I don't even know what this means LMAO (exam prep)

[eager cove]

.

[noble spruce]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[eager cove]

@opaque belfry

[opaque belfry]

ohh

I don't remember how to work with points that are in middle like this

imma try to seach and see

[eager cove]

it's so fine if u don't

ill wait for someone else

[opaque belfry]

Oooooo wait

we can do what we did before

place x and y coordinates in place of x and y

[opaque belfry]

lets start with this one

[eager cove]

I already did that 1

lmao

[opaque belfry]

$y = ax^3 + bx^2 + cx + d$

[trail tree]

Akki#gtfoanemia#2763'sOwner

[opaque belfry]

ayyy noicee

[eager cove]

I need the bottom two

[opaque belfry]

which one we gotta do now

oo oki imma see those

[eager cove]

but I didn't know if we had to use it

like idk what its even asking lmao

[opaque belfry]

hmm we don't have enough data to approximate it ?

we can have it like A and B are turning points

imma try graphing it

[eager cove]

I found what a b c d

all equal

and the local max and min

[opaque belfry]

oo no not that one, i mean the bottom 2

AX and YB

[eager cove]

I didn't know if they were relevant but thats whta the first two questions asked

[opaque belfry]

yee I don't think they were

OOOO wait

they were

because A and B are same

[eager cove]

well

[opaque belfry]

well still not alot of data

[eager cove]

a = 9/10000
b = -13/100
c = 17/4
d = 60

local max (20.869, 100.256)
local min (75.428, 27.174)

*3 decimal places

[opaque belfry]

what was the cubic function you found using it 👀

[eager cove]

the derivative = 27x^2 - 2600x + 42500 / 10000

well

[opaque belfry]

9x^3 - 1300x^2 + 4.25x + 60 is the function then

[eager cove]

well

I found a b c d

than found the derivative of y = ax^3 + bx^2 + cx + d

subbed in a b c and d into that

[opaque belfry]

that doesn't seem right

[eager cove]

and got this

the function was

[eager cove]

.

this isn't it

[opaque belfry]

oh

whats the function tho

not the derivative

[eager cove]

isn't that the function

y = ax^3 + bx^2 + cx + d

[opaque belfry]

yee

what is a,b,c,d

[eager cove]

.

a = 9/10000
b = -13/100
c = 17/4
d = 60

[opaque belfry]

ooo

[opaque belfry]

does that function come back down tho

[eager cove]

idk what u mean by that

waittt

[opaque belfry]

oo issoke issoke

got it

its right

[eager cove]

is that of any help

question 4 was the next 1 after that

[opaque belfry]

oo no no yours was right

got it

[eager cove]

this was question 3 and 4 idk if they are any help

[opaque belfry]

lets start with our equation for approximating first vertex

[eager cove]

https://cdn.discordapp.com/attachments/1166267628764024842/1166345030349369434/IMG_1868.jpg?ex=654a2674&is=6537b174&hm=9825455af1056ebb69cb46e81b10b4d1cc50004da4f65bb0cd341fe804a4af44&

[opaque belfry]

$y = px^2 + qx + r$

[trail tree]

Akki#gtfoanemia#2763'sOwner

[opaque belfry]

this is for approximating AX

local maxima of AX was

@eager cove do you remember coordinates of vertex of quadratic 👀

[eager cove]

what is vertex?

[opaque belfry]

maxima or minima

quadratics have only 1

[eager cove]

I just found the local max and min

[opaque belfry]

yee forget that part for now

[opaque belfry]

do you know how to find maxima / minima of this ?

in terms of p,q,r

[eager cove]

nope

[opaque belfry]

maxima/minima is at point where derivative is 0 right

[eager cove]

yep

[opaque belfry]

so differentiating it

2px + q = 0

x = -q/2p

right ?

[eager cove]

yep

[opaque belfry]

what would y at that point be

$y = p(\frac{-q}{2p})^2 + q(\frac{-q}{2p}) + r$

[trail tree]

Akki#gtfoanemia#2763'sOwner

[eager cove]

ngl

idk what any of this is

I have not learnt it lmao

[opaque belfry]

you have

[opaque belfry]

this part you have

[eager cove]

nope lmao

[opaque belfry]

this is from quadratic equations

[eager cove]

quadratic equatons

[opaque belfry]

yee

[eager cove]

never learnt it lmao

[opaque belfry]

oh

you might have learnt it in some other name then

equations that have maximum power of 2

ax^2 + bx + c = 0

[eager cove]

yeah ik thats a quadratic

[opaque belfry]

we just finding its maximum

[opaque belfry]

well did you understand it till here ?

@eager cove 👀

[eager cove]

kind of

[opaque belfry]

oo okie

now lets simplify it

[eager cove]

so what are we actually finding

AX

[opaque belfry]

maximum/minimum of a quadratic

just coordinates

and then we will use that to find a quadratic for AX

$y = \frac{pq^2}{4p^2} + \frac{-q^2}{2p} + r$

[trail tree]

Akki#gtfoanemia#2763'sOwner

[opaque belfry]

right

[eager cove]

k

[opaque belfry]

$y = \frac{q^2}{4p} + \frac{-2q^2}{4p} + \frac{4pr}{4p}$

[trail tree]

Akki#gtfoanemia#2763'sOwner

[opaque belfry]

is this right

seems right

yee probably

[eager cove]

idk lol

[opaque belfry]

$y = \frac{-q^2+4pr}{4p}$

[trail tree]

Akki#gtfoanemia#2763'sOwner

[opaque belfry]

that seems right

okie

soo

[trail tree]

Akki#gtfoanemia#2763'sOwner

Akki#gtfoanemia#2763'sOwner

[opaque belfry]

okie @eager cove did you get how we got till here ?

and do you know which x,y are these

the ones we found

[eager cove]

max?

[opaque belfry]

yup

or min

it could be any of those

now lets plaxe values of x and y

from

max first

cuz AX

20.869 = -q/2p

100.256 = (-q^2 + 4pr)/4p

$\frac{100.256}{20.869} = \frac{-q^2 + 4pr}{-2q}$

[trail tree]

Akki#gtfoanemia#2763'sOwner

[opaque belfry]

is that right

[eager cove]

im so lost

LOL

[opaque belfry]

you understood till here right ?

[eager cove]

I'm trying to finish my english as it's due in 27 mins and also try to understand this LOL

[opaque belfry]

did you understand this ?

[opaque belfry]

oo lets do this later then

[eager cove]

I appreciate that, can we do it tmr instead when I'm in a fresh mindset I've been doing hw all day

[opaque belfry]

yeee good idea

I will rest too today

[eager cove]

I do really appreciate the time you take to help me and make sure I understand you have a good rest of your night / evening

[opaque belfry]

thank youu, you too

[eager cove]

yo @opaque belfry lmk when your free

[finite pier]

is this the problem you were having issues with? it's alright if you were in the process of reposting it- but, i can try my best to work through it

[eager cove]

uploading atm

lemme fix them up

this is what i Don't know how to do

[finite pier]

ok sorry it's taking me a minute- my chromebook screen isn't the greatest resolution and i'm trying my best to read all these to get a gist of the problem :P

what level of calculus is this btw?

[eager cove]

idk it's just year 11 methods in australia

lmao

[finite pier]

ah gotcha
well then i gotta ask. when the last problem says "approximate" does it want exact answers, or will the teacher take answers that are close enough?

[eager cove]

which question

[finite pier]

4

[eager cove]

im guessing 2 decimal places

i don't really know

or exact answers

idk we have never done this before so idk what the teacher expects

[finite pier]

hmm... because to me, an "approximation" makes it seem like you have to come up with things that will work for the specific instance, and not some concrete answer.

my best expectation would be, you have two quadratics connected by a line. a quadratic from point A to point X, a line from point X to point Y, and a quadratic from point Y to point B.

and you want the transition between them to be smooth AKA differentiable, such that the derivative the quadratic AX going into point X equal the derivative of the line XY exiting X, and vice-versa for XY entering Y and YB leaving Y.

[eager cove]

idk 1 word u just said ngl

like I do but idk what to actually write

[finite pier]

ok- well, what i'm trying to say is-

wait does this server have tex typescript?

[eager cove]

yeah I think

[finite pier]

you think you can get me a better resolution photo of this page? i bet it's very similar to what i'm trying to do but i can't quite see what you did, nor what the problem is ACTUALLY saying

[eager cove]

umm I can try

Does this work ?

if you click on it, it will make it more clear

[finite pier]

yea it's like 300x400, maybe just internet things-

okay, regardless i'll try my best

thoguh i do see you computed an initial quadratic between A and B for question 2, correct?

[eager cove]

idk lmao, a friend helped me

[finite pier]

okay, well from what it looks like

you ultimately want to create piecewise equations that are similar to the cubic function

[eager cove]

idk lmao

[finite pier]

in question 2, it looks like you were able to create two functions f and g such that

$$\displaystyle \lim_{x \to \infty^- }f'(x)=\lim_{x \to \infty^+ }g'(x)$$

where f(x) was your cubic and g(x) was a quadratic

wait

not infinity

bru

$$\displaystyle \lim{x \to B^- }f'(x)=\lim{x \to B^+ }g'(x)$$

[trail tree]

BeRu_t

[finite pier]

i mean it didn't format right- ugh discord is weird with texscript

$$\displaystyle \lim_{x \to \infty^- }f'(x)=\lim_{x \to \infty^+ }g'(x)$$

WAIT THAT'S STILL WRONG- ok nevermind

sorry for my suffering pffff

[eager cove]

all good

[finite pier]

but yeah, with question 2, it looked as if you connected two unrelated graphs at a point, such that their derivatives (or gradients as your book calls them) were equal at that point

[eager cove]

all I know I did was in question 2 was get equations to find a b and k and subbed them in each other

[eager cove]

I guess in the mean time, how do I figure out if any of these lines are greater than 60 degrees

[finite pier]

$$tan(\frac{\mathrm{d}f(x)}{\mathrm{d}x})>\sqrt{3}$$

[trail tree]

BeRu_t

[eager cove]

wtf is that 💀

[finite pier]

okay, do you know how to find an equation for the slope of a line at a point x?

[eager cove]

I forgot 💀

y - y1 = m(x-x1) ?

[finite pier]

nah

$$\displaystyle \lim_{\Delta x \to \0} \frac{f(x+\Delta x)-f(x)}{\Delta x}$$

[trail tree]

BeRu_t
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[finite pier]

please, format correctly

that.... was close-

okay, so if this doesn't ring a bell

i'll explain the equation really quickly

basically, it's a limit. it's approximating a value, as a given value, in this case Delta x, approaches (what is supposed to be a) zero

it's taken from the slope equation, basically $$\frac{y_2-y_1}{x_2-x_1}$$ where the y is f(x) and &&x_2=\Delta x+x_1

[trail tree]

BeRu_t
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[finite pier]

bru i msessed up the compiling and it still worked, that's nice at least lol

[eager cove]

so what would I actually input to prove y=0.02x^2 {0 \le x \le 40}

[finite pier]

wait. i think i misinterpreted- what exactly are you attempting to prove with this question?

[finite pier]

okay, so we want to find when the angle of x is greater than 60 degrees, right?

well whe f(x) is

[eager cove]

well I have to prove that these lines

1. Are continuous at all transition points
2. Start and end at same height
3. Never gets as high as the previous point
4. Ascent or decent no more than 60 degrees from horizontal

[finite pier]

1. continuous at all transition points, basically means does f1(x) = f2(x) at x=n

[eager cove]

so how would I do the first 1

[finite pier]

so, the first point appears to be between your first and second equations, and they both meet at x=40

so, does f1(40) = f2(40)?

where f1 and f2 refer to the first two equations

[eager cove]

so I sub 40 into

or just write that?

f1(x) = f2(x) at x = n

f1(40) = f2(40) so it's continuous

[finite pier]

well, they want you to test it for all points where the two graphs meet

[finite pier]

yes, now do that between f2 and f3, f3 and f4, so-on until you can prove the whole thing.

you'll need to also show for f1(40) = f2(40) ==> 0.02(40)^2 = 32, then solve until you reach 32=32

and do this for all the spots where two graphs/equations meet

[eager cove]

so for 2 and 3

they meet at 45

so if I sub 45 into both equations

[finite pier]

indeed, that is what you gotta do

[eager cove]

ok

thanks

2. Prove it ends and starts at same point

would I make y = 0?

[finite pier]

if it helps, the formal definition of "Continuous" in mathematics is: "Two graphs who share an endpoint; for one graph f(x) ending at x=n and another graph g(x) beginning at x=n, f(x)=g(x) for x=n

[finite pier]

this one i'd say you go to the first and last equations. the whole series runs from x=0 to x=89.

so you'd need to test, does first function at x=0 EQUAL the last function at x=89

[eager cove]

hmm what u mean by that

like subbing in x = what

x = 0?

[finite pier]

2 says "Prove it ends and starts at the same point". by 'it' I can assume they mean the entire graph from x=0 to x=89.

'ends and starts' would mean the start and endpoints, x=0 and x=89.

'same point' likely means same y=value, and prove jsut means

f1(0) = f7(89)

show that they equal each other, basically

does that make sense how i got that?

[eager cove]

so like y = 0.02(0)^2 and y = -1.08(89-89)

[finite pier]

ye

[eager cove]

ok

[finite pier]

but does it make sense how i came to that conclusion?

[eager cove]

yea

[finite pier]

i will agree that these questions are worded pretty poorly, especially #3

[eager cove]

3. The rollercoaster never gets as high as the previous point

[finite pier]

this one, i'm going to re-write it.

3. Each section of the graph's endpoint is never as high or higher than the endpoint of the section previous to it.

[eager cove]

kk

[finite pier]

an example of this, would be f4 has an ending at x=64, and f5 has an ending at x=70. so prove that f4(64) > f5(70)

[eager cove]

wait i just realized I failed it

[finite pier]

huh?

[eager cove]

look it's the same height

[finite pier]

congrats; you've spotted something that you must say "we cannot prove this point"

so, point 3 cannot be proven true

[eager cove]

that would also fail

same as point 5

[finite pier]

exactly. so, looks like we have to say "This point cannot be proven true for the graph."

[eager cove]

ok

4. No ascent or decent can be steeper than 60 degrees from the horizontal

[finite pier]

alright, this is the hardest one

because it requires gradients/derivatives

[eager cove]

ok

[finite pier]

so, do you remember derivatives/gradients? if not, it might be a very good idea to go over a refresher for it

[eager cove]

yeah

[finite pier]

well, what we want to do with the graph then, is find the slopes, and put that through a tangent function

why? well, a slope is rise over run, change in y over change in x. y and x are perpendicular, and if you want the angle at the base of x, or the beginning of where you take the slope, you can do tan(x) = change in y / change in x

[eager cove]

hmm

[finite pier]

derivatives/gradients can get you slopes at points on a graph, which you can then use tan(n) = derivative of function, and find n. if it's +/- 60 degrees or greater/lesser, then it fails

[eager cove]

so for the first one

the derivative is 0.04x

[finite pier]

the nice thing about derivatives is you can also more easily determine which endpoint is gonna be greater, usually.

the first equation's derivative is positive linear, so you know the highest y value is the highest x value

so, what's the highest x value that f1(x) can achieve?

in this graph

[eager cove]

40?

[finite pier]

alright

so what's f1'(x) for x=40?

or, basically, 0.04 * 40

[eager cove]

1.6?

[finite pier]

yep. now, if we use tan(n) = 1.6

solve for n

[eager cove]

n = arctan(1.6)
n = 58.4 degrees

[finite pier]

is that greater than 60?

[eager cove]

nope

[finite pier]

then f1 satisfies the conjecture.

now onto f2.

its derivative will jsut be 0, so that's easy- 0 degrees- easy

now f3.

[eager cove]

i suck at derivatives

of functions like this

[finite pier]

there's an easy way to do this, something called u-substitution

[eager cove]

wait is it just -2/15(x-45) ?

[finite pier]

let's look at an example function, h(x) = 14(x+15)^3

we want to get h'(x). but normally, we'd have to expand out the x+15 cubed- which is not fun!

so, we can use a fun theorem that goes like this:

if we make x+15 = u(x), then h(x) = 14u^3 which is a lot more simple.

then when you do h'(x), you just do chain rule to get h'(u(x)) * u'(x)

[finite pier]

i would agree

[eager cove]

and the highest point is 45?

f'(45)=(-2/15(x-45))

wait what

[finite pier]

this one we wouldn't want to look at highest point necessarily- since at x=45, it's pretty evident that the slope is 0 just by observation alone.

the other end, however, is pretty sharp downwards.

the problem also said make sure nothing goes BELOW -60 degrees as well

so, let's look at tan(n) = f'(60) and hope n isn't less than -60 degrees

[eager cove]

k

so -2

tan(n) = -2?

[finite pier]

if that's what you got, yea

[eager cove]

it's greater than 60 ffs

I'm going to have to fix that

[finite pier]

nope, it's not wrong

the descent of that graph is lower than -60, so it can't satisfy that last condition

[eager cove]

k

can we do 1 more

[finite pier]

alright

[eager cove]

4. -1/20(60)^2 + 17

derivative = -2/20(60)

[opaque belfry]

I am free now but BeRu is doing a great job

[eager cove]

we skipped the questions lmao

[finite pier]

pfff thank you, but do feel free to join in

[eager cove]

but he is doing an AMAZING job at helping me with the other stuff

[opaque belfry]

Oooo

[slate vigil]

@finite pier has given 1 rep to @opaque belfry

[opaque belfry]

I love it when new helpers join 🤩

[finite pier]

yea, it had to do with differentiable estimates, and i don't think they went over that yet

[finite pier]

well i needed help with somethign of my own pfff

[opaque belfry]

Ooo

[opaque belfry]

make a post imma see if I can help

[finite pier]

thing is, it's not "help" related. it's an exploratory question. where do i post?

[eager cove]

4. -1/20(60)^2 + 17
   derivative = -2/20(60)
   = -6

tan(n) = -6
n = -80.54 ...

[finite pier]

also aussie, when u got the last question up i'll do my best for ya

[opaque belfry]

ooo #1018223258228760617

[opaque belfry]

I am on for awhile, ping me if you guys need help

[eager cove]

idk how that's greater than 60 degrees

am I being dumb rn

@opaque belfry

[opaque belfry]

oooo

wait lemme read this question

[finite pier]

posted

[finite pier]

wait, this is what you were solving?

which function was that

[eager cove]

4

[finite pier]

you forgot a -60

[eager cove]

?

oh

[finite pier]

it should have equated to -1/10 * (x-60). you had just x for the derivative.

and then u use x-64

[eager cove]

wait

1 sec

f'(x) = -2/20(x-60)

-2/20(64-60)

= -0.4

[finite pier]

keep going!

[eager cove]

tan(n) = -0.4

= -21.8

lmao oops

[finite pier]

WOOOO YOU GOT IT!

pffff its ok, we all make mistakes

[eager cove]

well

I should be able to go on with proving all the lines now for each criteria thanks a lot for the help @finite pier

[slate vigil]

@eager cove has given 1 rep to @finite pier

[finite pier]

you're welcome!

anything else ya need?

[eager cove]

I'll come back to the other questions later, but not atm

[finite pier]

alright, i'll be up a little while longer in case you change your mind

[eager cove]

idk if your still here

but I have to express it as a peicewise function

[finite pier]

which?

[eager cove]

the whole thing

https://cdn.discordapp.com/attachments/1166267628764024842/1166634811239190568/image.png?ex=654b3456&is=6538bf56&hm=29abf38d3c1099e637092913f3ef03cf79ad1832c78fbfbb4503cf80c739aaaf&

it says "it is appropriate to express the final function as a piecewise function"

[finite pier]

gimme a moment to remember how to type this thing out in latex

[eager cove]

All good

[finite pier]

$$f(x) = \begin{cases} 0.02x^2 & 0 \le x \le 40 \ 32 & 40 \le x \le 45 \ -\frac{1}{12}(x-45)^2+32 & 45 \le x \le 60 \end{cases}$$

[trail tree]

BeRu_t

[finite pier]

and just continue that down for all the functions

[eager cove]

Thx

[finite pier]

yw yw!

you might need to change the less-than-or-equal-to's on the 2nd term of each to just a less-than if your teacher is particular about the way the way it's written is

it shouldn't matter tho, only lose 1 point at most on a test for it

[eager cove]

wouldn't you want it less than?

yo @opaque belfry u here

[opaque belfry]

Sorry I was a lil busy

Lets do this tomorrow tho, it's midnight here and probably like 4 for you