help-10

[finite hawk]

How do I show that tana = sinA/cosA

Or the other one

[worn yoke]

it uses SOH CAH TOA

[finite hawk]

Yeah

What is question 2

Asking

[worn yoke]

so we call the opposite side "a", the adjacent side "b", and the hypotenuse "c"

[finite hawk]

How fo they get to question 2 I somewhat get q1

[worn yoke]

it is asking you to show, that for any angle A, it's true that [\sin^2 A + \cos^2 A = 1 ]

[warm shale]

pnoןɔ

[worn yoke]

and this is proved using the pythagorean theorem

[finite hawk]

So why does a/c^2 + b/c^2 =c^2

How do I even know what c^2 is equal to

[honest forum]

for any acute angle

[worn yoke]

the pythagorean theorem says that [ a^2 + b^2 = c^2 ]

[warm shale]

pnoןɔ

[finite hawk]

Where does the (c•sinA)^2 come from

[worn yoke]

SOH

write that out and solve for a

[finite hawk]

Isn’t soh A/C

[worn yoke]

by SOH we have $\sin(A) = \frac ac$

[warm shale]

pnoןɔ

[worn yoke]

so what is a?

[gloomy edge]

Adjacent

[finite hawk]

Oh sin A * c

[worn yoke]

yes

so we substitute that into the pythagoren theorem formula

[finite hawk]

So I’m subbing

Oooh

Okay

I think I get it now

Ty

I appreciate u

Thx

.close

[obtuse pebble]

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[obtuse pebble]

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[tiny spire]

What rules graph 2 out for this. They are quite similar I am having trouble seeing the differences here

[brave bramble]

For graph 2, bottom left looks nearly vertical

,calc e^(-2)(1 - (-1))^2

[warm shale]

Result:

0.54134113294645

[brave bramble]

But the slope should be pretty small

[tiny spire]

Alright that makes sense thank you : )

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[copper cape]

Hello

[copper cape]

Merp

Welp, i start to giving it up...

[gloomy edge]

Uhhhh, what?

Is there an English version of this?

[surreal forge]

i believe it is asking to find the point of intersection

[copper cape]

Line (blabla) is tangent to (blabla) in the point (a,b). For a < b, value of (a-b) is...

[copper cape]

Nope

I tried

[surreal forge]

wdym "Nope"

[copper cape]

Its tangent

[barren zenith]

Wait im indo ill do it

[surreal forge]

a tangent point is still a point of intersection...

[copper cape]

KudaBangke

Tangent doesnt cross

I googled it

It says i need to down grade the fynction

[barren zenith]

Iy harus dicari turunannya

[copper cape]

Fu(x)

[barren zenith]

Iy yang pertama jg harus diturunin relasinya jd

11+3dy/dx=0
dy/dx=-11/3

Nah dari sana 2 turunan ini harus sama pada titik (a,b)

Jadi kamu masukin ke fungsi yang sudah diturunkan

[barren zenith]

Ini varieblenya diganti a

= -11/3

Setelah ketemu a-nya tinggal lu masukin ke f(x) yg awal

Nanti a nya kalau di beresin bakal ketemu
a1=1 dan a2=3
Dr sini bisa dicari b nya
b1= -7/3 dan b2=5

Karena a<b
Yang dipake a2 dan b2
a-b=3-5=-2

[copper cape]

How about this...

Kalau emang rumus (persamaan1) = (persamaan2) itu berlaku universal

Berarti sama ajah kalau 2 buah garis punya titik temu entah itu perpotongan atau singgung?

[barren zenith]

There 2 persamaan at play here

The f(x)=g(x)
And f'(x)=g'(x)

Yang kita pake disini itu yang kedua

Karena tangent

Dan kebetulan angkanya enak jadi gw pake yg kedua

Secara umum

Kalau soal grs singgung

Lu akan pake 2 2nya

Udh gitu jd persamaan 2 variebel

Jd beda dengan perpotongan

Karena kalau perpotongan cuma pake persamaan yg pertama

I must ask u are indo right bcs i'll feel real stupid here if u werent and u were helpin a friend out or smth

Lmao

[copper cape]

Bentar bang 🗿

Memahami

[barren zenith]

Okay wkwkwkkwkwk

Siap

[copper cape]

11+3, turunannya -11/3

???

[barren zenith]

Bkn bkn itu tuh sbntr gw tulis dikertas

Jd gt

[copper cape]

Turunan...

Derivative

Saya punya byk pertanyaan, tapi Bro yg satu ini bukan helper pribadi

[barren zenith]

Santuy weh

Klo mau add friend boleh

[barren zenith]

Mungkin lu lebih paham klo dr awal persamaannya diubah
Jd
Dr 11x+3y-48=0
Jd y=-(11/3)x+16

Baru yg ynya diturunin

Jd lu y'=-11/3

[copper cape]

Koefisien tetap ada

Konstanta menghilang

Panggilan untuk istirahat...

[barren zenith]

Iyes

[barren zenith]

Wkwkwkwkwkw

[copper cape]

Tapi bener kan?

[barren zenith]

Iyh

[obtuse pebble]

@copper cape Has your question been resolved?

[obtuse pebble]

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[formal latch]

this question is probably simple, but how would I tell if the bottom function gets smaller than the top function as n -> infinitiy

like it crosses over

Is it just by degree

what if the degree has the variable in it

like would y = 5^x be greater than x^10

[elder inlet]

is it possible to use a limit comparison test here?

[formal latch]

that's true

[elder inlet]

but in terms of degrees, we can see that e^(4/n) approaches e^0 which would be 1, and e^4/n^2 will approach 0 as we go to infinity, quicker because of the degree

[formal latch]

I was just wondering if there was a rule to see if a function is smaller than the other

[elder inlet]

i would just imagine the limit of each sub-function like 4/n, n, and n^2 and see how they relate to the whole of the problem, however, it may not be enough for a proof

[obtuse pebble]

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[formal latch]

thank you

[obtuse pebble]

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[timid silo]

Find y x z

Explain it. Give me hint by hint. Dont just answear it.

[dusk widget]

okay, have you tried anything?

@timid silo

[timid silo]

Tired to find the big triangles, its a no.

[dusk widget]

I see

okay

[timid silo]

Since they arent the same size, still no.

[dusk widget]

well, the first thing to notice is that we have two small triangles in the big triangle

[timid silo]

yeah.

[dusk widget]

what can you say about those two triangles?

[timid silo]

All their sides have a square attached.

Also..

Pythagorian...

[dusk widget]

!

okay

[timid silo]

We have 2 sides

[dusk widget]

well

[timid silo]

2 90 degres angle.

[tawny copper]

similar.......

[dusk widget]

we can apply Pythagoras to many things here

[timid silo]

Rlly?

[dusk widget]

mhm

we have more than one right triangle

[timid silo]

But everywhere we have more than 2 unknown variables.

And nothing is same like 2x

[tawny copper]

Similar triangle

[tawny copper]

Use that to establish system of equations

[timid silo]

y2 + z2 = 5squared?

[tawny copper]

Yeah thats works too

[timid silo]

But the lengths arent the same.

[tawny copper]

?

Do u know whats a system of equations is?

[timid silo]

Brb I will keep trying.

[timid silo]

Oh,,

So put them in the same condition?

BRB, I will try by my self.

[tawny copper]

Good

Doing it by yourself is the best method is learn probably

[timid silo]

I will ping you soon, sorry for the disturb..

[tawny copper]

There's 3 unknows x y z

So u need to establish a system of 3 equations

[timid silo]

With 3 equations never done it before..

[tawny copper]

Its really easy

[timid silo]

3 of those?

[tawny copper]

Yep

But with z

[timid silo]

f

[tawny copper]

Like

A system but with 3 equations

[tawny copper]

There's 2 in this system

[timid silo]

Ik. I need 3

[tawny copper]

here's the color-coded, use pytha on 3 triangles to find 3 equations

[tawny copper]

Good job

Solve it

Here's how

For example this

[timid silo]

Solve for one, and fill others?

[tawny copper]

get x^2 term on one side, everything on the other side

[tawny copper]

Yep

[timid silo]

Ok let me try.

[tawny copper]

At the end, u should have 1 equation with 1 unknow

Which is solvable

[timid silo]

Xsq = ysq-1

Now fill?

[tawny copper]

Dont square it

U see this

x^2 term appear again

So just sub that in

[tawny copper]

sub this

[timid silo]

Ok

[tawny copper]

into this

[timid silo]

Now I have xsq and ysq

I will find zsq

[tawny copper]

Thats weird

[timid silo]

Lol, what did I do?

[tawny copper]

Where's the 4^2 goes?

[timid silo]

Oh let me check.

[tawny copper]

Hmm

Z and 2 looks alike, I think u should write Z like this

[timid silo]

Idk where to substitute the Xsq.

[tawny copper]

Z^2 = 20 then y^2 = 25 - 20 = 5

[timid silo]

Never seen these kinds of system.

[tawny copper]

U should get used to it

I will see it alot

[timid silo]

So in a system if it doesent solve in first try, we try again?

[tawny copper]

No?

The point of such system is that we can turn them into 1 equation with 1 unknown

[timid silo]

I mean in the second try it still have 2 unknown answears.

[tawny copper]

Yeah

Then we sub it again

So reduce the unknown

[timid silo]

Never before.

First time.

Let me try.

[timid silo]

Wait...

[tawny copper]

Dont use my answer

Try to get your own

[timid silo]

A question: when we have something like this: (zsq - 15)+ zsq = 25, do i always have to switch sides to get -()?

[tawny copper]

2 zsq = 25 + 15

Then zsq = 40/2 = 20

[timid silo]

I get it, but just for furture problems.

[tawny copper]

Its a 1 variable equation

U can do whatever u like

[timid silo]

So how to remove prentecese

[tawny copper]

prentecese?

[timid silo]

Yes.

A question: when we have something like this: (zsq - 15)+ zsq = 25, do i always have to switch sides to get -()?

[tawny copper]

where is the -() ?

[timid silo]

If orenteces are on the other side, i can because - changes signa.

[tawny copper]

U could just treat them the same like (x - 15) + x = 25
and find x

Thats all

[tawny copper]

This term cancel itself out

if u want to?

But then u get zsq = 25 - zsqu + 15

[timid silo]

Is there any easier away, if the prenteces are necessary?

[tawny copper]

Then u have to switch zsqu back

[timid silo]

(like when they cant just disappear.)

[tawny copper]

It will

If's not then there's no answer so either u make a mistake along the step or the teacher made a mistake when they made the question

Dont worry abt this @timid silo

If u not sure, u can use a calculator to solve it for u

[timid silo]

I am so sure about the concrete example.

And its done.

[tawny copper]

Almost all calculator has this functionality but u should stick to do it by hand first

[tawny copper]

Make sure u sqrt them

[timid silo]

But I was curious if I had an exact problem and would do something wrong...

[tawny copper]

U can use calculator to check

Take a pic of ur calculator

I will show u

[timid silo]

I use photo math plus.

[tawny copper]

No idea

[tawny copper]

Then thats it, u solve everything

Good job

[timid silo]

Photo math is an app that you take a picture of a math problem and it solves it.

[timid silo]

An example:

[tawny copper]

Yeah I dont like using app that do all the stuff for u

[timid silo]

Lol.

[timid silo]

Any advice before I thank you?

[tawny copper]

Hmm

No idea

Practice more?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[timid silo]

@tawny copper

[obtuse pebble]

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[tawdry vigil]

Can someone explain how this is differentiated using the product rule?

[last pilot]

why'd you want to use the product rule?

there's a very specific formula that can be applied here

[ember frost]

you COULD use product rule, but like ren said, why

[warm shale]

ren

[tawdry vigil]

You mean the quotient rule?

[tawdry vigil]

Ik, I was just curious

[last pilot]

yes but this is faster

[last pilot]

this is basically quotient rule with v = f(x) and u = 1

[tawdry vigil]

Oh ok

[tawdry vigil]

Do you mind explaining how it could be done using the product rule?

Just out of curiosity

[last pilot]

i mean sure

hang on

actually

yea no idt this is differentiable via product rule

[ember frost]

you can use prod rule, but you would still have to use quotient rule then

[last pilot]

^^

so purely product, no

[ember frost]

since youre differentiating the reciprocal of lnx

[last pilot]

exactly

inverse?

u mean reciprocal

[ember frost]

yes

[last pilot]

anyway, we get 2 log e / log x

then we gotta use quotient

bc we can't really differentiate log_x c w/o quotient

[tawdry vigil]

Oh ok

[last pilot]

mhm

[obtuse pebble]

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

[tawdry vigil]

Thanks

.close

[obtuse pebble]

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[last pilot]

np

[obtuse pebble]

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[meager mortar]

what is going on with this regression 😭
theres no other equation using the same variables

[meager mortar]

https://www.desmos.com/calculator/dgdbdzu5sv
this is the desmos link
(the equations and table under the first 3 entries are just some other regressions im trying with this section of data points)

[obtuse pebble]

@meager mortar Has your question been resolved?

[meager mortar]

nvm

.close

[obtuse pebble]

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[unreal marlin]

How do I do circles angles

[rocky shale]

??

[bronze nova]

circle theorems probably

[cinder basalt]

wat

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[unreal marlin]

This kind of stuff. This is one of my example but I don’t understand it

[cinder basalt]

use pythagoras

[unreal marlin]

How do I apply it?

[dim temple]

what angle are you trying to find

in there

[cinder basalt]

i think u trying to find the equation of some circle

[dim temple]

same thought

equation of the circle

[unreal marlin]

It did not tell me

[cinder basalt]

center (h,k)=(4,7) and r=4

[dim temple]

in that example, you are most likely trying to find the equation of a circle

[unreal marlin]

Yes

[dim temple]

since the center is (4,7)

and the raidus is 4

just substitute those values in the standard form of a circle

,,(x-h)^2+(y-k)^2=r^2

[warm shale]

Renz

[unreal marlin]

OH!!! Thank you

[dim temple]

h will be the x-coord of your center, which is 4

k will be the y-coord of your center, which is 7

for raidus, just count it in the graph

[unreal marlin]

Ok thank you!

.close

[obtuse pebble]

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[raw plinth]

I have a 3D physics engine for rectangles but I need to make it apply rotational velocity based on the collisions I detect with SAT. I have looked and found that to do this I need to get the contact points but I cant find how to get them between two 3D objects. Pls tell me if you know how I could get them or if you know a better way to get the needed rotational velocity. I have the vertexes for all cubes, their normals, the normal of the collision and the depth. as well as their center location

[obtuse pebble]

@raw plinth Has your question been resolved?

[obtuse pebble]

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[frozen geode]

Hi there, I have this integral which I'm sure should be simpler, but I've already tried different methods and I've always come to a dead end.

[warm shale]

Argón

[frozen geode]

The limits part for evaluating if it converges or diverges is fine. I'm having trouble with the integration

[fossil crag]

not sure if this function has a nice integral

[obtuse pebble]

@frozen geode Has your question been resolved?

[hybrid gyro]

@frozen geode what level of education are you in?

This question looks like it might be solvable with Feynman's Technique (but that was with sin(x)/x and I've never learnt or used it myself so I can't be sure).

Otherwise, the closest thing I can think of is to use complex analysis

where you try to do this as a chosen contour in the complex plane and try to use Cauchy's Theorem/Cauchy Residue's Theorem etc. to find this real integral. But this requires you've learnt a lot of complex analysis knowledge. If you've never learnt it then this method is not recommended

[obtuse pebble]

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[timid silo]

Decide which of the statements regarding the solution of the equation log3(6x−2)log3(x−3)=2
is correct.
Question 9Select one:

a.
None of the remaining possibilities is correct.

b.
The sum of all solutions is 12.

c.
The equation has two solutions and their product is 10.

d.
The equation has one solution.

e.
The equation has no solution.

[obtuse pebble]

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[obtuse pebble]

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[spiral void]

yuh hello, i cant wrap my head around this 😭

[spiral void]

also my wifi is messing up 💔

i cant think on what i should do

[dim temple]

alright so

what have you noticed in the diagram

[spiral void]

uhhh theyre parallel?

[dim temple]

particularly with these two

[spiral void]

yuhhh

[dim temple]

yeaa

they are parallel

meaning that these 2 angles must add up to what?

[spiral void]

yeah but the line isnt one intersecting line

i mean

[spiral void]

these ones

[dim temple]

they do

[spiral void]

180

[dim temple]

yea

[spiral void]

yeahh

[dim temple]

whats the angle on the top then

[spiral void]

104

[dim temple]

yea correcct

[spiral void]

subtractionnn

WOOOOO

[dim temple]

leme draw a new one

[spiral void]

okokok

[dim temple]

the pic is small to draw on

[spiral void]

😭

take your timee

thank you btw 🙏

[normal cairn]

while he does that, look at the upper triangle

its a special one

[dim temple]

^

[spiral void]

yeahh

isosceles

[normal cairn]

mhm, so what do we know about the angles in an isosceles triangle

[spiral void]

OHHH TWO ARE THE SAME

[normal cairn]

the ones at the base right?

[spiral void]

YEAHHH

[normal cairn]

combining that with the knowledge that angles in a triangle add up to 180, can you find the two angles in it?

[spiral void]

OH EM GEE

i can just divide 76 to 2 right?

[normal cairn]

yeah that works

[spiral void]

38

[dim temple]

these are 38, since they are comgruent like you said

since its an isoceles

now

[spiral void]

mhm

[dim temple]

if the red is 36

and the one in the middle is 62

[spiral void]

ohhhhhhhhhh

[dim temple]

and the bottom one is an unknown

[spiral void]

ADDITION

[dim temple]

what do they add up to?

they form a line, the green one

[spiral void]

THE BOTTOM MISSING ANGLE IS 80

but not the x

sooooo

76+80

then minus by 180

??

[dim temple]

yea

[spiral void]

WOOOOOOO

[dim temple]

180-(76+80)

[spiral void]

THANK YOUUU

🙏

[dim temple]

whats the final answer

[spiral void]

to bill too

24 💪

[dim temple]

correct

[spiral void]

which issss B

WOOOO THANK YOU

byee

.close

[obtuse pebble]

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[dim temple]

youre welcome

[obtuse pebble]

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[opaque mesa]

how to find the minimum of the function $f(x) = \sqrt{(8-0.5x)^2+(x-4)^2} \times 60 + 50x$

[warm shale]

General_Jacob

[surreal forge]

honestly i would just differentiate that

[versed stratus]

Differentiation

[opaque mesa]

and how to solve this?

$\frac{1}{2} \times (8+0.5x+2(x-4)) \times (\sqrt{(8+0.5x)^2+(x-4)^2})^{-1} \times 60 + 50 = 0$

[warm shale]

General_Jacob

[opaque mesa]

i'm not sure if it's correct

I got this from wolfram

[obtuse pebble]

@opaque mesa Has your question been resolved?

[obtuse pebble]

@opaque mesa Has your question been resolved?

[obtuse pebble]

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[rugged mountain]

Hi i need help please

[rugged mountain]

Is there a mistake at r?

[hybrid gyro]

what is the question?

[obtuse pebble]

@rugged mountain Has your question been resolved?

[obtuse pebble]

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[grizzled meteor]

ive tried to find the gradient of both functions and set them equal to each other but it said my answer was wrong. also that doesnt lead to two answers, so im really not sure what to do

[restive gorge]

Basically we are trying to look for such gradient that is orthogonal to the normal vector of the plane, right

[grizzled meteor]

im not sure, to be honest

because the final answer would be a point

or two points more specifically

[restive gorge]

the situation

[grizzled meteor]

yeah

i think i understand what the question is asking, jsut not how to go through with it

[restive gorge]

we need f_x f_y and f_z is my guess

[grizzled meteor]

yeah that would make the gradient, i just dont know what then to set it equal to

[restive gorge]

$\grad f = \begin{pmatrix} f_x \ f_y \ f_z \end{pmatrix}\$

$\ \begin{pmatrix} f_x \ f_y \ f_z \end{pmatrix} \cdot \begin{pmatrix} -4 \ 5 \ 4 \end{pmatrix} = 0\$

[warm shale]

𝔸dωn𝓲²s

[grizzled meteor]

OHHH let me try that hold on

[restive gorge]

This would be a linear system

[grizzled meteor]

where would the second equation come from

[restive gorge]

scalar product

If our gradient vector is orthogonal to the normal vector

then it's parallel to the plane

[restive gorge]

like here orthogonal to the black normal vector would go along the blue plane (parallel basically)

[grizzled meteor]

so i got -16x+10y+40z=0 for the first one

[restive gorge]

implicit differentiation?

we need to do partial

or implict partial

[grizzled meteor]

i just did the dot product of the gradient of the surface and the gradient of the plane

[restive gorge]

can yoou tell me first you derivatives

[grizzled meteor]

f_x = 4x
f_y = 2y
f_z = 10z

and then i did g(x,y,z) for the plane so
g_x = -4
g_y = 5
g_z = 4

[restive gorge]

then you get

-16x + 10y + 40z = 0

[grizzled meteor]

yeah thats what i put earlier

[restive gorge]

yea wait

hmm

what is next lemme think

my next step would be

to solve that for z

and plug that into 2x²+1y²+5z² = 1 for z

[grizzled meteor]

and then that would only leave x and y?

[restive gorge]

yea (hopefully)

you get solutions for x and y

[grizzled meteor]

i would still need a system of equations since there are two unknowns

or

could i sub it into both f(x,y,z) and g(x,y,z) and then that could be my system?

that would be really nasty though it doesnt feel right

i would have like x and x^2 and xy and y^2 all at the same time

[restive gorge]

hmm when we differentiate we get equations

4x = 0
2y = 0
10z = 0

[grizzled meteor]

the point (0,0,0) is incorrect just checked

[restive gorge]

ofc it's incorrect

i just differentaited

i was just pointing out we dont have f_x, f_y and f_z to begin with

[grizzled meteor]

ahhh i see

i might just have to take the L on this one honestly 😭

[restive gorge]

ahh

we have to treat z as afunction of f(x,y)

[grizzled meteor]

oh?

[restive gorge]

then it makes sense

yes

it's like

x² + y² = 1

here you treat y as a function of x

now analogically

z(x,y)

then solve for f_x YES

ok let's try it

[grizzled meteor]

so

[warm shale]

𝔸dωn𝓲²s

[restive gorge]

So

I have to think

[grizzled meteor]

do we do the same thing to the equation of the plane then?

[restive gorge]

yes

I hope it works out

[grizzled meteor]

ill tag you if i get it

[warm shale]

𝔸dωn𝓲²s

[grizzled meteor]

wait so from there

is there a way to get another equation to make a system

[restive gorge]

We get y = 8x/5 basically

[grizzled meteor]

i got y=8x/5

multiply both sides by z

and then you have (8x/5)-y=0

so y=8x/5

[restive gorge]

oh yes

[grizzled meteor]

but then how do we find x and z is my question

[restive gorge]

so

if we substitute y for 8x/5 in the gradient

and then maybe we can choose x

and z

then we got one gradient

and plug in x and z into the original equation to get y

i will check it

[grizzled meteor]

but the gradient is a vector so how would substitute

or rather how would substituting help

[restive gorge]

(0,0, 1/sqrt(5))

does this work?

[grizzled meteor]

unfortunately no 😭

[restive gorge]

weird

i think we need a second contraint

I subbed this into y=8x/5 intial equation and we get a circle solution

but that's not quiet it

yet

[grizzled meteor]

im honestly completely lost at this point LOL

[restive gorge]

we still have the derivatives

to get the tangent plane of f

$z = f_x(x-x_0) + f_y(y-y_0) + z(x_0,y_0)$

[warm shale]

𝔸dωn𝓲²s

[restive gorge]

$-4x + 5y + 4z = 9 \iff z = x-\frac{5y}{4}+\frac{9}{4}$

[warm shale]

𝔸dωn𝓲²s

[restive gorge]

we should tell that z(x,y) = 9/4 for some x and y

or no

gimme one try haha

[grizzled meteor]

okokok

i trust you 🙏

[obtuse pebble]

@grizzled meteor Has your question been resolved?

[restive gorge]

still on it

[grizzled meteor]

youre good! im working on some other problems lol

[restive gorge]

I found one point it's ugly

[grizzled meteor]

let me try it

X AND Y ARE CORRECT!!!!

z is wrong though

[restive gorge]

put a minus then before z

-4sqrt(905)/905

[grizzled meteor]

nope

[restive gorge]

ahaha why is this weird this task

how can x and y be correct then

[grizzled meteor]

no idea LOL

[restive gorge]

but the point is in the equation

also i got more than 2 points so that's even weirder

[grizzled meteor]

yeah i really dont know

oh wait

sorry that was my mistake

the original z was correct i just accidentally put 181 instead of 905

[restive gorge]

knew it

now for a second point

[grizzled meteor]

yup

[restive gorge]

switch signs

like + to - and - to +

since i got like 4 solutions

[grizzled meteor]

its right !

thank you so much omfg 😭

[restive gorge]

wait

I want to send my work so you can see what I did

i did a bit of research

but that wasnt enough

had to improvise

[grizzled meteor]

i really hope i never see a problem like that again but just in case id def like to see your work haha

[restive gorge]

it's not the problem but the fucking numbers

So our intial thoughts were correct, actually

to find the gradient of F we differentiate everything partially actually

Then now comes the trick

since we have the gradient

of F

the normal vector of our gradient has to be the same as the normal vector of our given plane to for some constant k

that constant k is very important

because if were to equate without a k like previously we wont find solutions

so we want to find such k so that everything is satisfied in our system

[grizzled meteor]

OHHHH

wait that makes so much sense

[restive gorge]

now the difficult part was k

in other problems sometimes x = 0 or y = z or some shit

so i didn't know what to do next since i got everything in terms of k

so i plugged instead k into F to get k lol

so i think what i really did was find such k so that i find the matching gradient

[grizzled meteor]

yeah yeah

[restive gorge]

then i got x and y which i need

and you get z by plugging in x and y from the gradient into F

and that's it

[grizzled meteor]

jesus christ

[restive gorge]

4 points actually

plug in plug out shit

[grizzled meteor]

its been 2 hours exactly by the way 😭

[restive gorge]

that's nothing

i had worse especially when i didnt was on the level i am now lol

or had a helper

but i also learnt something new

[grizzled meteor]

so should i close it now?

[restive gorge]

you can if that's it x)

[grizzled meteor]

that should be it for now hahaha

thanks for your help its been nice hanging out with you lol

.close

[obtuse pebble]

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[obtuse pebble]

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[cedar oracle]

how would i convert this parametric equation into polar form, where it's r = something

[latent walrus]

so x=5+2.5cos(t), y=3+4.5sin(t)
[(x-5)/2.5]^2+[(y-3)]/4.5]^2=1

then x=rcos(theta), y=rsin(theta)

could maybe work something from there

[cedar oracle]

well i got rid of the translation and ended up with this

but its slightly off

[cedar oracle]

i used r^2 = x^2 + y^2 to get to this

[latent walrus]

was t a stand in for theta?

[cedar oracle]

yes

[latent walrus]

ah, ignore my stuff then

[cedar oracle]

it wasnt working with theta

i dont thnik i need to include the translation so i think it would be fine to just have it as (3.5cos(t), 4.5sin(t))

[latent walrus]

pretty sure it matters

you need the 5+ and 3+

[cedar oracle]

oh

well i asked my teacher and he said that its fine if i just consider it without translation when converting

so if it was just (3.5cos(t), 4.5sin(t)) how would the converting work

[latent walrus]

how you did it

[cedar oracle]

because i just did the x = 3.5cos(t), y = 4.5sin(t), x^2 + y^2 = r^2

[cedar oracle]

yeah but its slightly off for some reason ☹️

[latent walrus]

try it without ignoring the translation

[cedar oracle]

no its like way off 💀

[latent walrus]

oh wow

[cedar oracle]

i think ignoring the translation is best

[cedar oracle]

this is what im confused about

because it should work right

any ideas?

[latent walrus]

just having a think

[cedar oracle]

alr

[latent walrus]

unsure of any mathematical reason

at least none i can see

[cedar oracle]

YEAH EXACTLY 😭

maybe desmos problem?

approximation

probably not desmos is pretty good

[latent walrus]

as you can see, the do align

[cedar oracle]

huhhh

how did you get it to be a point

ohhhh

nvm

you set the inside to t

oh it does align

oohhhhh

[cedar oracle]

wow tysm

ig it just isnt visible when still

ok thank you for everything

how do i close this

.close

[obtuse pebble]

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[obtuse pebble]

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[lusty spruce]

can someone explain how he got to T1 = 8sin50/sin70?

[tawny copper]

Triangle method

\

\

[obtuse pebble]

@lusty spruce Has your question been resolved?

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[knotty locust]

Hi, when doing mathematical induction, would it be possible to do P(k - 1) -> P(k) instead?

Base case would remain the same. But I assume the Induction Hypothesis would differ?

In this case, rather than 1^3 + 2^3 .... n ^3 = n^2 * (n+1)^2/4 , it would be (n-1)^2 * (n^2) / 4?

[knotty locust]

Here is my work if needed, not sure if its a valid proof

[astral silo]

it would be the same really, but then i would think u'd have to show 1 more of the sequence after P(1) or 0 or whatever the 1st one is but idk

[knotty locust]

kk thanks

[astral silo]

just do n^2(n+1)^2/4+ (n+1)^3 = RHS

i guess

[knotty locust]

ight be thanks :)))

[knotty locust]

but I think im proving LHS, so wouldnt RHS stay the same ?

so no (n+1)^3 ?

(i have almost no idea of what im doing)

[astral silo]

yea

wait whar

[knotty locust]

like why is there (n+1)^3 on the right ? I dont really get that part

[astral silo]

induction u need to use (assume) P(k) is true to prove P(k+1) is true, u need to use P(n) in some way

[knotty locust]

yea I tried doing that but it is more complicated with the binomials

[astral silo]

RHS is (n+1)^2(n+2)^2/4

[knotty locust]

yea but thats if you started with assumption of P(k) tho right

[astral silo]

LHS is p(n) + the last number which is (n+1)^3

yea

[knotty locust]

since I did P(k-1), then I would have normal "fomrula" on the right

ok so we chilling then I think

[astral silo]

probably yea

i dont think teachers will stress too much

except for exams

[knotty locust]

yea this is just practice so far

[astral silo]

exam markers want to see easy layouts that they can skim through

and they dont want to have to double check if ur other method is correct or not

[knotty locust]

oof yikes LOL

[astral silo]

cuz waste of time

but thats quite minor it should be fine

[knotty locust]

maybe ill just stick with k + 1 then

[astral silo]

yea better be safe than sorry

[knotty locust]

Huge help man, highly appreciated!

.close

[obtuse pebble]

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[astral silo]

np u are doing fine with induction

didnt need my help really

[knotty locust]

Just needed reassurance : >

[obtuse pebble]

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[flat rune]

am i right with everything here

i know the first pic is kinda blurry but its the best i could do

please let me know

<@&286206848099549185>

[worn yoke]

the whole process looks correct

[flat rune]

thank you so much

[obtuse pebble]

@flat rune Has your question been resolved?

[obtuse pebble]

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[granite dome]

need help verifying this

[broken tendon]

Do you know your trig identities?

[granite dome]

only some

[broken tendon]

This could be useful

[granite dome]

in class we were taught pythag identities, reciprocal identities, even odd identities

and we can only use those for now

[broken tendon]

I will be happy to help you tomorrow, because I need to go to sleep rn

Sorry

[granite dome]

it's due tomorrow morning 😭

its ok tho

i'll wait for someone else

[lapis quartz]

i gotchu blud

you know to rationalize?

[granite dome]

ya

[lapis quartz]

@granite dome

ok try rationalize the eqn

by multiplying and divideing by sint+cost+1

[granite dome]

so just multiply by conjugates on num and denom?

[lapis quartz]

ye

[granite dome]

to get squared functions

[lapis quartz]

take conjugate of denom

[lapis quartz]

ye

[granite dome]

👍

one sec

[lapis quartz]

use proper use of parenthesis generally during rationalizing you get a^2-b^2 format and (a-b)^2 or (a+b)^2 format

[granite dome]

would both the top and bottom be the same cuz

same factors

[lapis quartz]

not getting you

[granite dome]

because the conjugate of the numerator is equal to the denominator and the conjugate of the denominator is equal to the numerator would they end up being the same when u multiply it out

[lapis quartz]

ye u are right but once we get in terms of higher degree it might be easier to cancel each other out

i told you the method of rationalizing cuz both are conjugate of each other

[granite dome]

i see

i got this when i multiplied it out

not sure if that’s right

[lapis quartz]

uhh i see just one sec i will tell you about this in detail

ima show my work one sec

[granite dome]

tysm

[static furnace]

numerator should have 2 positive sin(theta)'s

wait

what are you multiplying

[lapis quartz]

you should multiply and divide by same conjugate

[granite dome]

oh

[lapis quartz]

you have just made the equation equal to one lol

[granite dome]

so which one should i use

the one in num or denom

numerator right?

[lapis quartz]

anyone will do try denom

[granite dome]

i might try num because two terms in num and one in denom

or is that flawed logic

[static furnace]

the conjugate should only change the sign of the one, not the trig functions yea?

[granite dome]

explain what u mean

[static furnace]

i could be wrong, but i did get the correct answer. conjugate of sinx+cosx-1 is sinx+cosx+1

just as the conjugate of -sinx-cos-1 would be -sinx-cosx+1

[granite dome]

isn't the conjugate just changing the signs in between terms

i'm not familiar with trinomial conjugates but i'm pretty sure it would function the same way as binomial conjugates where the signs just change

[lapis quartz]

you can take any conjugate

does not have to change only the sign of 1 alone

[static furnace]

as long as you only change 1 sign

[granite dome]

oh

[static furnace]

yes that

[granite dome]

did not know that

[lapis quartz]

you take take (sint+1)-(cost)

then conjugate it

whatever you wish

[granite dome]

wait what

that's the denom right

i'm just gonna try it

i'm gonna multiply by (sint+cost+1)

[lapis quartz]

sorry for bad handwriting

here i have taken conjugate of denom then multiplied and divided it

then on numerator i have written in terms of (sint+1)+cost

in denom (sint+cost)+1

[static furnace]

i think you have some missing terms

[lapis quartz]

i did this since its easier to use (a-b)(a+b)=a^2-b^2 property

[lapis quartz]

i dont think so

[granite dome]

ok i’m here now

[lapis quartz]

NICEU

[lapis quartz]

now do you see in the denom we have sin^2+cos^2?

[granite dome]

set equal to 1?

[lapis quartz]

ye

[granite dome]

then that will get rid of 1 right

[lapis quartz]

yes

[granite dome]

👍

[lapis quartz]

now next step is little hard to see but if we use some trial and error of substituing we can cancel out 2 on numerator and denominator

in numerator we can write sin^2t as 1-cos^2t

[granite dome]

ok then from there now what

[lapis quartz]

one sec

[granite dome]

not really sure how this leads to the answer

[lapis quartz]

just one minute

[lapis quartz]

you forgot one +1 term on numrator

[granite dome]

oh ya

[lapis quartz]

now add both cos^2t

you get -2cos^2t

[granite dome]

yup

[lapis quartz]

now take 2 common out

then cancel 2

or just divide all terms by 2

[granite dome]

simplifies to that right

[lapis quartz]

you again forgot that one

[lapis quartz]

here after you add +1 to num you get +2

[granite dome]

so what would it look like

now

[lapis quartz]

something like this

[lapis quartz]

just trace back to this step add +1 on num

[granite dome]

ok did it

[lapis quartz]

now

do you see 1-cos^2t?

[granite dome]

identity right

[lapis quartz]

ye

[granite dome]

sin^2t

[lapis quartz]

yep

[granite dome]

then sin in denom

cancels

[lapis quartz]

yes sir

got your answer?

happy?

[granite dome]

yes bro

tysm

[lapis quartz]

np

[granite dome]

after the last test my grade tanked from 91 to 86

[lapis quartz]

trignometric simplication is about trial and error

[granite dome]

i got a 66 on an inverse trig test 😭

[lapis quartz]

ahh its ok

try to solve more problems

thats the only way

[granite dome]

ya

[lapis quartz]

alright bro take care

[granite dome]

i'm p sure this one was the hardest on the whole worksheet tho

thanks so much man

[lapis quartz]

np

close it

[granite dome]

.close

[obtuse pebble]

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[obtuse pebble]

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[brave moon]

i need help understanding what a question is asking

[brave moon]

ill send it one sec

i know about span and stuff

like i understand what i means

i just dont know what this question is asking me to do

i got the right answer

i kinda guessed though

i just thought that its basically s times (-5 1 -5)

and put that in

[obtuse pebble]

@brave moon Has your question been resolved?

[brave moon]

.close

[obtuse pebble]

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[obtuse pebble]

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[sinful crest]

How would I take the curl of this expression

[thorn rose]

the curl is in terms of spatial derivatives, just chug that?

[sinful crest]

But to me I'd need to specify the components along which the wave travels?

[thorn rose]

\vec r is just the position vector no?

does the euclidean basis not work for this problem

[sinful crest]

Hmm yeah I think it would.

so just rewrite r in it's components then go as usual?

[thorn rose]

try it

[obtuse pebble]

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[obtuse pebble]

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[red badge]

how do you do this

[ember frost]

what have you tried

[red badge]

all i know is that i cna move it to 1/cos(x+45) =1

right

cause tis the recipricole

i usually work in radians its why im confused

.close

[obtuse pebble]

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[obtuse pebble]

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[south olive]

hi

[dim temple]

hello

[dim temple]

anything you need help with?

[obtuse pebble]

@south olive Has your question been resolved?

[obtuse pebble]

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Remember:
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Please don't occupy multiple help channels.

[surreal forge]

sooo

do you have work?

[obtuse pebble]

@timid silo Has your question been resolved?

[tawny copper]

sqrt(x^2) = abs(x)

[obtuse pebble]

@timid silo Has your question been resolved?

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[spiral void]

hi guys i have a question

[spiral void]

sooooo

[lapis quartz]

so what?

[spiral void]

when u add two sides of a triangle it should be larger than the third side right?

[lapis quartz]

yes