and found its +
#help-38
fossil cobalt
so i said increasing
so basically we are finding the slope of the slope
if that makes sense
rugged latch
yep exactly
fossil cobalt
so what does this mean
rugged latch
whether the "rate of change" is. increasing or decreasing"
fossil cobalt
why did my teacher evaluate it
rugged latch
because to solve for d^2y/dt^2 at t=1, u need dy/dt at t=1
look at
see how theres a dy/dt in there
thats why we need it
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brazen gazelle
how would you do question b) if it asked for displacement instead
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brazen gazelle
no
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cyan oxide
how would you do question b) if it asked for displacement instead
u just integrate velocity over time without worring about whether its positive or negative
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fallow marten
so any suggestion?
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fallow marten
<@&286206848099549185> any suggestion?please
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supple sierra
can anyone help with this
wraith hinge
Can someone draw for me the representation of this?
$arccos(z / hyp) * (180 / pi)$?
solid kilnBOT
Cut
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wraith hinge
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cyan knoll
(I'll send my working so far in a sec)
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cyan knoll
Assume that (0,0) is a root. Vector OM is half of OA (if A is root). Multiply OA by i, to rotate it 90 deg giving OB. OB+OA=OC. Therefore (0,0), OA, OB and OC are the roots to an equation.
I'm currently stuck trying to find an equation that will have roots at these, and I'm not sure whether to start with the polar form or the cartesian form.
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wraith hinge
hello
@cyan knoll you need help?
biquadratic equation means it will have 4 roots
so its safe to assume that the required eqn is biquadratic right?
now if you have the 4 roots and you want to find the eqn, you can use the general method right?
so start with the cartesian form
x^4-(sum of roots)x^3+(sum of roots taken 2 at a time)x^2-(sum of roots taken 3 at a time)x+product of roots
and then why not substitute polar coordinates
rcos(theta), rsin(theta)
bright quarry
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scarlet escarp
i would like some confirmation on this. if given y'f'(y) = 1, can we say that a necessary condition for y to be a solution is f(y) = t + C?
scarlet escarp
the lhs looks like it's just the derivative of f(y) wrt t, so it would seem that integrating both sides would give that condition
and similarly, can i say that y is a solution provided f(y(t)) = t - f(y(0))? this would be the result of integrating from 0 to t
wispy raptor
How to prove this?
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How to prove this?
simple haven
Both are well known identities equal to e
The first is using the compound interest formula, the second is using the series representation of e^x evaluated at x=1
wispy raptor
Without the pretense that both are equal to e, how to prove they are the same?
dapper swift
Without the pretense that both are equal to e, how to prove they are the same?
I recommend using https://approach0.xyz/search/
It searches for questions in Math SE and on AoPS
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austere cedar
Let e^x be the function that passes through (0,1) whose derivative is itself.
The left is the definition of the derivative for e^1
The right is the Taylor series for e^1
wispy raptor
Let e^x be the function that passes through (0,1) whose derivative is itself.
...
So like the derivative and taylor series should be same by definition?
austere cedar
Actually I guess I am assuming e^x is analytic
wispy raptor
Wats analytic 🙃
simple haven
Analytic means that a function can be represented by a Taylor series, essentially.
austere cedar
Does e^x actually match its Taylor series at x = 1? We don't know yet.
wispy raptor
https://math.stackexchange.com/questions/69806/prove-the-definitions-of-e-to-be-...
Anyways according to this i dont quite understand the circled part, can someone help explain it to me?
dapper swift
Anyways according to this i dont quite understand the circled part, can someone ...
Let n/x = u and so x/n = 1/u
Then you have (1 + 1/u)^u, where u also goes to infinity as n goes to infinity (we assume x is held constant)
So that limit becomes e^x
wispy raptor
I see
dapper swift
And now you can see why x can't be 0
wispy raptor
Anyways according to this i dont quite understand the circled part, can someone ...
What does the formal series n converge thingy mean tho?
dapper swift
Well we'll just have 1^n as n tends to infinity, no big deal
dapper swift
What does the formal series n converge thingy mean tho?
Basically we've formally shown the 2 series are the same
By taking the limit as n goes to infinity
wispy raptor
Well we'll just have 1^n as n tends to infinity, no big deal
So it can be 0 but it'll just be 1=1?
Which doesnt mean anything?
dapper swift
Formal series means what exactly
It's bad wording from someone who probably isn't a native speaker of English
dapper swift
So it can be 0 but it'll just be 1=1?
Yeah so it doesn't matter we couldn't let x = 0 when we substituted n/x = u
wispy raptor
It's bad wording from someone who probably isn't a native speaker of English
Oh lmfaoo
wispy raptor
Yeah so it doesn't matter we couldn't let x = 0 when we substituted n/x = u
Right thanks
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dapper swift
No worries man
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wraith hinge
If i am given, a=/0 and ax^2 + bx + c = 0
Will the equation have real roots if Discriminant is >= 0 ?
wraith hinge
If i am given, a=/0 and ax^2 + bx + c = 0
Will the equation have real roots if ...
here i know, if D = 0 then we get repeated roots which are real
low bison
If i am given, a=/0 and ax^2 + bx + c = 0
Will the equation have real roots if ...
If the discriminant of the quadratic equation ( ax^2 + bx + c = 0 ) is greater than or equal to zero, then yes, the equation will have real roots. A discriminant of zero means one real, repeated root, and a positive discriminant means two distinct real roots.
wraith hinge
If the discriminant of the quadratic equation \( ax^2 + bx + c = 0 \) is greater...
but
this dosent make sense
low bison
??
wraith hinge
what if a,b,c belong to imaginary numbers
astral ore
then it doesn't make sense to compare the number 1 + 2i with 0 for example
just use the quadratic formula to get the two solutions
it still works all the same
plucky grail
then you can't say anything about the solutions
wraith hinge
no you arent understanding
plucky grail
they may or may not be complex
astral ore
?
wraith hinge
the condition on a,b,c for the roots to be real if D >= 0
maybe i am phrasing it wrong mb
low bison
what if a,b,c belong to imaginary numbers
The roots will be real only if the discriminant’s imaginary part is zero and its real part is non-negative. Otherwise, the roots will be complex.
Was that more the kind of answer you were looking for..?
astral ore
the condition on a,b,c for the roots to be real if D >= 0
the condition is literally give by D >= 0 which implies b^2 - 4ac >= 0
what?
plucky grail
the condition on a,b,c for the roots to be real if D >= 0
this is IF the coefficients are real
wraith hinge
this is IF the coefficients are real
this is more of what i want
astral ore
you've trailed off your main question like multiple times
i don't think anyone knows what's actually going on
wraith hinge
you've trailed off your main question like multiple times
sigh
wait
forget this mess
let me rephrase
plucky grail
I think your question was clear enough 
astral ore
astral ore
Okay answer his question until he actually understands
without asking him to rephrase it lol
wraith hinge
ignore the very ovious grammar error
wraith hinge
this is so different from what you asked
😭
i am so sorry
should have done this first, my bad guys
fallen pawn
It is true.
wraith hinge
first
plucky grail
true if a, b, c are real
wraith hinge
but you can't answer False either
you can, theres a specific condition asked in the question
astral ore
if you want to play the a,b, and c are real/imaginary game that is
plucky grail
yeah this is all assuming the coefficients are real
wraith hinge
the condition given here(for D>0) for real roots is a,b,c belons to rational numbers
wraith hinge
yeah this is all assuming the coefficients are real
so for the answer, does real work or is rational a better answer?
plucky grail
Rational isn't wrong, since it's a subset anw. but real is more complete
wraith hinge
Rational isn't wrong, since it's a subset anw. but real is more complete
hm
so for this part
fallen pawn
the condition given here(for D>0) for real roots is a,b,c belons to rational num...
No. No need to be rational. Just imagine any quadratic and multiply its coefficients by root(2). D continues to be >0
wraith hinge
should it be real?
wraith hinge
No. No need to be rational. Just imagine any quadratic and multiply its coeffici...
so real numbers
fallen pawn
yes
astral ore
It's literally all the same
wraith hinge
Can we not predict if they belong to R?
plucky grail
Can we not predict if they belong to R?
we can
if you're confused about anything, just look at the whole quadratic formula. that works regardless of real/complex coefficients
fallen pawn
Okay. I think I get his question.
Given two real roots, is it possible to construct a quadratic equation with complex coefficients?
@wraith hinge is this it?
wraith hinge
no 😭
astral ore
Bruh
wraith hinge
guys
i will mess around with soem numbers and check what the confusion is
thank you for your time
and sorry 😞
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fading pebble
The stationary point (1,0) has eigenvalue 0 and eigenvalue 2
How do I find out what type it is ? I want to know if its minimum, maximum or saddle point
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rich breach
How to verify if the segments are parallel, i cant get the same number
crimson void
How to verify if the segments are parallel, i cant get the same number
Wdym by 'same number'
lusty delta
they just mean they don't get an equivalent ratio that proves parallel property
rich breach
Wdym by 'same number'
I tried solving it by gcf and dividing but they get two separate decimals
crimson void
Then that means AY is not parallel to WX
crimson void
Could you translate what the question says?
rich breach
Ok
rich breach
Could you translate what the question says?
It just says on this, decide if WX is parallel to AY
crimson void
Yes so the answer will be false
lusty delta
I tried solving it by gcf and dividing but they get two separate decimals
but will ask
what numbers do u get
like what numbers did u use to get that results
rich breach
but will ask
what numbers do u get
18 and 32 gcf and 15 25 gcf
I tried dividing but they get different numbers
merry coyote
can i just ask about this question
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merry coyote
I think it's uncountably infinite, because x and y go to infinity because they're just members of the uncountably infinite set of real numbers right?
Like there's nothing to limit them
so the graph will keep growing larger and wider as x or y increases
Am i correct in everything I've just said?
scarlet escarp
you're right, though there are probably nicer ways to phrase that
if you're allowed to use calculus/analysis there are some good proof of what you just said in nicer terms
merry coyote
if you're allowed to use calculus/analysis there are some good proof of what you...
i mean we were never told not to use anything
it's just the concept is fairly new to me so I'm trying to understand them
scarlet escarp
have you heard of the intermediate value theorem
merry coyote
i think I've seen it but we've never actually used it
Would it be like using f(a) ≤ x, y ≤ f(b) where a = -∞ and b = ∞
or something
scarlet escarp
this is a good place to use it, it gives you a nice solution to the question
it says that if f(a) < f(b), then for every number d between f(a) and f(b), there is some other number c so that f(c) = d
we can look at it in your specific case
rewrite the equation as
y = 2x^2 + e^x - 1
the right term is a continuous function
and you have that
y(0) = 0, y(1) = 1 + e
does this make sense so far
merry coyote
Yes
scarlet escarp
then you have that y(0) < y(1)
so y must cross through every value between 0 and 1 + e
so you have at least as many elements as [0, 1 + e]
scarlet escarp
well that the cardinality of [0, 1 + e] is uncountable, but that should be known
wraith hinge
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whole coral
Your volume proportion is wrong-
whole coral
Remember that lengths in B would be sqrt{25/64} of those in A, and how that relates to your volume proportions
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wraith hinge
Remember that *lengths* in B would be sqrt{25/64} of those in A, and how that re...
So I square 625/4096?
Would that mean the case has a volume of 22,728.33281?
That seems a bit unrealistically large
whole coral
So I square 625/4096?
No, you already are too large(!)
Square root the 25/64, then cube what you get from that
wraith hinge
i already did that for the sake of doing that
it’s 0.625
Is the answer 325.08
@whole coral
whole coral
Not quite no 
whole coral
it’s 0.625
As in this is what you get from (25/64)^(3/2)?
wraith hinge
As in this is what you get from (25/64)^(3/2)?
oh
why do you power it to 3/2?
originally you’d square it
then square it
whole coral
why do you power it to 3/2?
Square root to get the length proportion, cube to turn the length proportion into the volume proportion
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wraith hinge
Square root to get the length proportion, cube to turn the length proportion int...
What
whole coral
Do you know how the lengths, surface areas and volumes in similar shapes relate?
You were given how the surface areas are related, and you want how the volumes are related from that
whole coral
No it isn’t-
Remember that only surface areas have the squared units, volumes have cubed units (e.g. m^3, ft^3 etc etc)
whole coral
No you don’t - that would give you something to the power 4, not 3
whole coral
Square root (which will turn your cm^2 to cm) and then cube that (which will turn said cm to cm^3)
wraith hinge
yes that’s what I said right
So 3, 9, 27
as an example i mean
and we start with 9
so we square root then cube
whole coral
That’s how you’d do it sure (though you didn’t say that )
wraith hinge
so where was my mistake
whole coral
so we square root then cube
Well as per before, you basically start with 25/64 and you wanna do this
whole coral
So to get from SA to V we square it again
For the last one you said squaring it (which isn’t quite how you’d do it 
)
whole coral
Yep, that’s it 
wraith hinge
Then use that number as my factor?
whole coral
Yep, as the factor for the volume, the volume of B is that factor of the volume of A
wraith hinge
so 514.8 divided by 125/512
2108.6208
subtract 541.8 from that
you get 1,566.8208
Is that correct
whole coral
Well, more that you know that as the volume of B is 125/512 of the volume of A
Then the volume of A is 512/125 that of volume B
If I call V the volume of B, then in that case, you know that (512/125)V - V = 541.8, and you can solve that for V
wraith hinge
can you just solve it
it’s so late
ive been working with a group of friends to solve this 💀
also this came on my mock paper
and i got like 1/5
whole coral
We can’t, not allowed to-
wraith hinge
😕
whole coral
If I call V the volume of B, then in that case, you know that (512/125)V - V = 5...
This is pretty close to solved though, you just need to find V from that
wraith hinge
is x 175
whole coral
Then the V is your answer
wraith hinge
Then the V is your answer
175?
whole coral
Yep that’s it 
whole coral
There you go 
whole coral
No problem 
A tiny bit, you gotta know how the proportions of volumes work, and knowing that you can write everything in terms of the volume of B makes life much easier to work with
Something that trips people up is relating surface areas to volumes and vice versa
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calm marsh
hi. i need to find volume of it but i have no idea
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hi. i need to find volume of it but i have no idea
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calm marsh
<@&286206848099549185>
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edgy willow
hi. i need to find volume of it but i have no idea
try the polar coordinates integral of that height z, you have to convert the z=.. with x=rcos(theta) and y=rsin(theta), then just integrate that times r (because polar) from r=0 to sin(theta) on the inside and theta=0 to pi/2 on the outside
calm marsh
okay
can you write in integral equation . i am having a little bit trouble understanding it due to my lack of proeffiency in English also in math
edgy willow
like this
calm marsh
thanks
edgy willow
since it's like dxdy = r*drdtheta
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turbid crow
when we're referring about frequency we mean how the original signal its splitted up with different signal's ?
I.e u[n] is declared for n>0 with amplitude 1.
And it’s z transform is 1/1-z^-1
If you substitute at z=2, you get z=0.5 , what does this mean ? What it makes it uniquely and differentiate from the time domain ?when we're referring about frequency we mean how the original signal its splitted up with different signal's ?
I.e u[n] is declared for n>0 with amplitude 1.
And it’s z transform is 1/1-z^-1
If you substitute at z=2, you get z=0.5 , what does this mean ? What it makes it uniquely and differentiate from the time domain ?
why frequency domain aka z domain is useful
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atomic totem
I’ve proven that for any solution the gcd will be divisible by b, but I can’t do it the other way
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wraith hinge
Hello i have 2 questions regarding how do i prove that log_a(x)/log_a(b) = log_b(x) , and when do i know when to use differential equations what is its concept expect for the fact that it combinea a function and its derivative
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wraith hinge
Thank you !🌸
lost minnow
Help
blazing geode
,rotate
solid kilnBOT
lost minnow
yes
blazing geode
3
lost minnow
huh
blazing geode
pi
lost minnow
huhhh
north latch
not that good at math but isn't it just 3sin(x) + some offset
blazing geode
y=3sin(1/2x)
fast ginkgo
no
lost minnow
can u explain?
fast ginkgo
there's no vertical or horizontal shift here
blazing geode
the period is being modified as opposed to the regular one sinx=y
blazing geode
2pi
fast ginkgo
that's PERIOD
lost minnow
yo let me answer
fast ginkgo
plz don't dispense incorrect advice
blazing geode
I'm sorry jack
lost minnow
its alr
we make mistakes
now
can someone tell me THE STEPS TO MAKE AN EQUATION FROM THE GRAPH PLS uwu 🙂
fast ginkgo
whats the amplitude for sinx?
^ answer this
then check the amplitude of the graph on yr question
lost minnow
the amplityue is the max + min / 2
dapper swift
the amplityue is the max + min / 2
No
lost minnow
yes
lost minnow
(max - min)/2
Thats what I mean
dapper swift
Basically the amplitude is 3
lost minnow
oh ya
cuz the upper bound and lower bound is just 3 and negative 3 but amp is always |B| so its just 3
dapper swift
Yep... this is the easy case
lost minnow
what about horizontal stretch/shirnk
fast ginkgo
the factor responsible for stretch is sin (ax)
lost minnow
how do i find that
i dont get what u mean by comparing with 2pi
one sec brb
alr I am back
@fast ginkgo @dapper swift
fast ginkgo
how long is the period in a normal sine wave sinx typically
2pi, what's the period here
note that the stretch factor is inversely proportional
dapper swift
how do i find that
The period is when you have one full cycle
So it goes through one min and one max very cycle
fast ginkgo
note that the stretch factor is inversely proportional
e.g. if the period is 2 times the usual then the factor is 1/2
fast ginkgo
a period is the length of one full wave
lost minnow
mhmmm
fast ginkgo
or from peak to peak
lost minnow
so its pi?
fast ginkgo
correct
so the period is 1/2 of sinx which has a period of 2pi
does that give you any indication of how to find the stretch factor
lost minnow
holup
fast ginkgo
you can graph these things on desmos fyi
fast ginkgo
i dunno how you drew that conclusion
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lost minnow
@fast ginkgo
fast ginkgo
why ping
fast ginkgo
i don't understand the point of your question
you literally plotted it on desmos no
naturally youd have a better idea of what the period is than i would by estimating from a picture
lost minnow
the graph is cos(x)
fast ginkgo
then wdym by asking if it's the period of a cos graph
fast ginkgo
do you know how to count one cycle
i think you may have some terms confused because your questions are not making a whole lot of sense right now ibr
lost minnow
pls explain
the perird is the time it takes for one cycle to complete
I am pretty sure that right
fast ginkgo
*the length
lost minnow
?
fast ginkgo
okay sure the duration of a cycle
fast ginkgo
you can count number of cycles from anywhere
e.g.
it doesn't really matter as long as a full wave is accounted for
lost minnow
so was blue line from left to the blue line a full wave?
fast ginkgo
you only have 1 blue line here
if you had another around x=11ish that would look like a full wave
but again idt youll be asked to visually estimate periods very often (???)
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how do you convert y = x into polar equation?
naive geyser
i know that x = rcos(theta) and y = rsin(theta)
but the thing is
when i put rcos(theta) = rsin(theta) into desmos, it doesn't work
hallow cliff
Thats because your parameter is theta
left oriole
what went wrong?
naive geyser
what should i do then?
left oriole
you can easily solve it by hand
naive geyser
left oriole
r = 0 is one solution, and for nonzero r you can divide both sides by r
left oriole
r cos(theta) = r sin(theta)
plug in r = 0, you get 0 = 0 which is true
to get the rest of the solution (the case where r is nonzero), divide both sides by r
hallow cliff
They are trying to find the polar equation for y=x
naive geyser
yeah
so graphing works with y=x
but i don't know how to do it with polar equation
left oriole
r cos(theta) = r sin(theta)
divide both sides by r
it's like the third time i've said that
did you try it?
left oriole
you can do this by hand, ignore the software
if you divide both sides of r cos(theta) = r sin(theta) by r, what is your new equation?
naive geyser
it would be cos(theta)=sin(theta)
naive geyser
i still don't know how to graph cos(theta)=sin(theta) on the paper
its kinda confusing
left oriole
do you know a trig identity that involves sin and cos?
naive geyser
oh tan?
left oriole
in particular, what happens if you divide both sides of cos(theta) = sin(theta) by cos(theta)?
naive geyser
um if you divide both sides by cos(theta), then you would get 0 = tan(theta)?
left oriole
you divide cos(theta) by cos(theta) and you get 0?
naive geyser
um theta could be pi/4, 3pi/4
left oriole
pi/4 yes
naive geyser
wait no
left oriole
3pi/4 no, that would be -1
left oriole
yea
left oriole
yea
naive geyser
the line only invovles
left oriole
theta = pi/4 or 5pi/4
naive geyser
the 1st and 3rd quadrants!!!
left oriole
or those plus any multiple of 2pi
and r can be anything
as you vary r you move along the line
naive geyser
theta = pi/4 or 5pi/4
wait but that's not a line is it?
i know those are 2 points of the line y = x
but how is it a line?
left oriole
theta = pi/4, that's a half line, right?
the 1st quadrant part of the line y=x
and theta = 5pi/4 is the other half, the part in the 3rd quadrant
technically the origin isn't included in either of those, since the theta is undefined at the origin
that's why i mentioned r=0 separately earlier
tepid hamlet
🤔
graph of theta = pi/4 is a full line i think, not just a half line
r can take on any value including negative ones, and 0
naive geyser
tepid hamlet
theta = pi/4 and theta = 5pi/4 are both valid polar equations representing the line y = x
but you don't need both
left oriole
r can take on any value including negative ones, and 0
yea depends on the convention in force here
whether we're restricting r to be >=0 or not
tepid hamlet
i gotta admit i didn't read the full chat history, but would there be a reason for r to be >= 0
left oriole
tepid hamlet
if you're just talking about generic polar coordinates \ equation that should definitely include any value for r right?
left oriole
depending on context that could mean r >= 0 or not
only the OP can tell us which is the case in their context
tepid hamlet
when is r restricted to >= 0 in polar coordinates? 🤔
naive geyser
wait now that i think
both
sin(theta)=cos(theta)
and
tan(theta) = 1
seem acceptable?
left oriole
sure, they are equivalent
naive geyser
they all just mean y = x
tepid hamlet
both
sin(theta)=cos(theta)
and
tan(theta) = 1
seem acceptable?
tan theta = 1 for theta = pi/4 or theta = 5pi/4
tepid hamlet
ya but like i don't think you actually ever DEFINE polar coordinates like that, in some contexts you just might need to have a unique representation of each point so you restrict r to a subset
naive geyser
tan theta = 1 for theta = pi/4 or theta = 5pi/4
mhm
naive geyser
tan theta = 1 for theta = pi/4 or theta = 5pi/4
but im also thinking that pi/4 is both 1st and 3rd quadrants?
left oriole
ya but like i don't think you actually ever DEFINE polar coordinates like that, ...
i'm fairly agnostic about it personally, but i have no idea whether whoever's grading chimken's work would accept just theta = pi/4 as they might have the r>0 requirement.. whereas theta = pi/4 and theta = 5pi/4 works regardless of convention
naive geyser
because it can be (n, pi/4) as well as (-n, pi/4)
left oriole
also, if they want the full solution it would have to be pi/4, 5pi/4, plus any integer multiples of 2pi
tepid hamlet
i'm fairly agnostic about it personally, but i have no idea whether whoever's gr...
put theta = pi/4 into any graphing calculator
left oriole
put theta = pi/4 into any graphing calculator
i'm not gonna engage this further, it seems like a pointless debate without knowing the op's convention
tepid hamlet
fair, but my argument is that it's a widely known / accepted convention, there is not really variation here
r can be any real number, so coordinate (r, pi/4) includes points like (-2, pi/4), (0, pi/4) etc.
naive geyser
put theta = pi/4 into any graphing calculator
this is the answer key i have
looks like they treat pi/4 as a line
tepid hamlet
yes, thats how it should be
naive geyser
because it can be (n, pi/4) as well as (-n, pi/4)
like, (-n, pi/4) is the same as (n, 5pi/4)
right
tepid hamlet
uh 5pi/4 you mean
tepid hamlet
yes, so in cartesian coordinates each point has a unique representation, it's convenient
the thing with polar coordinates is there are many representations of the same point, in fact infinite representations
so (-n, pi/4), (n, 5pi/4), also (-n, 9pi/4) etc. you can keep adding 2pi to either of those two and its the same point
naive geyser
so (-n, pi/4), (n, 5pi/4), also (-n, 9pi/4) etc. you can keep adding 2pi to eith...
yup
i see the convention
tepid hamlet
so just keep in mind that the point is the actual object and a set of coordinates is just a representation of it and not unique, it can cause confusion with polar coordinates
naive geyser
theta = pi/4 is the same as y=x
it could've been 5pi/4 too
but just for the convention
mathematicians said
lets use pi/4?
tepid hamlet
but like bungo said, each point will have a unique representation if you restrict r >= 0 and 0<= theta < 2pi
except origin i guess, not sure what the convention for that is
if you think of those restrictions
naive geyser
yeah i think i get the idea at this point
it will definitely help me on the test 
i appreciate your help!
tepid hamlet
in general, origin can be represented by any theta as long as r = 0
naive geyser
.close
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neat crystal
how do i prove this series converges or diverges?
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how do i prove this series converges or diverges?
neat crystal
trail ingot
for large n, (n-1)/(n^2-n+1) is approximately 1/n and arcsin(1/n) is approximately 1/n
that should motivate a solution
neat crystal
for large n, (n-1)/(n^2-n+1) is approximately 1/n and arcsin(1/n) is approximate...
what do i use to turn (n-1)/(n^2-n+1) into 1/n? if i divide everything by n^2 wont it turn into 0?
trail ingot
you don't need to "turn it into anything"
what i wrote above was supposed to motivate limit comparison test
neat crystal
ok
hexed bone
summer yarrow
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plucky grail
note that $-8(1-2x)^3 = 8(2x-1)^3$
solid kilnBOT
nebula40
plucky grail
you dont have to expand, just factorize and solve the inequality like normal
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fast gull
Hi MY mind is blanking for part b anyone care to guide me
fast gull
I forgot im still in grade 9 math
lmfao
frozen plover
Hi MY mind is blanking for part b anyone care to guide me
do you mean g?
fast gull
do you mean g?
yeah i do the lettering throws me off cos these are the only parts
solid kilnBOT
²⁰Ne
fast gull
i just wrote it as the conjugates of alpha... and left it at that
vagrant marsh
fast gull
wait omg i'm so stupid guys im sorry its part f because its part of all these questions i was doing prior. LMFAO its actually the 7th root of unity. im going to close it and retry with this knowledge bc i might get it
.close
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fast gull
wait im back
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fast gull
i found a = 1 but now im stuck on getting b after because i forgot to easily get the modulus squared of alpha (which was how i was going to find b)
fast gull
Hi MY mind is blanking for part b anyone care to guide me
this was the og question btw
let me know if theres a simpler way of finding b than what i did lol
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digital musk
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little garden
Hello I want to know something about gauss divergence theorem can somebody help me
little garden
<@&286206848099549185>
compact plover
Hello I want to know something about gauss divergence theorem can somebody help ...
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spare helm
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split chasm
show your work
split chasm
from exponent laws
$$\frac{p^x}{q^x} = \br{\frac pq}^x$$
solid kilnBOT
ℝαμΩℕωⅤ
split chasm
since the power is the same, you can do division on the bases, the power will remain
split chasm
apply
$$\frac{p^x}{q^x} = \br{\frac pq}^x$$
to
$$\frac{36^{2m}}{108^{2m}}$$
solid kilnBOT
ℝαμΩℕωⅤ
spare helm
Like after that
split chasm
wdym
split chasm
well, the bases aren't the same, powers aren't the same
so you can't do much directly
spare helm
So what would be next
split chasm
simplify the 36/108
spare helm
1/3
split chasm
you may also recognise that 216 is a power of 6
spare helm
This correct so far?
split chasm
yes
split chasm
recall what i mentioned earlier about 216 being a power of 6
split chasm
powers don't get cancelled like that at the very bottom
spare helm
OHHH
I FORGOT
Right
According to the rule you sent
But the answer was like
1/9^m
Omds
Nvm
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split chasm
yeh
spare helm
How would I reach that from here
low sand
Hellow guys dose anyone knows how to do this
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Hellow guys dose anyone knows how to do this
warm vortex
Hellow guys dose anyone knows how to do this
dear
turbid leaf
Hellow guys dose anyone knows how to do this
turbid leaf
should be (secx + tanx)’
warm vortex
btw ||dy/dx=secx|| only view to confirm
low sand
don't forget chain rule
I did chain rule and I got the 1/1+sinx 😭😭
warm vortex
I did chain rule and I got the 1/1+sinx 😭😭
no you forgot to differentiate
(u(v(x)))'=u'(v(x))v'(x)
you wrote down v(x)
low sand
low sand
don't you know d/dx(secx)=secxtanx
I didn’t know about this
warm vortex
I didn’t know about this
there should've been a look up table in your book for such basic derivatives
anyway
you do now
so apply
zealous summit
is this rocket scinece or what. how do i even prove that they are equal to each other. should i just put the two equations equal to each other and solve? but solve for what variable?
ionic pendant
you're showing a trigonometric identity (true no matter what), so you take one side and manipulate it until it looks like the other side
i would recommend you start with the angle sum identity
zealous summit
.reopen
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zealous summit
wait. i dont get it
i tried it but it didnt work
since theyre practically all variables
what am i supposed to do after i implent the formula (the first one, right?)
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zealous summit
is this right?
stiff locust
yo
stiff locust
if a question says something is dropped from a building that is 1362 feet tall
and you find the instanteous velocity at a point
would your unit be feet per second
lament reef
yes
latent axle
Depends if you are taking it with respect to seconds, minutes, hours
Probably seconds though yea
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frank abyss
Hello people, I’m facing yet another math problem. So here is the issue - we’re writing a test tomorrow and I couldn’t come to math class cause I was sick. Now, I’m trying to review the material before the test and I simply don’t even know how one should approach the problem I’m facing. Here is the problem:
You are given a circle k, with center S and radius r. Then you are also given a circle m, with center O and radius v. Find all line segments XY, such that X belongs to k and Y belongs to m, XY is parallel to SO and also |XY| = 1/2|SO|, where |x| represents the length of the segment.
Any hints would be greatly appreciated
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Hello people, I’m facing yet another math problem. So here is the issue - we’re ...
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frank abyss
<@&286206848099549185> if anyone feels like it pls 🙏🙏
ok here is a picture with all the information given ig
Yea I'm thinking maybe since it's parallel, there wouldn't be that much solutions
tender yacht
first off check this value: distance between S and O minus (v+r). this value should be less than 1/2 of SO
if it is greater, you dont have a solution
frank abyss
yea that makes sense
tender yacht
if they are equal, you get one solution. that is the line segment that lies on SO
if it is less, then you get only 2 solutions
frank abyss
can't there be more symmetric solutions
oh there can't
cause there are at most 2 pairs of points w/ distance 1/2 of SO
tender yacht
yepp
frank abyss
however, how would you find that?
frank abyss
first off check this value: distance between S and O minus (v+r). this value sho...
oh let me look at thjis
tender yacht
now im not sure if my method is the most efficient one, but i would say parameterise the distance in terms of theta
frank abyss
now im not sure if my method is the most efficient one, but i would say paramete...
WHAT did you just say??
parameterise
tender yacht
if I have a circle of radius r, with centre at origin
then for all points on that circle, x= rcos theta and y = r sin theta
does this make sense?
frank abyss
oh wait let me draw this
tender yacht
actually nevermind
frank abyss
I think I know what you mean
tender yacht
im sure there is a simpler solution
frank abyss
its like a right angled triangle with that radius
tender yacht
someone else will probably be able to help better. in case i find the solution, I'll let uk
frank abyss
still tysm for trying to help!
i'll work with the observations you mentioned in the meantime
frank abyss
then for all points on that circle, x= rcos theta and y = r sin theta
ok I get this now
you just kind of made a vector out of it
tender yacht
yepp
so assume the origin is at O
first circle's coordinates are v cos theta and v sin theta
the other circle's coordinates are r cos alpha and SO + r sin alpha
frank abyss
ok i'll draw this one out
tender yacht
for our line to be parallel, v cos theta = r cos alpha (same x coordinates)
and v sin theta + r sin alpha = 1/2 SO
if you solve these two equations, and get either alpha or theta
frank abyss
yea like a relation
what does the other angle have to be when the first angle is fixed
tender yacht
for our line to be parallel, v cos theta = r cos alpha (same x coordinates)
get the other angle from this relation
frank abyss
for our line to be parallel, v cos theta = r cos alpha (same x coordinates)
they don't necessairly have to be the same tho?
oh but in this case yes
frank abyss
but in this exact case the x coordinate has to be the same regardless
cause then the lines wouldn't be parallel
but if the lines were like slanted
you could still use the same relation
tender yacht
x coordinate will be same but the angles alpha and theta will be different
tender yacht
but if the lines were like slanted
then the x coordinates wont be the same
frank abyss
then the x coordinates wont be the same
yea but it doesn't matter much, cause you can still sort of transform the problem
tender yacht
mhm
frank abyss
or just calculate out the second point
cause you just have to add some constant
which is the |x1 - x2| of the first line segment
tender yacht
hmm i guess so
frank abyss
yea okay lets assume both are on the same x axis, I can solve these edge cases by myself
but what you are essentially saying is that we have a system of two equations
that are both solvable
tender yacht
yes
frank abyss
v cos theta = r cos alpha
tender yacht
yea okay lets assume both are on the same x axis, I can solve these edge cases b...
its a fact not an assumption
cuz we have taken s directly above o
and any line parallel
will have both its points having the same x coordinate
frank abyss
cuz we have taken s directly above o
yea in my drawing but if the line segment SO were to be slanted, to draw a parallel line distant by sth we just have to add some constants that can be calculated to the equation
tender yacht
yep... but the question doesnt mention any specific coordinate system
so for similar questions, we can always take our system such that the centres have the same x coordinates
frank abyss
cause like in this special case it is a fact that |SO| + rsin alpha - v sin theta = 1/2|SO|
tender yacht
makes life much easier
frank abyss
thats actually very smart
and that trick with moving stuff to the origin aswell
im a competetive programmer so i don't know much about real geometry problems lol
tender yacht
ah its just experience
frank abyss
i always just cross/dot product it xd
tender yacht
thats cool!
frank abyss
ok thank you for your help
i value it that you have taken your own time
and this trick with parameterising trick looks neat
i will definitely (ab)use it in the future!
tender yacht
glad I helped!
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bright patrol
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$$
\begin{aligned}
\text{Surface Area: }2\pi\int_a^b f(x)\sqrt{1+f'(x)^2}
\end{aligned}
$$
$$
\begin{aligned}
f(x)=x^3\
f'(x)=3x^2 \
f'(x)^2=9x^4\
\end{aligned}
$$
$$
\begin{aligned}
2\pi&\int_0^5x^3\sqrt{1+9x^4}\
&u=1+9x^4\qquad du=36x^3
\end{aligned}
$$
solid kilnBOT
AWACS Sky Eye
bright patrol
here's where i'm stuck. i started to do u-sub because i was watching the video tutorial and then it stopped making sense
wraith arch
which part is confusing you?
solid kilnBOT
AWACS Sky Eye
wraith arch
alright so you understand how you got to the second part there right
where we just plugged in y’ into the formula
bright patrol
no, not at all.