help-13

[crimson sedge]

I found the value of c which is 3 then plugged it into f(4c) and got f(36) and plugged in 36 into f(x) which leads to 10

[open nacelle]

Yea you got it

[crimson sedge]

thank you

❤️

.close

[cedar kiln]

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[ebon finch]

calc help derivatives f(x)=66-30e^-3.5x

[ebon finch]

any tips on how to go about this

[dense basalt]

You can use standard derivative rules

Get derivative of each, and subtract

Also, e^(-1.35)t is the same as e^(-1.35) all to the power of t

So you can use the standard formula for (a^x)'

[ebon finch]

you lost me ngl

[neon moon]

You may let u = -3.5t, then apply chain rule.

[cedar kiln]

@ebon finch Has your question been resolved?

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[crimson sedge]

How would I look at a graph like this?

[upper abyss]

When does g(u) = -1?

[crimson sedge]

Like this?

Would I be looking at x first?

Then f(x)

[thick cipher]

Two questions :

1. for which values of y does g(y) =-1
2. for which values of x does f(x) =y

[crimson sedge]

1- 4

2- 2

[thick cipher]

So you have your answer

[upper abyss]

You've got it, really. I'll write it in full so you can see the whole picture:
g(4) = -1
f(2) = 4

So that means
g(f(2)) = -1

[crimson sedge]

ok

thx

.close

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[gentle egret]

is this right?

[dire geode]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[gentle egret]

f(a)=(-7)^3+3(-7)^2-6(-7)-1= -155
f(b)=(-4)^3+3(-4)^2-6(-4)-1= 7

[gentle egret]

am i doing it right?

i think my solution is wrong in f(b)=(-4)^3+3(-4)^2-6(-4)-1= 7

can anyone help pls?

[dire geode]

Why do you think these are wrong

Do you know the statement of the intermediate value theorem

[gentle egret]

i think im supposed to put 4 instead of 7 in f(b)

[dire geode]

Why

[gentle egret]

idk

[dire geode]

Show this statement and apply it to your problem. Which part are you unsure of

[gentle egret]

i just need confirmation from real person to know if its right. im very sceptical of anything i solve

[cedar kiln]

@gentle egret Has your question been resolved?

[dire geode]

Where did 4 come from

[dire geode]

You used a calculator, right?

[gentle egret]

i just solved a similar question and got f(b) wrong

then i looked at the intital question and thought it would be simliar tothat

[dire geode]

,calc (-4)^3+3(-4)^2-6(-4)-1

[wraith dagger]

Result:

7

[dire geode]

That doesn't explain where 4 comes from

[gentle egret]

i thought it comes from here

[dire geode]

f(b) = f(-4)

Where would 4 come form

[cedar kiln]

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[cedar kiln]

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[tropic oxide]

(z+i)^10 = -(z-i)^10

take the absolute value of both sides

what secret equation laws

we are not solving the equation

we are just showing that one equation implies another

"did something" is vague

multiplication by x is legal

except depending on your goals there might be something to account for

[vagrant elbow]

If two complex numbers are the same, then their moduli are also equal

[crimson delta]

the problem is only ever if you want to go back. so eg if you multiply by x and then see that x=0 is a solution, that might not be valid in the original equation

[vagrant elbow]

Just think about it

[burnt vapor]

But the question says they are

[crimson delta]

the equal sign quite literally says that they are equal

[tropic oxide]

but these two dont look equal .-.
i made the original equation into that...

your equation was (z+i)^10 + (z-i)^10 = 0

i subtracted (z-i)^10 from both sides

do you have any objections to that

[burnt vapor]

You can write z = a+bi and use |z+i| = |z-i|

Using the formula |α+βi| = sqrt(α^2 + β^2)

Yes, that's what we proved before

Yes, you just need to square both sides of the equation

Yes

[cedar kiln]

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[bright sorrel]

Need help solving "The sum of the first n members of an arithmetic progression is Sn=2n^2-n. What does the third member equal to?"

[bright sorrel]

so I'm guessing I have to go like this

S/3/=2(3)^2-3=15

a1+a2+a3=15

an=a1+(n-1)d

where d is the difference of the arithmetic progression

so a2=a1+d
a3=a1+2d

a1+a1+d+a1+2d=15
3a1+3d=15
a1+d=5

am I doing good so far?

[dreamy sleet]

Yes, also notice that a1 + a2 = 2*(2^2) - 2 = 6

So you can get a3 quickly

[dreamy sleet]

These steps are correct so far

[merry axle]

I need help

[dreamy sleet]

!help

[cedar kiln]

Please read #❓how-to-get-help

[merry axle]

WHT

[bright sorrel]

oh S/2/=6

[dreamy sleet]

Please read #❓how-to-get-help

[dreamy sleet]

$$ S_2 $$ you mean?

[wraith dagger]

even order group => solvable

[bright sorrel]

yeah

[bright sorrel]

so I get
a1+d=5
and
2a1+d=6

I get a1=1

d=4

a3=a1+2d=1+8=9

yup, that's it

thank you

.close

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[crisp flint]

If a side of a square is x and the middle of the square is empty and only the perimeter exists, why is it wrong to use x^2-{x-2}^2 to find the area

[floral obsidian]

Can you explain why (x-2)²?

[crisp flint]

Because the centers doesnt exist and only frames with 1 thickness

So it is area of full square - area of center

[floral obsidian]

Oh thickness is 1. You did not mention that before.

@crisp flint So like this?

[crisp flint]

Yeah

[floral obsidian]

I don't see it being incorrect.

It should be correct.

Maybe you should simplify the expression more.

[crisp flint]

I think it is because \sqrt{x^2-{x-2}^2} \neq x

[floral obsidian]

What square root?

And who told you the answer is incorrect?

[crisp flint]

Just incorrect

[floral obsidian]

Try 4x-4

Is this correct?

@crisp flint

[crisp flint]

I dont think so

Because 4 is constant

Also \sqrt{2x-4} \neq x

[floral obsidian]

Explain this

[crisp flint]

Because it is one side of square

[floral obsidian]

Send the question you are trying to solve.

Screenshot

[floral obsidian]

That is not the question you asked.

There is no square or rectangle with empty middle here.

Anyways to solve this, notice that the side Length of the square is x

But we need x+2 as the width of the new figure

Can you think of a way where we take 1 square and 1 of any other shape to make a rectangle of width x+2 (length does not matter for now)

[crisp flint]

It is square filling with 1 difference

[cedar kiln]

@crisp flint Has your question been resolved?

[simple bane]

so is it about finding the perimeter of the 'square'?

[cedar kiln]

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[hazy flicker]

gah I am stressing so bad

[hazy flicker]

these problems have me so mad

this one FIRST, i have one other which is very easy but my homework keeps counting it wrong, I've even used calculators on the second one

everything makes sense but BECAUSE the radical x is in the denominator it's throwing me off so hard, like there's 4-5 steps affected by that movement and literally I don't know what to do on any of them

[surreal cave]

this is a definite integral, you still have x in your answer?

[hazy flicker]

I did this when I was exhausted 😭 , I gave up and put that just to save my answer, because I didn't wanna try to put in numbers just to get it wrong

that answer was an extremely rough estimate

[surreal cave]

alright, that's fine, so first off yo did the indefinite integral correct

[hazy flicker]

damn

I doubted myself 🥺

[surreal cave]

now just use $\int_{b}^{a}f(x)\dd x=F(a)-F(b)$ :)https://www.wolframalpha.com/input?i=int+1%2F(sqrt(x)(9%2Bx))

[hazy flicker]

I'm trying to redo the problem now and IM STUCK??

[surreal cave]

wolfram link in case you don't believe me lol

[hazy flicker]

could you help walk me through the steps of the indefinite part?

[wraith dagger]

XxMrFancyu2xX

[hazy flicker]

like cause now that I'm actually awake somehow im confused

[surreal cave]

alright

[hazy flicker]

du = (1)/((2)(x)^(1/2))

and dx with that

[surreal cave]

can you send the integral? 😅

OH are we still doing the first one?

[hazy flicker]

yes

yeah yeah

[surreal cave]

aight

so if $u=\sqrt{x}$ then $\dd u=\frac{1}{2\sqrt{x}}\dd x$ yes I agree

[wraith dagger]

XxMrFancyu2xX

[hazy flicker]

yeah perfect

so dx = 2sqrt(x)DU

which makes sense of course, but WITH RADICAL X in the denominator, doesn't that change things??..

[surreal cave]

hang on

$\int\frac{1}{\cancel{\sqrt{x}}(9-u^2)}\cdot 2\cancel{\sqrt{x}} \dd u$ :)

[wraith dagger]

XxMrFancyu2xX

[hazy flicker]

oh wow

I guess I'm just dumb

if the radical x was in the numerator what would you do though?..

[surreal cave]

then we'd have $\int \frac{u}{(9-u^2)}\dd u$ :)

[wraith dagger]

XxMrFancyu2xX

[hazy flicker]

woahhh with the two moved outside right?..

[surreal cave]

oh right I forgor it my bad 💀

[hazy flicker]

wow so if you ever have a situation with x ontop of like a normal inverse trig function, WHAT do you do? like in that one you just sent the u over 9 - u^2

[surreal cave]

$\int\frac{u}{9-u}\dd u=\int\frac{9}{u}-\frac{u}{u}\dd u=\int\frac{9}{u}\dd u-\int\dd u$

then you ened up with some logarithm business

oh wait it was u^2

well it remains the same

use partial fraction decomp

albeit it changes because of the u^2

[hazy flicker]

no it's okay lmao I appreciate it, and then I have this goofy problem which I've literally calculated with 5 different sources :

I've put in 1.057

and it was WRONG??

[surreal cave]

hyberbolic cotanget?!

man I have not seen that in math problem like ever

,w coth(ln(6))

[hazy flicker]

these thigns are so annoying

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by XxMrFancyu2xX#6050

[surreal cave]

try 37/35

or does it want decimals?

[hazy flicker]

decimals I guess

[surreal cave]

mayeb try 37/35?

[hazy flicker]

I've tried

[surreal cave]

that's a good question, maybe ask your teacher what went wrong idk man

[hazy flicker]

bro the fraction worked

thanks sm

.close

[cedar kiln]

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[surreal cave]

yw!

[cedar kiln]

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[crystal mason]

"A liquid consists of a mixture of four liquids with different densities. When the mixture is left in a test tube, four layers form after some time. By measuring the heights h_i, i = 1, ..., 4 of the layers using a ruler, one can estimate the proportions of the respective liquids. However, due to measurement inaccuracies, these estimations are flawed. Suppose the heights h_1 = 4.3 cm, h_2 = 3.5 cm, h_3 = 1.4 cm, and h_4 = 0.7 cm are measured with an error margin of 0.1 cm each. Provide an estimation and an error margin for the proportion of the third liquid."

(This is supposed to be solved using multidimensional analysis, but idk how I could do that)

[dire geode]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

@crystal mason Has your question been resolved?

[cedar kiln]

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[fervent pike]

For this question, how is it sqrt3/2X?

[dire geode]

where's the question

[fervent pike]

I mean how do you know sqrt3/2 is multiplied by x component in X

[dire geode]

where's v?

[dire geode]

did you calculate this?

[fervent pike]

oh waitt, i think i know what you mean

[dire geode]

it's helpful to read left to right and understand step by step each equal sign

[fervent pike]

its gonna be 0 w/ y=1

[fervent pike]

thank you

At this point, the math I have no problem its my problem solving abilities

.close

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[patent creek]

Waves are travelling across a rope attached to a fixed end. An initial wave with an amplitude of 5 cm is sent across the rope, then a second wave with an amplitude of 7 cm is sent across once the first wave reaches the end. When the two waves meet each other, what will be the amplitude of the interference wave?

[patent creek]

I think its 12

right

[cedar kiln]

@patent creek Has your question been resolved?

[slate lintel]

seems right to me

[patent creek]

.
Waves are travelling across a rope attached to an open end. An initial wave with an amplitude of 5 cm is sent across the rope, then a second wave with an amplitude of 7 cm is sent across once the first wave reaches the end. When the two waves meet each other, what will be the amplitude of the interference wave?

what about this

one is open end one is fixed en

what is the difference

@slate lintel

[slate lintel]

will the two waves even meet each other if there's an open end?

[patent creek]

no

but

its asking me to put in a number

[slate lintel]

that's weird

[patent creek]

it says once the first one reaches the end

[slate lintel]

once the first one reaches the end it'll die surely

[cedar kiln]

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[crystal mason]

"A liquid consists of a mixture of four liquids with different densities. When the mixture is left in a test tube, four layers form after some time. By measuring the heights h_i, i = 1, ..., 4 of the layers using a ruler, one can estimate the proportions of the respective liquids. However, due to measurement inaccuracies, these estimations are flawed. Suppose the heights h_1 = 4.3 cm, h_2 = 3.5 cm, h_3 = 1.4 cm, and h_4 = 0.7 cm are measured with an error margin of 0.1 cm each. Provide an estimation and an error margin for the proportion of the third liquid."

(This is supposed to be solved using multidimensional analysis, but idk how I could do that)

[crystal mason]

oh oops, i shouldve resent my question

[crystal mason]

I have no work.. I mean my first thought was just simple calculations like (h_3)/(h_1+h_2+h_3+h_4) (which was also suggested by a helper here) and such but that has nothing to do with my subject so i came to this server for help

[cedar kiln]

@crystal mason Has your question been resolved?

[crystal mason]

<@&286206848099549185>

[crystal mason]

Analysis II

Lol

[cedar kiln]

@crystal mason Has your question been resolved?

[crystal mason]

<@&286206848099549185>

[twin scarab]

.

[twin scarab]

yes?

[crystal mason]

check out my pinned message

[cedar kiln]

@crystal mason Has your question been resolved?

[crystal mason]

<@&286206848099549185>

[crystal mason]

😭

.close

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[crimson sedge]

How would I do this question?

[wraith dagger]

You must be a bot manager to use this command!

Jun 22, 2023, 9:50 PM

[cedar kiln]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[crimson sedge]

1

[ancient lodge]

Make the directions of the inequalities the same then ||add them||

[crimson sedge]

What do you mean by direction?

[ancient lodge]

As in $<, \leq$ vs $>, \geq$

[wraith dagger]

civil_service_pigeon

[crimson sedge]

ok

what is that

[crimson sedge]

can someone help me solve 2.91+52.90

[crimson sedge]

#help-0

And delete your message

[ancient lodge]

#❓how-to-get-help

[open nacelle]

Whatd you do to get this

[crimson sedge]

He said to make them in the same direction so I just multiplied the bottom by -1 on both sides to flip the inequality.

[open nacelle]

Did you forget -2?

[crimson sedge]

Yup haha

Done

[ancient lodge]

Can you see any opportunities for cancellation?

[crimson sedge]

Add

But if we add

what will happen to the inequality

one has the equal to

[ancient lodge]

The equality will go away (if you want to justify this for yourself, consider some cases with concrete numbers, such as 1__<__1 and 2<3 -> 3<4)

In general, even though the two sides in the first inequality can be equal, because this isn’t the case in the second inequality, the equality goes away

[crimson sedge]

so it will just be greater than or less than

[ancient lodge]

(We can add because the directions of the inequality signs are the same)

[ancient lodge]

Yeah it’s one of those

[crimson sedge]

Answer is D

i divided by 3

yay thank you so much

i appreciate it

.close

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[foggy mason]

for multivariable functions, does the region have to be bounded to find an abs max or min?

[flint plinth]

not necessarily

[dire geode]

No

[flint plinth]

consider a constant function

[foggy mason]

oh

well i suppose ignoring that

Just in all of my practice problems

if it was not bounded it was never stated wether a point was an abs max or min

[flint plinth]

plenty of other counterexamples exist

[dawn junco]

f(x, y) = e^(-(x^2+y^2))

[flint plinth]

e^(-||x||^2)

[foggy mason]

hmmm

well i suppose i can see there are counter examples

why cant we find abs extema on a function like f(x,y) = 2x^3 + 2y^3 -9x^2 +3y^2 - 12y

(that was pulled straight from my booklet lol)

even though it is possible to find all the critical points

[dire geode]

critical point doesn't imply extrema

[foggy mason]

right but

if we found like

all the critical points

[flint plinth]

let x and y be very large (both positive or both negative), then the x^3 and y^3 terms dominate... therefore?

[foggy mason]

then like one of them should be a max right

[foggy mason]

oh

mann

hmmm

i think as a general rule ima say if its not bounded then i probably can't find an abs min or max

[flint plinth]

if it's not bounded above, then there's no abs max
if it's not bounded below, then there's no abs min

[foggy mason]

i doubt they will give me something super troll

aint no way

gotcha

thats actually just a better way and easier way to check

[flint plinth]

odd order polynomials are not bounded above or below

[foggy mason]

checking if this thing is bounded tho what the heck

only thing i can see in there is a circle

and i mean thats bounded but e^that is idk

[flint plinth]

all you need to know here is that x^2 + y^2 >= 0, so you're taking e to a negative power, which gives you something between 0 and 1

[foggy mason]

oh right

wait sorry this might be silly

but even numbered polynomials are bounded?

wait

i guess either above or below

[flint plinth]

they can be bounded above or below (but not necessarily) but not both

[foggy mason]

i can see its one or the other

[flint plinth]

for example in one variable, f(x) = x^2 is bounded below

[foggy mason]

right

[flint plinth]

and in two variables, f(x, y) = x^2 + y^2 is bounded below, but f(x, y) = x^2 - y^2 is not bounded above or below

[foggy mason]

hmmm oh boy

okay yea

thats good info

my booklet thing here just also says

"to find the absoloute extreama on a closed and bounded region R"

so thats kinda inferring that my prof wont do anything nasty

but just incase

i know some edge cases

[flint plinth]

yeah closed and bounded region matters here, because there's a key theorem which says that a continuous function on a closed and bounded region has an abs max and an abs min

[foggy mason]

ah right

is that like IVT

nvm

[flint plinth]

no

[foggy mason]

hahahaha

[flint plinth]

it's somewhat related but deeper

[foggy mason]

ah okay

well, that helps a bunch

thank you

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[glad shore]

Hey

[glad shore]

this is a pretty easy question about calculus, but I'm a bit confused

So I've got to determine the equation of the normal line and tangent to the curve:

y = (1 + 2x)^2 in the point (1,9)

[neon moon]

What are you confused about then?

[glad shore]

So I know that the derivative of the function will be the slope of the function

the derivative being 4 + 8x

So then, I must operate the equation with the point

but then what's the y = 9 for?

It's got to have a function

[deft gull]

the y=9 is just there to let you know that it’s a point on the curve i think

[glad shore]

What am I supposed to do

[deft gull]

and you can use it to write your equation in point slope form

[glad shore]

Soooo... do I just check the equation with the x = 1?

[solemn pawn]

U just find the derivative then 1/- of it then sub in x value

[deft gull]

for the slope, yes, then write the equation of the line

[glad shore]

wut

[solemn pawn]

Find the derivative and then sub in x for the gradient of tangent

[glad shore]

so you're telling me to use the equation y - y1 = m(x - x1)

[solemn pawn]

And then do point gradient for equation

[glad shore]

and then I must use y1 as 9?

[deft gull]

you can

[solemn pawn]

And for Normal it’s negative reciprocal as gradient and do point gradient for it too

[glad shore]

welp that's what I'm doing, isn't it?

[deft gull]

yes because you know the line passes through the point (1,9), and point slope is for a line passing through a certain point (x1, y1), so yes y1 = 9 works as long as x1 = 1

[glad shore]

soooo, the equation would be

y = 13x - 4?

if 4 + 8(1) equals the slope

wait no

Its y = 12x - 3

right?

[deft gull]

yep that’s the tangent line

[glad shore]

aight, and then, what about the normal line?

that's the confusing part

[deft gull]

like Warner Highsenberg said, the slope of the normal line is basically the negative reciprocal of the tangent line’s slope

basically, the normal line is the line perpendicular to the tangent line that passes through the same point

[glad shore]

Right, so what do I calculate?

Srry I'm not very good at it

[deft gull]

yeah so u know the slope of the tangent line is 12 right

[glad shore]

Yep

[deft gull]

you use that slope and take the negative reciprocal

[glad shore]

-1/12?

[deft gull]

yep!

so then to make the line equation

you just do the same thing, except instead of 12 you use -1/12

[glad shore]

so, using the equation y = mx + b, the tangent line's b is the same as the normal line's b?

[deft gull]

not always

i mean you use the same point slope form

and then you go from there

[glad shore]

ooooooh got it

[deft gull]

so like you would have the y-9 and x-1 again

yep

[glad shore]

using point gradient, alright

so, the normal equation is y = -1/12x + 109/12?

Lovely

beautiful

seems to be working

so yeah guys @deft gull @solemn pawn thanks for your help, it is very appreciated

have a great day guys

[deft gull]

thanks you too

[glad shore]

: )

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[primal lodge]

The question is find the coefficient of (2x+3y-4z+w)^(9)

i understand the solution mathematically but where did 9!/(3! 3! 3!) come from

is that a permutation or something where you get rid of duplicates?

I know it says "that there are this many terms" but need some context

[neon moon]

You may want to read about "Multinomial Theorem".

https://en.m.wikipedia.org/wiki/Multinomial_theorem

[primal lodge]

Oh okay that makes sense

This will hold for binomial as well right

I guess obviously it should 🤦‍♂️

Could you also tell me

what I would do when products are involved?

@neon moon

is there a slick trick for this like the one i posted for multinomial

instead of just trying to expand (2x-1) and then multiply by (x+3)

this is an easy question

if it was harder it'd require more effort

i want a generalized method

[neon moon]

It is a product of two binomial, so an x^5 term must be formed by a product of two term, e.g. a x^2 term in the first product and x^3 in the second product.

Hence your goal should be

1. Write 5 as sums, from 0 + 5, 1 + 4, to 5 + 0.
2. Examine (x^0, x^5), (x^1, x^4) to (x^5, x^0) in the two products
   And so on.

[primal lodge]

that's the long way

having to examine 6 terms wouldn't always be feasible

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[jolly sable]

So Jimmy said use limit in it

I didn't understand how it works

[grave inlet]

nice question

asymptotes isnt tangent right?

[grave inlet]

hmmm

i do beleive that x+y=-a isnt a tangent

/graph

@tropic oxide what the cmd to draw a graph here?

[tropic oxide]

there is none

you go to http://desmos.com/calculator and screenshot

[grave inlet]

it wil approach it till infinity

aight

@jolly sable yup its correct

[cedar kiln]

@jolly sable Has your question been resolved?

[jolly sable]

Yes. I found it. But my question is how to find an asymptote of this equation by using limits?

@grave inlet

[grave inlet]

ohh

[weary vessel]

I have been taught a way to find asymptotes of functions in general

I tried applying it for this example, but it did not work out

Do you still want to know the method?

[jolly sable]

Yes

I want to know the method differentiation

[weary vessel]

It did not work for this example because this function is implicit

[jolly sable]

Ahhhhhhhhhhh

[weary vessel]

It works for usual explicit functions

[weary vessel]

Differentiation?

[jolly sable]

Let's wait for someone if they know it

Like use of calculus directly

Limits differentiation

[weary vessel]

Those are two different ways to find the asymptotes

I know them

There is differentiation and another one

I will show you differentiation

Assume we got a function f(x) which has an asymptote at x -> +oo

[jolly sable]

Hmm sure

We can try it.

[weary vessel]

The asymptote is a straight line, and since the function behaves like it

The function is "like a straight line at x -> +oo"

y = ax + b

Therefore, the slope can be found by

$a = lim_{x \to \infty}$ $f'(x)$

[wraith dagger]

LUNA

[jolly sable]

🤔

Hmm okay

[weary vessel]

Looks odd, but that's it

😋

Try it on this example

$f(x) = e^{-x} +2x + 1$

[wraith dagger]

LUNA

[jolly sable]

Here slope will be $a=-e^{-x}+2$

[slate lintel]

curly braces

[weary vessel]

That's the derivative

Now take the limit

[wraith dagger]

arjunn5589

[jolly sable]

A=2

[weary vessel]

Yes, the asymptote of the function at +oo has a slope of 2

Now to find the y-intercept of the asymptote

[jolly sable]

I see

So?

[weary vessel]

At infinity, we have the function being approximately the asymptote

if not it lol

So, at infinity, the difference between the function and the asymptote is zero

$lim_{x \to \infty}$ $f(x) - (ax+b)=0$

[wraith dagger]

LUNA

[weary vessel]

$b = lim_{x \to \infty}$ $f(x) - ax$

[wraith dagger]

LUNA

[jolly sable]

Absolutely

[weary vessel]

And that's how we find b!

These 2 steps are always performed

[jolly sable]

Which y intercept

[weary vessel]

The equation of the asymptote is y = ax + b

a is found by

$a = lim_{x \to \infty}$ $f'(x)$

[wraith dagger]

LUNA

[weary vessel]

and b is found by

$b = lim_{x \to \infty}$ $f(x) - ax$

[wraith dagger]

LUNA

[jolly sable]

Interesting

[weary vessel]

This is all you gotta remember, forget the proof if you want

Apply b's formula

for the example

Tell me what you find

[jolly sable]

Slope will not be negative?

[weary vessel]

What?

Apply both formulas to this

$f(x) = e^{-x} +2x + 1$

[wraith dagger]

LUNA

[weary vessel]

Just to practice it

Give me the equation of the asymptote

[jolly sable]

Hmm sure

$y=2$

[wraith dagger]

arjunn5589

[weary vessel]

y=2?

[jolly sable]

$y=2x-2-2x=-2$

[weary vessel]

What are you doing xd

[wraith dagger]

arjunn5589

[weary vessel]

I gave you the formula for the slope a and the y-intercept b

Of the asymptote of a function f(x)

All you do it apply them

And you will have your asymptote equation

[jolly sable]

Y=ma+b will be asymptote
I put a and b value in

[weary vessel]

y = ax + b

asymptote for the function

$f(x) = e^{-x} +2x + 1$

[wraith dagger]

LUNA

[weary vessel]

Just find a and b

Using the two formulas

[jolly sable]

We have already find a=2

[weary vessel]

This is the differentiation technique

Yes

[jolly sable]

For b = e^{-x}+2x+1-2x
B=1

So y=2x+1

[weary vessel]

Right!

That's it

And look, it truly is its asymptote at +infinity

[jolly sable]

That's true

[weary vessel]

So, I summarize

Given a function f(x)

If it has an asymptote at infinity, the slope of the asymptote can be found using

$a = lim_{x \to \infty}$ $f'(x)$

[wraith dagger]

LUNA

[weary vessel]

and the y-intercept of the asymptote can then be found using

$b = lim_{x \to \infty}$ $f(x) - ax$

[wraith dagger]

LUNA

[weary vessel]

This is the differentiation technique

Just remember the formulas

[jolly sable]

Let me try this for our given question

[floral obsidian]

What's the problem?

[weary vessel]

That's the thing, in this problem the function is implicit

It means you cannot write it in the form

y = f(x) = ...........

[jolly sable]

a=lim x to infinity 3x^2-3ay

[weary vessel]

you will always have y on both sides

[jolly sable]

@floral obsidian

[jolly sable]

What about partial differentiation?

[weary vessel]

You cannot isolate y

You cannot isolate dy/dx

ideally, we'd love to have y = something in x only

or dy/dx = something in x only

But you keep getting something in x and y

So, you cannot apply xd

[jolly sable]

Please find any idea for this

Implicit

[weary vessel]

im helpless

the technique works wonders for normal functions tho, you can memorize and use it

I give you another example, try to work it out and tell me the equation of the asymptote

$g(x)=\frac{2x²-x+1}{x-1}$ ; $x \in \mathbb{R}$-(1)

[wraith dagger]

LUNA

[jolly sable]

I'll upload after my lunch

[jolly sable]

I am stuck at f'x

@weary vessel

[weary vessel]

Return to the one after you canceled terms

Actually no, just take the limit

The form you arrived to makes it easy

[jolly sable]

Which one?

[weary vessel]

Take the limit of the last one

it's the best!

[jolly sable]

A=2

[weary vessel]

Noice

[jolly sable]

B = 1

[weary vessel]

Lovely

Again, it works very well

You can memorize it and use it

[jolly sable]

Wowwww

[weary vessel]

The negative side too

[jolly sable]

True

,w d/dx of x^3+y^3=3axy

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by arjunn5589#0

[weary vessel]

Finally, I found it

It's actually (b)

The asymptote is y = -x-a

[jolly sable]

Yes

How did you get it?

Please teach me😀

[weary vessel]

It's such a long process

But I used the same technique

derivative for a, then limit for b

i gotta go unfortunately

I will show you later today

Have a nice day!

[jolly sable]

You too

You can upload the picture

Whenever you get time

[cedar kiln]

@jolly sable Has your question been resolved?

[cedar kiln]

@jolly sable Has your question been resolved?

[cedar kiln]

@jolly sable Has your question been resolved?

[wraith dagger]

Riku

[limpid plume]

Which simply means the shortest distance between the curve and the line is approaching 0 as x approaches infinitt

but the problem lies elsewhere

[wraith dagger]

Riku

[limpid plume]

There is no staright forward way to write y entirely in terms of x, though it can be done

One such way (albeit arduous) is to solve the cubic equation, y as the variable.

That gives an f(x) to us

[limpid plume]

I don't usually solve these so I'm not confident enough but there seems to exist Tartaglia-Cardano method of solving cubic equations.

[wraith dagger]

Riku

[jolly sable]

I don't know anything about these terms. How to tackle it

[limpid plume]

This just means the slope of the curve apporaches the slope of the asymptotic line when x approaches infinity

[jolly sable]

Fine

What is m here?

[limpid plume]

the slope of the asymptote line

[limpid plume]

honestly I don't know Cardano's method either

So I'd recommend video lectures on it

[jolly sable]

Yes. Please send the link

The problem is we can't find f'x here

[keen drum]

what's the issue?

Do you need to find the oblique asymptote?

[limpid plume]

https://www.youtube.com/watch?v=msNEUxpSO9w It's in Hindi also, so it might be easier for you to follow.

[jolly sable]

Yes we have find the oblique asymptote by x=1 y=m method but I am trying to learn differntiation/limit use of it ti get asymptote where luna told me a nice way but it's not working with $x^3+y^3=3axy$ because it's f'x

[wraith dagger]

arjunn5589

[jolly sable]

I can find the roots of any cubic equation. @limpid plume

How is this video related to this question?

[keen drum]

So assume it to be y=mx+c, and as x approaches infinity, the lim of the y value will be equal. So just substitute the y=mx+c into y^3+x^3=3axy. And for that equation to be linear the coeff of higher degree of x(>1) should be zero.

[limpid plume]

We will use that to write the equation in form y = f(x)

[keen drum]

this should work ig

[limpid plume]

this idea seems to be working but I don't know how accurate it would be

[jolly sable]

No, this method is very long. I know it. Divide by highest order

There is no use limit/differentiation in it

[jolly sable]

It works for all

[limpid plume]

$x^3 + y^3 = 3axy \implies y^3 - 3axy + x^3 = 0$

[wraith dagger]

Riku

[limpid plume]

solve this cubic equation to get y in terms of x and a

then you can write the equation as f(x)

[jolly sable]

How to get it? In the video they are talking about function only in x order

😐

[limpid plume]

treat x and a as constants

🙂

When your equation is in y, take everything else as constants

$y^3 - (3ax)y + (x^3) = 0$ which is now similar to $x^3 + px + q = 0$

[wraith dagger]

Riku

[vestal wolf]

Is there a general rule for cubic equations with that form

[limpid plume]

Cardano's Method should work

[vestal wolf]

[limpid plume]

Where it involves writing x = u + v

[vestal wolf]

I want to cry

[limpid plume]

and then reforming a quadratic equation with u^3 and v^3 as roots to find those, from those to find u + v, and then we get a root.

Rest of the roots can be found after we long divide the original polynomial with the corresponsiding factor, and then simply solve the resultant quadratic equation

[limpid plume]

honestly same

at this point I'd just tell wolfram alpha to do the computation for me

[keen drum]

dude then why don't you apply the usual method of multivariable functions which involves finding phi of different degrees, replacing x with 1 and y with m

[limpid plume]

I think he mentioned that method but he wants to do this the limit way instead

[keen drum]

ok

[jolly sable]

See

@keen drum

[limpid plume]

which is why I told to make the equation explicit instead of the implicit form it currently is

[jolly sable]

I want to learn other short methods to solve problems related to asymptote

[limpid plume]

and that does require solving the equation

[limpid plume]

The problem with these questions are - there are usually no short methods. And this is the only reason why, I hate any type of coordinate geometry in competitive exams

You have to do mindless computations for no reason at all

[vestal wolf]

But isn't that the point of competitive exams

[limpid plume]

well olympiad exists

it's definitely not this

[vestal wolf]

Idk suneung was probably like this according to my experience

[limpid plume]

Well I guess then I'm not attempting that honestly waste of my already declining brainpower sorry if that sounds harsh

[vestal wolf]

Or that!s just because I was an average korean highschool student who wasn't too interested at math

Ofc suneung does not cover asymptotes like this

[jolly sable]

I guess this will be long and confusing. I guess i have to remember only the traditional method

X=1 y=m

Thank you all

.close

[cedar kiln]

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[prisma ether]

This is the process I followed:

[limpid plume]

this is unrelated but do you use dark web reader extension?

[prisma ether]

yes

[limpid plume]

Firefox gang 🤝

[prisma ether]

Opera

[limpid plume]

anyways let's see the question

[prisma ether]

$\int :\frac{f'\left(x\right)}{f\left(x\right)}=\int :\frac{x}{x^2+9}$

[wraith dagger]

Chuti | Argentina

[limpid plume]

oh still nice

[prisma ether]

After this, I got

[wraith dagger]

Riku

[prisma ether]

for the moment what I shared seems correct though

still writing latex

1 moment

[limpid plume]

i mean yeah, i was just saying what you would need to solve it

[prisma ether]

$ln\left|f\left(x\right)\right|+c_1=\frac{ln\left|2x^2+1\right|}{2}+c_2$

[wraith dagger]

Chuti | Argentina

[prisma ether]

$ln\left|f\left(x\right)\right|=\frac{ln\left|2x^2+1\right|}{2}+c_2-c_1$

[wraith dagger]

Chuti | Argentina

[prisma ether]

still writing

🤣

[limpid plume]

[\int \frac{x}{x^2 + 9}dx = \frac{1}{2}\int \frac{2x}{x^2 + 9}dx = \frac{1}{2}\ln(x^2 + 9) + c]

[wraith dagger]

Riku

[prisma ether]

$e^{ln\left|f\left(x\right)\right|}=e^{\frac{ln\left|2x^2+1\right|}{2}+c_3};:c_3=:c_2-c_1$

[wraith dagger]

Chuti | Argentina

[limpid plume]

i don't get how did you get this

[prisma ether]

$e^{ln\left|f\left(x\right)\right|}=e^{\frac{ln\left|2x^2+1\right|}{2}}\cdot e^{c_3}$

[wraith dagger]

Chuti | Argentina

[prisma ether]

its wrong?

[limpid plume]

.

[prisma ether]

yes you are right

so this:

[limpid plume]

seikai

i mean

correct

[prisma ether]

alr so whats the deal with the

$e^{c_3}$

[wraith dagger]

Chuti | Argentina

[limpid plume]

that's just a random constant

keep it K

[prisma ether]

k

$f(4) = 3$

[wraith dagger]

Chuti | Argentina

[prisma ether]

$\frac{3}{5}=e^{c_3}$

[wraith dagger]

Chuti | Argentina

[prisma ether]

😬

ah I just replace 3/5 and end of story right?

ok I think I got it, cool

[limpid plume]

yes

absolutely

nicee

[prisma ether]

[limpid plume]

[cedar kiln]

@prisma ether Has your question been resolved?

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[fleet terrace]

In this integral, I get the answer that Wolfram gets, but one step before the last, how did they get to the last step?

[fleet terrace]

„which is equivalent for restricted x values to“

[gritty viper]

Using sin = cos*tan and log rules

Though there's no good reason to actually do that

[fleet terrace]

Hm

So my answer counts?

[cedar kiln]

@fleet terrace Has your question been resolved?

[cedar kiln]

@fleet terrace Has your question been resolved?

[gritty viper]

Yeah your answer is better

[cedar kiln]

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[clever patio]

when deriving the 3rd equation of motion, i don't understand the integral part

how it goes from the third line left to the top line right

... unless the lecturer forgot to put the bounds v, v_0, and x,x_0. However, are we allowed to do this? v is not equal to x and v_0 is not equal to x_0

[crystal raptor]

doing the integration leaves you with a constant of integration, you can plug in t = 0 so see what that constant should be

[clever patio]

thanks for the reply. so you would get v^2/2+c=a(x+c). how would you plug in t=0?

[crystal raptor]

first you want to write it as v^2 = ax + c

moving one constant to the other side and taking it away from the other is still just a constant

then the notation for v(0) is v_0 and x(0) = x_0

note this is just the same as adding in the bounds

[clever patio]

so when t=0 I'd get v_0^2=ax_0+c. C=v_0^2-ax_0

[crystal raptor]

eh i made i typo earlier

[clever patio]

then I end up with v^2-v_0^2=a(x-x_0), different from the equation in the png

[crystal raptor]

should still be v^2/2

[clever patio]

ah i see

okay id get the equation above then

[clever patio]

bounds v, v_0, x, x_0 right?

[crystal raptor]

yes

[clever patio]

even though it works, im not sure why we can integrate both sides with different bounds

like if you have a function 2x=x^2 and i inegrated 2x with bounds 0 and 2 and integrated x^2 with 1 and 3. What's the difference here?

actually that was a bad example, but you get what i mean?

[crystal raptor]

well yeah whats written is very sloppy

really you should be integrating both sides with the same bounds and then doing a usub on one side

but you cant really see to do that with the notation theyve gone with

[clever patio]

ye pretty sure v is not equal to x and x_0 is not equal to v_0 (in general)

[clever patio]

https://www.youtube.com/watch?v=tcZX1TmTxJ4&ab_channel=BRICS%3AWORLD

this guy seems to be doing the same though

as in with those bounds

[crystal raptor]

yes because the majority of people are sloppy with differentials

[clever patio]

ah

[clever patio]

how would you do this?

I'm trying to figure it out with no success

[crystal raptor]

if you have $\int_0^T v\frac{\dd v}{\dd t}, \dd t$

[wraith dagger]

ΣΑC

[crystal raptor]

then cancelling the dt's is a result of a substitution, hence the bounds change

[clever patio]

so the T becomes v and 0 becomes v_0?

what you are saying that if the variable you are integrating to changes, than so does the bounds?

[crystal raptor]

definite integration by subsitution involves changing the bounds of integration

[clever patio]

I understand that

[clever patio]

but what are you substituting from here?

[crystal raptor]

u = v

[clever patio]

I'm not really seeing this

if u=v then du=dv

[crystal raptor]

which is dv/dt * dt

and you have to just take on faith that you can treat differentials as fractions

[clever patio]

but aren't you substituting v for u?

[crystal raptor]

du = dv/dt * dt

[clever patio]

agreed

[crystal raptor]

so the integrand is just u du

[clever patio]

yes

[crystal raptor]

which you can rename to v if you want

its a dummy variable

[clever patio]

and the upper bound "T"

[crystal raptor]

the point is you are integrating something times its derivative

this is exactly the set up that u-sub is designed for

T is just an arbitrary time, we cant use t because we're using that in the integral

[clever patio]

so from that equation you'd get u^2/2 with limits of 0 and T

so you would get T^2/2

[crystal raptor]

you have to change the limits

u(0) = v(0) = v_0

u(T) = v(T) = v

[clever patio]

oh I see now

because v=v_0 + at

if time is 0

and time is T

thanks a lot for clearing things up

[crystal raptor]

no worries:)

[cedar kiln]

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[orchid oriole]

GUYS PLEASE

[orchid oriole]

HOW DID I DO PART A WRONGGG

I worked it out using alternate anbkes

Angles

Pls help

[livid hound]

wdym using alternate angles

[orchid oriole]

As in I used 3x equals 80 degrees bc they’re alternate

[livid hound]

those angles are technically alternate
however you don't have parallel lines here so they aren't congruent

[orchid oriole]

Oh, so I only do they when they are parallel?

[livid hound]

yes

[orchid oriole]

What should I do instead

[cedar kiln]

@orchid oriole Has your question been resolved?

[orchid oriole]

.close

[cedar kiln]

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[clever jacinth]

im confused as to how they reached the answer

[runic garnet]

Which step specifically

[clever jacinth]

im good with the 5-3x=ln10 part, since ln(e^x)=x

but after that, wouldn't 5 become a -5 when put on the other side?

also with the -3x suddenly becoming positive

and at the end, where to get x alone, wouldn't you have to divide 3 with everything? where did that 1/3 come from?

[surreal cave]

$5-3x=\ln10$\
$-5+3x=-\ln10$\
$3x=5-\ln10$ :)

[wraith dagger]

XxMrFancyu2xX

[surreal cave]

just skipped a few obvious steps, no funny business going on here

as for the 1/3 is being mutliplied by everything so it has the same effect as dividing everything by 3!

[clever jacinth]

ohh youre so right

[surreal cave]

also hello Stephen!

[clever jacinth]

it would be the same thing as writing everything /3

[surreal cave]

just looks nicer :)

[clever jacinth]

it rlly does

i never thought of it that way, thank you!

[runic garnet]

Hola amigo

[clever jacinth]

.close

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[crimson sedge]

How do I do iii?

[carmine whale]

You have v in terms of u
Notice that d is the total sum of the magnitude of horizontal distances covered till collision by both the projectiles

[crimson sedge]

Yes

So do I find the time for max height separately

Plug them into x distance

And add them?

[carmine whale]

Yep

[crimson sedge]

Oh okay

Thank you

[carmine whale]

Not max height
Rather the height where the collision took place

The time where both projectiles r colliding... U have to find that time

[crimson sedge]

But the question says both of the particles are at max height at p

So how do I do it again? @carmine whale

[dreamy sleet]

Notice that the projectiles are always at the same height

[crimson sedge]

Yes

[dreamy sleet]

so you only need to find the horizontal distance

[crimson sedge]

I tried

<@&286206848099549185>

[cedar kiln]

@crimson sedge Has your question been resolved?

[dense hornet]

you are given the displacement from the origin wrt t, so in order to collide at the time t, what condition is needed? ||Hint: the vertical displacement has to be the same for both vector functions, what about the horizontal displacement?||

[cedar kiln]

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[crimson sedge]

.reopen

[cedar kiln]

✅

[crimson sedge]

Idk they add to give d?

[dense hornet]

no

[crimson sedge]

I have no idea

[dense hornet]

if they are colliding at the point P at the time t, then that means their displacement should be equal to the displacement of point P correct?

so in order for that to happen, what would that imply for the horizontal displacement of particle B?

[crimson sedge]

Why?

Isn’t it distance not displacement

[dense hornet]

distance here means nothing. You have to add a preposition to add context. The displacement given by the function r_A and r_B shows you the distance from their current position to the origin

so if say at t=0, then the particle A would have the displacement given by r_A(0)=0i + 0j, this implies particle A is at the origin, for particle B, the displacement is given by r_B(0)=di + 0j. This means that the particle B is at the point where it's d metres horizontally from the origin and 0 metre vertically from the origin

in similar fashion, if at the time t=Usqrt(3)/20, both particle are at the point P, then their horizontal and vertical displacement shows the horizontal distance and vertical distance to the origin

now, if both particle are at the point P at this time t, then does that mean their horizontal displacement have to be equal to each other?

[crimson sedge]

No

[dense hornet]

the answer is yes

how are the particle colliding if their horizontal displacement is not the same

lets say particle A is at 2m horizontally from the origin at time t, but particle is 4m away from the origin horizontally. It's imposible for these to to be colliding

and this is not considering whether or not they are at the same height too

[crimson sedge]

Ohhhhh waittt

I get it now

So I just equate the displacements?

[dense hornet]

yes

even better, you only need to equate the horizontal displacement, i.e the displacement in the i direction

since you know that the height will be equal to each other anyway

[crimson sedge]

Is this correct?

@dense hornet

[dense hornet]

yes, looks good

[crimson sedge]

Thank you

.close

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