help-42

[fringe reef]

u sub for x+b/2

and then it's a standard integral

[quick tiger]

what about the d

[brazen elbow]

trig sub

[quick tiger]

the problem is you'll get u^2 /d^2

[fringe reef]

v-sub lol

just make another substitution

[quick tiger]

yeah let me show you

i know it will be arctan something

[fringe reef]

yes

[quick tiger]

yeah but the d^2 is a problem

[brazen elbow]

leave it be

[brazen elbow]

if there is no information for d just leave it be, its just some constant

[brazen elbow]

mspaint lol

tex it out

[quick tiger]

some shaky hands ....

[brazen elbow]

$\frac1{d^2}\int\frac{\dd u}{\frac{u^2}{d^2}+1}$

is this what you meant

pay attention to the d, the upright d is different from normal d

[quick tiger]

no

$\frac{1}{d^2} \int \frac{du}{\left(\frac{u^2}{d^2} + 1\right)}$

[potent lotus]

RulzerFly

[quick tiger]

i meant this

[potent lotus]

∮μ²ƞdS

[brazen elbow]

why take 1/d^2 out?

you are making it harder

put that back in

[quick tiger]

why there's no other way

[brazen elbow]

just do it

$\int\frac{\dd u}{u^2+d^2}$

[potent lotus]

∮μ²ƞdS

[brazen elbow]

this is what you should have

like artametra said, do another substitution

[quick tiger]

what about the d^2

it's not arctan(u)

[brazen elbow]

d is just a constant

treat it like so

whats the formula for trig sub that you learned?

[quick tiger]

what's that you mean variable changing ?

[brazen elbow]

yes

[quick tiger]

they gave us this in the lesson

$\arctan(u) = \frac{1}{u^2 + 1}$

[potent lotus]

RulzerFly

[quick tiger]

no no

$\arctan(u) = \frac{1}{1 - u^2 }$

[potent lotus]

RulzerFly

[brazen elbow]

well you dont exactly need that

heres a formula you should use

[quick tiger]

you mean the one with the tan

$\frac{d}{du} \arctan(u) = \frac{1}{1 + \tan^2(\arctan(u))}
$

$\frac{d}{du} \arctan(u) = \frac{1}{1 + \tan^2(\arctan(u))}$

[potent lotus]

RulzerFly

[brazen elbow]

no not that

wait

[quick tiger]

okay

[brazen elbow]

$\int\frac{\dd x}{(x^2+a^2)^n} \to x=a\tan u$

[potent lotus]

∮μ²ƞdS

[brazen elbow]

use this

[quick tiger]

a arctan u no ?

[quick tiger]

look what if we took u (1/d) * (x + b/2)

1/d is constant

[brazen elbow]

no

[quick tiger]

so du = 1/d dx

What the f i mixed everything together.....

[quick tiger]

$it's \frac{1}{d} \arctan\left(\frac{x + \frac{b}{2}}{d}\right) $

$\frac{1}{d} \arctan\left(\frac{x + \frac{b}{2}}{d}\right) $

$\frac{1}{d} \arctan\left(\frac{x + \frac{b}{2}}{d}\right)$

[potent lotus]

RulzerFly

[quick tiger]

i should be this no ?

[brazen elbow]

seems to check out

yeah thats correct

[quick tiger]

finally ....

[quick tiger]

thanks for help (i know that i've given you headaches 💀 )

[brazen elbow]

not really lol

but yeah np

[calm coral]

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[sturdy haven]

Hi, i am a bit confused about the process on the right side- i understand why you want to multiply magnitude by sin and cos because the hypotenuse on the bottom will cancel out and you will simply have the x and y components, but how does this work for the 300N vector (the work done on the right) because there isnt really a triangle made here (and i know you could just look at and know its -300,0 but i kinda wanna understand how it works)

[pearl dune]

Better to think in terms of the angle, like the angle of F1 is 60 and the angle of F2 is 180, but if you think of it as a triangle it's effectively a triangle where the 'height'/'opposite' is 0 so the 'hypotenuse' and 'adjacent' merge into a single line

Like imagine what happens to the sides of this triangle as theta gets smaller and tends to 0

In the context of that diagram in your picture, their angle is measured from the +ve x, so is just 180-theta where theta is the angle in the triangle I drew

So as theta becomes 0, the angle in your diagram is 180-0 so just 180

Hence you arrive at the derivation of F2

[sturdy haven]

ok

yeah i think that makes sense ty

[pearl dune]

Np!

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[crystal axle]

<@&286206848099549185>

[remote mural]

do you know what a circle is

[crystal axle]

uhh no

yes i do

[remote mural]

what about the center of a circle

[crystal axle]

the radius?

[remote mural]

no, the center

[crystal axle]

no

i dont know

[remote mural]

@crystal axle ^^

[crystal axle]

the center looks like thats the radius

oh nvm

so what is the center

[sinful pasture]

the point in the center of the circle

[crystal axle]

its 12 am here i js want the answer 😭

anyone gonna help or what

[sinful pasture]

what do you think the answer is

[crystal axle]

i dont know what the asnwer is

[sinful pasture]

look closely

[crystal axle]

can you just tell me step by step of what to do

[sinful pasture]

the red thing on the left is 2.5 ft

what's that red thing called looking at the second picture

[crystal axle]

radius?

[sinful pasture]

yes

radius = 2.5 ft

[crystal axle]

ok so whats the next step?

now that i know the radius

[sinful pasture]

the diameter is 2*radius if you look at the image

[crystal axle]

so 5?

[sinful pasture]

yes

[crystal axle]

ok so do i do 5*Pi

aka 5 x 3.14?

[sinful pasture]

no

you are just being asked for the radius and the diameter

[crystal axle]

okay...

sorry im just confused

and tired

how do i find the diameter

[sinful pasture]

you just found it

d = 2*r

d = 2*(2.5)

d = 5

[crystal axle]

2 x 2.5?

[sinful pasture]

yes

[crystal axle]

5

oh i just did this

so what the answer to the radius

[sinful pasture]

.

[crystal axle]

i got the answer

Radius: 2.5
Diameter: 5

[sinful pasture]

yes

[crystal axle]

.close

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[dusky flax]

diameter is already shown in diagram (1/4 or 0.25 yd)

[dusky flax]

radius is half of diameter = 1/4 * 1/2 = 1/8 yd

[calm coral]

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[white mountain]

145

[white mountain]

,rotate

[white mountain]

Where do I begin?

[sinful pasture]

you want to maximize the area

and the dimensions are x and f(x)

[white mountain]

Yes

[sinful pasture]

Ok so find a maximum for A = xf(x)

[white mountain]

I say x*y
Letting y be e^(-x^2)

Is that the right path

[sinful pasture]

yes

its more like an optimization problem

[white mountain]

Now what

Take derivative?

[sinful pasture]

you want a maximum for the Area, so yes

[white mountain]

e^(-x^2)+xe^(-x^2)*-2x = 0

How do I find x?

[sinful pasture]

$e^{(-x^2)}+xe^{(-x^2)}(-2x)=0$

ok let me type it correctl woops

[potent lotus]

svc

[sinful pasture]

you can factorize e^(-x^2)

e^a is never 0 so you can already get rid of that term

[white mountain]

(1-2x^2)(e^-x^2)

[sinful pasture]

yes

so (1-2x^2)=0 or (e^-x^2)=0

[white mountain]

And only 1 of those could be 0

[sinful pasture]

ab = 0 is the same as a = 0 or b = 0

set both equal to 0 maybe to avoid skiping step

(for example (x-3)(x-2)=0 is the same as x=3 or x=2)

[calm coral]

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[abstract snow]

Ask a question

this is Arithmetic Progression

no wait

I read it as A_1 mb

https://www.quora.com/The-sequence-An-is-defined-by-A1-2-and-An+1-An+2n-what-is-the-value-of-A100

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[astral mural]

Hello

[astral mural]

What does root mean?

Why are we using 1/n?

[civic dirge]

3rd root. ³√

[astral mural]

Why not use this equation?

[civic dirge]

Like, the third root of 8 is 2.

[astral mural]

So we can't use that equation for even roots?

[civic dirge]

That equation is being used

[astral mural]

Ahhh I see

Tysm

.close

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[atomic heron]

I want to make a graph

[atomic heron]

where basically the mens 100meter world records

at on the y axis

and the date they were made on is on the x axis

but the thing i wanna do is

I have a theoretical limit of 8.97

so I want an asymptote

at 8.97

and I want the equation so that I can like predict future world records and when the theoretical limit will be reached

[graceful dust]

hmm it will be hard to fit something “useful” here

[atomic heron]

wdum

[graceful dust]

before anything i would want to see what the points look like, with race time plotted against date

[atomic heron]

this is what it looks like for swimming

I want something similar for running, but an asymptote

[graceful dust]

yea i get what you’re asking for

[atomic heron]

what platform should i use

[graceful dust]

i dunno, desmos or something if you just wanna see it, but eventually you’re probably going to need to put the data into some coding language with curve fitting support

[atomic heron]

is it doable on geogebra

[graceful dust]

or maybe wolframalpha will do it but you might want more customization than it offers

for seeing the points sure

[atomic heron]

does geogebra have the ability to create the function

[atomic heron]

hmm, is it not possible

[graceful dust]

is what not possible?

[atomic heron]

doing the plotting

and the software automatically just giving me a best fit line equation

[graceful dust]

well no software would use the plot, it would use the numerical datapoints

you need to be able to specify the conditions you want on the curve though

[atomic heron]

hmm

[graceful dust]

like as an example you might ask it to find the “best” fit among curves of the form y = ae^(-bx) + 8.97

so a and b are something the software would compute

or you can ask it for some other type of curve with an asymptote, not necessarily an exponential function like that

[atomic heron]

oh

got it

makes sense

thats yeah

ok

makes sense

[graceful dust]

you can also try just plotting the points in desmos and playing around with a and b until you find ones that look ok if you want a lower effort “solution” lol

[atomic heron]

lmao

its ok

need it to be accurate

ill try this

thanks

.close

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[grand shoal]

find all the value for a and b so that u v w are linearly independent.
I tried using ERO but still can't get the value for a and b respectively

[marble olive]

.reopen

.open

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[grand shoal]

.close

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[marsh blade]

something confuses me here

why did we do sin theta = y / 15

i am confused

[calm coral]

@marsh blade Has your question been resolved?

[near cosmos]

Opposite over hypotenuse is the sine

[zinc crater]

we took sin theta= y/15 since sin theta = opposite side / hypotenuse

[near cosmos]

Also if your wondering why we only took sin theta it is because we need to compare the theta with something that is variable in the problem. Since the height is varying we took sin theta so that theta can be compared to the rate of change of the height.

[calm coral]

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[rich ore]

Im getting 54 for 5d is it correct?

[rich ore]

<@&286206848099549185>

[proven pelican]

Tell ur way

As we can't just tell if ur ans is correct or not

[rich ore]

So there are 3 numbers at start to make it greater than40000

So 3 at start

Then 4 num left

[proven pelican]

Hmm

[rich ore]

Out of which 3 can make it even

[proven pelican]

As u want to get even

Put max even first

[rich ore]

For each of the 3 there will be 3! Different prrmutations

3x3!x3

[rich ore]

What does that mean

[calm coral]

@rich ore Has your question been resolved?

[rich ore]

<@&286206848099549185>

[calm coral]

@rich ore Has your question been resolved?

[torpid spade]

I would suggest considering three cases

i) Those who start with a 4
ii) Those who start with a 5
iii) Those who start with a 6

And then you can add up the amount of solutions for each of the particular cases

[rich ore]

Wont that be lengthy making each permutation individually might even miss some

[torpid spade]

No, it will be in fact easier

[calm coral]

@rich ore Has your question been resolved?

[calm coral]

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[proud egret]

Guys, does anyone know how to find the volume of solid of revolution bounded by two functions, if one of the functions is negative on some interval and positive on the other?

[proud egret]

To be more concrete, here's the problem I was trying to solve:

But I'm worried this solution may over or under count the actual volume

Maybe computing it as
$$
\pi \int_a^b \left| f^2(x)-g^2(x) \right| , dx
$$

will fix the issue?

[potent lotus]

Sweet Tea 🧋

[crystal forge]

so if that part of the shape isnt contributing any volume (basically it would be going over part of the volume traced out by the upper part of the function) then you can just ignore it altogether

[proud egret]

Hmmm. I think I can see that (at least if we were given only the parabola without the line), but I'm not sure if that works when we have 2 functions as well

[crystal forge]

if you wanted to be technically correct you can show that | lower function | <= | upper function | on the given interval

[brazen elbow]

calculate the shape as usual, but subtract the redundant volume

[proud egret]

Yeah, yeah. I get the idea. But again, I'm not sure abt this: #help-42 message

[proud egret]

which redundant volume and how as usual?

[crystal forge]

so you can just consider the volume traced by this region and set your bounds accordingly or do what fungus suggested and subtract the redundant volume, i guess whichever you think would be easier to set up the integral

[brazen elbow]

if you revolve the graph around the x-axis, you can see that there is a portion that is revolved twice

thats the redundant volume

calculate that too

[crystal forge]

for this ^ way (as in picture) you can do the intgral of line - parabola from first interseciton to x = 0 then just integral of the line function between 0 and the 2nd x intercept, then integral of line - parabola on the right

like these 3 regions will have different setups

[proud egret]

Oh, I see! Thanks!

So this is usually done just by eyeballing it?

Like, no 'universal' formula?

[crystal forge]

eyeballing what?

[proud egret]

I think
$$
\int_a^b \left|f(x)-g(x)\right| , dx
$$
for computing the area bounded by f and g on [a, b] will always work, for example

[potent lotus]

Sweet Tea 🧋

[proud egret]

I mean, the regions that we'll split our function into

[crystal forge]

well that still applies on the interval [left intersection of the functions, 0] and [2nd zero of the parabola, 2nd intersection of the functions] and on the one in the middle its just area under a single graph which you know the formula is just a basic integral, but yeah you just to consider what is bounding the region i guess that's all

you can also do what fung suggested, both setups are pretty easy to work with i would say

[proud egret]

wait, how do you mean that still applies?
those are two different things

In the screenshot we are asked to find the volume of the solid of revolution
While the formula int_a^b |f(x)-g(x)| dif x gives the area (notice the absence of pi and f^2 and g^2)

[proud egret]

I'm referencing this formula I've just sent

[crystal forge]

erm yeah sorry not area, use the volume equations

like pi (upper function^2 - lower function^2) dx isn't it?

[proud egret]

yeah

[crystal forge]

for the red and green

and purple is just pi (line function)^2 dx

[proud egret]

lemme ask a separate (but still related) question then\
Why doesn't the formula
$$
\pi \int_a^b \left|f(x)^2- g(x)^2\right| , dx
$$
work in this case?

[potent lotus]

Sweet Tea 🧋

[proud egret]

ie why do we need to separate

(notice the abs value here)

[crystal forge]

well that piece under the x axis like you say will be counted twice

since it traces out the same volume as its reflection above the x axis

which is already included

so what fung suggested is to just do the integral anyway, not caring about it being counted twice

[proud egret]

hm, interesting. Bc I saw that on wiki and it didn't make any assumptions regarding whether f,g are positive or not

[proud egret]

gotcha. I like yours method more tho

[crystal forge]

can you show me where you are reading that on wiki

[proud egret]

one moment

[crystal forge]

and then you would find the volume the rotation of this region creates and subtract it from the total

[proud egret]

https://en.wikipedia.org/wiki/Solid_of_revolution
(It's described for revolution around y-axis, but I guess it also works when we revolve around x-axis)

[crystal forge]

this is the problem

[proud egret]

Ohhhhhh. I didn't see it

Gotcha

[crystal forge]

actually thats maybe about something else

[proud egret]

Also, do you know of a few (like one or two) exercises for this idea (when a function crossess the axis), preferably with answers
Just want to make sure I got it 100%

[crystal forge]

ok that quote was a bit about something else, but yeah this discussion is when the generating curve is not crossing the axis

the generatrix i guess is the curve used to generate the volume so like your function basically, in your case one of these generating curves is in fact crossing the axis of rotation

so you can see why that generates issues :d

[crystal forge]

i dunno, are you learning from a book? otherwise maybe you can google stuff, i dont really have any examples at ahnd myself

@proud egret you could have something like this where if you are rotating these regions over the x axis and considering volume, well you wouldn't have any issues because neither of the regions has a reflection over the x axis so they would both just be contributing their volume as normal, what you have to consider is just if the area you are rotating has a corresponding area reflected over the axis of rotation, thats what you need to consider

[remote mural]

<@&268886789983436800>

[calm coral]

@proud egret Has your question been resolved?

[proud egret]

I'm not closing yet because I'm still working on an example I just came up with to test my understanding
I hope to continue the discussion from here in case I have any difficulties

[calm coral]

@proud egret Has your question been resolved?

[proud egret]

Ok, so, here's an example:
I'm trying to properly find the V(sol. of rev. of the green shape when rotated around x-axis)
The functions are given by f(x) = x/2 - 5 and g(x) = -1/10•(x-10)^2 + 5

I'm a bit confused how to approach it.
Do I like draw y=|f(x)| and y=|g(x)| first and try to eyeball it from there?

the thing I'm a bit confused it that it isn't always that easy to determine what function will have «greater magnitude»

In this example, we'll have
$$
|f(x)|\ge |g(x)|, \quad \forall x \in [0, 5]
$$

but then it changes:
$$
|f(x)|\le |g(x)|, \quad \forall x \in [5, 15]
$$

[potent lotus]

Sweet Tea 🧋

[proud egret]

I've drawn the abs values of the functions (red is $|g(x)|$ and blue is $|f(x)|$) \
And it doesn't look that easy to actually find out when we just have to do $\pi \int_a^b h^2$ and when $\pi \int_a^b h^2 - u^2$ (where u is just a function, I've ran out of letters xDD)

[potent lotus]

Sweet Tea 🧋

[proud egret]

<@&286206848099549185> ||just in case you only get a notification only when a new help channel is opened||

[pulsar vigil]

What do u need help with

[proud egret]

Calculating the volume of the solid formed by the shape shaded in green, when revoled around x-axis

[proud egret]

In particular, I'm a bit confused how to apply the disk method
$$
V=\pi \int_a^b f^2-g^2
$$
when either the functions cross the x-axis or when it isn't true that one of them is always (on the interval $[a, b]$) greater than the other

[potent lotus]

Sweet Tea 🧋

[calm coral]

@proud egret Has your question been resolved?

[crystal forge]

@proud egret hm for this one seems like it'd be easiest to just like find the volume overall including the overlapping part and then subtract what is counted twice, i think the picture you drew earlier is helpful

like subtract this part here that is the volume generated by these two regions, the red is a bit easier since its a triangular region that will generate a right circular cone which you dont even have to integrate, you can use the volume formula for a cone, then you can find the volume generated by that blue region using the quadratic

ping me if you're back / still need help later

[proud egret]

Oh, thank you for the reply!
Honestly, I don't have intuition of what you are trying to communicate to me

[crystal forge]

its just same idea as previous problem we discussed, thats the part of the region being rotated that has a reflection on the other side of the x axis so both parts arent necessary to create the whole volume

[calm coral]

@proud egret Has your question been resolved?

[proud egret]

.close

[calm coral]

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[calm coral]

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[light patrol]

Anyone able to help with beginner scripting in python? Or know a place to get help?

[brazen elbow]

ask your question

we answer anything lol

[light patrol]

It’s not a question it’s wanting me to write code and idk where to start or what to do. I’ll send pics

[hasty coyote]

I can suggest freecodecamp.org ............... it's free

[light patrol]

I’ll check that out!

[hasty coyote]

you're choosing a harsh path bro, but i'm sure you can do it, KEEP UP

[light patrol]

It’s for a college course that’s kicking my ass lol

I’m a geology major, I get why they’re having me learn it but I’m drowning out here

[hasty coyote]

wtf

[light patrol]

FR

[hasty coyote]

GEOLOGY !!!, I'm CS student and yet this rubbish is swallowing me

you've got it hard bro

F

[light patrol]

It’s not even that I don’t think it’s interesting, I just have a small lizard brain that doesn’t understand math lmao

[light patrol]

Drowning together

[hasty coyote]

that's fine, even though i study cs, 90% of my study is math and electronics, and 10% left is filled with a bit of useless pascal programs

anyways close the channel or the mods gonna get mad

[light patrol]

o7

.close

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[frozen vault]

can anyone prove that sinx + cosx*√3 = 0

[left canopy]

,w graph sin(x)+cos(x)*sqrt(3)

[potent lotus]

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[left canopy]

,w graph 0

[potent lotus]

Results provided by WolframAlpha

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Requested by akishbrabus

[exotic bane]

Do you need to prove this or find x where this eq holds true

Cause I don't think this equation can be proven for all x

[frozen vault]

find x where the equation is true

I'm actually kind of stuck in the middle of a math and I know the answer to this should be zero?

[exotic bane]

Can I see your work?

[frozen vault]

yea hold up

This is the math actually which should be equal to -3/4*sin3x

I've already got the answer but there's this part left so i thought it should be equal to zero

[calm coral]

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[calm coral]

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[remote mural]

I'm asked to show condition work/state conditions met or not met etc.

bit confused, I know that the conditons are

f(a) exists
limf(x) exists
f(a)=limf(x)

how do I find the value at which it is continution/not continuous

for a it is quite obvious being 3

but how would I prove the 2nd condition

f(3) DNE fine

wait lmme try smth

.close

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[remote mural]

When our bottom is equal to zero, the function is not continuous or discontinuous

It becomes a vertical asymptote

[calm coral]

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[calm coral]

@full mirage Has your question been resolved?

[calm coral]

@full mirage Has your question been resolved?

[calm coral]

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[remote mural]

I'm noticing the transition points -2 and 0

so what I tried was

$2=\lim_{x\rightarrow -2}(ax^2+b)$

[potent lotus]

Remlis

[remote mural]

2 because -(-2)=2

from the first eq.

[cloud nexus]

Yep

[remote mural]

$2=4a+b$

[potent lotus]

Remlis

[cloud nexus]

And we also need b=6

[remote mural]

this I don't understand

why

[cloud nexus]

To satisfy ax^2+b=6 at x=0

The quadratic term goes to 0

So we are left with b=6

[remote mural]

6=ax^2+b when x=0

b=6

[cloud nexus]

Yes

[remote mural]

okay i think I get it

that's it?

[cloud nexus]

Basically we want all the boundary connections to be smooth

Thats the goal of these types of problems

[cloud nexus]

And then we solve for a

[remote mural]

mm

$2=4a+6$

[potent lotus]

Remlis

[remote mural]

$-4=4a$

[potent lotus]

Remlis

[remote mural]

$a=-1$

[potent lotus]

Remlis

[cloud nexus]

Yes

[remote mural]

perfect

thank you very much

[calm coral]

@remote mural Has your question been resolved?

[calm coral]

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[nimble hill]

@vivid surge

[nimble hill]

They closed that channel

[vivid surge]

oh

[nimble hill]

Is this right ?

[vivid surge]

yeah it is

[nimble hill]

The formal is 0, 1

[vivid surge]

it's wrong ;-;

[nimble hill]

I solved like one hour ago 😢

[vivid surge]

this is correct

[nimble hill]

Thanks

[vivid surge]

i think the answers given might be a typo

[nimble hill]

Yeah maybe thank you 😊

[vivid surge]

[nimble hill]

. Close

. close

IDK how close the channel

[lofty nacelle]

no space between . and close

like .close

[raw dirge]

!done

[calm coral]

If you are done with this channel, please mark your problem as solved by typing .close

[raw dirge]

copy paste it

[drifting seal]

.close

[calm coral]

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[stable raven]

how do i do a

[runic rune]

if f'(x) is in the positive area what does that tell us

[stable raven]

ohh that its increasing?

[runic rune]

yep

[stable raven]

what about b?

is it concave down because as x --> infinity the values reach 0?

[runic rune]

it is concave down

[stable raven]

but why

[runic rune]

f'(x) is decreasing.

if f'(x) was increasing it would be concave up

[stable raven]

ohh okay thank you

ohh okay thank you

.close

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[mighty dagger]

can i use here l'hopital ?

and a>0

[pseudo moss]

no

[unkempt drift]

No, you have a -infinity/0 form

https://en.wikipedia.org/wiki/L'Hôpital's_rule#Derivative_of_denominator_is_zero

[mighty dagger]

ye ok just asking cuz the teacher used l'hopital to solve this

tysm

[unkempt drift]

Npnp

[mighty dagger]

.close

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[pale panther]

I don't understand how they got the domain and range of the bottom one

[calm coral]

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[mighty dagger]

how did he go from first step to 2nd

like ik how he removed the 2 down there by taking out 1/2

but how did he move the e^x up and made these stuff

[mystic obsidian]

if you are studying integration then you just know it

[modern peak]

basic rules of exponents?

[lucid torrent]

1. when a constant is a coefficient of all the variables you can factor it
2. dividing 1/e^x is same as multiplying the reciprocal

[mighty dagger]

oh wait a min

its (all the thing up)/2/e^x

so i can move the e^x up

[modern peak]

$\int \frac {e^{1.5x + 3.5} + e^{2x}}{\frac {2}{e^x}} \text { d}x= $0.5\int e^x \cdot (e^{1.5x + 3.5} + e^{2x}) \text { d}x$

$\int_{ }^{ }\frac{\sqrt{e^{3x+7}}+\left(e^{x}\right)^{2}}{\frac{2}{e^{x}}}dx=0.5\int_{ }^{ }e^{x}\left(e^{1.5x+3.5}+e^{2x}\right)$

[potent lotus]

ren

[mighty dagger]

ohhhhh

damn ok gotcha

ok tysm

appreciate it

.close

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[molten moat]

Hello

[molten moat]

My answer is SDZBGDQ

And there is no such option

From the first example of student

I thought it just changed the letter and put the letter preceeding it

S is replaced by R because it is before it and T by S and so on

And hence C should have been replaced by B

[graceful dust]

[hot creek]

observe that the 4th letter changes successively

[bronze adder]

this question is...

[molten moat]

oh shit you're right

I only went for 3 letters and stopped

Thank you

then the fourth answer is correct

Thank you

.close

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[molten moat]

I mean 3rd

.close

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[shy light]

What should I do after this

Stuck with it from yesterday

I've got 4 equations but it turns out to be a loop

[marble escarp]

Question? 🙂

[shy light]

find the angle CFD

Here it is

Here is something I've tried

[remote mural]

Does this work

[shy light]

N9

I've already tried it

And the equations come out to be a loop

[remote mural]

Ohhh

[shy light]

Check this

<@&286206848099549185>

[dull wagon]

why does noone read what i say

[marble escarp]

Did you write anything above?

[shy light]

I've read

[marble escarp]

About?

[shy light]

I've made some constructions

And I've read adventitious triangle

[marble escarp]

Oh the instructions

Didn't 'bot' say it though?

[shy light]

what are you saying I don't get anything

[marble escarp]

The meaning of this

[shy light]

Oh

It was a problem from yesterday

[marble escarp]

Oh, my bad

[shy light]

And he suggested us to read adventitious triangle

but that problem and this problem is a bit different

and I'm stuck in the midway

[calm coral]

@shy light Has your question been resolved?

[calm coral]

@shy light Has your question been resolved?

[calm coral]

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[remote mural]

Why they did (A-7I) and not

[remote mural]

$\frac{Ax}{7}$

[potent lotus]

Eigenvalues

[left sorrel]

i haven't done eigenvalues but what use is this? Lol

to show that there's nontrivial solutions, you have to show that Ax = 0 for nontrivial x's

so $Ax = \lambda x \rightarrow (A - \lambda \mbb{I}) x = 0$

[potent lotus]

nyxie9151

[left sorrel]

this "form" does exactly that

you show that there's a nontrivial x that solves (A - lambda* I)x = 0

which is equivalent to showing that the determinant (A - lambda*I) = 0 or the columns/rows are linearly dependent

[calm coral]

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[potent lotus]

mathematician8256

mathematician8256

mathematician8256

[cloud nexus]

why are you deleting messages

??

the fuck?

[calm coral]

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[brittle siren]

wth

[calm coral]

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[midnight thistle]

could anyone explain this terrible markscheme to me please 😭

[calm coral]

@midnight thistle Has your question been resolved?

[alpine gust]

if they intersect at a point, then y of the line equals y of the curve

y = k - x = 6 - (x-3)^2

after some simplification and rearranging u will get a quadratic equation that must have two roots

cuz he says that it intersect at two different points

[midnight thistle]

the roots of the quadratic I get don't give me the answer though

[alpine gust]

show me

also the roots will have the variable k in it

[alpine gust]

thats why he make sure that
b^2 - 4ac > 0

[midnight thistle]

i rearranged k-x = 6 - (x-3)^2
to k-x = x^2 -6x - 3
to
0 = x^2 - 5x - 3 - k

[alpine gust]

there is a negative sign in front of the ()

6 -( x^2 - 6x +9)

[midnight thistle]

oh i see so it should be -x^2 - 6x - 9

[alpine gust]

+6x *

[midnight thistle]

i see so then that simplifies to x^2 -7x + (3+k) but then how do i find the roots of this

[alpine gust]

in order to have two roots, the value of b^2 - 4ac should be greater then zero

[midnight thistle]

i see so now i have worked out the first value of k for 37/4 but how do i work out the lower bound

[alpine gust]

thats what im trying to figure 😂

one way is to say k = x + y

= 3 + 2sin(t) + 4 + 2cos(2t)

[midnight thistle]

thats quite hard to solve lol

[alpine gust]

calculus is one way if u know it

[midnight thistle]

i do understand calculus to a degree

[calm coral]

@midnight thistle Has your question been resolved?

[calm coral]

@midnight thistle Has your question been resolved?

[calm coral]

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[winter herald]

isnt L1 = 2???

[pure kayak]

what?

[winter herald]

and arent Lucas numbers defined by Ln + Ln+1 = Ln+2

[pure kayak]

L1's value is defined as part of the equation

[inland vortex]

the 0th lucas number is 2

[winter herald]

i think the indexing starts from 1

but okay lets just assume it starts from 0

[inland vortex]

in the question youve given, yes, which is why we have L_1 = 1

https://en.wikipedia.org/wiki/Lucas_number

on wikipedia we see n = 0 is defined as 2

[winter herald]

why are the fibbonacci numbers in the DEFINITION of lucas numbers

[inland vortex]

i agree tho this is the standard definition

but since theyre equivalent, they chose to state the fibonacci version as "the definition"

[winter herald]

yea okay so i jsut want to prove both direction right?

[inland vortex]

theyre only asking you to prove one direction

[winter herald]

but i wont be satisfied then '^'

lol

ok

[inland vortex]

i think the proof will probably give u both directions for free

since ur showing that the two sequences coincide after all

[winter herald]

yea eya

okey

let me try for some time

[winter herald]

i just wanted to clear this doubt

[inland vortex]

yea sure

[winter herald]

okay

i proved one direction i guess

and the L1 = 1 is probably a typo

[winter herald]

okay L3 = F4 + F2 = 3 + 1 = 4 is just false

OK

that is if L1 = 2

nah

the book is a bit badly written i guess

no problem

got it

kay thanks @inland vortex

.close

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[dusky crescent]

Can anyone help with this question?

[pallid halo]

compare the real and imaginary parts of the LHS and the RHS

[ocean cypress]

z^3/4 = i
z^3 = i^4 = 1

take the cube root. you will get 3 solutions

some of them are extraneous

you will have to plug them back in

[dusky crescent]

THis is the similar example we had in class but with a real number

I think I have to do a similar process to this

but how would I know the argument

[pallid halo]

you have 1 = r^(3/4) sin(3/4 theta)
you know r = 1, so this means sin(3/4 theta) = 1

you can solve that last equation in the usual way

[dusky crescent]

How do we know r gives 1

I think I am missing something

[pallid halo]

look here

the magnitude of the LHS is 1, so the magnitude of the RHS is 1

thus r^(3/4) is 1

so r is 1

[dusky crescent]

Ohh okie. How would i find the argument though to solve?

[pallid halo]

sin(3/4 theta) = 1

so 3/4 theta = pi/2 + 2pi n
where n can be any integer

now solve for the values of theta in say [0,2pi) that satisfy this

[dusky crescent]

Where does this come from'? I know r is 1 but why does this make sin(3/4 theta) = 1

[pallid halo]

take the imaginary part of this equation:

what do you get?

[dusky crescent]

The part before the sin?

or the LHS

[pallid halo]

the whole equation

what's the imaginary part of the left hand side?

[dusky crescent]

1

but wouldn't that mean 1(sin(3/4theta)

How do you arrive that 3/4 theta is pi/2 + 2pi n ?

[pallid halo]

1 = sin(3/4 theta), right?

sine of what angles gives you 1?

[dusky crescent]

Hmm I don't see how I thought that 1 was just for the imaginary part

I think I am forgetting some rule or something

[pallid halo]

can you clarify?
are you ok with the equation 1 = sin(3/4 theta) or are you confused about where it came from?

[dusky crescent]

Yeah I am confused about where it came from

Like why multiplying the imaginary part by the angle gives 1

[pallid halo]

suppose i have an equation of the form:
a + bi = c + di

where a,b,c,d are all real

that mean:
a = c
and
b = d

now apply that to this equation

the left hand side can be written as 0 + 1i

and the right hand side can be written as $$r^{3/4}\cos(3\theta/4) + i r^{3/4}\sin(3\theta/4)$$

[potent lotus]

Bungo

[pallid halo]

so:
a = 0, b = 1

c = r^(3/4)cos(3 theta/4)
d = r^(3/4)sin(3 theta/4)

b = d gives you...

[dusky crescent]

1

ohh that makes so much more sense now!!

Thank you : )

Does that mean I can just write the equation as r^3/4cos(3/4 theta) + 1

@pallid halo

[pallid halo]

if you want to use the cosine term, that's the real part of the RHS

so you'd want to equate that with the real part of the LHS

i.e.

you would be doing
a = c

in this case
a = 0, and c = r^(3/4)cos(3 theta/4)

so your equation would be
0 = r^(3/4)cos(3 theta/4)

which will give you the same solutions

[dusky crescent]

When my prof did this example (a) she didn't get rid of the sin

@pallid halo

Is there any way I can do that for this question as well?

[pallid halo]

she did get rid of it

that's what is happening when going from the top equation here to the two bottom equations:

[dusky crescent]

where did she get the 2n+1 from

@pallid halo

[pallid halo]

she wants r^15 = 2
and cos(15 theta) = -1

cosine is negative at pi

plus any multiple of 2pi

in other words, cosine is multiple when the angle is pi + 2pi n for any integer n

which you can rewrite as pi(2n+1)

[dusky crescent]

okay so for this question how would I figure that piece out ? @pallid halo

like with 3/4 theta

[calm coral]

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[calm coral]

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[glacial raptor]

Need help on probability and statistics. Distribution of sample mean

[cloud nexus]

which question

[glacial raptor]

all of them

[hasty fiber]

a should be obvious, the question tells you the parameters explicitly

[glacial raptor]

yeah just realized it...

overthinked it

[hasty fiber]

b should also be clear, you just need to take into account that your sample size is 14

[glacial raptor]

so find the answer for b, it would be finding the standard deviation x bar?

I'm thinking to hard on this...

[hasty fiber]

You have to find the mean and std dev yes

[glacial raptor]

yeah so far

X = 64, STD = 20, N = 14

X x-bar = 64, STD X-bar= 5.34

[hasty fiber]

I don't like X=64

[glacial raptor]

isn't the 64 million the mean?

[hasty fiber]

It is

But X is a random variable, not the mean

[glacial raptor]

ah ok

[hasty fiber]

Anyway, carry on

[glacial raptor]

so overall I got STD and N correct but not X

[hasty fiber]

64 looks fine to me

[glacial raptor]

64 million as the mean, STD 20 million, N as the random sample is 14

[calm coral]

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[trim ermine]

when i find the oblique asymptote, for example for the equation

$\frac{3x^2-2x-17}{x-3}$

i'd get $3x+7$

if a question asked "determine whether the curve approaches the OA from above or below"

with limit notation, how would I show the sides the equation approached the OA from

$\lim_{x->\infty}f(x) = (3x+7)^+$

would that be valid?

[potent lotus]

notnick

[brisk pelican]

Now I don’t think so

At least I’ve never seen that notation or heard of that defined

What you can do is show that the equation can be written as 3x+7 + (something over x-3)

Maybe 4/(x-3)? Not sure

But anyway that goes to 0 when x is big

But this shows that the equation is always a bit above the asymptote

[calm coral]

@trim ermine Has your question been resolved?

[trim ermine]

no not really, when you evaluate it goes to like 30000007

but yeah im not sure either

my teacher is a bit strange

as well as this textbook

like i could easily write "f(x) approaches the OA from above as x -> infty"

but that is a bit long

[brisk pelican]

No I mean

The extra factor goes to 0

[trim ermine]

extra factor?

no im talking about oblique asymptote

[brisk pelican]

Dawg 😭

[trim ermine]

the OA is 3x+7

[brisk pelican]

You gon make me use latex on mobile

Ok gimme a second

[trim ermine]

um 😭

[brisk pelican]

$\frac{3x^2-2x-17}{x-3} = \frac{3x^2-2x-21}{x-3} + \frac{4}{x-3} = 3x+7 + \frac{4}{x-3}$

[potent lotus]

coak

[brisk pelican]

Wow I was sure that wasn’t gonna a compile

Let’s get it

[trim ermine]

ahhh

OH

um

im still not sure what a proper answer for 8a would be tho

[brisk pelican]

Well you just have to ask yourself if 4/(x-3) is positive or negative for big x

Obviously it’s positive

So as x goes to infinity, the function will be just barely greater than 3x+7

But in the limit the difference is negligible and it goes to 0

Make sense?

[calm coral]

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[snow bough]

Why is this false? I can't find anything wrong with it except for the right integral missing a dx

[calm coral]

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[calm coral]

Please don't occupy multiple help channels.

[eager forge]

How do i do ths?

[eager forge]

mb

[calm coral]

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[warm warren]

Well if 14y = 13k then surely y is a multiple of 13

[ocean cypress]

gcd(13,14) = 1

in fact:

gcd(n, n+1) = 1

always!

25 - x is an integer

let that integer = k

14y = 13k

then divide both sides by 13

14y/13 = k

the right side is an integer because k is an integer

so the left side must also be an integer

then invoke euclid's lemma

13 is a prime, and if 13 | 14y, then either 13 | 14 or 13 | y

13 does not divide 14 because of what i said above about n being coprime to n+1

so 13 must divide y

25 - x is an integer

if a prime p divides the product ab, then p divides a or p divides b

here's an example:
3 divides 12 * 2, so either 3 divides 12 or 3 divides 2

is that not given?

[warm warren]

So a natural number

If it’s true for integers then its then it’s true for counting numbers because counting numbers are a subset of integers

[calm coral]

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[white mountain]

why does lhopital only apply to indeterminate cases

[runic rune]

if the limit is determinate, then you have your solution

[white mountain]

But lhopital does not work for determinant form

Why

[runic rune]

.. to prolly simply put it, its because its not needed. its better to just direct substitution at that point

[white mountain]

But why can it not be used

What makes it not work

Because it doesn’t work for determinant form

It’s not a matter of easier or harder

[pale blade]

If you're interested in understanding why L'H doesn't work on conventional limits, start by understanding why L'H works for indeterminate ones, from a physical perspective

[white mountain]

And why is that

[pale blade]

Well if you don't know how the limits of a function's derivatives relate in a compartive (inf/inf or 0/0) case, its tough to figure out why it (might) fail when the limit is determinate.

What happens when you apply l'h to a limit that has already been solved by it? You get the wrong answer

[white mountain]

Okay, but why

if I were to lhopital lim x->2 (2x+4)/x I would not get an undefined answer, I would get the wrong answer

What is so special about f(x) and g(x) being 0 that makes it work

[pale blade]

If you decompose a regular l'h limit into its two consitituent functions, the visual intuition of l'h is that even though both f(x) values are zero, they have different rates of change. Since you have no initial "displacement", these rates of change dictate the nature of these two functions (or their ratio) in the neighbourhood of x=a

To illustrate:

Now, what if f(x) and g(x) don't match?

[white mountain]

How does this show lhopital

I get the f(x) and g(x) being seperate functions but how does the df and dg give you lhopital

[pale blade]

Because we want to get the limit of f/g

since f=g=0 at the point where they intersect, we use their derivatives to model what will happen right afterwards

aka their derivatives respectively, df and dg

get their quotient

and that lh for you

,rotate ccw

[pale blade]

You end up with something like this if f is not equal to g at the point of interest

now the ratio of "dg" and "df" have no meaning, since the increment from f=0 and g=0 is nonzero, meaning f hits zero first, before g

to correctly apply the same process, we would need the two "beams" to start from the same place, where the two functions are equal to zero

All of this requires a comparison to be made from the same starting point

and beyond that, from the same starting point where f=g=0, since starting at any other value (f=g=a) while trivial (lim = 1 duh), would cause l'h to fail, due to the added "increment" that differentiates the intended "limit of derivatives" to be used for l'h and simple limit of derivatives one would try to use for it.

All of this is a bit hand wavy (and way beyond the level of calculus 1). The formal proof uses the mean value theorem IIRC

[white mountain]

Why can’t they both just equal 1

And work from there

[pale blade]

Try to draw it out. You'll have a sort of "increment" that screws up the dg/df ratio

[white mountain]

Wdym

[pale blade]

im assuing you mean f=g=1?

[white mountain]

I’m lost

[white mountain]

Sure

[pale blade]

Try to think about it this way

We can get the relative change of f and g when they both equal to 1

But that change is infinitesimal

Compared to the big green box (from y = 0 to 1), whatever difference in their rates of change there is is completely irrelevant

ADDENDUM: This ratio is thus only relevant when there is no big green box (ie. when f=g=0).

It’s ultimately adding an infinitesimal to a number, 1+0.0…2 = 1

Sure, red grows faster than blue, but 1.0000..2/1.0000..1=1 still

In all honesty though, I wouldn’t overthink it unless it’s out of personal interest. The “formal” way of expressing and communicating this is beyond Calculus 1. You’ll get more exposure to it if you end up taking Analysis

[calm coral]

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[tall moon]

assume there is an isoceles triangle ABC where AB=AC, point E is the midpoint of line AC, and point K is the intersection point of the line that bisects CAB and the circle that touches point A, B, and E, prove that the diameter of the circumcirle is $\frac{4}{3}$ times AK

[potent lotus]

Skill_Issue

[tall moon]

it would look something like this

i have no idea where to start this

[calm coral]

@tall moon Has your question been resolved?

[dusky flax]

hm

[calm coral]

@tall moon Has your question been resolved?

[calm coral]

@tall moon Has your question been resolved?

[calm coral]

@tall moon Has your question been resolved?

[tall moon]

<@&286206848099549185>

[calm coral]

@tall moon Has your question been resolved?

[left canopy]

Okay, I suppose I have a somewhat computational proof to get some Intuition

You can compute the circumradius of AEB. Let O be the center of this circle. From there, you can compute the angle of the arc AOE and EOK which should give you the angle AOK. You can use this angle and the circumradius to compute AK. Then compare that value to the circumradius of ABC.

[calm coral]

@tall moon Has your question been resolved?

[tall moon]

how do i compute the arc for aoe and eok tho?

[left canopy]

It's double the angle EAK and, AOE you can find using the length of AE

[tall moon]

im sorry, i dont know that much about the coorelation between the triangles and their circles, but is arc AOE equal to arc EOK?

if so how?

oh wait nvm

do i use trig for AOE?

[tall moon]

arc EK=$\pi r(\frac{EAK}{90})$
arc AE=$\pi r(\frac{sin^{-1}(\frac{AE}{2r})}{90})$

[potent lotus]

Skill_Issue

[tall moon]

um

EOK=4EAK $\quad$
AOE=$4sin^{-1}(\frac{AE}{2r})$

[potent lotus]

Skill_Issue

[tall moon]

wait

arc AK=$\pi r(\frac{EAK+sin^{-1}(\frac{AE}{2r})}{90})$

[potent lotus]

Skill_Issue

[tall moon]

AOK=$4EAK+4sin^{-1}\frac{AE}{2r}$

[potent lotus]

Skill_Issue

[tall moon]

AK=$2r sin(2EAK+2sin^{-1}\frac{AE}{2r})$

[potent lotus]

Skill_Issue

[tall moon]

D: this is absurd why would the olympiad have this question