i forgot how
 
Completing the square is how I learned in school
I don't know if everyone learned it that way though
 
unrelated to math: Sciborg, is it currently a bobblie or a blobcrown?
 
4:19 AM
It looks close enough now that it has graduated to bobblie status, it just needs some pointies
 
I'll ask my teacher on Monday. I'm confused on how to do it so
 
I'm trying to think of how to explain, I learned it a while ago and it's blurry
 
I'm sorta jealous that you can ask your teacher. Mine don't really have time for one-on-one help as much now.
 
@bobble Pfffft. I'm laughing so much right now
She doesn't teach which is why I'm basically asking here
 
but god i don't remember it being this weird and complicated
 
4:21 AM
@bobble But she at least goes over homework
I got to go. Thanks for all the help!
 
I think I got it! Remember, you said that the sides should be equal length for max area. So set the two terms (9 * x^2 / pi^2 and sqrt(4pi^2 - x^2)) equal to each other. Solve for x
 
Bobble you're a genius
 
that was from the puzzling ability North's bobblie has siphoned off
 
Of course
That's way easier than completing the square nonsense
 
4:24 AM
there's going to be x^4 and x^2 terms, which is nasty
 
Honestly if I had time I would teach you derivatives since it makes a lot of graph and algebra problems much easier :p
 
substituting y = x^2 and then quadratic formula should work
 
 
18 hours later…
10:31 PM
*cries*
 
me here
there there
pats North's bark
 
This is coming from the 2019 handbook for AP physics btw
"A police car of mass m moves with constant speed around a curve of radius R. (The car is, from your point of view, coming out of the page and is in the process of turning towards the left side of the page.) The car is moving as fast as it can without sliding out of control on the flat roadway to respond to an emergency. This maximum safe speed isv 0 . The coefficient of static friction between the car’s tires and the roadway is μ s ."
"Derive an expression for the maximum safe speed that the car can take the turn in terms of μ and R."
 
How far did you get?
 
No where. I have zero clue where to start
 
(also, my class is past this point so I already solved the problem)
 
10:33 PM
All I know is that (hold on flipping back)
v_1=v_0, right?
Because of the relation between for static friction and kinetic friction
 
Do you know what the difference is, and what the equations are, respectively?
 
Oh, I think I know where to start
 
.... then let the physics begin
 
@bobble The equation/relation is F_fr(static) =< F_fr(kinetic)
I think this is my first equation for derivation
 
Um, you need to relate friction to F_norm
Relating the two kinds of friction to each other isn't much use - only one is present
 
10:36 PM
Uh, I need to grab my equation sheet
 
In case you want to copy it, the coefficient of friction usually has the variable mu: μ
 
Oh smacks head
|F_fr| =< μ|F_norm|
 
restrains self from asking if trees have heads
@PrinceNorthLæraðr great! So how can we get F_norm?
 
Can you derive an expression for F_norm now?
 
10:40 PM
|F_fr|=<|μ|F_grav|
 
Can you re-write F_grav in terms of physical quantities/constants?
(and you did write down the intermediate steps, right?)
 
Hold on
Hurm..
I got to |F_fr|/|mg|=<mu. I'm not sure whereelse to go
Also, can you check my derivation, I'm not sure I'm doing it correctly because I've used up 5/6 spaces
 
> constant speed around a curve
^^ think about this while I check your work
 
omega=omega_0+at
 
> constant speed
 
I wouldn't have used quite so many absolute values (just declare signs!), and you should keep F_frict isolated
 
and hint: constant tangential velocity means there is uniform centripetal acceleration
 
a_c = ?
 
10:55 PM
Oh, what does c stand for?
circle?
 
centripetal acceleration
acceleration_centripetal
 
Oh, that's what that equation was oh my
a_c=v^2/r
 
can you use this?
 
Yes, because the car is turning at a constant speed around a curve
 
F_c will probably be more useful, but it requires a_c
 
10:58 PM
Wait let me derive this out
Now that you've told me about F_c
Ohh, okay
So F_c=a_c*m
F_net=F_c-F_fric=a_c*m
 
F_c is not a force by itself, it is a net force
 
Oh whoops
sorry
 
common mistake
 
Oh, is it F_c=(a_c*m)-F_fr?
 
1 min ago, by bobble
F_c is not a force by itself, it is a net force
 
11:03 PM
Hng
 
What force(s) make(s) up F_c?
(which force(s) point in that direction on a FBD?)
 
am reading the transcript and laughing at deus's "here is a garbage picture"
 
@bobble Uh... uh...
 
Back up a bit. What forces exist?
 
Force of friction, force of gravity
 
11:05 PM
and?
 
Urm.. normal force?
 
and which directions do they point?
 
Wait, that's it?
 
that's all the ones that matter
I mean, there's also air resistance, but you can usually just ignore it
 
Huh. That's all I need on the FBD? I guess I misunderstood how to draw it
Let me fix the FBD
Normal goes up, gravity goes down, friction goes right
 
11:08 PM
Not exactly right, it goes inwards
 
How do I draw that on a fbd?
 
Well, on the FBD it's right
 
but going inwards matters when you consider which force(s) contribute(s) to F_c
hint hint
 
Okay processing
Oh, since the only force that's acting is friction, the body will turn to the left. Trying to explain it in a way that makes sense
It's not moving to the right, but inward. That means that it's really acting on itself to turn?
 
11:12 PM
correction: since it's turning to the left, friction is to the left on FBD, since that is inwards
 
Oh
So, F_c=F_net=F_g+F_norm+F_frict?
 
F_c is only composed of forces that point inwards/outwards
Which forces point inwards/outwards, and which point in other directions?
 
F_fric, inwards, towards the left
 
So F_c = ?
 
11:14 PM
yes
 
14 mins ago, by Prince North Læraðr
So F_c=a_c*m
Can you combine that with this?
And the inequality F_frict <= μmg
 
F_fric=a_cm; (mu)|mg|<=a_cm. It also asks the relationship between the radius though so
 
did formatting mess up your equations?
 
(mu)|mg|<=(v^2/r)*m
Not really
I forgot the "<"
 
11:17 PM
a_c = v^2/R, remember
 
Ah, right
Wait so, I have only six spaces. How do I put ALL of that?
 
mv^2/R <= μmg is what I'm trying to steer you to
 
Urm, that's what I did ^^
 
ah I got confused with your equations
 
But like there are only six spaces on the answer sheet. This is like 10 steps or something
 
11:20 PM
for right now, finish it off by isolating v on one side
 
v<=sqrt(μRg)
Oh wow, that's neat
The masses cancel out
 
Okay, it's a terrible picture
But that's how I fit it into 6 rows
 
Ohh, you put all the variable moving in one box? That's smart
 
Can you tell what's happening?
 
Kind of. Nice handwriting
 
11:22 PM
sarcasm?
 
Yes. I thought you said you had good handwriting?
 
If I care. The teacher grades this for completeness, not accuracy. So I rush the handwriting
 
Ah
Okay, I'm going to do my derivation and look over the other questions and see if I'll need help
Thank you~
 
see you later!
 

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