yes yes yes, thank you đ
#help-33
drowsy sundial
marsh citrusBOT
If you are done with this channel, please mark your problem as solved by typing .close
molten moss
wahoo
wise oxide
marsh citrusBOT
wise oxide
Is the answer 7/5?
glossy flint
Yep
marsh citrusBOT
@wise oxide Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
still temple
iâm not sure how to start this.. i missed a day of lecture and iâm confused. ):
marsh citrusBOT
iâm not sure how to start this.. i missed a day of lecture and iâm confused. ):
quasi patrol
,rotate
elfin berryBOT
sterile rock
well
quasi patrol
well, its easy to see that this is the same as (probability it is atleast 7)-(probablity that it is atleast 8)
bc that limits it to exactly 7
try working from there
sterile rock
I think there is an easier way to do that
still temple
on the calculator?
wary kite
yes
sterile rock
||you get it correct 7 times, incorrect once.||
||(0.2)^7 * (0.8) * 8C1 (for the possible orders this may occur)||
I think that works
quasi patrol
||you get it correct 7 times, incorrect once.||
||(0.2)^7 * (0.8) * 8C1 (for the...
8C1 is the worlds worst way to write 8
sterile rock
||you get it correct 7 times, incorrect once.||
||(0.2)^7 * (0.8) * 8C1 (for the...
Do not look at this until you have struggled on it a bit
sterile rock
8C1 is the worlds worst way to write 8
I wrote it like that so you can apply similar logic to other questions
still temple
i still donât understand what i should do first..
sterile rock
like if it was 6 correct, 2 incorrect
sterile rock
i still donât understand what i should do first..
try working out the probability that you get:
CCCCCCCI
C= correct
I = incorrect
and work from there
still temple
since thereâs 8 trials there would be one incorrect?
sterile rock
||consider ICCCCCCC and CICCCCCC and all permutations of this sort||
sterile rock
since thereâs 8 trials there would be one incorrect?
yes but also notice that ||order matters||
sterile rock
7 correct and one incorrect
yes
still temple
on the bookâs answers page itâs: 0.0000819
quasi patrol
on the bookâs answers page itâs: 0.0000819
ok
i dont really see how that affects the matter
sterile rock
on the bookâs answers page itâs: 0.0000819
i got that using my method
still temple
this chapter is on discrete probability distributions
quasi patrol
this chapter is on discrete probability distributions
alr
so wait
we can do this another
way
whats the probabiltiy its atleast 7
sterile rock
isnt this an SAT question? Areyou allowed a calculator? (I dont know myself)
still temple
we are allowed to use calcs
quasi patrol
isnt this an SAT question? Areyou allowed a calculator? (I dont know myself)
this isnt an sat question
sterile rock
we are allowed to use calcs
oh
quasi patrol
SAST test?
thats the scenario
still temple
can you guide me through the calc? mine is a ti-84 plus
quasi patrol
alright
lets phrase it another way
so
you get 7 questions right
and 1 question wrong
still temple
yes
quasi patrol
so whats the chance of getting a question right
quasi patrol
whats the chance of getting a question right
quasi patrol
because the answer is big
the answer is small you mean but yeah
still temple
1/7 since thereâs 7 trials so 7 correct and 1 incorrect?
still temple
the answer is small you mean but yeah
i meant the numbers total for the answer but yeah
quasi patrol
whats the chance of getting an individual question right
still temple
1/7? wait iâm confused
quasi patrol
1/7? wait iâm confused
look
each question
imagine theres only 1 question for a second
whats the chance of getting that right
quasi patrol
out of how many choices though?
read the question
still temple
it says 8 multiple choice questions
quasi patrol
it says 8 multiple choice questions
read the question carefully
each individual question
1 question
imagine theres 1 question
whats teh chance of getting that right
still temple
iâm lost
still temple
.close
marsh citrusBOT
still temple
found a helpful video
marsh citrusBOT
Available help channel!
main garnet
how is it the case that if it holds for sequences in D, it must hold for sequences in D \ {a} ??
marsh citrusBOT
how is it the case that if it holds for sequences in D, it must hold for sequenc...
main garnet
that dont make sense to me
i understand how if it holds in D \ {a} then it must hold in D but how the other way
devout mauve
well it holds for all sequences in D
some of them will be sequences in D\{a}
and so it also holds for these
main garnet
well it holds for all sequences in D
ohhh im getting it now
it's saying that for all sequences that are in D \ {a}
i was thinking it was saying that all sequences contained in D are also contained in D \ {a}
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
vague verge
marsh citrusBOT
potent fog
have you drawn a diagram yet
vague verge
Yup
mystic ledge
trigonometry might help
fallow fossil
do you know your sin, cos, tan?
vague verge
potent fog
is that a 9 or an a
vague verge
A
marsh citrusBOT
Available help channel!
molten ferry
why does $\int_a^b\dv{y}{x}\dd{x}\neq\int_a^b\dd{y}$
marsh citrusBOT
why does $\int_a^b\dv{y}{x}\dd{x}\neq\int_a^b\dd{y}$
elfin berryBOT
Jash
slim spire
I think you can fix it if you integrate from y(a) to y(b) on the right
you do this for separation of variables so it should work
molten ferry
I think you can fix it if you integrate from y(a) to y(b) on the right
oh ok
ty
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
light estuary
ello can someone help me study for my upcoming test
marsh citrusBOT
ello can someone help me study for my upcoming test
light estuary
do you guys do physics?
cobalt sentinel
send a sample question u need help w
marsh citrusBOT
@light estuary Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
keen snow
Evaluating a integral from 0 to 2 -10x^3 * sqrt(x^2 + 4) dx. Use trig-substitution
marsh citrusBOT
Evaluating a integral from 0 to 2 -10x^3 * sqrt(x^2 + 4) dx. Use trig-substituti...
amber birch
!status
marsh citrusBOT
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 was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
lofty hull
i am in grade 6 so pleaz dont say this is easy and you should be able to do it
left wadi
i am in grade 6 so pleaz dont say this is easy and you should be able to do it
!occupied
marsh citrusBOT
i am in grade 6 so pleaz dont say this is easy and you should be able to do it
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #âhow-to-get-help for instructions).
amber birch
You're not that far off: what's tan^2 theta + 1?
It's one of those useful trig identities
keen snow
thats just sec^2 right?
amber birch
Yeah
amber birch
So yeah the next step is to factor a 4^(1/2) out
And then you need another trig sub unfortunately: this one uses (sec x)' = sec x tan x
keen snow
oh lord
gilded smelt
So yeah the next step is to factor a 4^(1/2) out
And then you need another trig...
no??
your final integral is of sin/cos^4
which is a simple u sub
unless I am sorely mistaken
keen snow
the way our teacher wants, is to use trig sub not u sub
gilded smelt
I mean yes.... but one is drastically easier than the rest
you've already used the trig sub
now it's a question of how to evaluate tan * sec^3
amber birch
I'm thinking just (2 times) integral of tan theta sec^3 theta
Yeah so write it as (sec theta tan theta) (sec^2 theta)
Well you can just spot what the answer will be probably, but technically it does rely on u-sub
gilded smelt
u sub is a lie, it's just a simplification
like, you can u sub anything for any purpose
amber birch
like, you can u sub anything for any purpose
And sometimes it doesn't work out, I know
gilded smelt
regardless i maintain integrating sin/cos^4 is easier than tan * sec^3 (or (tan*sec)(sec^2) )
whether you rely on a u sub to assist you in finding the integral or not
amber birch
the way our teacher wants, is to use trig sub not u sub
Your teacher probably wants you to use a trig sub, instead of integration by parts
You've already used a trig sub so it doesn't matter now
gilded smelt
^ ie you could've done tabular integration no subs at all
keen snow
so after pulling that four out (multiplying it by the -160) and using the trig identity for tan^2 theta + 1
is where i am looking good, or did I go off the rails?
gilded smelt
you messed up
you had sqrt(tan^2+1)
and you turned it directly into sec^2
instead of sqrt(sec^2)
keen snow
so after taking that square root of sec^2,
you have sec, multiplied by sec^2, which would result in tan^3*sec^3
right?
gilded smelt
yup
hint: ||pythagorean theorem, sin^2 = 1-cos^2|| if you can't figure out how to solve from here
marsh citrusBOT
@keen snow Has your question been resolved?
keen snow
still solving
gilded smelt
lol fair enough
keen snow
kinda, forgetting my trig things, tan^3 can be simplified to sin^3/cos^3
then sin^3 can be sin^2 * sin and cos^3 can be cos^2 * cos
from there i can use trig identies to simplify
and sec^3 can be 1/cos^2*cos
gilded smelt
you're overcomplicating things but you might know what you're doing lol
I strongly recommend putting it all in terms of sines and cosines
keen snow
thanks, our calc 2 prof doesnt really teach us
my boy and i have been teaching ourselves
gilded smelt
rip
keen snow
ong
so, i have converted everything in terms of sine and cos, and got (sin^3/cos^3)*(1/cos^3)
keen snow
lmaoooo
gilded smelt
from there, you only need one step to get a simple integral
keen snow
was going to do that
gilded smelt
hint (again): ||p y t h a g o r e a n t h e o r e m||
keen snow
ngl little confused with what to do with that đ
gilded smelt
right now you have sin^3/cos^6 yes?
which is an... unhelpful fraction
however, you can convert a sin^2 into 1-cos^2
keen snow
split the fraction how?
gilded smelt
like, into two fractions
keen snow
like 1 - cos^2 * sin and 1/cos^6
keen snow
1/cos^6 and cos^2 sin / cos^6
gilded smelt
the first is incorrect, and reduce the second
keen snow
sin/cos^4
gilded smelt
whoops second is also incorrect but now the first should be right
keen snow
is it a -sin/cos^4?
gilded smelt
nope
[\frac{\sin^3\theta}{\cos^6\theta}\to(\sin\theta)\frac{1-\cos^2\theta}{\cos^6\theta}\to\frac{\sin\theta}{\cos^6\theta}+\frac{-\sin\theta}{\cos^4\theta}]
the sin multiplies the whole thing
keen snow
yeah i just realized that would get us back to where we started....
elfin berryBOT
Astral
gilded smelt
we started far left
we sub in 1-cos^2 = sin^2 in the middle
we expand in the right
do you follow?
keen snow
i follow,
quick question though
where does that 1 go?
seems to just vanish
nvm
i saw that plus sign as an arrow
therefore that on the far right is our answer?
keen snow
thats what i meant
gilded smelt
do you know how to integrate it then?
keen snow
we are able to integrate by using calculators on our homework,
but as to integrate that thing by hand
not really
gilded smelt
P: do a u-sub
keen snow
i assume u would be sin/cos^6
keen snow
yes
keen snow
do we sub u for cos?
gilded smelt
yup
keen snow
now, in our first fraction its sin / cos^6
rn i have it as sin/u^6 which could be turned into sin*u^-6 if i so pleased
the second fraction would be du/u^4 which could be u^(-4)*du
for the first because rn it is sin/u^6 could i add a -1 to the numerator and add the recripical of (-1) outside the integral to account for it?
gilded smelt
you forgot to do the du for the first one
what you have right now is
[(\frac{\sin\theta}{\cos^6\theta}+\frac{-\sin\theta}{\cos^4\theta})d\theta\to\frac{\sin\theta}{u^6}d\theta+\frac{du}{u^4}]
elfin berryBOT
Astral
keen snow
if our u = cos
and du = -sin
doesnt -sin just turn into du?
rather than -du?
also shouldn't theta go to both terms resulting in something like:
,, [(\frac{\sin\theta}{\cos^6\theta}+\frac{-\sin\theta}{\cos^4\theta})d\theta\to\frac{\sin\theta}{u^6}d\theta+\frac{du}{u^4}d\theta]
elfin berryBOT
aleyy
gilded smelt
u=cosø
du=-sinødø
keen snow
right.
gilded smelt
actually whoops you are correct
the fact remains you forgot to sub properly for the first term
keen snow
thats what i figured after you sent that correction
but more of how would i right that first term in terms of u sub
if the du is negative sin, how would i go about that?
gilded smelt
[(\frac{\sin\theta}{\cos^6\theta}+\frac{-\sin\theta}{\cos^4\theta})d\theta\to\frac{-(-\sin\theta{d\theta})}{u^6}+\frac{(-\sin\theta{d\theta})}{u^4}]
elfin berryBOT
Astral
keen snow
sorry, keep forgetting that
gilded smelt
it's an important detail
otherwise we could say u^2=cos^2=1-sin^2 = 1-du^2
which is a mutilation of mathematics
keen snow
but adding that negative in there, i would have to multiply a negative one outside the intergral right?
gilded smelt
**Astral**
as per here
keen snow
right, but we multplied a negative one to get that
wouldnt i have to multiply a negative one outside the integral because we did that inside the integral?
i remember my calc 1 prof said that if you multiply a constant in an integral, you must multiply its reciprocal outside the intrgral
keen snow
right
gilded smelt
[\int\frac{\sin\theta}{\cos^6\theta}d\theta+\int\frac{-\sin\theta}{\cos^4\theta}d\theta\to-\int\frac{-\sin\theta}{\cos^6\theta}d\theta+\int\frac{-\sin\theta}{\cos^4\theta}d\theta]
keen snow
correct
gilded smelt
so we can simply do each one individually if it makes things easier for you
in this case though
elfin berryBOT
Astral
gilded smelt
if that makes things clearer for you
gilded smelt
good good, so long as we are on the same page
keen snow
so now that i can simply each integral into u and du
gilded smelt
yes
keen snow
plug u back in and evaluate from 0 to 2 i assume?
gilded smelt
plug u back in, then plug ø back in
keen snow
wait, since that negative was infront of that first integral
gilded smelt
you started with an integral in terms of x
keen snow
doesnt that make my first term negative?
gilded smelt
wait, since that negative was infront of that first integral
yup you forgot all the coefficients too lol
both here and from the original
elfin berryBOT
aleyy
keen snow
quick question, what coefficients, i only seem to have the -640 that we pulled out from in the beginning
keen snow
yall gave a 2
yep
so therefore -320 would get multiplied to each term?
giving us (cancelling all of the negatives of course)
,, -320\frac{u^-5}{5}+320\frac{u^-3}{3}
elfin berryBOT
aleyy
elfin berryBOT
Astral
gilded smelt
you need to convert back to x from u, which means going from u to theta then theta to x
gilded smelt
literally, just replace every (u) with (cosø)
it's that simple P:
we already did the hard part
gilded smelt
so now we have
keen snow
but do i just plug the x's for theta?
gilded smelt
[-320|_{x=0}^{x=2}[\frac{1}{5}\sec^5\theta-\frac{1}{3}\sec^3\theta]]
elfin berryBOT
Astral
gilded smelt
that is not the sub you made
you said x = 2tanø
however, you can do it differently now
you can say x/2 = tanø
so ø = arctan(x/2)
and evaluate from ø = arctan((x=0)/2) to ø = arctan((x=2)/2)
make sense?
keen snow
yep
was originally stuck on how you got sec in that equation
it made sense
and then we said that theta is equal to arctan(x/2) after we said that x = 2tanø which was then tanø = x/2
got it, then we just plug in and evaluate at those intevals
then multiply our differences by -320
right?
gilded smelt
yup
keen snow
i got that ø = -14400
gilded smelt
?? how
keen snow
i did -320*[tan^-1(2/2) - tan^-1(0/2)]
gilded smelt
you're skipping steps
we have found F(ø), the antiderivative of our integral
we know we have to integrate from x=0 to x=2
x=0, but x=2tanø, so tanø=x/2, so ø = arctan(x/2), so ø(x=0) = arctan(0) = 0, so our first theta is 0 when x=0
what is the second coordinate we integrate to? corresponding to x=2
keen snow
so doing that arctan(2) is about 63.4349
gilded smelt
incorrect
keen snow
does this sound correct?
gilded smelt
let's put it this way
x=2tanø
2=2tanø
what is ø?
you forgot the 2 infront of the tan
aka the /2 on the x
keen snow
45
keen snow
why would it be wrong?
kinda messed up so many times, didnt really trust my instinct
keen snow
and find the exact form haha
also, 2 quick question, what is the exact form?
keen snow
oh i see, our teacher is fine with inexact form
keen snow
real quick
my friend wanted to ask
was this a "hard question"
or is this kinda average for calc 2
gilded smelt
eh it had a fair number of steps but they forced you to do a trig sub which are always terrible
I have never once needed a trig sub to do an integral
they only make things harder imo
otherwise it was fairly varied, ie the way we solved it required two substitutions and changing the bounds of the integral once
keen snow
so u sub is usually the play then?
gilded smelt
u sub is just a mental shortcut for doing integrals but it is one of the most common methods of solving an integral, yes
gilded smelt
so harder than usual
keen snow
thank you so much
gilded smelt
but not as hard as it could be
ie integrals involving partial fractions can get very nasty for the unexperienced
particularly if you layer steps
keen snow
oh my word
my boy and i are going to need several redbulls right about nowđ
but fr thanks for helping us for like 3 hours man, absolute legend
thanks, from rowan university calc 2 students
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
steel rapids
hi! could you please help me with finding why this is true and how to calculate it?
marsh citrusBOT
hi! could you please help me with finding why this is true and how to calculate ...
steel rapids
ik you are supposed to calculate the limit with n going to infinity of ln^7(n)/sqrt(base=3)(n) and its supposed to give you 0 for this to be true
but how to calculate this limit?
slim spire
it's annoying to work with ln(n) so you can try a change of variables that'll cancel it, like n=e^c
it'll turn into c^7/e^(c/3) which is like comparing a polynomial with an exponential, and can be proved with the ratio test or taylor series
steel rapids
alright thank you
i also found on the internet that limits looking like this where the upper side of the fraction is growing very very fast divided by something growing slowly is 0
is this true? does it imply in this case? @slim spire
steel rapids
oh yeah sorry i meant the opposite
slim spire
yea that's true
marsh citrusBOT
Available help channel!
solar mesa
In differential equations, even if I get a constant like "-C", will it always be placed as positive? "+C"
because what I did first was the thing on the left and when comparing with the answer it is something else
marsh citrusBOT
In differential equations, even if I get a constant like "-C", will it always be...
solar mesa
this is the problem, when I solved it they gave me logarithms, is there no problem if I work with either of the two? if I use the first one, I get what is on the left, and that is why I came to ask about the constants
marsh citrusBOT
@solar mesa Has your question been resolved?
marsh citrusBOT
@solar mesa Has your question been resolved?
marsh citrusBOT
@solar mesa Has your question been resolved?
marsh citrusBOT
@solar mesa Has your question been resolved?
marsh citrusBOT
@solar mesa Has your question been resolved?
oak gulch
In differential equations, even if I get a constant like "-C", will it always be...
Yeah that's how it works, but usually you can denote -C as K where C and K are both constants
oak gulch
this is the problem, when I solved it they gave me logarithms, is there no probl...
Both are the same, since ln|1-p| = ln|p-1|, and for the value of log(x), x must be > 0, and that's why there's an absolute bracket there to denote that the value must be positive
worn plank
It doesn't matter
Choose C < 0 and so -C would be greater than 0
C = -5, --> your expression -(-5) = your expression + 5
marsh citrusBOT
@solar mesa Has your question been resolved?
elfin berryBOT
nyes1375
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
nyes1375
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
late geode
split fraction
lethal bridge
What's that?
late geode
$\frac{a+b}{c} = \frac ac + \frac bc$
elfin berryBOT
âιΟΊâĎâ ¤
marsh citrusBOT
marsh citrusBOT
Available help channel!
timid wasp
marsh citrusBOT
timid wasp
dont understand where to start even after looking at mark scheme
late geode
what's the mark scheme
timid wasp
timid wasp
what's the mark scheme
^
late geode
which part don't you understand
timid wasp
how do i start on the question
late geode
the worked solution indicates that
divide both sides of the equation by cos(x)cos(y)
which leads to getting the expressions tan(x) and tan(y)
timid wasp
.close
summer cradle
marsh citrusBOT
summer cradle
is this even doable
stoic slate
Have you tried anything?
late geode
did you copy down the question correctly?
stoic slate
The exercise is very straightforward
urban shard
stoic slate
What are you trying to find?
late geode
its strange that they have both a and x present here
stoic slate
The last equation doesnt make too much sense to mw because u can replace immediately with 20
So the system solves by itself but you cant get a single value for everything
summer cradle
is it like simultaneous equations
stoic slate
It is called system of equations
summer cradle
hows that work
stoic slate
So for example y+z = 20
So you can replace that in the last equation
a+20=?
But even with this simplification you cannot get single values for all the five variables
marsh citrusBOT
marsh citrusBOT
Available help channel!
flint axle
Hi !
flint axle
I need some help
marsh citrusBOT
@flint axle Has your question been resolved?
marsh citrusBOT
@flint axle Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
loud flicker
okay
loud flicker
$f\left(x\right)=\frac{x+1}{x-4}$
elfin berryBOT
eclipseryu
elfin berryBOT
eclipseryu
elfin berryBOT
eclipseryu
loud flicker
after three tries i get this
it says the answer is this
i seriously dont understand why
could anyone possibly tell me why this is the case?
im gonna try it a fourth time and see if i can recreate the answer
nevermind
i seriously hate having adhd
.close
marsh citrusBOT
solemn plover
11?
chrome olive
i'm not sure the way you're doing induction is correct or rigorous, that's a bigger thing
for one you're assuming the conclusion during the inductive step
you can assume $2^{k+1} > k^2$ for some $k \geq 3$ holds and you follow it by manipulating the inequality to get $k+2$, so something like $2^{k+2} > k^2 \cdot 2$
elfin berryBOT
jack
chrome olive
you can't simply make the assumption that $2^{k+2} > (k+1)^2$ since this is the very thing you're trying to prove
elfin berryBOT
jack
marsh citrusBOT
@split forum Has your question been resolved?
wary yew
how to determine the factor ring $\mathbb{Z}[i] /<1-i>$?
marsh citrusBOT
how to determine the factor ring $\mathbb{Z}[i] /<1-i>$?
elfin berryBOT
Dubs
marsh citrusBOT
@wary yew Has your question been resolved?
wary yew
please check
marsh citrusBOT
Available help channel!
main mica
Rombuses perimeter is 40 need to find the acute angles cos if one of the diagnals is 12
marsh citrusBOT
Rombuses perimeter is 40 need to find the acute angles cos if one of the diagnal...
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
@main mica Has your question been resolved?
marsh citrusBOT
marsh citrusBOT
Available help channel!
wary yew
Determine the factor ring $\mathbb{R}[x] /<x^2 +2x+1>$
marsh citrusBOT
Determine the factor ring $\mathbb{R}[x] /<x^2 +2x+1>$
elfin berryBOT
Dubs
wary yew
my answer: $\mathbb{R}[x] /<1-i> = {ax +b + I : a,b \in \mathbb{R} }$
elfin berryBOT
Dubs
marsh citrusBOT
@wary yew Has your question been resolved?
marsh citrusBOT
wary yew
.reopen
wary yew
<@&286206848099549185>
tulip marsh
marsh citrusBOT
tulip marsh
for this one i do not understand why we are solvving for A
what key word in the probelm tells us we are solving for a
fervent rampart
we have to find where the graphs intersect to find the bounds of integration
tulip marsh
yes but what key word in the probelm tells us we are solving for a
static quarry
"enclosed by the graph[s] of..."
tulip marsh
well shouldnt we be finding the x and y values of this point instead? to fin the "enclosed by the graph[s] of..." part
fervent rampart
a is the value of x at that point
fervent rampart
we are finding the value of x where y (the output of both functions) is the same
tulip marsh
so it would all be in respect to x
since our limits on teh integral are also X values then
okay
uj
(reposting for reference)
so when teh question says "find the area of the region" it is really saying solve for the upper limit?
those 2 and 2 dont make sense
fervent rampart
if you want to use an integral to find the area of the region, then you can't do that until you have limits of integration
tulip marsh
if you want to use an integral to find the area of the region, then you can't do...
but i was thinking we find upper limit yes but then after we find uppper limit then solve fo the area of the region and that is the answer.
in this question i think they just made the upper limit be the answer to the problem
fervent rampart
in this question the area coincidentally happens to be a number that is close to the upper limit
according to the solution the upper limit is a = 2.754 whereas the area is 2.784, whcih is close but not the same number
tulip marsh
ohhhhhhhh, so they do actually find upper limit then solve fo the area of the region with the upper limit there
guys?
fervent rampart
ohhhhhhhh, so they do actually find upper limit then solve fo the area of the ...
yes
fervent rampart
a is the intersection of the two graphs, so you set the two functions equal and solve for x
fervent rampart
i think they used a calculator to find an approximate solution
marsh citrusBOT
marsh citrusBOT
Available help channel!
potent nova
Hello can you help me to prove it part a i think we can not use archimedean property here can you help me or give me a hint , about part b is it true yn must be converge
marsh citrusBOT
Hello can you help me to prove it part a i think we can not use archimedean pro...
gilded sage
you're on the right track for part a, you do need to use archimedean property
marsh citrusBOT
@potent nova Has your question been resolved?
potent nova
you're on the right track for part a, you do need to use archimedean property
How i can prove it?
potent nova
for every real number (r) we can found a natural (n) s.t n>r
marsh citrusBOT
@potent nova Has your question been resolved?
devout mauve
or: for every positive real r you can find a natural number with 1/n < r. that form is often more useful
marsh citrusBOT
@potent nova Has your question been resolved?
potent nova
or: for every positive real r you can find a natural number with 1/n < r. that f...
yes but how can i use this idea to prove the above question ?
devout mauve
x<y means that y-x is positive. motivated by that you might rearrange the inequality you want to show to 1/n+1/m < y-x
potent nova
x<y means that y-x is positive. motivated by that you might rearrange the inequa...
we can not use archimidean property here
devout mauve
why not
marsh citrusBOT
@potent nova Has your question been resolved?
devout mauve
-x could be negative so I cant do that
y-x is positive so we know there is a natural k with 1/k < y-x
can you now find natural numbers m,n with 1/n+1/m < 1/k ?
potent nova
y-x is positive so we know there is a natural k with 1/k < y-x
maybe if we take m=n=k , but then ?
devout mauve
is 1/k +1/k < 1/k ?
potent nova
no i make mistake
potent nova
is 1/k +1/k < 1/k ?
n>m>k ?
devout mauve
no thats not enough
potent nova
no thats not enough
then how ?
devout mauve
for simplicity you could assume that n=m
so then 1/n+1/n <1/k
can you solve that for n?
potent nova
so then 1/n+1/n <1/k
n>2k
devout mauve
you are supposed to show that some n and m exist
you now showed that some n and m exist
so you are done
you are allowed to assume that they might happen to be equal
potent nova
you are allowed to assume that they might happen to be equal
i got it thanks
can you see part b plz
potent nova
Hello can you help me to prove it part a i think we can not use archimedean pro...
here
marsh citrusBOT
@potent nova Has your question been resolved?
fallow fossil
0 is not a natural number where u learn right?
well i mean it is useless too mb
u use infinity ig
marsh citrusBOT
marsh citrusBOT
Available help channel!
jade hazel
Find the parameter m so that for every x that is a real number the equation is satisfied. Please someone help
marsh citrusBOT
Find the parameter m so that for every x that is a real number the equation is s...
desert socket
okay
so f(x)=mx^2-2(m+2)x+m+1
and, f(x)<0 for all real x
what can you conclude?
@jade hazel
marsh citrusBOT
@jade hazel Has your question been resolved?
jade hazel
and, f(x)<0 for all real x
that m must be less than 0?
I really dont know how to calculate this
terse turtle
(LHS is a quadratic -- in what case is a quadratic always <0, i.e. not touching the x-axis?)
terse turtle
try finding D then
jade hazel
OMG
I am so dumb I was doing that but forgot to add one more m
ok thank you
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
midnight estuary
Titu's lemma seems natural because of the form but it ain't helping much, simplifying LHS isn't working either, any ideas?
marsh citrusBOT
Titu's lemma seems natural because of the form but it ain't helping much, simpli...
@midnight estuary Has your question been resolved?
amber birch
Titu's lemma seems natural because of the form but it ain't helping much, simpli...
Nah, Titu's lemma is the way to go
$\frac{(2(a + b + c) - 3)^2}{a + b + c} \ge a + b + c$
$\implies (2u - 3)^2 \ge u^2$
elfin berryBOT
south
amber birch
This is true for $u \ge 3$
elfin berryBOT
south
amber birch
$\frac{a + b + c}{3} \ge (abc)^{1/3} = 1$
elfin berryBOT
south
amber birch
Yeah lol
midnight estuary
Thanks
amber birch
Np
midnight estuary
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
glossy harness
I donât understand. Why did they put a negative sign there?
marsh citrusBOT
I donât understand. Why did they put a negative sign there?
late geode
definition of abs value
sqrt(n^2) = |n|
for negative n, that = -n
similarly
sqrt(x^6) = |x^3| = -x^3
for negative x
marsh citrusBOT
@glossy harness Has your question been resolved?
novel juniper
why is the centre of the form (-a,-a)
marsh citrusBOT
why is the centre of the form (-a,-a)
quasi patrol
why is the centre of the form (-a,-a)
Bc its in the third quadrant
novel juniper
no, why are the abcissa and oordinate the same
fathom ridge
touches both axes
livid maple
Because it touches both axes
fathom ridge
so it must be equidistant from both axes
marsh citrusBOT
marsh citrusBOT
Available help channel!
limpid oxide
how do I do this question
I get [(ι^3 + β^3 + γ^3 + δ^3) + (ιβγ)^3 + (βγι)^3 + (γιδ)^3]/[(ιβ)^3+(βγ)^3+(γδ)^3+(δι)^3] - 30(ιβγδ)^3 but everything after that gets too lengthy for me and I screw up everytime
marsh citrusBOT
how do I do this question
I get [(ι^3 + β^3 + γ^3 + δ^3) + (ιβγ)^3 + (βγι)^3 + (...
@limpid oxide Has your question been resolved?
limpid oxide
<@&286206848099549185>
livid maple
What does summation sign mean here?
amber birch
What does summation sign mean here?
So it's a cyclic sum
brazen jungle
Ig you need to expand and apply vieta + identities
marsh citrusBOT
marsh citrusBOT
Available help channel!
azure karma
So the question here is to write a statement so that the function f is shown in the graph a and not in graph b, would this be enough?
marsh citrusBOT
So the question here is to write a statement so that the function f is shown in ...
azure karma
Also, I intitially wrote this, would this be enough in this case?
marsh citrusBOT
@azure karma Has your question been resolved?
azure karma
<@&286206848099549185>
marsh citrusBOT
@azure karma Has your question been resolved?
marsh citrusBOT
@azure karma Has your question been resolved?
slim spire
So the question here is to write a statement so that the function f is shown in ...
yea that's correct
the second one is the right idea but technically you don't know the units along the x axis so maybe 0.1 is too big
azure karma
Thanks
marsh citrusBOT
Available help channel!
weak eagle
marsh citrusBOT
weak eagle
Iâm confused on number 15
Right now I know they both touch at 2pi/3 and 4pi/3
But Iâm not sure how to set up the integral
marsh citrusBOT
@weak eagle Has your question been resolved?
weak eagle
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
lost birch
I dont understand how to do this question
marsh citrusBOT
I dont understand how to do this question
lost birch
shadow nest
what have you tried
gilded pivot
Do you know what kind of calculations youâre allowed to do in a system of equations ?
@lost birch
lunar terrace
hey what's going on
marsh citrusBOT
marsh citrusBOT
Available help channel!
brisk totem
21
marsh citrusBOT
@brisk totem Has your question been resolved?
solid breach
Supposed to use finite sums
can you show what you tried
don't really need finite sums
hint: ||split it up into triangles from the centre of the circle||
brisk totem
hint: ||split it up into triangles from the centre of the circle||
im supposed to use finite sums
its the lesson
marsh citrusBOT
@brisk totem Has your question been resolved?
marsh citrusBOT
@brisk totem Has your question been resolved?
marsh citrusBOT
@brisk totem Has your question been resolved?
marsh citrusBOT
@brisk totem Has your question been resolved?
inner jay
.close
elfin tree
(x/a) + (y/b) = 1
marsh citrusBOT
(x/a) + (y/b) = 1
marsh citrusBOT
marsh citrusBOT
Available help channel!
still temple
marsh citrusBOT
still temple
Hi, not sure what Iâm doing wrong here. Any help would be appreciated
amber birch
Hi, not sure what Iâm doing wrong here. Any help would be appreciated
d/dx (a^x) = a^x ln a only works when a is a constant
You need to write $t^{1/t} = e^{(\ln t)(1/t)}$
elfin berryBOT
south
amber birch
Then you'll see you need the product rule
still temple
Ohhh ok ok thank you
amber birch
d/dx (a^x) = a^x ln a only works when a is a constant
So why this works is cause $\frac{d}{dx} e^{(\ln a) x} = e^{(\ln a) x} \ln a$, as the derivative is nice here
Nw
elfin berryBOT
south
marsh citrusBOT
marsh citrusBOT
Available help channel!
uncut island
Hi! If i have some function like a 1 sheet hyperboloid given by: 1 = x^2 - y^2 +z^2 . How would I go about finding the vector function that creates the space curve
marsh citrusBOT
Hi! If i have some function like a 1 sheet hyperboloid given by: 1 = x^2 - y^2 +...
uncut island
I was reviewing and I noticed a lot of times they just provided a vector function and just had me parameterize it
but what if we just have the function of a surface, how do we then find the vector function
amber birch
but what if we just have the function of a surface, how do we then find the vect...
Well there's more than one way to parameterise it, but using the identity $\cos^2 \theta + \sin^2 \theta$ is probably the easiest
elfin berryBOT
south
amber birch
So we can let z^2 = cos^2 theta and x^2 - y^2 = sin^2 theta
Now recall sec^2 x - tan^2 x = 1
So one possible parameterisation would be $(x, y, z) = (\sec \phi \sin \theta, \tan \phi \sin \theta, \cos \theta)$
elfin berryBOT
south
marsh citrusBOT
@uncut island Has your question been resolved?
uncut island
So we can let z^2 = cos^2 theta and x^2 - y^2 = sin^2 theta
why are we using sec^2x-tan2x - 1
i assume thats for z
wouldnt x = sin^2(theta)
amber birch
So we can let z^2 = cos^2 theta and x^2 - y^2 = sin^2 theta
@uncut island Do you get this part?
<@&268886789983436800>
amber birch
sin^2(theta) = 1-cos^2(theta)
Yes that is true
If x^2 - y^2 = sin^2 theta and z^2 = cos^2 theta
Then x^2 - y^2 + z^2 = 1
That's why I did that
uncut island
1 != x^2 ?
uncut island
If x^2 - y^2 = sin^2 theta and z^2 = cos^2 theta
Then x^2 - y^2 + z^2 = 1
I dont see how this works
amber birch
I dont see how this works
sin^2 theta + cos^2 theta = 1
uncut island
yeah
uncut island
sin^2 theta + cos^2 theta = 1
but isnt that x^2 + y^2 = 1
amber birch
but isnt that x^2 + y^2 = 1
Ah, well here x isn't sin theta
y isn't cos theta
Here, x, y, and z have nothing to do with the unit circle
amber birch
Hi! If i have some function like a 1 sheet hyperboloid given by: 1 = x^2 - y^2 +...
This is a surface in 3 freaking dimensions
uncut island
yeah
amber birch
So x and y absolutely aren't points on the unit circle, which is in 2 dimensions
They're points on the hyperboloid
Once you understand why cos^2 theta + sin^2 theta = 1 for all theta
You don't need the unit circle anymore
You can just use the identity
x and y can be anything, well anything that lies on the hyperboloid
uncut island
i thought it was because cos^2theta and sin^2theta were lenths of each side of that triangle in purple
amber birch
i thought it was because cos^2theta and sin^2theta were lenths of each side of t...
It's cause of the Pythagorean theorem
We define cos theta to be the base of the triangle, the x-coordinate
uncut island
yeah
amber birch
And we define sin theta to be the height of the triangle, the y-coordinate
uncut island
thats wat i mean
amber birch
So since the radius is 1, the hypotenuse is 1
uncut island
i get that
amber birch
x = cos^2x
Okay this doesn't make sense firstly
You can't define x in terms of itself
amber birch
Yes
uncut island
so from here where do we get sin^2(theta). = x^2 - y^2
amber birch
so from here where do we get sin^2(theta). = x^2 - y^2
We're just defining x^2 - y^2 to be sin^2 theta
uncut island
yeah but why and how is that true
amber birch
x^2 = cos^2(theta)
x and y in x^2 - y^2 aren't the same x and y in here
amber birch
So I'm not saying that x = cos(theta) and y = sin(theta)
amber birch
We're just defining x^2 - y^2 to be sin^2 theta
I'm not talking about the unit circle when I say this
uncut island
ok I can see that. But i dont see what your doing to make that statement work
amber birch
If x^2 - y^2 = sin^2 theta and z^2 = cos^2 theta
Then x^2 - y^2 + z^2 = 1
OK back to this
Do you get this part?
uncut island
no i dont get the x^2 - y^2 = sin^2 theta
i dont see how we can just say they arbitrarily = sin^2 theta
how do u know to set that = to sin^2(theta) and not something else
amber birch
how do u know to set that = to sin^2(theta) and not something else
Cause we're using the identity sin^2 theta + cos^2 theta = 1
uncut island
yeah
amber birch
how do u know to set that = to sin^2(theta) and not something else
And yes this is an arbitrary choice
You could also let z^2 = 1 - a and x^2 - y^2 = a for some number a
But notice how this is very similar to the idea we use behind spherical coordinates
uncut island
right
amber birch
Yeah it's the same principle: sin^2 + cos^2 = 1 and use another angle for x^2 + y^2 to be r^2 sin^2 theta
So that z^2 = r^2 cos^2 theta
uncut island
ok
so when working in 3 dimensions we cant use polar?
because of a 3rd angle
2nd*
so if were to parameterize it in polar coords
x^2-y^2+z^2 = 1 goes to. x^2 = (rsin(theta)cos(phi))^2 , y^2 = -(rsin(theta)sin(phi))^2 , z^2 = (rcos(theta)^2
pretty confident I'm doing something wrong here
amber birch
so when working in 3 dimensions we cant use polar?
Yeah, we need another angle to describe 3 dimensions
For 3 dimensions we need at least 3 pieces of info
uncut island
You could also let z^2 = 1 - a and x^2 - y^2 = a for some number a
(rsin(theta)cos(phi))^2 - (rsin(theta)sin(phi))^2 = 1
amber birch
x^2-y^2+z^2 = 1 goes to. x^2 = (rsin(theta)cos(phi))^2 , y^2 = -(rsin(theta)sin...
Ah cause cos^2 x - sin^2 is not 1
So you can't use cos x and sin x
You need sec^2 x - tan^2 x = 1
amber birch
(rsin(theta)cos(phi))^2 - (rsin(theta)sin(phi))^2 = 1
Also you don't need r here cause the RHS is just 1
uncut island
x^2-y^2+z^2 = 1 goes to. x^2 = (sin(theta)cos(phi))^2 , y^2 = (sin(theta)sin(phi))^2 , z^2 = (cos(theta)^2
so is this now parameterized?
amber birch
so is this now parameterized?
There's a problem
The problem is with y
Cause if you square root both sides you will get y = i * plus/minus sin theta sin phi
uncut island
is it the negative
uncut island
x^2-y^2+z^2 = 1 goes to. x^2 = (sin(theta)cos(phi))^2 , y^2 = (sin(theta)sin(phi))^2 , z^2 = (cos(theta))^2
amber birch
x^2-y^2+z^2 = 1 goes to. **x^2 = (sin(theta)cos(phi))^2 , y^2 = (sin(theta)sin(...
Nope, cos(phi)^2 - sin(phi)^2 is not 1
So you won't get x^2 - y^2 = sin^2 theta
amber birch
You need sec^2 x - tan^2 x = 1
This
uncut island
but sec^2x = 1/cos^2x
i dont see where that is coming from
i understand
where u get the sec^2x from its an identity buti dont get how thats relevent
amber birch
where u get the sec^2x from its an identity buti dont get how thats relevent
So if x = cos theta sec phi and y = cos theta tan phi
Then x^2 - y^2 = cos^2 theta sec^2 phi - cos^2 theta tan^2 phi
= cos^2 theta (sec^2 phi - tan^2 phi)
uncut island
So if x = cos theta sec phi and y = cos theta tan phi
where are we getting the sec?
amber birch
= cos^2 theta
uncut island
isnt is x = sin(theta)cos(phi)
amber birch
isnt is x = sin(theta)cos(phi)
It's not cos phi
We could have sin theta or cos theta
But we must have sec phi to use the identity sec^2 phi - tan^2 phi = 1
uncut island
im lost. the spherical conversion says x = sin(theta)*cos(phi)
amber birch
im lost. the spherical conversion says x = sin(theta)*cos(phi)
THE SPHERICAL CONVERSION DOESN'T WORK FOR EVERYTHING
The spherical conversion was just an example
It uses the same ideas but it's definitely not the same
uncut island
gotcha
oobh
i think i see it
nvm
like i can see how sec*2 phi - tan^2phi = 1 and x^2 = sec^2phi and y^2 = tan^2(phi)
1-x^2+y^2 = z^2
not sure what do with z
z^2 = 1 - sec^2 (phi) - tan^2phi(phi)
but that just z = 0
uncut island
.close
marsh citrusBOT
marsh citrusBOT
Available help channel!
weary quarry
The mark scheme totally ignored any change in the i direction even though there is a change present
marsh citrusBOT
The mark scheme totally ignored any change in the **i** direction even though th...
weary quarry
Why?