help-49

[last slate]

sent a picture btw

<@&286206848099549185>

[midnight plank]

@last slate Has your question been resolved?

[wet pollen]

Give me a moment while I help someone else first.

[last slate]

okkk

[wet pollen]

Ok, so when you encounter a situation as you did with y = 0 and lambda = 1, treat them as two possible cases.

[wet pollen]

.

[last slate]

Alrr

[wet pollen]

That seems to be where you got stuck anyways, am I correct?

[last slate]

I think so

How do i go about solving after i determine that though?

[wet pollen]

So you have the case of y = 0. Given all of the equations you came up with and were given, which ones could you use that with?

[last slate]

The constraint equation?

[wet pollen]

Yes.

[last slate]

But wouldnt that only give me that -xz = 64

[wet pollen]

Sorry for the wait, I just needed to verify what I was doing.

[wet pollen]

Now use that result in one of the other two lambda functions you found to solve for either x or z.

[last slate]

Ohhhh i see

I think i understnad now

Thank you!

[wet pollen]

yw

[last slate]

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[turbid spire]

DE

[turbid spire]

arcsin(y'') = x(y')

why is degree not defined

can we just say
if the argument of trig funcs got derivative inside , the degree is always negative

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[lethal path]

f'(0) is defined because from the left, we have 2(0) + 2 = 2 and from the right, we have 2

So the limits from the left and the right are equal

We're using properties of continuity here since we are given that the function represents f'(x) already

(and when x = 0, we have x >= 0 for the top branch, so the limit equals f'(0) = 2(0) + 2)

For the second image, the slope of the graph is clearly different from the left and right of each jagged point

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[last slate]

If a set of eigenvalues containing n eigenvalues has some non-simple elements, does that mean it is not possible to form n linearly independent eigenvectors from that set?

[last slate]

by "simple" I mean each eigenvalue having an algebraic multipicity of 1

[nova yoke]

you mean linearly dependent eigenvectors, i assume?

[last slate]

yeah edited sorry

[nova yoke]

no, it does not necessarily imply that

for example, the identity matrix has only 1 for an eigenvalue, repeated n times

and every nonzero vector is an eigenvector

so you can take the standard basis for example

what you're asking about is the geometric multiplicity of the eigenvalue, as opposed to the algebraic multiplicity

you always have GeoMult <= AlgMult for each eigenvalue
GeoMult is the dimension of the eigenspace associated with the eigenvalue

so the situation you're asking about is when GeoMult is strictly less than AlgMult

which can happen

for example: $$\begin{pmatrix}1 & 1 \ 0 & 1 \end{pmatrix}$$

[grand pond]

Bungo

[nova yoke]

1 has AlgMult 2, but GeoMult 1

[last slate]

right, so i guess in a way like

you can have an eigenvalue repeated 5 times for example

but maybe u will only produce 3 linearly independnet eigenvectors from those eigenvalues

[nova yoke]

yep that can happen

[last slate]

so that eigenvalue has a geo multiplicity of 3 and the eigenspace (the space formed by the span of th eigenvectors) will be of dimension 3

[nova yoke]

correct

[last slate]

like do they span R^3 or is it entirely different

[nova yoke]

they span a 3 dimensional subspace of whatever R^n they live in

at some point you'll learn about the jordan canonical form

which puts the matrix into a form where you can see at a glance the whole structure

you can see what are the eigenvalues, and what are their algebraic and geometric multiplicities, at a glance from the JCF

[last slate]

oh woah thats cool

okay thank you i think i understand this now

[last slate]

But wait just to conclude

For my original question

considering our set, assume we have only 1 element with an algebraic multiplicity m, with the rest being simple, and no knowledge of what the geometric multipicity of that element is

then we can assure ourselves of there existing at least n - (m + 1) linearly independent eigenvectors right?

since i assume each eigenvalue needs to have at least one linearly independent eigenvector

[nova yoke]

well if there are a total of k distinct eigenvalues then you're guaranteed at least k linearly independent eigenvectors

[last slate]

right

[nova yoke]

since eigenvectors associated with distinct eigenvalues are linearly independent

(if the matrix is symmetric then you can even choose them to be orthogonal)

[last slate]

right

[last slate]

and what about the case i mentioned? like with onlu one non-unique eigenvalue

[nova yoke]

haha let me do the arithmetic in my head, i'm working with 2 sips of coffee so far

1 eigenvalue with algmult m, the rest are simple, so you have n-m other eigenvalues, and a total of n-m+1 eigenvalues including the first one

so yea

at least n-m+1 LI eigenvectors

n-m+1, not n-(m+1)

[last slate]

haha no worries sorry about that

okay gotcha! i think i understand this topic now

much appreciated bungo

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[nova yoke]

sure

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[regal dome]

I have to write it as one log but im not sure if my answer is correct. My answer is 3log(9x^2)

[slender walrus]

can you show your work

[regal dome]

yes give me a second

[slender walrus]

what laws are you applying to get from the second to third line

[slender walrus]

you don't quite have that here

[regal dome]

oh yeah youre right

hmm what should i have done?

how do i get rid of the 2

[slender walrus]

apply power law to the
$$\frac{\log_{3}(y)}{2}$$

[grand pond]

ℝαμΩℕωⅤ

[regal dome]

How does the powerlaw apply here

[slender walrus]

what does the power law for log say

[regal dome]

isnt that just for multiplying

[slender walrus]

wdym

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[regal dome]

idk my brain is fried at this point

could i do 3log(y)^-2

[slender walrus]

no

before you do something, state the specific law you are trying to apply

also state the power law i recommended you apply earlier

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[umbral rampart]

correct or nah

[turbid spire]

correct

[umbral rampart]

thx

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[echo perch]

are there a total of 4 possible answers to this question?

depending on which pivot you choose

or are there infinitely many solutions

or maybe there is only 1 solution? i went with x1 as the obvious choice for the pivot, but i haven't learned about column operations yet, only row operations (RREF)

I suppose RREF is not even needed here, with only 1 row

so i'm gonna guess infinitely many basis, not just 4, but what do you think?

and if that's true, i wonder how WebWork can check this, if there are infinitely many answers

[midnight plank]

@echo perch Has your question been resolved?

[robust isle]

indeed there are infinitely many bases here

[robust isle]

hopefully they have some custom computation behind the scenes to check that your answer is actually a basis

you could try and see if the site is good or not

try and replace the first vector by [-1 1 1 0] (added the first and second vector together, the span doesn't change)

[echo perch]

interesting

to be "proper" basis it doesn't need to have a single "1" for each pivot? for each vector

in terms of the actual math (not WebWork), it doesn't matter, it's still the same basis, just a different way to write it?

[robust isle]

well no it's a different basis obviously

they describe the same vector space though (ie the vectors (x1, x2, x3, x4) for which 7x1 + ... = 0)

[echo perch]

oh right, different basis but to describe the exact same vectors

i look at basis as conversion system, like an oven that is F and an oven that is C

both get the same result if you follow instructions for temp and time

so changing the basis is like putting a skew filter in photoshop onto a vector grid system. the grid lines still match up to the vectors, they just look different to the naked eye

[tidal token]

3Blue1Brown has a great video on this

[echo perch]

yes, that's exactly where i got this analogy from 🙂

helpful to understand basis

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[sage knoll]

hello !!
i kinda need some help on this cuz i really dk how to approach it

[sage knoll]

got this from somewhere as well so yea

<@&286206848099549185> a

i guess i found the area to be 18 using geogebra but how i actually prove it

i know it forms 3-4-5 triangle but is it just a coincidence??

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[mystic flower]

hi

[last slate]

Ello

[mystic flower]

need help with the top half

if you can it would be appreciated

[last slate]

2x^2 +3x - 2 = 0

No gcf so you factor then

Idk #3 I’m only in grade 11

[mystic flower]

this is grade 9

[last slate]

I never learnt about a zero product

[mystic flower]

oh

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[mystic flower]

thanks

[last slate]

Oh I get it

@mystic flower

[mystic flower]

.reopen

[midnight plank]

✅

[last slate]

Aren’t u just finding the zeros of the factored equation?

[mystic flower]

yes

[last slate]

So what do you not get?

[mystic flower]

idk its confusing

[last slate]

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[mystic flower]

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[woven citrus]

i think it is d

but i am not too sure

[still nacelle]

that's not correct

[woven citrus]

i thought with jump disscontinuity the x does not exist

[sleek gorge]

The plus means look at right side

[still nacelle]

that is if the question asks for the limit as x->1, not 1+

[sleek gorge]

If it was minus it would mean end behavior on left side

[woven citrus]

is it 3

c

[still nacelle]

yup

[woven citrus]

k makes more sense

thanks

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[normal plover]

howd they get theta = 60 and 180

nvm

.clse

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[strange river]

What aspect of this question do you need help with?

[signal rock]

i solved it just now

but i have another question

i have got 10(rt3 -rt3/3)

[livid wyvern]

maybe some factorization will advance the problem

[signal rock]

how do i get this i have got 10(rt3 -rt3/3)
to their answer

ok i got it

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[strange river]

Use your triangle identities

Oh

Too late

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[plain gazelle]

How do I even start?

[last slate]

Look for dilapidated idols

[fallen vigil]

The algebraic rule is that we add (b/2)^2

That is how you answer 1.

[plain gazelle]

so we add 6?

by 6

so 36

[fallen vigil]

Yeah

[plain gazelle]

ahhhh

[fallen vigil]

Here is the geometric reason

The reason we do b/2 is so we can form the two rectangles, which gives the edge lengths of the bigger square we add

[plain gazelle]

i think i did this one wrong are you meant to multiply 2 with -4?

[fallen vigil]

,rotate

[fallen vigil]

wdym

if ur doign the slide and divide thing yeah

[plain gazelle]

is my answer correct?

wait nvm it is

if its like can be divided by the first number then it will be fine but if its not then i would multiply the first number by the end number

[fallen vigil]

ye

You need to multiply by 2 though if u only have (x-2)(x+1) as the final answer

since that is only x^2-x-2

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[twin kernel]

Where did r/h from

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[latent elk]

Hint:

Cut it into triangle

Then get one triangle's area then multiply with 6

Fun fact

[tacit blade]

Just use sss after finding angles

[latent elk]

@last slate

[tacit blade]

Each angle is 120

Which means you have a 60 60 60 triangle

[latent elk]

Dont hand it out for him

He should figure it out himself

Then whats the height?

[tacit blade]

Do you know how to find area via trig?

Or only using side av

[latent elk]

Here's another hint

Well done

[latent elk]

.

Area of Octagon = 6 * one triangle

No?

Show the work of how u calculate the height

Use pytha theorem is much faster

Google use radian

not degree

So sin60 = sqrt(3) / 2

Is it?

Even that

Google calc gives out decimal number

Well, u can use whatever u like

Wdym

[latent elk]

Look

See the green triangle

Use pytha on it

U can only use pytha for right triangle

No

Whats the pytha theorem states?

Whats c?

Whats a then

💀

Dude

You need a lot of practice

That is a very terrible way of doing math

[latent elk]

Any teacher who endorse this way of thinking shouldnt be a teacher\

b is the height

Did ur teacher only teach Pytha theorem to calculate the hypothnuse

get a new teacher then

[latent elk]

b is the height, so set b as a variable

plug in c and a

Then find b

GOOD

Now get b

Is a calculator too expensive for u? Cause having one is a must if u want to be good with math

Yep

now get one triangle area

*6

Then u get the final answer

Thats weird

Oh yeah thats correct

My calculator = 216V3

= 374.1229744

[latent elk]

Nope

@last slate

Theta is in radian

Not degree

U can convert from deg to rad by :

Rad = Deg * pi / 180

So 120 deg = 2pi/3 rad

Its just a number in an unit

If u haven't learnt it

U could think that Area of the arch = x * area of Circle

Where x is just a number, and u need to find it

Hint: ||It has to do with the angle 120 deg||

No

Whats the angle of the whole circlce

Ok

here's th way to find x

If 360 deg = the whole cirlce (100%)

Then 120 deg will take up how much percent of the whole circle

Hint: ||Triangle method||

Yep

So that area will = 33.333% of the total area of circle

33.3333% is just 1/3

Multiply

33.3333 percent

No 33.33333

33.33333 percent = 0.3333333 = 1/3

Yep

Good job

Remember the unit

cm^2

[latent elk]

.

what is this?

This is the correct answer

u just need to plug cm^2 at the end

Oh

No

Sorry

U can use the same method to find the arch length

Since 120 deg take up the same percentage of the whole circle arch so just 1/3 * whole arch

Yeah

[midnight plank]

@last slate Has your question been resolved?

[viral dagger]

angle 3x+10 and 6x-28 create a line, or a 180 degree angle, which means 3x+10+6x-28=180

do you know how to continue it?

[midnight plank]

@last slate Has your question been resolved?

[viral dagger]

$$3x+10+6x-28=180$$
$$9x-18=180$$
$$9x=180+18=198$$
$$x=22$$

[grand pond]

Skill_Issue

[viral dagger]

do you know how to continue this for y?

[viral dagger]

former

[grand pond]

Its Cadet Time

[viral dagger]

no

you can see that its an icoseles triangle, with the base angle being 3x-35

so 2x+2+3x-35+3x-35=180

and continue from there

yes since its a triangle

[grand pond]

Its Cadet Time

[viral dagger]

no afain

[viral dagger]

use this

here i labeled it for convenience sake

AB=CB, which mean BAC=BCA, the inside angles all add up ti 180
ABC+BCA+BAC=180
we know that ABC=2x+2
BAC=BCA=3x-35
which means 2x+2+3x-35+3x-35=180

well BA=BC=18

do it yourself :p

show your work

it should've been $$ 2x+2+3x-35+3x-35$$

[grand pond]

Skill_Issue

[viral dagger]

cool

ya correct i think

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[feral sedge]

consider the cases for x > 1 and x < 1

[rain wasp]

the lhs is just absolute value

[feral sedge]

you could but this is not good idea

consider what happens to sqrt((x - 1)^2) when x > 1 and when x < 1

only when x > 1

misunderstanding here

sqrt((x - 1)^2)

say u just allows this to equal (x - 1)

then for x = 0 or something, this would be -1

so you are essentially claiming that sqrt((0 - 1)^2) = -1

but a square root can't be negative, so there is something wrong with this reasoning

also no

to get around this, we take note of the following: (x - 1)^2 = (1 - x)^2

does it make sense why this is true

consider: (x - 1)^2 = (x - 1)^2 * 1 = (x - 1)^2 * (-1)^2 = ((-1)(x - 1))^2 = (1 - x)^2

cool

uhhh

probably?

[feral sedge]

anyway basically that is how we split the expression into our x > 1 and x < 1 cases

and to complete the problem, you solve two inequalities (one for x > 1, one for x < 1)

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[tepid atlas]

how do i solve this?

i cant find the factors for it

[latent elk]

Let f(x) = x^3 + 5x -10

f(x) is continous throughout R

Then x0 is the root of f(x) if a < x0 < b and f(a)*f(b) < 0

In other sense

[tepid atlas]

wdym by this... kinda complex notation that we havnt gone over

in my class

[latent elk]

This is way more easy to visualize

If x0 is the root

[tepid atlas]

ah ok

(x,0) u mean?

[latent elk]

yeah

Then both side must have different sign

[tepid atlas]

yes

ok

[tepid atlas]

howcome you know that the cubic is on the positive side of the x axis?

[latent elk]

This is just a visualize,

[tepid atlas]

oih ok

[latent elk]

The point is, this is always truth for any graph

[tepid atlas]

so what next?

[latent elk]

bruh

i thought u get it

[tepid atlas]

i do

but how do i find the points

oh

just checked the answers

only one is opposite signs]

[latent elk]

These are a and b

[tepid atlas]

b

[latent elk]

this is f(x)

[tepid atlas]

is there any other ways of working it out?

nvm the answer is

D

[latent elk]

Yeah, find the actual root

[tepid atlas]

how can i do that

i cant factorise it for some reason

[latent elk]

Its a x^3

3rd degree poly

[tepid atlas]

yeah

[latent elk]

The point is, your teacher probably set it so hard to find to root, thus force u to use this

[tepid atlas]

oh ok

so how do i use that

u said the signs must be opposite

but the answer is not b

[latent elk]

For a)

[tepid atlas]

so keep doing that until you get opposite signs

[latent elk]

So D is the correct answer

[tepid atlas]

so its something like that

[latent elk]

YEP

[tepid atlas]

sicne one y coordinate is positive

one is negative

[latent elk]

Yep

[tepid atlas]

therefor in between is the x int

[latent elk]

yep

[tepid atlas]

ahh ok

bit abstract this question

nothing like the others

thank you

[latent elk]

Not abstract

its new for u

[tepid atlas]

thats what i meant

this kinda thinking

[latent elk]

but its pretty easy to understand

[tepid atlas]

it is

thank u for ur help

.close

[midnight plank]

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[midnight plank]

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[pastel tree]

Linearize the function around the point A. Then use it to calculate an approximate value of f(B), where B lies on
distance 0.17 l.e. from A in the direction (8, −15). (Please compare with f(B) calculated with a mini calculator).
Also state the equation for the tangent plane to the surface z = f(x, y) at point A.

[pastel tree]

@tribal temple

[midnight plank]

@pastel tree Has your question been resolved?

[midnight plank]

@pastel tree Has your question been resolved?

[midnight plank]

@pastel tree Has your question been resolved?

[pastel tree]

............................................................................................

[tribal temple]

[pastel tree]

😭

[tribal temple]

Hehe I was away back now

[pastel tree]

Please tell me how you learned multi variable calculus:(

[tribal temple]

“With great difficulty”

[pastel tree]

I’m currently dead in the water trying to find somewhere where I might learn what is necessary to be able to solve these hand in assignments that we are given

The course literature is so hard read

Not sure if it’s a me problem or the book

I can't even find the appropriate section where it covers Linearizing functions to be able to solve the questions

[tribal temple]

Most likely either they might have used different terms (my notes don’t call this linearisation) or that they’re just terrible, as before this is probably one of the harder topics overall

Not even sure if I know of any good sources overall

Anyways, yesterday we got to $f(6 + h, 5 + k) \approx 332 - 85h - 102k$, and figured that $h = \frac8{100}$ and $k = -\frac{15}{100}$

[grand pond]

@tribal temple

[pastel tree]

Yes but, if i'm being honest with myself, i have no idea what we did?

This feels awful not being able to understand anything whatesoever

I'll never be able to pass the test at this rate

[tribal temple]

Awwww

[tribal temple]

Are you at least happy accepting that we need to use this as the formula for now?

[pastel tree]

I wish i was

i dont know what the formula is or what it does

[tribal temple]

Well, erm, basically it's like the "linear approximation" for multivariable functions I guess

Like as per yesterday, similar to single variable functions, you can approximate a function near a point using the function value and the deriviative(s) there

[pastel tree]

Idk chartbit i'm kinda losing it mentally atm

It's been nearly three hours and i have literally zero clue or idea what i'm doing?

I don't have any sources to learn from

and i'm already like 2 weeks behind

[tribal temple]

hmmmm

Alright say that you were given a function, and you knew nothing about it, other than you were given a point, and then you know the function output, and how the derivatives looked like

And you wanted to estimate the function output at some other point that's close to the point you were given

Do you have an idea of how you'd go about doing that?

[pastel tree]

No...

I have literally no idea

[tribal temple]

[pastel tree]

Like i know absolutely nothing

I have no source of learning material at all

No teacher, no book, no friends

So idk how i'm supposed to be able to solve this

at single variable calculus i could at the very least solve it by asking gpt or here

but it doesn't seem many study multivariable calc here

[tribal temple]

More that it's horrible to explain

Anyways, you're happy that $\pdv{f}{x}$ tells you how the function $f$ changes in the $x$ direction, and similarly $\pdv{f}{y}$ tells you how $f$ changes in the $y$ direction? Or no?

[grand pond]

@tribal temple

[pastel tree]

No i did not know that

i know that derivates shows the incline/decline of a given function

[tribal temple]

-that wasn't explained?

Bruh

Anyways that's kind of the idea of it, that's why you hold all the other variables constant when you're doing the partial derivative with respect to one variable

Because you want to see how the change in that one variable changes the output of your function

[pastel tree]

Ok

[tribal temple]

Cool, so are you at least happy with that for the time being?

[pastel tree]

But like he has covered

all of this already and i have literally zero zero zeor clue on what they do

or how to use them

or the theory behind them

or what the signs means

Shouldn't i be able to solve the assignments that he gives us if i understand the formulas and the theory behind them?

For example for this question that we are doing right not it covers linearlization (this image^). Shouldn't i essentially be able to just utilize this formula to get the answer to the question?

[tribal temple]

Well yea and that's what we're doing

[pastel tree]

Oh ;-; sorry :c

[tribal temple]

Us using that formula is how we got this here

[pastel tree]

ik i'm a pain but math really gets to me emotionally :( ||(tired of crying :))||

[tribal temple]

Wait are you happy at least with all we did yesterday, at least with how we found f(a,b) and the partial derivatives?

[tribal temple]

(and you're not a pain )

[pastel tree]

Welll what we did yesterday was that we were given a function and a point

and we utilized the middle part of this formula

But why is there a lot of other info on the formula?

what does the triangle mean? delta?

[tribal temple]

There are other equivalent forms to use this in, for now you can ignore this part (but it's talking about the directional derivative of f in the direction of v)

[tribal temple]

And yea, the change of f

"The change of $f$, which is $f(a + h, b + k) - f(a, b)$ is approximately $h f'_x(a, b) + k f'_y(a, b)$"

[grand pond]

@tribal temple

[pastel tree]

oh oki

"For differential functions of two variables, the linearization equals the change in the tangent plane"

It this true?

What does this even mean xD

For functions of two variables, the linearization of it is the same as the change in the tangent plane

idk what a tangent plane is (i maybe know what linearization is, but not 100% sure.. )

[tribal temple]

You're happy in the single variable case with what a tangent line is right?

[pastel tree]

Yes. A single variable case tangent line shows the incline/decline of a given function at a given point

[tribal temple]

Cool, and of course you notice that there's basically only one direction you consider in the single variable case

[pastel tree]

Right

[tribal temple]

But when it comes to two variables, then basically as you have effectively two dimensions in which you can move in, that at a certain point, you'd have more of a tangent plane, a two dimensional object, if you give me a second

something that looks like this, see how you have that plane that doesn't (locally at least) intersect the surface

[pastel tree]

right

[tribal temple]

Anyways, basically we have that this here effectively is the equation for the tangent plane at that point, you of course use the function value and the derivatives

[pastel tree]

Okay

makes sense

so instead of a tangentline we get a tangentplane

Why do we get new variables h and k?

[tribal temple]

Well, those are basically describing how much (in the x and y direction respectively) you change from (a, b)

[pastel tree]

I tried visualizing it but, wtf is this abomination?

[pastel tree]

It looks nothing like this

[tribal temple]

Well that one was more of a particular example rather than how exactly it would look, but (give me a moment )

[pastel tree]

Then use it to calculate an approximate value of f(B), where B lies on
distance 0.17 l.e. from A in the direction (8, −15).

What approximate value do they mean?

[tribal temple]

Basically they're asking you to work out what f(B) is

But as per before, B is basically (6, 5) + (8, -15)/100

[pastel tree]

But how do you know that?

I'm given a Line from A to B

with a distance and a length

[tribal temple]

They tell you:

B lies on distance 0.17 l.e. from A in the direction (8, −15)
and (8, -15) has length 17

It's a direction vector for here

[pastel tree]

Right

why length 17?

What exactly is f(B)?

[tribal temple]

well you can work it out manually that sqrt{8^2 + (-15)^2} is 17

[pastel tree]

Makes sense

[tribal temple]

And as in its exact value, or what it's supposed to mean?
For the former, I'd need to work it out
the latter, it's the value of f at the new point B, where as per before, there are those steps to work out what B actually is and all

[pastel tree]

So f(B) is a value?

It's so unclear to me what they are asking

[tribal temple]

Yep, it's the function value we want to know

But we're just estimating it for now

[pastel tree]

Estimating the function?

aaaaaaaaaaaaaaaaaaaaaaaah i donj't understasnd HJAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

I don't even understand the question.

[tribal temple]

That's the whole linearisation thing we're doing now

[pastel tree]

the problem is that i don't even understand the question

calculate an approximate value of f(B)

what is f(B)

we neither have a function

or B

[tribal temple]

f is the function in the question, and...

[tribal temple]

...B is that point as per before, a point that's "distance 0.17 [...] from A in the direction (8, −15)"

"Start at A, the point (6, 5), and then point in the direction (8, -15), but move a distance of 0.17"

[pastel tree]

Oh ok

but it doesn't specify that f is the funciton in the question?

That's what got me so confused

Okay so we are looking for the point B to be able to determine f(B)

Can we use the formula of a line?

[tribal temple]

Notice how the function is called f

[pastel tree]

We start at A and end at B

that gives us a vector

and also a line L

and the variable t in the line L we set to 1/17

thus giving us the point B?

[tribal temple]

You could but as per before it's 1/100 you're moving, the line direction vector is length 17 after all and you wanna move a distance of 0.17, in other words, 1/100th of the direction vector

[pastel tree]

but

okay wait

So the directional vector becomes
B - A = (8,-15) - (6,5) = (2, -20)

And the Lines equation becomes :

L : (x,y,z) = (6,5) + t(2,-20)

[tribal temple]

The direction vector is (8, -15)

[pastel tree]

oh crap right

Ok so

L : (x,y,z) = (6,5) + t(8,-15)

and now we want the point B

They say it's 0.17 l.e

shouldn't t be 0.17?

[tribal temple]

That would only be the case if (8, -15) had length 1

[pastel tree]

if t = 1 then we get the point where the directional vector ends, right?

the tip

Can't i set t = 1, figure out the point. and then use t = 0.17?

[tribal temple]

Well sure, that would be how you've moved (8, -15) from (6, 5)

[pastel tree]

:)
then i have no idea

[tribal temple]

And as per before, you want to consider the length of the direction vector

[pastel tree]

t = 0.01 ?

[tribal temple]

Unless you want to change the direction vector, you could do that, you'd need to make it have length 1 if you really wanted to

[pastel tree]

That was way to hard lol

[tribal temple]

Awwww

[pastel tree]

B = (6,08 . 4,85)

based on the line L : $(x,y,z) = (6,5) + 0,01*(8,-15)$

[grand pond]

Merineth

[pastel tree]

and now i input it into the function?

[tribal temple]

Yep that's fairs

[tribal temple]

And we want basically the h and the k, which is how much you've changed to get from (6, 5) to B, which of course is just h = 0.08, k = -0.15

[tribal temple]

For which, put those h and k into here

[pastel tree]

huh

$f(B)=f(6,08.4,85) = 332 + ln(31-(6,08 * 4,85))\sqrt{228+6,08^2 +4,85^2}$

[grand pond]

Merineth

[tribal temple]

"Linearize the function around the point A. Then use it to calculate an approximate value of f(B)"

[pastel tree]

Is this wrong?

:)))))))))))))))

[tribal temple]

It isn't wrong, but it's not what they asked you to do

[pastel tree]

okay so we use

-85h-102k

where h = x and k = y

[tribal temple]

Well more that h is the change in x, the k is the change in y

[tribal temple]

hence this here

[pastel tree]

but why do you decrease

x and y to

0.08 and k -0.15

[tribal temple]

Because it's not the x and y values, h and k are the change in x and y: how does the x value change (for h)? How does the y value change (for k)?

[grand pond]

Merineth

[pastel tree]

ok..

I clearly don't understand what i'm doing

[tribal temple]

Notice how in the formula

It's f(a + h, b + k) that they have in there?

So here, we had that B was the point (6.08, 4.85), but that is (6 + h, 5 + k)

In other words, 6.08 = 6 + h and 4.85 = 5 + k

That's why as per before you don't really need to find B exactly, you just needed to divide that (8, -15) by 100, as that gets you the h and k instantly for you

[pastel tree]

😔 oki

$-85k-102k = -850.08-102-0,15$

[grand pond]

Merineth

[pastel tree]

= 8,5

?

[tribal temple]

Yep, that's fine, you have that $f(B) - 332 \approx 8.5$

[grand pond]

@tribal temple

[tribal temple]

Noting that f((6,5)) = 332

[pastel tree]

what does that value indicate?

332

and 8.5

the z axis coordinate?

[tribal temple]

8.5 is the approximate change in the value of f between B and A

The 332 is the value of f(A)

[pastel tree]

oki

and lastly i want the equation for the plane?

iirc the planes equation was something like this
x+y+z = t

[tribal temple]

Yep, you want the plane equation, but that's quite simple-

[pastel tree]

i want the normal vector?

[tribal temple]

Well actually you've basically almost found the plane equation

[pastel tree]

It's very hard to keep track of what i'm doing when i don't have an image in my head

but rather only numbers

(8,-15) isn't this the directional vector along the plane?

no?

meh

How am i supposed to visualize this

[tribal temple]

Right, see this here

We have as per before that $a = 6, b = 5$, and we already found that $f(6,5) = 332$ and that $\nabla f(6, 5) = \pmqty{f'_x(6, 5) \ f'_y(6, 5) } = \pmqty{-85 \ -102}$

[grand pond]

@tribal temple

[pastel tree]

right!

what does x and y mean?

[tribal temple]

Basically it's the thing you found before, but you replace h with x - a, k with y - b

[pastel tree]

just variables?

[tribal temple]

Yep, any x and y value you want to be

[pastel tree]

okay so

$z = 332 + (-85, -102) \cdot (x-6, y-5)$

[grand pond]

Merineth

[tribal temple]

Yep

[pastel tree]

so

$85x+102y+z= 1352$

[grand pond]

Merineth

[pastel tree]

$z= 332+(-85)(x-6)+(-102)(y-5) \
z= 332 - 85x+510 -102y+510 \
85x+102y+z= 1352$

[grand pond]

Merineth

[pastel tree]

[tribal temple]

Yep pretty much

[pastel tree]

thank god i have you charbit

honestly idk what i would've done if i didn't have you???

(catropeemote)

[tribal temple]

Awww

A certain cat here demanding treats

But yea other than that, pretty much linearising a function and finding the tangent plane are the same thing

[pastel tree]

pet him IMMEDIATELY!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

GOD what a cutie

i'll close for today ;-; i hope i see you tomorrow <3

tysm charbit :(

literally my saviour

.close

[midnight plank]

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[tribal temple]

They’ve all gotten more than enough treats and attention, this one too
See ya later might be a bit busier tomorrow

[midnight plank]

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[maiden river]

can someone help me figure out what base case i use ? is it n = 0?

[carmine void]

Yes

[maiden river]

Could you explain why its n = 0 as a base case?

is it cause of i = 0 ?

[carmine void]

Yes, exactly

[maiden river]

Ahh i see! Thank you 🙂

[carmine void]

You're welcome

[maiden river]

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[lyric wraith]

wouldnt the two adjusted equations be same ?

[vague seal]

no

one is saying

take f(x)

and turn it entirely negative

and the other is saying

take the x in f(x) and make that negative

[lyric wraith]

oh 😭 thank you

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[normal plover]

.close

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[quasi cedar]

Shouldn't this be moved to MATH HELP?

[midnight plank]

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[normal plover]

what do i do if i want to solve for theta

[zenith stone]

do u know what tan(x) means?

well before that

[normal plover]

oh

yes

[zenith stone]

if 12=12k

what is k?

[normal plover]

1

[zenith stone]

yes

so 1=sinx/cosx

do u agree?

[normal plover]

yes

thak you sorry i didtn realize

[zenith stone]

so u get it?

[normal plover]

yes

[zenith stone]

okie

[normal plover]

how do we know that tan is 45 degrees tho when its 1?

[zenith stone]

sinx/cosx = 1
sinx = cosx

now for what angle is that true?

sin(45) = √2/2
cos(45) = √2/2

do u know the unit circle?

there is one more solution

[normal plover]

I dont have it memorized but yes

[zenith stone]

in the third quadrant

5π/4 = x

both sinx and cosx are -√2/2

so that must be a solution as well

[normal plover]

ohh

so would the teacher expect us to know the trig circle by heart?

or is there a simpler way

[zenith stone]

well u only need to know the first quadrant

[normal plover]

that my calculator can do

oh

[zenith stone]

ur calculator will only give u one solution

its good to know one quadrant of the unit circle

[plush prism]

since the radius is 1 on the unit circle, we get this:

and for sin(theta)=cos(theta), we want these 2 lines to be the same length

which is the case on the diagonal

[normal plover]

ohh

[plush prism]

so you get these two solutions

they also have the same length on the other diagonal ( f(x)=-x ), but then you have opposite sign

[normal plover]

ohh

[plush prism]

notice that usually, the sides would be sin(theta)/r and cos(theta)/r, i didn't write r because r=1

[normal plover]

oh ok

ty guys

[plush prism]

👍

[normal plover]

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[candid rampart]

What do you know so far

[pale wyvern]

I might know that I have to do the equation first and the answer I plug it into a graph?

Or am I wrong

Been a bit since I've done these types of problems though

[rich haven]

first find the slope

if a line with slope -1/3 is perpendicular to another line, whats the slope of the other line?

[pale wyvern]

-1/3?

Wait no

[rich haven]

that would be parallel

[pale wyvern]

Right

3?

[rich haven]

yep

we can set a variable c for the constant

so y = 3x + c

[pale wyvern]

Now i solve for x right?

[rich haven]

yes by plugging in (6, 8)

[pale wyvern]

How would that look like quick question?

[rich haven]

6 is x, 8 is y

8 = 3*6 + c

[pale wyvern]

Oh

So c = -10?

[rich haven]

yep

[pale wyvern]

sweet

[rich haven]

so your final equation is

y = 3x - 10

[pale wyvern]

So I solve for x or y?

Or both

Brain fart

So I got 10/3

x = y/3 + 10/3

[rich haven]

wdym

you had y=3x + c

and found that c = -10

so its just y = 3x - 10

[pale wyvern]

Oh right

Do I leave it at that?

[pale wyvern]

Am I done with the equation? Or are there any other stuff I need to do?

[rich haven]

thats it

[pale wyvern]

Yay, thank you for your help!

[midnight plank]

@pale wyvern Has your question been resolved?

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[mint oriole]

If each x_i is strictly between 0 and 1 is this matrix invertible

[midnight plank]

@mint oriole Has your question been resolved?

[final monolith]

should be yea

each column is linearly independent, since they're all almost (1 1 1 1) but point into different dimensions a bit

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[mint oriole]

.reopen

[midnight plank]

✅

@mint oriole Has your question been resolved?

[midnight plank]

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[zealous tinsel]

here is my work progress for the following question

[zealous tinsel]

1a

and this is the correct answer

i’m not sure how they got this answer and what it means

[last slate]

They didn't ask you to differentiate it?

[zealous tinsel]

i dont think so

thats all the instructions im given

[last slate]

They only then why did you do so

You just have to write in $au^{n}$ form

[grand pond]

fredthebread69

[zealous tinsel]

i just found the derivative but im not completely sure what that means

[last slate]

What are you unsure about, your derivative or the solution of the question

[zealous tinsel]

both….

i thought my derivative was right but i dont know if i was even supposed to derive or what the answer means in general

[last slate]

Ok

They have taken u=2x-1

[zealous tinsel]

where does that come form? is u just the inside?

[last slate]

So the expression
$1/(2x-1)^{2}$

[zealous tinsel]

like inside of parentheses?

[grand pond]

fredthebread69

[last slate]

Can be written as?

[zealous tinsel]

(2x-1)^-2

[last slate]

$1/u^{2}$

[grand pond]

fredthebread69

[last slate]

Which is just u^(-2)

Correct?

[zealous tinsel]

what is u? like what does it symbolize

or is it just a variable

and yes i agree

just curious if u has some sort of significance

[last slate]

It's just a variable

That is dependent on x

Because that's what you let

[zealous tinsel]

ohh okay okay

[last slate]

Clear?

[zealous tinsel]

got it

yes!

[last slate]

Or in other words, u is a function of x

u=f(x)=2x-1

Agree?

[zealous tinsel]

yes

[last slate]

Now you wish to differentiate it, so it's differentiation of u^(-2) with respect to dx

So we get $-2u^{-3} du/dx$

[grand pond]

fredthebread69

[last slate]

Agreed?

[zealous tinsel]

yes

[last slate]

You seem to have done it right

[zealous tinsel]

ohhh so

the -2 is like

the outside basically

like the chain rule

that puts the -2 at front (i dont know how else tk explain it)

[last slate]

Pardon

[zealous tinsel]

and u is whatever’s inside?

like

how in the 2nd step

the -2

went to the front of the parentheses

is that where it’s coming from?

[last slate]

Say I differentiate

g²

All I know g is a function of x

I would get
$2g dg/dx$

[grand pond]

fredthebread69

[last slate]

You differentiate your term normally then multiply dg/dx

Does that answer your question

[zealous tinsel]

yes

thank you for the help!

[midnight plank]

@zealous tinsel Has your question been resolved?

[midnight plank]

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[midnight plank]

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Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[clear tusk]

How can I justify the second one on the top

[sinful trout]

,rccw

[clear tusk]

Umm. Does someone know this

I keep trying random methods that don't work

They didn't taught how to do this

!!

<@&286206848099549185>

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[clear tusk]

Sure
Sorry for being impatient

[sinful trout]

no problem

what have you tried for (ii)?

[clear tusk]

I just tried random stuff like trying dividing em

I know thats not the way

[manic bison]

i think the question might be wrong as you cant divide a quadratic polynomial by a cubic

[clear tusk]

Oh , but it's in the NCERT example

[sullen kayak]

I think they mean the other way round

[clear tusk]

Yeah it could be

Let me just try it

[frosty veldt]

i think thats the point maybe

[clear tusk]

Oh ok

[frosty veldt]

q = 0 and r= ax^2 + bx + c