It is currently 7:20 PDT. I will be back in around 30-40 minutes

I'm here, and will be for a while.

@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

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?

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

the actual values of a and b don't matter

huh?

There are two factors that act separately: a * x^2 and sqrt(b - x^2)

Right

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

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?

Hold on

@bobble this looks sketchy

nvm edited

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?

sure

let me thin

If you make it very long and skinny, with same perimeter, does that increase or decrease area?

decrease

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)?

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?

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?

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

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!

yeah

So what does your original graph look like?

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)

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...

It peaks at theta=5.13 according to desmos

waa

because increasing x^2 decreases sqrt(b - x^2) by less

????

because of the "b" offset for that term

oh

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

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

x axis in terms of pi

thank you, desmos

oh?

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

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?

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

i think this is it maybe? mathsisfun.com/algebra/completing-square.html

but god i don't remember it being this weird and complicated

@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

Ah!

Nice

Bobble you're a genius

that was from the puzzling ability North's bobblie has siphoned off

Of course

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

*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)

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

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?

F_grav

Can you derive an expression for F_norm now?

|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

Oh, okay

and hint: constant tangential velocity means there is uniform centripetal acceleration

thinking

a_c = ?

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

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?

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

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

Not exactly right, it goes inwards

How do I draw that on a fbd?

Well, on the FBD it's right

Oh, okay

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?

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 = ?

F_fric?

yes

Oh wow

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 "<"

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

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

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!