you have your cuboid thing
#help-28
glass crystal
that changes size over time
dV/dt represents the change of volume over time
dx/dt is the change of size of the sides
over time
glass crystal
oh its a typo
torn jolt
ohh ty
glass crystal
sides*
but no those are rates of change
so infinitesimal stuf
well derivatives
glass crystal
so from the initial size to the x=4
i was answering this
torn jolt
is dx/dt 15x^2?
torn jolt
ohh right
glass crystal
you said it there yourself
glass crystal
also i dont like the d/dx notation for maths class
glass crystal
its useful for physics but its very confusing
you could do this whole exercise this way
glass crystal
i'm writing the process
torn jolt
ok ty
glass crystal
torn jolt
how does it become 1/15x^2?
glass crystal
overheating problem
torn jolt
wait
ahhh im sorry it taking so long
ohh i understand
but why are you doing that division?
are you trying to isolate dx/dt?
glass crystal
dont be sorry
if i want to leave I can at any time
i'm not trapped with you
relax
torn jolt
mhm
i think ive done so much maths that i was starting to confuse stuff
thank you for your help. have a good evening (im assuming its evening)
bitter vine
hi
safe trench
Just ask
hot herald
make sure to first read #❓how-to-get-help and #rules
bitter vine
hi
im a high school pupil i didnt finish studiying every notion in maths
is that ok
safe trench
Depends what you want to do after high school
bitter vine
engeneering
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@bitter vine Has your question been resolved?
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Available help channel!
indigo patrol
hi
scenic wren
these are the curves
so we make a vector that goes from one point on r1 to one point on r2
and the points are arbitrary
then we need tangents of the curves
which i do not know how to do
nova basin
the tangent is only here to give the direction, so compute dr/dlambda and dr/dmu
linear terms matter
indigo patrol
oh sorry yea
nova basin
So then you can solve for when their dot product is 0
Like finding the global max of any function
Global max = max(local max)
So the biggest local maximum, which has the advantage of being the max of a finite set usually, rather than an uncountable one
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meager saddle
I am so confused on how to do this
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I am so confused on how to do this
smoky ocean
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@meager saddle Has your question been resolved?
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@meager saddle Has your question been resolved?
supple token
i dont understand why we multiply by tan
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i dont understand why we multiply by tan
scenic yew
i dont understand why we multiply by tan
It's just a trick for division. You can just do it at any division. Multiply the same number (it can be anything) on both the numerator and the denominator
It likes 1/25. To make it as a decimal easier, we multiply 4 on both the numerator and the denominator to make the denominator 100
it would become 4/100
and 1/25 is equal to 0.04
toxic wren
question is from amc (australian math competition) juniors 2022
the question is as follows: The first number in a list is 2. After that, each number is calculated by adding the digits of the previous number together and squaring the result. What is the 2022nd number in the list?
if I havent misunderstood, the beginning of the list would be [2, 4, 16, 49, 169...].
i understand the question, but not really how i would calculate the nth number in this list. i couldnt find any pattern within it either
remote veldt
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rare dock
think about 0s and 1s 🙂
remote veldt
Yes but a,b,c must be different
rare dock
ya I mean make one of them 0, one of them 1, and the other something else lol
remote veldt
I see that a=1 and c=0 but b can take on any integer
rare dock
so problem solved?
rare dock
that's not what that means lol
remote veldt
Bruh what
rare dock
0,1,2 are distinct integers and 1^2 = 1^0
remote veldt
But you can also have 0,1,3 or 0,1,4 and so on
rare dock
that's just not what the question is saying
0,1,2 are distinct
it doesn't say unique or anything
anywhere
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wise tartan
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tame marten
C
tame marten
its c
wise tartan
yeah thats what I had
tame marten
what do u need help with?
sacred trout
I'm need help understanding this
full forumBOT
I'm need help understanding this
sacred trout
How do I find the inequality of f'(x) > 0?
the function is this:
I already found the derivative to this function which is this:
I just need to find the inequality of f'(x) > 0 for x ∈
which is this function above
torn jolt
which is this function above
well, you already have found a critical point
sacred trout
which is 5
torn jolt
so sub in numbers greater or lesser than that critical point
yes
see what happens then
substiute f'(4) and f'(6)
torn jolt
yes, thats the point of what i was saying
i just said sub in both to see the difference in signs
scenic yew
I just need to find the inequality of f'(x) > 0 for x ∈
Hey I think it's incomplete
element of what?
sacred trout
thats supposed to be the answer mb
torn jolt
did u figure out what f'(4) and f'(6) would give you though?
sacred trout
(5,inf) right
torn jolt
you dont have to calculate it, just observe how the signs would be
torn jolt
(5,inf) right
yes
sacred trout
you dont have to calculate it, just observe how the signs would be
I don't?
torn jolt
you dont really care about what the result would be, you just need to know when it is positive and negative
scenic yew
I don't?
You see, it's also a fractional exponent
So not only would be a negative, it would also be an imaginary number or a complex solution
torn jolt
So not only would be a negative, it would also be an imaginary number or a compl...
would it?
it is an odd powered root, the negativity should not matter
torn jolt
it has a cusp at a x = 5 which makes the derivative undefined there, which is your only critical point
but if x = 5 was also undefined in the original function, then you don't consider it at all btw @sacred trout
sacred trout
it has a cusp at a x = 5 which makes the derivative undefined there, which is yo...
undefined?
torn jolt
if your original function f(x) was something like 1/(x-5)^2 and ur derivative was 1/(x-5) (this doesnt make sense, but just for the sake of example)
x = 5 would make the derivative undefined, which indicates it could have a cusp or a turn at that point
but since x = 5 is also not defined at f(5) then that point is just a vertical asymptote
it can not be a critical point
i just said that for future reference
merry atlas
How do I find the primitive function of $sin(2^(1/2))$ ?
full forumBOT
How do I find the primitive function of $sin(2^(1/2))$ ?
torn jolt
oh are u still stuck with it haha
glossy valveBOT
Kienai
merry atlas
oh are u still stuck with it haha
Hi again, yeah I slept on it to no avail
glossy valveBOT
EndTimes
vapid barn
is that what you mean?
merry atlas
Yes
torn jolt
**EndTimes**
okay so for this, take u = $\sqrt{x}$ and $dx = 2\sqrt{x} du$
glossy valveBOT
♡LexQa♡
torn jolt
this should be a good start
vapid barn
yeah let t = 2sqrt(x), so t^2/4 = x, then dx = t/2 dt
torn jolt
yeah let t = 2sqrt(x), so t^2/4 = x, then dx = t/2 dt
yep
vapid barn
then you just have the integral:
merry atlas
okay so for this, take u = $\sqrt{x}$ and $dx = 2\sqrt{x} du$
Is this not what I did? I just wrote it as dx= 2tdt
vapid barn
$\frac{1}{2} \int t \sin(t) dt$
glossy valveBOT
EndTimes
torn jolt
then integrate by parts
torn jolt
Is this not what I did? I just wrote it as dx= 2tdt
well its better for us to start from the beginning
merry atlas
**EndTimes**
Ok, I got 2 there, not 1/2 so obviously wrong immediately
merry atlas
Oh ok
torn jolt
both would be correct, eventually
merry atlas
Right, cos I also had sin(2t)
Can you point out where in my attempt I went wrong then?
vapid barn
probably applying the double angle formula
but you can avoid that with the correct substitution
merry atlas
Ok so you substituted the whole of the internal part to avoid that?
vapid barn
yea
merry atlas
Interesting
vapid barn
constants are free
merry atlas
Yeah, I see what you mean
vapid barn
if you can get rid of a constant do it
merry atlas
Back at my desk now, gonna give it another go with your substitution then
merry atlas
yeah let t = 2sqrt(x), so t^2/4 = x, then dx = t/2 dt
Why is dx = t/2 dt here? Is the derivative of 2sqrtx not 1/sqrtx?
vapid barn
also another helpful trick
avoid taking derivatives of roots, let me write it out in full
merry atlas
Yeah they're not exactly convenient
vapid barn
t = 2sqrt(x)
t/2 = sqrt(x)
(t/2)^2 = x
t^2/4 = x
2t/4 dt = dx
t/2 dt = dx
merry atlas
ooh, yeah
vapid barn
so instead of taking the derivative of sqrt(x) you can take the derivative of t^2 which is a lot less prone to error
vapid barn
yeah that's what i did
differentiation is a linear operation, which means that multiplicative constants get factored out
merry atlas
but the quotient rule at least leads to squaring the denominator, no?
vapid barn
so d/dt(t^2/4) = 1/4 d/dt(t^2)
merry atlas
the derivative of the denominator leads to 0 so nothing happens with that part but the squaring still happens, no?
vapid barn
but the quotient rule at least leads to squaring the denominator, no?
no quotient rule at all here, we don't have a quotient of functions
merry atlas
That doesn't apply when the denominator is a constant?
vapid barn
i mean it technically does, but all the terms in the derivative of the denominator become 0
so it ends up being the same thing as just factoring out the constant and taking the derivative of the function
merry atlas
Ah right, cos when you multiply the denominator with the derivative of the nominator it cancels out the square of the denominator
Never mind me, it's been one of those days
Ok, back to trying it your way, sorry
vapid barn
are you confused on how (t/2)^2 becomes t^2/4
merry atlas
Oh no, I get the whole dx susbstitution now
vapid barn
there's not any calculus going on in that step, just algebra
merry atlas
I can't get there for some reason. I just get a circle of partial integration that never ends
I'm probably integrating the wrong parts..
vapid barn
that happens
you have to solve for the original integral
like let's say x = 1 - x, then you know that x = 2, same thing but this time your 'x' is the entire function that you're trying to integrate
let me just walk you through it
merry atlas
Yeah, like knowing when to derivate the trigonometric function and when to integrate it has me beat
glossy valveBOT
EndTimes
merry atlas
There I integrated t
vapid barn
how can we integrate that, knowing that:
$$ u dv + v du = d(uv)$$
$$ \int u dv + \int v du = uv$$
$$\int u dv = uv - \int v du$$
glossy valveBOT
EndTimes
vapid barn
just a quick derivation of ibp formula, i cannot remember it right unless i rederive it every time
merry atlas
Haha, yeah that form was not taught to me. $\int f(x)g(x) = F(x)g(x) - \int F(x)g'(x) dx$
glossy valveBOT
Kienai
vapid barn
it's really just the product rule for derivatives
but alright, let's try letting:
$$\sin(t) = v$$
$$t dt = du$$
glossy valveBOT
EndTimes
vapid barn
so what is u, and what is dv?
merry atlas
So I take the integral of t, get $ ((t^2)/2)sin(t) - \int((t^2)/2)cos(t)dt $
Bot didn't like that
vapid barn
we should end up getting:
$$ \int t \sin(t) dt = \frac{1}{2} t^2 \sin(t) - \int t \sin(t) dt$$
glossy valveBOT
EndTimes
merry atlas
nocturne creek
**Kienai**
The superior form
merry atlas
,rotate
glossy valveBOT
merry atlas
Here's the circle I was talking about
vapid barn
**EndTimes**
look at this
merry atlas
What?
You didn't take the derivative of sin
or the primitive of t?
I'm so confused'
vapid barn
yep i made a typo there, let me sort it out
merry atlas
Ok so, so far we should have the same then? Except in my handwritten one I did include the 1/2 you wanted to ignore earlier
I get why, it's annoying
vapid barn
sin(t) dt = dv, cos(t) = v
t dt = du, u = t^2/2
so we should end up getting:
merry atlas
again why is t not primitive after the integral there?
merry atlas
Ok, but then we definitely have the same thing so far
vapid barn
i'll actually just write this out on paper, latex is not being kind today
merry atlas
Yeah this bot version does not work well for me either
But in the next step, I want to derivate t^2 and take the primitive of cos (t)
vapid barn
alright, i'll just send a pic
merry atlas
But that leads to the circle I got stuck in so I guess that's wrong
vapid barn
vapid barn
But that leads to the circle I got stuck in so I guess that's wrong
Actually a lot simpler than it seems, maybe you chose the wrong u and dv, or just made a little error along the way
merry atlas
Hmm, this udv stuff is very confusing. I've not seen it and don't know how to use it either
That formula looks to me like it skips half the operation
vapid barn
Haha, yeah that form was not taught to me. $\int f(x)g(x) = F(x)g(x) - \int F(x)...
this formula isn't right
vapid barn
uv = (t)(-cos(t)) = -tcos(t)
merry atlas
No, after the integral
the formula is minus the integral, but the substitution is v=-cos(t)
vapid barn
yeah that should be a plus sign there, did it in my head but didn't write it down, the final answer is right though.
merry atlas
The last line looks right to me though
merry atlas
this formula isn't right
What part isn't? That's straight from my lectures
merry atlas
I missed the dx parts
glossy valveBOT
EndTimes
merry atlas
Yeah that is exactly the form I got from my lectures
See why this udv stuff is so confusing?
I see it gives the right answer in this case, but it's like it skips half the operation by ignoring the derivative on the right, somehow
vapid barn
nothing is being ignored
merry atlas
That's also what led to my circle earlier though so idk, if I can wrap my head around it this udv stuff seems better
vapid barn
i showed you the derivation earlier, it's a straightforwards application of the product rule
merry atlas
Then where is the derivative of t after the integral in the second to last line on your notes?
That's the part I don't get
vapid barn
it's there. u = t, so du = dt
merry atlas
so in my version g' = dg ?
vapid barn
i suppose so, let me see if that works out
merry atlas
That would explain it in my mind at least... I don't quite grasp this whole dx stuff either. Nobody ever explained it, it just appeared
Sometimes it's like you actually multiply with it, sometimes it just disappears after integration. So is it there or is it just a note on what is being derived? I have no idea
vapid barn
Let me try using your formula, which admittedly is a bit weird but I made it work:
We have that:
$$\int fg = Fg - \int Fg'$$
Let f = sin(t), then F = -cos(t)
Let g = t, then g' = 1, so we should get:
$$\int t\sin(t) dt = -t\cos(t) - \int -\cos(t) dt$$
glossy valveBOT
EndTimes
vapid barn
yeah that works out actually.
merry atlas
Hmm, so you can take the derivative of a substituted variable like that and use the 1 to multiply with?
vapid barn
Sometimes it's like you actually multiply with it, sometimes it just disappears ...
it's a differential, you can think of 'dx' as a very tiny change in x
merry atlas
even though t= 2\sqr{x} ?
vapid barn
Hmm, so you can take the derivative of a substituted variable like that and use ...
what are you referring to
vapid barn
a lot of the time the shorthand obscures the notation
when i said to let g = t, that really means to let g(t) = t
merry atlas
It clearly gives the right answer
vapid barn
and so we can take:
dg(t)/dt = dt/dt = 1
vapid barn
we're not concerned about the derivative of g(t) with respect to x, but only with respect to t
merry atlas
We never use the dx/dy form for derivatives either so I'm not at all familiar with it
vapid barn
yeah that is strange
merry atlas
it's all f'(x)
vapid barn
you need to be taught how to treat differentials
merry atlas
So getting a one like that is like witchcraft to me xD
Ok, well this clears up a lot actually
vapid barn
and the cool thing about them is that they mostly respect basic arithmetic.
how were you taught chain rule?
merry atlas
Yeah once I have the primitive of something it is really amazing to use honestly
Hang on
Chain rule is written as $$ D(f(g(x))= f'(g(x))g'(x)$$
So outer times inner derivative
glossy valveBOT
Kienai
merry atlas
That works for me at least, but when you never use dx and it just appears in integrals it is very confusing
vapid barn
you can also write that as:
$$\frac{d(fg(x)}{dx} = \frac{df(g(x))}{dg(x)} \frac{dg(x)}{dx}$$
merry atlas
But can I ask about this udv stuff. How do you know to pick which for which variable?
glossy valveBOT
EndTimes
merry atlas
It comes down to a similar problem as picking which part to integrate and derivate in my form, which I also don't quite grasp.
vapid barn
merry atlas
you can also write that as:
$$\frac{d(fg(x)}{dx} = \frac{df(g(x))}{dg(x)} \frac{...
the /dx here is read as "with respect to x"?
vapid barn
the /dx here is read as "with respect to x"?
yeah, pretty much. notice how the dg(x) just cancels, like in a fraction
merry atlas
Right. The only thing my lecturer said about this is "it's not a fraction"
when he wrote up dx/dy as an example
vapid barn
it's not a fraction because it's not a number
but we can algebraically treat it 'like' a fraction, if that makes sense
vapid barn
yeah, that's because when we write:
dy/dx, we mean: d/dx (y)
so dx/dx is just d/dx (x) which is 1
you have to be careful because you're not working with numbers, but rather linear operators, but normal "rules" of arithmetic still apply mostly:
d(f+g)/dx = (df + dg)/dx = df/dx + dg/dx
alright i have to go, read that link i sent for which functions to pick for "u"
acoustic pebble
Hi again
acoustic pebble
The probability of me getting a queen or a king in a deck of card is 2/13. If I use the classical probability formula I get this which is correct.
Can permutation and combination be used here ??
If yes then how
I believe it can
But idk how
acoustic pebble
Yes
onyx glen
right
in that case
if you REALLY want to barf combinatorial functions over this
you can go $\frac{\binom{8}{1}}{\binom{52}{1}}$
glossy valveBOT
Ann
severe basin
Yes
severe basin
${}^nC_k=\binom{n}{k}$
glossy valveBOT
QuantumBee
onyx glen
you're drawing 1 card
severe basin
8C1 means I'm taking a card out of 8 cards right??
There are 4 kings and 4 queens, you can get a king or a queen in 8c1 ways.
onyx glen
there are 8 cards which are successes for you
acoustic pebble
So, 8c1 is the number of ways I can get king or queen from a deck of cards
severe basin
Yes
merry atlas
I need to find the primitive function of this, and I got $$ \frac{1}{2}(x+sin(6x-14)) $$ by using the trigonemtric identity cos^2x= 1/2(1+cos(2x))
merry atlas
But something is wrong, what is it?
glossy valveBOT
Kienai
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
merry atlas
Should it be 1/12? instead
merry atlas
Hello, still dumb here xD
torn jolt
Should it be 1/12? instead
yes
you are on the right track
your answer is missing one thign
merry atlas
Okay what?
torn jolt
are you sure it is just x?
torn jolt
yes, but u r doing this for 'u' remember
glossy valveBOT
torn jolt
man im more used to integrating by reduction than with the half angle formulas
well uhh
torn jolt
set u = 3x-7 and then do the half angle formula
torn jolt
\frac{sin(6x-14)+6x}{12}
merry atlas
half from the half angle, half from 2 u
torn jolt
$\frac{sin(6x-14)+6x}{12}$ + C
glossy valveBOT
♡LexQa♡
torn jolt
Or wait, shouldn't it be 4 under the sin?
no?
torn jolt
yeah u did lmaoo
merry atlas
Again
merry atlas
Yeah this subject is so damn hard
I have so many more but I'll have a crack at them myself first
torn jolt
Yeah this subject is so damn hard
haha integrals are a pain for everyone so yes
u=3x-7
followed by
cos^2(x)=(1+cos(2x)) /2
torn jolt
haha integrals are a pain for everyone so yes
** everyone ** 
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torn jolt
you got it
wait wha
that's not the correct identity I think
cos^2(x)=(1+cos(2x)) /2
not (1+sin(2x)) /2
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torn jolt
that's not the correct identity I think
thats after
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thats after
torn jolt
I get x/2 + (sin(6x-14))/2
also i just realised
i thought u wrote 12 but it is 2
why is it 2?
torn jolt
ic
No jokes on you i just created my OWN trigonometric identity
get BAMBOOZLED
Anyways uhm
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torn jolt
<:catGiggle:912459497954500659>
👎🏻👎🏻👎🏻
ofc depends on the question
ok solve $\int e^{-x^{2}}$ right now
glossy valveBOT
♡LexQa♡
glossy valveBOT
torn jolt
😔
torn jolt
😔
how isn't erf*sqrt(π) /2 not satisfactory
because i said so 🙄
we don't know the answer define a answer
we don't know what gives i^2+1=0
create a number that does
don't do this for 0/0 just
ok sure just a sec
creating a number
yes
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elfin mica
When a particle is moving in the positive direction or negative
direction, do we just make a sign chart of the velocity and find out where its pos and neg
elfin mica
but if we are trying to figure out whether the particle is speeding up or slowing down, we compare both a(t) and v(t) to see
misty topaz
You should just need velocity
delicate idol
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delicate idol
.close
torn jolt
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torn jolt
can someone help please
winged fern
Hey I am working on a coding problem where I have an equation I am unable to solve:
I got this 2 equations:
s1 = k^(−1) * (z1 + r1 * d) % n
s2 = (a · k + b % n)^(−1) * (z2 + r2 * d) % n
and I want to get d out of it but it has to be a whole number so i have to use the inverse multiplicative modulo but I have no idea how to do that
If someone could help be that would be great (:
limber flicker
$s1 = k^{(−1)} * (\frac{z1+r1*d}{n})$
glossy valveBOT
Clarkie
limber flicker
$s2 = (\frac{ak+b}{n})^{(−1)}* \frac{z2+r2*d}{n}$
glossy valveBOT
Clarkie
limber flicker
@winged fernignoring the odd looking brackets are those the equations?
winged fern
sorry the % signs are modulo so it should be s1 = mod(k^(−1) * (z1 + r1 * d), n)
limber flicker
right but that would be a line of code orr?
only modulo knowledge I have in python is just something about remainder or diving the left by the right
idk how to interoperate the stuff well enough
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@winged fern Has your question been resolved?
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@winged fern Has your question been resolved?
winged fern
i need a whole number between 0 and n at the end and the way i understood is that in order to get a whole number at the end i have to prevent fractures so the only way i know is to use the inverse multiplicative modulo for fractions and then modulo n everything at the end but i don't get to the right solution
full forumBOT
@winged fern Has your question been resolved?
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@winged fern Has your question been resolved?
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rugged harbor
What is the question?
I don't think I can help, I was about to go to sleep
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@torn jolt Has your question been resolved?
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Available help channel!
modern urchin
part 2 of this
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part 2 of this
modern urchin
I am considering the values of lim f as xy approaches 1 from either side
it's going to +inf and -inf
i dont know what to make of it
modern urchin
it's going to +inf and -inf
can i just say that since it's not approaching the same value the limit dne?
cloud shore
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@modern urchin Has your question been resolved?
barren iris
I'm doing #12 and it seems like I'm doing something wrong here. I tried to isolate x and then y which was fine(got big scary numbers). Then, I tried to put those into one of the systems and my z's canceled..
atomic blade
You can isolate z I guess
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glass crystal
dont you have a sign error here
tepid basin
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tepid basin
can i get some help on this?
i dont rlly know where to start
well i know i start with deriavative
quotient rule can be applied
wise widget
yup was just about to say that! yup!
keen spruce
sign chart ❤️
wise widget
^^
find when the derivative is zero or undefined
or in other words when the numerator = 0 or the denominator = 0
tepid basin
- get derivative
- set equal to zero and solve for x
- use numbers solved for x in sign chart?
then use domain things from sign chat
wise widget
- set numerator or denominator = 0 solve for x
keen spruce
look for a sign change
tepid basin
okok let me try that rq
neat bay
Yeah gotta check when derivative is undefined
For example |x| switches from decreasing to increasing at x = 0, which is exactly where the derivative is undefined
tepid basin
2) set numerator or denominator = 0 solve for x
what does this mean by
numerator or denominator?
both give the same answer?
@neat bay
neat bay
Same answer? No
Numerator is for when derivative equals 0
Denominator is for when it’s undefined
(literally, check when it’s dividing by zero)
wise widget
^^
neat bay
Denominator is for when it’s undefined
Have you set the denominator to zero? What happens?
tepid basin
tepid basin
to solve for x?
neat bay
but what does the x you get as a result actually represent
tepid basin
good question
tepid basin
where some curve occurs?
neat bay
here let’s start a little earlier
You understand derivative is the slope of the tangent line, right
tepid basin
correct
neat bay
So in a sense it measures the rate of change at a point
tepid basin
i do
neat bay
So one thing that’s cool is that it “detects” when a function is increasing or decreasing
If the slope is positive, it’s pointing up, so the function is increasing there
and if slope is negative, then it’s decreasing
cool
what if the slope is zero?
neat bay
Yep exactly
tepid basin
on the x axis
tepid basin
i'm not sure
neat bay
It can mean it’s at the top of a hill, like this
tepid basin
ah
neat bay
In other words, a (local) maximum point
local because it’s higher than the points immediately left and right of it
tepid basin
i see
neat bay
It could also mean it’s a local minimum point
But there is a special case which we need to check
tepid basin
red is f(x) and blue is the derivative graph right?
neat bay
Blue is the tangent line
The derivative is a little different when you graph it for all values of x, don’t worry about that
Look at this
Here the derivative is zero, so a flat line, but it’s not a minimum or a maximum
It’s basically like you’re running, slow down, stop, and then start running again
So that’s kind of an edge case
Anyway
For this problem, they’re asking you when the function is increasing or decreasing, right
neat bay
Ah
Yeah
Oh whoops another thing
A maximum/minimum can occur without the derivative equalling zero
Look at this
Here is |x|
There is a minimum at zero, but the derivative is undefined (why?)
So to check for min/max, it’s also useful to check when the derivative is undefined
neat bay
Yes exactly
Sorry denominator of the derivative
Think about it
How could the derivative be undefined?
This could only happen if you were trying to, say, divide by zero
neat bay
Or you’re trying to take the square root of a negative number
(But that’s less common)
So
What did you get for the derivative of this function?
tepid basin
numerator = 36 - 4x^2
denominator = (x^2 + 9)^2
neat bay
nice
ok let’s do min/max
find where it equals zero
and then find where it’s undefined (there might be no points like this)
tepid basin
so the first step would be set the numerator equal to zero?
i got x equals plus minus 3
neat bay
Nice
tepid basin
now we check to see when the denominator is undefined?
and set denominator equal to zero?
neat bay
Yeah, so x = 3 and x = -3 are possible min/maxs
Again keep in mind, you might not get an answer for denominator
neat bay
yep
tepid basin
because square roots cant have negatives
neat bay
It’s imaginary
tepid basin
meaning its defined
tepid basin
ahhh
okay
and now i do my sign chart
hmmmm
but what if my denominator was undefined
what would happen then?
neat bay
If it was undefined, let’s say at x = whatever, then you would add “whatever” to your sign chart
tepid basin
ohh ok
neat bay
Here is |x|
(It’s a possible min/max value, like this example)
tepid basin
so additional points
neat bay
yeah
tepid basin
so techncially u set both parts equal to zero
and the denominator might not always give a valid value
neat bay
Yeah
Well actually
Numerator may also not give a value; what if the derivative was 1/x?
Then you’d only get one possible point, x = 0
the numerator can’t ever equal zero, because the numerator equals 1, and 1 does not equal 0
Lol
tepid basin
ahh
neat bay
Ok so you have x = 3 and x = -3 on your sign chart
That means the derivative is flat at those points
So that means that in between those points, the derivative is gonna be either positive or negative
tepid basin
so i would plug those x values into f(x)?
and determine if y is positive or negative?
neat bay
Those in-between values are gonna be the intervals for increasing/decreasing
No you plug them into the derivative
That’s the slope
neat bay
Yeah, in between two zeroes, it’s gonna be all positive or all negative
So you only need to test one number
And that’s gonna be the positive/negative for the entire interval
tepid basin
i got
inc: (-3,3)
dec: (-infinity,-3)U(3,infinty)
now determining the relative max and min is tricky
tepid basin
neat bay
Yeah
So look at the left side
It goes negative, then zero at x = -3, then positive
In other words, decreasing, then flat, then increasing
This means that x = -3 is a local minimum
Think about it
neat bay
Yep nice! Yeah
tepid basin
but
how do we know there isnt a minimum from 3 to infinity?
well
if it goes to infinity it never reaches another min?
it jsut keeps going and going?
neat bay
how do we know there isnt a minimum from 3 to infinity?
Because that would mean the derivative is either zero or undefined somewhere from 3 to infinity
But we found all the points where it’s zero or undefined
And it’s only at x = -3 and x = 3
There literally cannot be a max anywhere else since we found all the potential values
neat bay
yeah fs
tepid basin
sometimes i kinda just follow formulas without understanding the why
neat bay
yeah
I like helping people see the why
now the rest of the problems should be trivial; you’re just repeating the exact same procedure lol
tepid basin
.close
lofty holly
Hey! Can you help me with my question? (Matrix)
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Hey! Can you help me with my question? (Matrix)
lofty holly
.close
void swift
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short crest
Ok so first of all, do you know how to find the gradient of a line that is perpendicular to another line?
short crest
Wait... i might've skipped some steps
ok so first of all
the line you have is
2y + 4x = 12
turn it into slope-intercept form
y = mx + b
where m and b are constants
short crest
are you sure?
void swift
wait
short crest
close... but there might be some miscalculations
void swift
nvm
void swift
-2
short crest
yes
so for 2 perpendicular lines.. the product of the 2 line's gradient is always -1
for example...
y = 3x + 3
the gradient/slope is 3
to find the gradient of a perpendicular line, it's just
m1 * m2 = -1
3 * m2 = -1
m2 = - 1/3
so the equation of the perpendicular line will be
y = -1/3 x + b
where b can be any value
since you have
m1 = -2
find m2 (the gradient/slope of the perpendicular line)
short crest
You're welcome
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tall juniper
Do you know the rule for parallel lines?
void swift
no
tall juniper
The rule is that the slope of line 1 = slope of line 2 (m₁= m₂)
tall juniper
If m₁ is 3 then m₂ is?
void swift
1/3?
tall juniper
no m₂ is 3
tall juniper
The rule is that the slope of line 1 = slope of line 2 (m₁= m₂)
Just like
void swift
oh
tall juniper
So the equation of the line (a) should have a slope value of 3
What does the equation of the line look like now?
tall juniper
Yeah
void swift
b=-11
tall juniper
Exactly
void swift
is that for b?
tall juniper
So the equation now looks like
y = 3x-11
void swift
yes
tall juniper
That is it, we have found the equation of line (a)
torpid stag
I think he means (b)
tall juniper
Do you know the rule of perpendicular lines?
void swift
no
tall juniper
line₁ is perpendicular to line₂ if m₁= -1/m₂
torpid stag
Why do y'all use m for the gradient in the US?
void swift
idk
torpid stag
line₁ is perpendicular to line₂ if m₁= -1/m₂
-1/m2*
tall juniper
Oh right
void swift
3 = -1/3
tall juniper
You can't substitute 3 for both m₁ and m₂
void swift
oh
tall juniper
You only substitute that for m₁ so you get 3 = -1/m₂
void swift
how do I find m2?
torpid stag
3 = -1/3
Remember that "=" is a relation that says something has equal value. Using the symbol as a "step"-symbolizer is not the correct usage
void swift
oh
torpid stag
how do I find m2?
Just say m2 = 3 instead of m1 = 3, then it gives you the gradient immediately instead of having to solve for m2
So for m1 = -1/m2, we have if m2 = 3?
void swift
how to find m2?
torpid stag
So for m1 = -1/m2, we have if m2 = 3?
^
void swift
oh
tall juniper
for the rest of the solution the equation of line (b) is the equation of the line with y=m₁x+b₁
And the original line in the picture is the line with equation y=m₂x+b₂ where m₂=3 and b₂=-7
void swift
y=3x-7
tall juniper
Basically: We are trying to find m₁ and b₁ now
If you substitute in m₂=3 into the equation m₁= -1/m₂ what do you get?
void swift
b=-1
tall juniper
Exactly
void swift
oh so it would be y=-1/3x-1
tall juniper
Almost
It would be y= (-1/3)x-1
(-1/3)x is different from -1/3x because the first means
$\frac{-1}{3}x$ while the second means $\frac{-1}{3x}$
void swift
oh ok
glossy valveBOT
jafar/جعفر
void swift
(-1/3)x is different from -1/3x because the first means
$\frac{-1}{3}x$ while th...
Thank you
I got all of the questions correct
tall juniper
Presto
tall juniper
I got all of the questions correct
Don't forget to .close the channel
torpid stag
And make sure to wait around 15 minutes after stating your question before pinging Helpers
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@steady lagoon Has your question been resolved?
tidal reef
Given that $A={2011, 2022, ..., 2021}$. Three subsets, $A_1, A_2, A_3$ are created from $A$ with the conditions:
\begin{itemize}
\item $A_1\cup A_2\cup A_3=A;$
\item $A_2$ has no intersection with other subsets;
\item $|A_1\cap A_3|=1$;
\item Every subset must have a prime number.
\end{itemize}
The number of subsets $(A_1, A_2, A_3)$ that satisfy the conditions above are...
tidal reef
how would i approach this question?
glossy valveBOT
nichoals
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@tidal reef Has your question been resolved?
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Available help channel!
twilit cloud
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tidal reef
damn
twilit cloud
12 c
deft zodiac
o
twilit cloud
i sneed help with that
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@twilit cloud Has your question been resolved?
glossy valveBOT
deft zodiac
i think u forgot +1 in ur eq
native flicker
G
twilit cloud
o rite
deft zodiac
so hard i gon lol
little fiber
Its about trigonometry specifically
sum identity of sine
how does this explain that
sin (a+b) = sin(a)cos(b) + cos(a)sin(b)
thanks
