help-23

[fallen thunder]

Hi, could someone give me a hint on how to do this?

[hardy lion]

idk if its a troll problem

if x is valid, so is -x

[fallen thunder]

i thought of this as well

not sure if i'm missing something lol

is it just 0?

weird

[opal oar]

cant u just simplify it

[shadow glade]

that seems like it would be it, yeah

[fallen thunder]

okay okay thanks

[opal oar]

yeah its 0

[shadow glade]

did you check what i said in the other channel about there being an infinite # of vectors perpendicular to another one in R3? : )

[opal oar]

i assumed they had to originating at the same point

which is why i said 2

[shadow glade]

yes, but it's still true with that 🤔

[opal oar]

chipi chipi chapa chapa dubi dubi dapa dapa

[shadow glade]

i...ok well i guess that means you aren't interested

[opal oar]

why is it not 2

[shadow glade]

hrm i will try to draw another bad but better picture

[fallen thunder]

[shadow glade]

so if you have a blue vector along z axis (starting at origin), you can draw vectors in all sorts of directions (assume they are all in the xy plane) and they are all perpendicular and they all start at the same point as you required, they can all be unit vectors etc.

[opal oar]

butthose arent perpendicular

[shadow glade]

check analytically: any vector in the xy plane has coordinates (x, y, 0) and the blue vector has coordinates (0, 0, z), what is the dot product of those?

a vector parallel to the z axis is perpendicular to any vector in the xy plane...

take a pen...stick it vertically on top of your desk...now walk your hand in any direction along your desk away from the pen...is it not perpendicular?

[opal oar]

but that doesnt originate at the same point

[opal oar]

nor are any of these lines perpendicular

[shadow glade]

ok well just think about it for a while or discuss it with someone else, i don't know how else to convince you 😄

[radiant crypt]

it's just the direction that matters

[shadow glade]

the point is where the pen meets the desk, they DO originate at same point, i don't understand what you mean. in the drawing, the origin is the origin point of all the vectors

[opal oar]

if i move a pen around a desk by definition of "move" it does not originate at the same place

[shadow glade]

no i didn't say move the pen sigh

[opal oar]

walk

sorry

[shadow glade]

imagine a line from the base of the pen along the desk without moving the pen, thats the perpendicular vector

are you sure you aren't trolling me at this point?

[opal oar]

maybe i have the geometric definition of perpendicular wrong

[shadow glade]

so you agree a vector can be perpendicular to a plane right?

[opal oar]

yeah

[shadow glade]

ok and say it meets that plane at some point O

[opal oar]

O

[shadow glade]

you can draw many lines that coincide with the plane through O right?

(sorry bad typing) lol

[opal oar]

what

like by rotating the flat angle

2 pens

one sticking up

one forming a 90 degree laying on the desk

yes you can spin it

[shadow glade]

yes, and each one of those configurations is perpendicular isn't it?

[shadow glade]

exactly

[opal oar]

yeah i had my definition wrong

i was thinking complete opposites

instead of forming a 90

[shadow glade]

ok well sorry if i was rude by giving you the thumbs down, hope you didnt take it that way, but thanks for sticking around for following my argument 😄

[opal oar]

preciate it ❤️

[toxic stratus]

damn kanna got her channel taken over

[opal oar]

They got their answer before we started

[plucky elk]

damn kanna how does it feel to be conquered

[toxic stratus]

you should still be discussing in your own channel

also @shadow glade

[shadow glade]

yes, sorry, but I believe the question was answered already?

[plucky elk]

.

[toxic stratus]

kanna wont be able to close this channel while you're still discussing here

[shadow glade]

yes, fair point, will open new channel next time 👍

[fallen thunder]

[shadow glade]

!done

[safe radish]

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

[fallen thunder]

yeah i was waiting for you guys

anyway thanks~

.close

[safe radish]

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[rain fiber]

hi i just have a simple question about radicals
8rad8 - 9rad2
Im confused on how you simplify them to be able to subtract them

[rain fiber]

<@&286206848099549185>

am i not allowed to do that

oops

i just saw someone else do that

[junior smelt]

You're supposed to wait 15 minutes

[opal oar]

8rad8 = 8rad(4*2) = 8rad(2^2 2) = 8 2rad(2)

Assuming it’s square root

[rain fiber]

im confused on what this means i might just be stupid but like i looked itup and they simplified both of the inner parts of the radicals to be 2 so they could subtract

and i dont know how they did that

[rain fiber]

mb

[opal oar]

It kinda depends what root it is

Can you show the example

[rain fiber]

sure

[opal oar]

Do you agree that sqrt(8) is the same as sqrt(4*2)

And that 4 is the same as 2^2

[rain fiber]

yeah

[opal oar]

So sqrt(8) = sqrt(2^2 * 2)

[turbid pumice]

sqrt(8) is nothing but sqrt(2*2*2)

[opal oar]

And u can separate the right side

Sqrt(2^2 * 2) = sqrt(2^2) * sqrt(2)

[turbid pumice]

so 8 * 2 * sqrt(2)

[turbid pumice]

that is 16sqrt(2)

[turbid pumice]

now you can go ahead and subtract

[rain fiber]

im still confused how they went from 8sqrt8 to 16sqrt2

sorry

so like

did they divided by 4

to get that inside part to be 2?

[opal oar]

Or just ignore everything we both said that works too

[turbid pumice]

exactly

[rain fiber]

ohhh okay seeing it visually like that helped more thanks

[rain fiber]

man im just confused trying to get some help and didnt understand what u said

thats all thank you

.close

[safe radish]

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[mystic bolt]

I need help knowing how to start

[halcyon marsh]

We first need to find tangent equation

[mystic bolt]

It seems

[mystic bolt]

So x + 1/2

[halcyon marsh]

You know in a equation dy/dx tells about slope of a tangent at that point

What is dy/dx

[mystic bolt]

Of the tangent equation?

Or the parabola

[halcyon marsh]

Ya we need a slope and a point to get tangent equation

[halcyon marsh]

Parabola

You know that for a line equation we need slope and a point y=mx+c

[mystic bolt]

Yes

[halcyon marsh]

What dy/dx for parabola

[mystic bolt]

One sec

3 - 2x/2?

Or 3 - x?

[halcyon marsh]

So this is slope of tangent according to point x

For different points on parabola

Now you have to assume your point Q

[mystic bolt]

How do I do this

[halcyon marsh]

Like assume it is (x1,y1)

[mystic bolt]

Oh yes

[halcyon marsh]

So what is slope of tangent at x1,y1

[mystic bolt]

Do I set equal to 0 and f(0) to get points?

[halcyon marsh]

No not that simple

Slope of tangent at x1,y1

[halcyon marsh]

Here

It is 3-x1

Are you following

[mystic bolt]

I’m not sure

I have the 3 - x for the derivative

[halcyon marsh]

No it's right i know

[mystic bolt]

So it would be the same slope at Q

[halcyon marsh]

Ya

[mystic bolt]

👍

[halcyon marsh]

Slope at Q is 3-x1

[mystic bolt]

Yes

[halcyon marsh]

And it is passing through (0,1/2) also according to question

[mystic bolt]

Correct

[halcyon marsh]

So what is equation of line

[mystic bolt]

Oh I see

y - 0 = 3 - x(x - 1/2)?

[halcyon marsh]

X1 there

They are different

[mystic bolt]

Yes edited

[halcyon marsh]

Y-1/2=(3-x1)(x-0)

This is right

Check

[mystic bolt]

Oh yes I swapped the x’s and y’s 🤦

Thank you for the help

[halcyon marsh]

Now what is slope of a line passing through 2 points

[mystic bolt]

Y-1/2=(3-x1)(x-0)

[halcyon marsh]

We need now x1

[mystic bolt]

I don’t know

How do I find that

[halcyon marsh]

See point is also on parabola

It's also satisfy parabolic equation

[mystic bolt]

(6,0)?

[halcyon marsh]

No that's not satisfying parabola

Point Q is on parabola so it must satisfy it's equation

[mystic bolt]

What does this mean?

Sorry I am not following

[halcyon marsh]

Mean point Q is lying on parabola right

See

So it satisfy parabolic equation

We assumed Q is x1,y1

[mystic bolt]

I thought we said that was (0, 1/2) to get the equation of tangent

[halcyon marsh]

No (0,1/2) is another point which is on tangent

See question

[mystic bolt]

Yes so how do i find the x1

[halcyon marsh]

So Q is on parabola so it satisfy it's equation

So x1,y1 satisfy parabolic equation

So y1=3x1-(x1^2)/2

Get it

You understanding or we start again

[mystic bolt]

I’m confused about this. Why is there 2 x1

[halcyon marsh]

You just put x1 ,y1 in parabolic equation

There is no 2 x1

[mystic bolt]

Oh I see the equation now

So what must be done

[halcyon marsh]

You got a equation in x1,y1 right

[mystic bolt]

Yes this one: y1=3x1-(x1^2)/2

[halcyon marsh]

Ya good

Now this point also on tangent line

[mystic bolt]

Is it point Q?

[halcyon marsh]

Ya

What is equation of tangent line

[halcyon marsh]

Point Q on tangent line

[mystic bolt]

y - 1/2 = 3 - x(x -0)

[halcyon marsh]

There is 3-x1

Now put x1,y1 point in this equation

Because x1,y1 point is on this tangent line

[mystic bolt]

y1 - 1/2 = 3 - x1(x -0)?

[halcyon marsh]

x to x1 one more x

[mystic bolt]

Both are x1?

[halcyon marsh]

Ya

[mystic bolt]

Ok got it

[halcyon marsh]

Now we have two equation in x1,y1 can you solve them

[mystic bolt]

Ok I’ll do it and text back

[halcyon marsh]

Ok

[mystic bolt]

I don’t think I did it right

Y1 = 3 - x1^2

[halcyon marsh]

We need value of y1 and x1 not another equation

Our equations is

[mystic bolt]

How did you get the last line of math you did?

[halcyon marsh]

I put y1 equal

[mystic bolt]

Oh mb i didn’t see the =

[halcyon marsh]

Y1=1st equation
Y1=2nd equation
1st equation=2nd.equation

[mystic bolt]

👍

[halcyon marsh]

You try to solve the equations

We get your x1 and y1

X1=1, y1=5/2

https://tenor.com/view/high-five-top-gun-tom-cruise-gif-9817538

[mystic bolt]

-1^2 is 1 though?

So 4.5

[halcyon marsh]

Ya but we forget it because it is causing problems

😂

[mystic bolt]

Oh fr?

[halcyon marsh]

No not fr

[mystic bolt]

So 4.5 or 5/2 😂

[halcyon marsh]

It is correct because we are working in first quadrant

Maybe there is a another tangent passing through (0,1/2) and cutting parabola

[mystic bolt]

So we are doing 5/2?

[halcyon marsh]

Ya

There are two tangent one cuts at (1,5/2) and one cuts at (-1,-7/2)

Which is in 3rd quadrant

[mystic bolt]

And Q is the 1, 5/2

[halcyon marsh]

And pass through (0,1/2)

[halcyon marsh]

Ya

[mystic bolt]

W

Thanks for the help

[halcyon marsh]

I think we are done

https://tenor.com/view/michael-scott-wink-yes-no-problem-gif-16355545

[mystic bolt]

Would you be able to help with the last part?

😅

[halcyon marsh]

Which part

Step 1 First we assume point Q

Step 2 Then we compute slope at Q and find the equation of tangent line

Step 3 we put Q in parabolic equation and tangent line equation

Step 4 we solve equations and find Q

[mystic bolt]

It’s one that wasn’t pictured sorry

It’s the same slope

[halcyon marsh]

Can you send question picture again

[halcyon marsh]

So tree is top of the hill

[dapper flame]

what

[mystic bolt]

Correct

But +2.5 ft

[halcyon marsh]

We have to find f'=0 point

[mystic bolt]

For the intersection

Oh ok

[halcyon marsh]

Which is 3

Right x=3

[mystic bolt]

3 is the x

Yes

F’(3)

?

[halcyon marsh]

At x=3 we need a y on tangent line

Put x in tangent line equation

What we get

[mystic bolt]

3 in as the x yes?

[halcyon marsh]

Ya

This will get us point of intersection of laser and tree

[mystic bolt]

5.5

[halcyon marsh]

Now we need foot of tree

Put x=3 in parabola

[mystic bolt]

4.5

So 1 foot up

[halcyon marsh]

So laser hit 1 foot above from its root

Get it

[mystic bolt]

That was much easier than I thought

Thanks

[halcyon marsh]

2.5 foot tree

[mystic bolt]

I just didn’t know how you knew how to start it

[halcyon marsh]

So answer is 1.5

It's simple you. Just have to understand concepts

[mystic bolt]

But isn’t the difference between 5.5 and 4.5 just 1?

[halcyon marsh]

But we need distance from the top of tree

1 is distance from down of the tree

[mystic bolt]

Oh I see

Not from the 5.5

But the tree height

[halcyon marsh]

Good 👍

Ya 2..5

[mystic bolt]

Thank you so much

.close

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[pearl pollen]

To count the radius, do I have to count the lines on the x and y plane with no indicated numbers like for example, if I only count the lines on the plane in Circle A then the radius will be 3 but if I don't it will be radius 6 which means I'm counting a decimal aswell

[safe radish]

@pearl pollen Has your question been resolved?

[dapper venture]

radius is 6 squares which is 3 units

[pearl pollen]

wdym 3 units?

[dapper venture]

I mean unit length

They labelled 1 unit length every 2 sqaures

[pearl pollen]

Woah so this is wrong

,w rotate

[flat frigate]

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[pearl pollen]

Wrong command

So this is wrong?

[safe radish]

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[ripe dune]

Can someone help me with this

[rotund crest]

you have all the x values

so put them into the equation

thats given

[ripe dune]

thank you

[rotund crest]

np

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[twin solar]

Hey, I am supposed to solve this differential equation
$$
y''(t)-y(t)=e^{-1}\mathcal{H}(t)
$$, so using a fourier transform I got it into this form
$$
\hat{y}(\omega)=\frac{1}{(-\omega^2-1)(1+i\omega)}
$$ and from that i got this integral
$$
y(t)=\int_{-\infty}^{\infty}\frac{e^{i\omega t}}{(-\omega^2-1)(1+i\omega)}d\omega
$$, is it correct?

[flat frigate]

konxmok

[twin solar]

where $\mathcal{H}(t)$ is the Heaviside step function

[flat frigate]

konxmok

[safe radish]

@twin solar Has your question been resolved?

[twin solar]

<@&286206848099549185>

[twin solar]

[safe radish]

@twin solar Has your question been resolved?

[safe radish]

@twin solar Has your question been resolved?

[safe radish]

@twin solar Has your question been resolved?

[safe radish]

@twin solar Has your question been resolved?

[safe radish]

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[safe radish]

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[flat frigate]

woods

woods

[safe radish]

@smoky minnow Has your question been resolved?

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[fading iris]

How do I find the area in Square meters of the composite figure

[fading iris]

Like idk what to do

<@&286206848099549185>

Idk

Idk the steps

Like actually

Idk how to find area

A shape

Yeah

Yeah

[lean otter]

Area of a circle is pi r^2

[fading iris]

Ok

Uh

Which one is the radius again?

10

Ok

So

Pi10^2

I got 314

Uh

Ok

Um

[lean otter]

What

[fading iris]

Dude idk

[lean otter]

What's the area of a rectangle

[fading iris]

I was held back

Length times with

[lean otter]

Yes

Now do that for the entire rectangle

[fading iris]

Ok

[lean otter]

What do you get

[fading iris]

Hold up

12800

[lean otter]

No

[fading iris]

Oh wait

3200

[lean otter]

Yes

[fading iris]

Oh so it’s 80 x 40

[lean otter]

Now what is the area of the small rectangle

[fading iris]

600

[lean otter]

Good job

[fading iris]

Yessir

[lean otter]

Now that you know the big rectangle equals 3200

And the small equals 600

How can you find the red part

[fading iris]

Huh

Add

[lean otter]

subtract

[fading iris]

Oh sorry

3200-600

2600

Ok

Area

Right?

20x20

400

400-20pi

D?

<@&286206848099549185>

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@fading iris Has your question been resolved?

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[west hedge]

Natural log isn’t not part of the curriculum in my region.

[lean otter]

The derivative of lnx is 1/x

[west hedge]

Mickeydawg

I never thought Mickey could be another animal than a mouse

[west hedge]

Should I just memorize it?

Or the proof for it is really simple and intuitive

[lean otter]

Be surprised

[lean otter]

Yes

You should've memorized a lot of the derivatives previously

[west hedge]

Is this the definition of natural log?

I just found it on my desk

I wonder what it is used for

[wild mountain]

for magic

[lean otter]

You mean the derivative?

[west hedge]

I digress

Lets refocus on the original question

I’m about to make some claims

Correct me if I’m wrong

I need some break

I’m stressed

.close

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[raw thistle]

im stuck after number 5

ive been on this worksheet for way too long lmao

[light shoal]

solving for x?

for 6, try multiplying by the denominators

btw, what do the colors mean

[raw thistle]

theres a mosaic on the back that my teacher wanted us to color in

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@raw thistle Has your question been resolved?

[raw thistle]

.close

[safe radish]

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[lean otter]

uh hi

[lean otter]

sorry idk how this works but i had a question

abt sinusoidal functions

[severe pond]

ask

!da2a

[safe radish]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[lean otter]

The main problem i have is getting what their concept is

like usually the way i learn things is like breaking them down by what is it that this math is for

[severe pond]

like what the trig functions represent?

[lean otter]

yes

and what they are for/ whats happening

[severe pond]

what have you heard they represent

[lean otter]

waves

[severe pond]

but are you aware of right triangles

and what the ratios

of the sides mean

and how we relate them with these trig functions

like opposite sides adjacent sides and the hypotenuse

[lean otter]

oh?

i mean i remember something abt that

[severe pond]

https://youtu.be/dUkCgTOOpQ0?si=YShX1lULzq8wV0hH

[lean otter]

thank you that looks good already

[severe pond]

the different trig functions just represent the ratios between the lengths of the sides of triangles given an angle

[lean otter]

yo

[severe pond]

yo

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@lean otter Has your question been resolved?

[safe radish]

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[pulsar cosmos]

can anybody explain what the steps for this assignment are asking me?

[pulsar cosmos]

like for part 1, what do they mean at 1 to start with a true statement?

[oak wraith]

state a fact

[pulsar cosmos]

my fault for not asking the teacher when i had the chance, but if anyone could explain that would be great

[oak wraith]

this seems related to trig so maybe something like

hmmm

[pulsar cosmos]

so like anything? evenlike 1 + 1 = 2

oooo

[oak wraith]

the sum of the interior angles of a triangle is always 180?

[oak wraith]

lol probably want it related to trig

[pulsar cosmos]

haha yeaah

[oak wraith]

oh

maybe pick one of these

[pulsar cosmos]

hmm okay so i could just do like cos^2(x) + sin^2(x) = 1

thats a true pythagorean statement right

i kinda was guessing thats what that part might mean but then for 2 it's saying to apply identities to it

which i do not understand lol

[mortal arrow]

yes

[pulsar cosmos]

🙏

[mortal arrow]

the applying identities part is using those 4 identities to change it up i think

[pulsar cosmos]

ahh okay

so take any parts from other identities too

[mortal arrow]

kinda like the double angle identity where cos(2a) = sin^2 - cos^2 = sin^2- cos^2** + cos ^2 - cos^2** = 1 - 2cos^2

[pulsar cosmos]

ohh yeah

[mortal arrow]

suppose so

[pulsar cosmos]

so i could take the id that says 1 + tan^2(x) = sec^2(x)

and isolate the 1 so its 1 = sec^2x - tan^2x

then put the one into the sin + cos one

so it would be like sin^2x + cos^2x = sec^2x - tan^2x

or am i misunderstanding it

[mortal arrow]

i suppose

you could probably simplify a lil more tho

[pulsar cosmos]

simplify where? at the right hand side?

idk it says to have something complex

anyway i think i get it way more now :) thanks guys!

i will come back if i am confused again 👍

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[split fulcrum]

I need some help with starting for part a)

[delicate bobcat]

If you draw a segment from O to A and O to B

They would both be radii of the circle

Therefore they would be equal in length

And they would form an isosceles triangle with the chord of the circle

[split fulcrum]

They would be each 6cm in length right?

[delicate bobcat]

Yes the two radii would be

OA and OB

[split fulcrum]

And I’m trying to find angle O with this information

[delicate bobcat]

Yes

So you can make a triangle with those 3 lengths involving angle O

You would also know all three of the side lengths

[split fulcrum]

right

[delicate bobcat]

That would be sufficient to figure out the angle

Depending on what trig stuff you've learned in your class there are a couple ways to do it

I'm assuming/hoping you've at least done some right triangle trigonometry

Otherwise I'm unsure how to help from there

If you know law of cosines you can find the angle directly with that

If you've only covered right triangle trig relationships then you have to manipulate the figure a bit more

If you have not discussed trig functions at all I'm not sure how to proceed

[split fulcrum]

Yeah I’m familiar with trig stuff I can probably do it

[delicate bobcat]

If you have not yet seen law of cosines

Cut the isosceles triangle in half with a segment that goes from O to the chord

[delicate bobcat]

This will cut the triangle into 2 congruent right triangles

Oh yea you did know it already cool

[split fulcrum]

Yeah

For part B how should I go about it?

[delicate bobcat]

Well the way i can reason out without special formulas of some sort would be like this:

Once you know the central angle AOB, then you can determine the area of the sector "behind" the isosceles triangle

So you'd do 360 degrees - measure of angle AOB

To get the reverse central angle

Then you could also find the area of the isosceles triangle

The shaded area is:

Area of Entire Circle - Area of Isosceles triangle - Area of the big sector behind the triangle

[split fulcrum]

Hm

Is there a way I can calculate the area of the sector

and then just get that and subtract the area of the circle

Wait

That wouldn’t do anything

Nvm

[delicate bobcat]

Well actually i over complicated it lol

You could find area of the small sector

And subtract the triangle

I did a roundabout way of finding it

You already know the central angle for the sector

The shaded area is just the sector with the triangle removed

So you'd do

(Angle/360 degrees ) * area of circle to find the sector area

[split fulcrum]

Oh yeah sector of AOB and then subtract that with the area of triangle AOB

[delicate bobcat]

Yea exactly

And you have a couple ways to get the triangle area

[split fulcrum]

I’ll try it now

[split fulcrum]

How’s this?

Not sure if I made any mistakes

[delicate bobcat]

Well it's a good sign that everything ended up being rational (besides pi)

[split fulcrum]

haha yeah

[delicate bobcat]

I will leave the calculations to you though lol

[split fulcrum]

I think this is the simplest I can make it

[delicate bobcat]

You can divide out the 3

36 and 27 both divisible by 3

36pi/3 is really 12pi

[split fulcrum]

Oh shit that didn’t even occur to me 36 is divisible by 3 lmao

[delicate bobcat]

Lol yea arithmetic is the hardest part afterall

That's why i don't do it lol

[split fulcrum]

haha

So that leaves me with 12pi + 9

Which sounds pretty good I think

[delicate bobcat]

Yea that looks neat so that is a great sign

[split fulcrum]

okay, thank you for the help

[delicate bobcat]

Yw gl with your schoolwork

[split fulcrum]

thank you!

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[trail otter]

let $V$ be vector spece over $\mathbb{F}, v \in V, W$ is subspace of $V$. Set $v+W$, whose all ements are sum of emelts from $v$ with some element from $W$, is called affine subspace in V. How can i proove that $v+W$ is subscpace of $V$ when $v\in W$ ?

[flat frigate]

Slowaq

[final halo]

What happens when you take a subspace, and add an element of that subspace to all other elements

[trail otter]

itll still be subspace wont it ?

[final halo]

Yeah, which subspace will it be

[safe radish]

@trail otter Has your question been resolved?

[trail otter]

Well since we took element from sub space W and added it to sebspace W I think it’ll be W once again

[final halo]

Yep it will! Maybe try and prove this rigorously, i.e. if you fix v in W you can write any element w in W as v + u for some u in W

[trail otter]

Hm I am having hard time with those rigorous proofs should I add anything to this statement you wrote?

[final halo]

Well you need to prove the statement I wrote

If you add an fixed element of W to everything in W, youre going to stay inside W since it's a subspace, that much is clear

The question is whether you can reach everything in W by doing this, as only then will v + W = W

[trail otter]

Ugh unfortunately I don’t have a clue on how to prove this could you dhow me please

[final halo]

Youre looking to solve v + ? = w

[trail otter]

since W is subspace of V then $\forall r,s \in \mathbb{F}, \forall v,w\in W$ holds that $rv+sw \in W$ thus we can get any $w$ from $W$ by doing linear combination of two elements from W for example $v+u=w$

[flat frigate]

Slowaq

[trail otter]

is this somewhat close to needed proof? @final halo

[final halo]

You need to specifically tell me what u is

The game is: I give you an element of W, w, and you tell me what I have to add to v to get to w

[trail otter]

is it some element from W which is eqaal to w-v?

[final halo]

That's exactly what u needs to be yes

And we know w-v is in W so v + (w-v) is in v + W

[trail otter]

nice so just to recap when to some element v from W we add another element u from W we span the whole subspace W and since v+u=w thus u has to in in form of u=w-v. we never left W because any linear combination of elemnts from W yields another element from W thus we stayed in W and W is subscpace of V thus v+W is subspace of V when v is from W

is this correct? @final halo

[safe radish]

@trail otter Has your question been resolved?

[safe radish]

@trail otter Has your question been resolved?

[safe radish]

@trail otter Has your question been resolved?

[upper nymph]

https://tenor.com/view/tesla-albert-einstein-madwolfofōkami-gif-13692030474614117380

help

!ship

bknb kj

onegai... naruto

https://tenor.com/view/na-naruto-sennim-gif-18628840

[trail otter]

.close

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[void musk]

guys, i struck on drive 10 from this process

[void musk]

anyone?

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@void musk Has your question been resolved?

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[dapper shell]

Is this work correct

[dapper shell]

is the formula in red correct?

[safe radish]

@dapper shell Has your question been resolved?

[dapper shell]

<@&286206848099549185>

[graceful heron]

Are you trying to integrate by parts?

[dapper shell]

yes

[graceful heron]

Then it is wrong.

For your f(x) and g(x),

You need to choose which one is more suitable to differentiate and which one is more suitable to integrate.

[dapper shell]

is the formula correct atleast

[graceful heron]

No, you shouldn't divide.

Its multiplication

And on the first term there shouldn't be a dx

Big F means primitive of f

[dapper shell]

i dont understand

[graceful heron]

Primitive = antiderivative

[dapper shell]

so integral?

[graceful heron]

integral of f

Yes

[dapper shell]

is this still correct

[graceful heron]

First term is correct, second one is not

it should be u'(x) * integral of v dx

the vdx should not be inside those parentheses

[graceful heron]

almost! now you removed the big integral sign

[dapper shell]

yea

[graceful heron]

oh wait, you're right here

Yes you're correct.

[dapper shell]

oh wow thank you

[dapper shell]

goodnight o/

[graceful heron]

gn!

[safe radish]

@dapper shell Has your question been resolved?

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[quick cloud]

What's the square root if the opposite of I?

[hard crest]

as in like

$\sqrt{-i}$?

[flat frigate]

ヘイリー

[safe radish]

@quick cloud Has your question been resolved?

[quick cloud]

Yes 👍

[visual kraken]

Well complex numbers have a really nice property where multiplying two of them multiplies their lengths and sums their angles

[quick cloud]

Oh, that's sick!

[visual kraken]

So since -i has length 1 and angle 3pi/2 (or 270 degrees), you should find the number that has length a and angle b such that a*a = 1 and b + b = 3pi/2

[quick cloud]

So is that that answer?

[visual kraken]

Well that's how to find the answer

There's a pretty specific answer in terms of a+bi (Different a and b than before)

[quick cloud]

Ah ok

So it would be a complex number right?

[visual kraken]

Yep!

[quick cloud]

Ah ok

Ty!

[visual kraken]

No prob

[quick cloud]

!done

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[visual kraken]

Real quick

[quick cloud]

Ye?

[visual kraken]

Something else you might find useful is this identity

$e^{i\theta} = \cos(\theta) + i\sin(\theta)$

[flat frigate]

NotABot

[quick cloud]

Oh yeah!

I've seen this!

[visual kraken]

Yeah it's a really cool and useful identity

[quick cloud]

Ty for showing me!

[visual kraken]

If you can find which value of theta gives -i, it's pretty easy to use that identity to find sqrt(-i)

No prob 🙂

[quick cloud]

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[tardy garden]

cosec(x) = -sqrt(2) and x is in Q II, what is the value of cotan(x)

i tried doing it like this, by trying to find x first then just plugging that in cotan

1/sin(x) = -sqrt(2)
sin(x) = -sqrt(2) / 2

but doesn't this show that x can't be in the second quadrant?

[tardy garden]

for sin to give a negative value x must be in quadrant III or IV, no?

[plucky elk]

Can you screenshot the original question

[tardy garden]

it's in french

[plucky elk]

Do it

[tardy garden]

.rotate

!rotate

[plucky elk]

,rotate

[tardy garden]

"if cosec(x) = -sqrt(2) and x is in QII, what is the value of cotan(x)"

[plucky elk]

Yea quadrant 2 is between pi/2 and pi

Maybe your teacher made a mistake

[tardy garden]

she makes a lot of mistakes so that is possible

ok then

[plucky elk]

Probably meant quadrant 3

But only she knows

[tardy garden]

thanks

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[worthy frost]

Literally don't understand what im doing wrong

can't i do this transformation?

the same here

why I can do it at the bottom, but not at the top?

[hard crest]

that's probably -3 / -0

[worthy frost]

I don't understand

[worthy frost]

is the transformation here wrong?

[hard crest]

no, that transformation is fine

but when you end up with -3 / 0, it's not clear whether that's positive or negative infinity

[worthy frost]

is this an indeterminate form?

i only see [0/0] is indeterminate, not [a/0]

[hard crest]

0/0 is indeterminate because it could be anything

-3 / 0 is infinite in one way or another

but like

$\lim_{x\to\y} \f{-3}{-e^{-x}}$ is clearly $\y$

[flat frigate]

ヘイリー

[hard crest]

even though naïvely it would be -3 / 0

[worthy frost]

oh i see

I don't see how i could change this transformation though

[hard crest]

i mean you really don't have to do any of that

it's pretty clear that it's increasing since it's built out of increasing functions

[worthy frost]

oh yeah

what am i doing

so i cant just say x+sqrt(x^2+3) is infinite, since its [inf+inf], but I can do lim x->inf x + x*sqrt(1+3/x^2) = 2x = infinity

[hard crest]

think it's just x, not 2x, but whatever

because it'll reduce to x * sqrt(1)

[worthy frost]

won't it be x + x*sqrt(1)?

[hard crest]

oh sorry i got lost in the text. yes

[worthy frost]

oh okay

thanks for the help

sorry, my brain wasn't working

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[pallid fractal]

The answer is 0 and negative numbers, but how do I illustrate that in this notation?

[pallid fractal]

Is it just x<1

[true zephyr]

well,

x<1 would include .5 as well

and |.5| isnt -.5

you want to use $\leq$

[flat frigate]

znsn

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[lofty mesa]

doing calc ab work with related rates and i dont understand the coffee pot problem at all 😭😭😭

[lofty mesa]

.close

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[hard crest]

write $\sqrt{x}$ as $x^{\sfrac12}$

[flat frigate]

ヘイリー

[indigo parcel]

I tried that

This is one of the ways I've tried to work it out

[hard crest]

instead of doing product rule between x^6 and x^1/2

try combining them

x^13/2

like you can use the product rule and that's valid, but it's more effort

[indigo parcel]

Thank you! Trying that now

I have a tendency to overcomplicate things

[hard crest]

(you can do a similar thing for that other term)

[indigo parcel]

?

No it looks like that's wrong too

Aaaaaa okay I can do this

Maybe?

Nope

Apparently this is the answer

Any idea how that happened?

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[indigo parcel]

Oh whatever

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[vocal kiln]

Need help with some U-sub

[vocal kiln]

Lemme convert to latex

$2\pi \int_{0}^{3} \frac{x}{1+\left(x^2\right)} ,dx$

[flat frigate]

KAVA

[vocal kiln]

Im not sure how to u-sub this

[stoic dune]

Sometimes it's easier to start with the derivative you should get, and then work backwards

[vocal kiln]

1/2x?

[terse lichen]

the denominator has 1+x^2, think about the derivative of that

[stoic dune]

In this case, you have two choices:
du = x dx
Or
du = 1/(1 + x²) dx

One of those is much easier to work with, and ends up working out well

Let's say du = x dx
What would u = ?

[vocal kiln]

Its been a while Im not certain 😅

Still shaking off the rust

u would just be constant yes?

[stoic dune]

Let's say, for fun:
u = x²
Then what would du be?

[vocal kiln]

Oh its anti derivative

du = 2x

[stoic dune]

Close. du = 2x dx

Note the very important implication that dx is "multiplying" the 2x here

[vocal kiln]

What does that do to the function?

[stoic dune]

We can even rearrange that:
du/2 = x dx

[vocal kiln]

I see yes

[stoic dune]

Btw, there's an x dx in the integral. We can now replace x dx with du/2

As they are the same

[vocal kiln]

Oh true

[stoic dune]

2π ∫ 1/(1 + x²) du/2

Now, we don't want a mixed integral. Having an x and u at the same time is terrible

[vocal kiln]

Agreed

[stoic dune]

Remember, we have these two equations:
u = x²
du = 2x dx

It would be nice if we could convert all x to u

[vocal kiln]

Yes

[stoic dune]

I'll let you decide what should be replaced here

[vocal kiln]

Probably good to sub in the u

1/1+u

is it proper form to have du/2 though?

[stoic dune]

It's just a multiplicative constant. If you like, you can move the 1/2 outside the integral

[vocal kiln]

Oh dang

Im gonna try to work it from here

Then return

if I have issue

But close for now

Thank you for your help

.close

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[calm snow]

i need help

[hard crest]

.close

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[modest cloak]

I'm gearing up to start Calc 2, and I need to make sure I'm well-prepared. Any advice on what specific topics and concepts from Calc 1 I should review to set myself up for success in Calc 2? Any recommended resources?

[modest cloak]

functions, limits, derivatives, integration? what else?

[shadow glade]

yeah probably just be comfortable in computing derivatives (easy) and integrals (takes more practice), be familiar with the big theorems like the mean value theorem (used in many proofs) and fundamental theorem of calculus, the major trig identities, review delta epsilon concepts to make sure it has clicked

be clear also on the definition of integrals as riemann sums since definitions of more complex integrals in terms of riemann sums will also happen

[modest cloak]

thank you

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[steel crypt]

I've used L'hopitals to arrive at this ugly looking equation. How do I got about solving it? n > 0, and a,b are integers >= 2

[obsidian stream]

k

one min

it's easy

1st

divide by ln(2)

[frozen marlin]

...

u sure u dont mean multiply

[obsidian stream]

in num and denominator

[obsidian stream]

bruh no

[frozen marlin]

?

dude that'll make it ln^2 (2)

[obsidian stream]

huh

[frozen marlin]

;-;

[obsidian stream]

see this

[frozen marlin]

yk what idk how to solve limits anyway never really done em

imma leave u to help this dude

peace

[obsidian stream]

Wait

Divide my ln(2 on num and denominator)

ln(2) will dissaperar in denominator

[lean otter]

split the fraction in this fashion (a+b)/c = a/c + b/c

[stray socket]

Since it's an infinite limit, you only need to focus on end behavior. You should know that as x -> infinity, 1/xln(a) = 0, so we don't care about that part. Everything else can be done analytically using end behavior, once again

[obsidian stream]

answers 2^b

[frozen marlin]

damn

[steel crypt]

I was thinking about ignoring that term, but wasn't sure if I was able to

[stray socket]

Well i mean

Its the best way imo

Because if you simplify it it's legit just $\lim_{x\to \infty}\frac{\ln(2) + 2^x2^b}{\ln(2) + 2^x}$

[flat frigate]

Umbraleviathan

[stray socket]

You know ln(2) = O(2^x) so you only focus on the 2^x term, effectively just erasing ln(2) from the top and bottom

[lean otter]

...

[steel crypt]

Its been a while since I've taken a calculus course, and its popping back up again in my algorithm analysis class, so I've been trying to remember how to go about these

[stray socket]

If you have a function F(x) = f(x)/g(x), and you evaluate an infinite limit for F(x), you only really need to analyze to whatever dominates f(x) and g(x) for whatever infinity

[steel crypt]

How is that done?

[stray socket]

Let's consider something simple for now like

(x^4 + 6x + 7)/(6x^2 + ln(x)), and we evaluate for x -> inf

[steel crypt]

Thanks to everyone else who helped, btw

[stray socket]

We can say that f(x) = x^4 + 6x + 7 and g(x) = 6x^2 + ln(x)

We gotta look at the highest power term for each

Which is x^4 and 6x^2 respectively

[stray socket]

So this is equivalent to evaluating the limit as x -> inf for x^4 / 6x^2

Likewise for your problem, we identified what term is dominant as x -> infinity

[steel crypt]

So if the limit was a constant, like 2^b, using this approach would still allow us to find the answer = 2^b?

[stray socket]

well

[steel crypt]

Meaning for this problem, instead of L'Hopital, I can just ignore loga(n), and simplify (2^n * 2^b)/(2^n) = 2^b?

[stray socket]

Establishing end behavior, the leading term for the numerator is 2^x 2^b and the denominator is 2^x

[steel crypt]

Should be n -> inf, sorry

[stray socket]

So you get (2^x 2^b)/2^x

Which sinplifiies down to 2^b

Which is a constant

[stray socket]

You can ignore log(n)

[steel crypt]

My brain is a bit fried right now, but why is it we can ignore a term that would go to infinity (log(n)) in order to find 2^b?

[stray socket]

That's because log(n) = O(2^n). In other words, to extreme values, 0 < log(n) < C 2^n where C is some constant

So log(x) kinda becomes nothing

[steel crypt]

So we can also apply log < exponential in a limit, and ignore the log

[stray socket]

Like a pebble to a whole mountain

[stray socket]

Pretty much

It's a bit discrete analysis but i mean

[steel crypt]

Well that makes this question a lot easier

[stray socket]

It does kinda rely on you knowing what certain parent functions look like

[steel crypt]

I was able to prove loga(n) + 2^(n+b) was big-theta(2^n) using the c1g(n) <= t(n) <= c2g(n), but had to do it again, only using limits

So wasn't sure exactly what I was and wasn't allowed to do when proving with limits

[stray socket]

I mean that also works

Colloquially, just something that indicates growth

[steel crypt]

I had to show using both approaches, so they MAY want me to actually use L'hopitals and do the calculations to get lim = 2^b

Thanks, and thank you @obsidian stream and @lean otter

[stray socket]

Splitting the fraction ends up working for the first one

You don't need to do L'H

Limits are a linear operator

[steel crypt]

I tend to make things more complicated than they have to be.

[stray socket]

It be like that

The more you know the more you think lmao

[safe radish]

@steel crypt Has your question been resolved?

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[vagrant oriole]

For question #6 I got the part where the boats move north and west but how would I go about finding the bearings?

[safe radish]

@vagrant oriole Has your question been resolved?

[uncut jungle]

Hello, the bearing is the angle that is swept when you imagine a vertical upwards (north) line from the first point, and rotate it clockwise until it hits the second point

So for example bearing from North to East is 90degrees, South to North is 0degrees, North to West is 270 degrees

[vagrant oriole]

Would I just write 270 degrees for the bearing of Boat A (21 mi) to Boat B (18 mi)?

[uncut jungle]

Sorry on reflection, those examples are wrong, but the first point is what i meant

It would be the angle 180 degrees + the top angle of the triangle

Because you imagine a north line at the top point, sweep it until it hits the second, so it is the 180 sweep to the right side of the triangle, and then the top angle more to sweep all the way to B clockwise

Does that make sense?

[vagrant oriole]

I am understanding what ur saying but for the top of the triangles angle would I have to find the hypotenuse first and then use Cos?

[uncut jungle]

Just use tan, you have the opposite and adjacent

But yes, that would work as well but longer

Here's a drawing of the bearings so the concept is clearer

[vagrant oriole]

For B how come the direction used is North and not West?

[uncut jungle]

The bearing is always the angle swept from north

[vagrant oriole]

Ooh ok

I found A to B (220.6 degrees) but for B to A im having trouble finding the bearing

Would I also have to find the angle on the inside first in order to find the outside on?

[uncut jungle]

There is a relation between angles here because of parallel lines

(alternate angles)

[vagrant oriole]

Oh so it would be 40.6 if I am understanding you correctly.

As that is the angle I found on the top with tan

[uncut jungle]

Yep 😄

[vagrant oriole]

Thank you very much, now I will go struggle on my other problems 🫡

[uncut jungle]

No problem, best of luck!

[vagrant oriole]

.close

[safe radish]

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[violet gale]

given the graph of f(x) de function goes trough point (1,18) and has a minimum/maximum at (-1,2). give the function of f(x)

[violet gale]

this is the graph that it is supposed to represent. i know how to do calculation when you get 2 coordinates and the y coordinate of the y axis

but not this method

[sullen lotus]

this isnt a good qn

but assuming f(x) is quadratic it should be manageable

do u know how to solve?