60/-4 = -15
#help-23
lean otter
X-5x
X equals 1?
sly plank
1 - 5 = -4
when a variable has no coefficient, it is defined as 1
x = 1x
The value is exactly the same
lean otter
when a variable has no coefficient, it is defined as 1
Thanks homie
Yeah its makes more sense to me now.
Thanks for your help
lean otter
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night ingot
what should my first step be? (i get the concept, just confused how to start)
obtuse plover
id multiply numerator and denominator by sqrt3
night ingot
so after i do that i should get ((1+sqrt3)/sqrt3))/((-1+sqrt3)/sqrt3) ?
srry idk how to format that better 😭
ohh wait i think i got it
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mighty ridge
Whats the formula or how would I do this question?
safe radishBOT
Whats the formula or how would I do this question?
mighty ridge
so 90 degrees
twilit plover
yes
twilit plover
yes
void mirage
mighty ridge
oh wow it was that easy
twilit plover
math is funn :>
mighty ridge
tysm guyss
twilit plover
ur welcome
mighty ridge
I have another question
twilit plover
what
mighty ridge
umm yes
180-135?
oh my bad
actually no its not making sense when I say it out
okk
ohh great ty
yess
90?
noo
45?
ohhhh = 45
yes right
the remaining part?
ohhh yess
I understand noww
TPQ is 45
because
T is 45 and Q is 90 so P would be 45 to add up to 180?
Oh okok
Tysm tyler
okk sure
omg yesss I got it on my test today
I can say I did in fact got it wrong
What if a circle is between them is it the same thing
yes please
ohh
on the test was a full circle
yess
Ill try to draw it if I remember correctly
give me a min
so It wasnt a full circle
it looked like b3
ohhhh
Thx again ik ill get questions like this nextweek I hope I understand them😔
really?
Ok thx
yeahh okk
I have another question I just want to knw if its correct
this would be x= 37.35?
I used sine
samee 😭
i will draw it then
yh
yk im actually looking at it its actually easy
my bad😭
well greatt im proud
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unreal rune
Sorry to come back again but idk how to aproach this one at all. Determine the following infinite summation: $\sum^{\infty}_{i=0} (2i+1)(\frac{1}{3})^i$
safe radishBOT
Sorry to come back again but idk how to aproach this one at all. Determine the f...
flat frigateBOT
Zenflip
unreal rune
I know that it is converging yes
although navigating how to actually find the sum, do I just need to estimate what it's approaching? that feels inneficient
safe radishBOT
@unreal rune Has your question been resolved?
unreal rune
is there a way to split the sum? or maybe a way to simplify this that I just am not getting?
safe radishBOT
@unreal rune Has your question been resolved?
undone burrow
is there a way to split the sum? or maybe a way to simplify this that I just am ...
yea splitting the sum is a good idea
then you can use the fact that d/dx x^i = i x^(i-1)
unreal rune
wait why would I need to use that?
and how would I split the sum as you can't just multiply sums like that
undone burrow
split it as 2i(1/3)^i and (1/3)^i
unreal rune
oh you meant that way
undone burrow
the derivative thing is useful for the 2i(1/3)^i one
unreal rune
dang I never expected I would have to start bringing in my calculus for discrete math
uhg
time to look back through my old notes to remember the rules of derivative summation
wait so when is does the starting interval need to add 1 again when doing a derivative of a summation?
undone burrow
I don't think you need to change the bounds for this one
(sorry, not too experienced with sums, I just know this method)
unreal rune
it wouldn't = 5/2 would it?
I'm looking at the sum of a telescopic function
wait I'm just wrong nvm
uhg I forgot how tedious this was
undone burrow
how're you approaching it?
it shouldn't be that tedious
unreal rune
well I'm reading in my notes to see how it would work mechanically the issue that I'm seeing is that the derivative of the sum doesn't mean I can find the actual solution of the sum
they aren't equivalent no?
undone burrow
the method is rewriting $\sum_{i = 0}^\infty i(\frac13)^i$ as $\sum_{i = 0}^\infty \frac13 \frac{d}{dx}x^i$ then at the end, evaluating at x = 1/3
flat frigateBOT
Sepdron
unreal rune
what I'm getting is tthat I need to substitute out i*1/3?
wait no
is it the integral you want as I have 2i*(1/3)^i
the integral is 2*(1/3)^(i+1) + c
I don't know how that is useable but it is something
ig the c isn't necessay as it is always between a range
but then using the sum of that kind of series I get a 1
undone burrow
no you don't need to do the integral
bc that'd be 2*(1/3)^(i+1)/(i+1)
unreal rune
you're right
my bad
it's late so that's probably why I'm not registering the use of the derivative right now
unreal rune
the method is rewriting $\sum_{i = 0}^\infty i(\frac13)^i$ as $\sum_{i = 0}^\inf...
because as much as I would love to know what you're getting at here, my brain is not understanding why at all
alright imma close this and think about it when I have brain power, my head is killing me
thank you
.close
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hazy field
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hazy field
My answer is basically right but I’m not sure when to add the arrows
which keeps bugging me
“😡” this is me rn
lowkey
hard crest
what's the y-value of this for x = -3?
hard crest
okay, so the 3/5 equation shouldn't be happening left of 0
hazy field
could u explain why
hard crest
well because it's only for 0 ≤ x ≤ 5
hard crest
when it doesn't have an end
like instead of 0 ≤ x ≤ 5 it's just 0 ≤ x
then you'd have an arrow on the right side
hazy field
and vice versa?
hard crest
yeah
hard crest
nah yeah
hazy field
I thought the kangaroo reaction was a yes
hard crest
yeah nah
hazy field
Nah yeah?
hard crest
nah yeah nah
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tiny terrace
a quick question if i wanted to know the domain of, would it be any number higher or equal than 0 right?
safe radishBOT
a quick question if i wanted to know the domain of, would it be any number highe...
lean otter
well not always
it depends on what you are rooting it to
okay one sec I'll give you a better generalisation
this isn't particularly helpful
tiny terrace
for even roots the argument must begreater than or equal to 0 and for odd there is no such thing
is that?
flat frigateBOT
Edmund Cloudsley
lean otter
for even roots the argument must begreater than or equal to 0 and for odd there ...
precisely
lean otter
**Edmund Cloudsley**
this is what you just said represented mathematically ⏫
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trail otter
Why polynomial of degree 2 isn’t sub space of R_[x]?
safe radishBOT
Why polynomial of degree 2 isn’t sub space of R_[x]?
trail otter
Why it doesn’t have zero vector?
rough storm
note that it says polynomials of degree 2, not polynomials of degree less than or equal to 2.
@trail otter
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@trail otter Has your question been resolved?
rocky hemlock
in these we have to just use the addition substraction or UV or chain rule right? im confused on what to do because the things are not given to be = to y
safe radishBOT
in these we have to just use the addition substraction or UV or chain rule right...
rocky hemlock
and i was absent on the day this was given
safe radishBOT
@rocky hemlock Has your question been resolved?
rocky hemlock
and hoiw to do the 10th one
flat steppe
Man do you know some trigonometry
onyx haven
in these we have to just use the addition substraction or UV or chain rule right...
To graph the functions, you often let these expressions like $x^3 \cdot e^x = y$.
If you think about applying a "derivative operator" to both sides, you can do:
$\frac{d}{dx} (x^3 \cdot e^x) = \frac{d}{dx} (y)$
The way these questions are framed, you are asked:
Given an expression $x^3 \cdot e^x$, find $\frac{d}{dx} (x^3 \cdot e^x)$. (I pressume, since you are asking about the chain rule)
You don't have to have them equal to $y$. Just think of it as implicit.
flat frigateBOT
abfish
onyx haven
d/dx (y) i meant there
rocky hemlock
To graph the functions, you often let these expressions like $x^3 \cdot e^x = y$...
is doing them like this correct
rocky hemlock
d/dx (y) i meant there
?
onyx haven
just a correction i made to the text i wrote
rocky hemlock
just a correction i made to the text i wrote
the way i solved in the image is that correct?
rocky hemlock
seems like you are doing it correctly
i was just confused beevause they didnt say that whatever is given = y
onyx haven
You can also write $\frac{d}{dx}(x^2 e^x)$ like this:
$(x^2\cdot e^x)' = 2x \cdot e^x + x^2e^x$
flat frigateBOT
abfish
onyx haven
Which is a little shorter notation
onyx haven
**abfish**
Anyways, I hope this explanation makes sense @rocky hemlock
They are sort of "implicitly" equal to y
makes sense to set them equal to y, when you want to graph them
rocky hemlock
Which is a little shorter notation
ik i did it for y` but my maths teacher i am nnot sure he would allow it so just to be safe
and how to do the 10th and 11th one? i cant figure them out
onyx haven
It's again just applying the product rule and addition rule.
Addition rule:
$$
\frac{d}{dx}(f(x) + g(x))=\frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))
$$
flat frigateBOT
abfish
rocky hemlock
It's again just applying the product rule and addition rule.
but when i apply one rule y is already dy/dx so after product rule it is y`=whatever is the answer os the product rule
does it matter if we do product rule first or addition rule first
onyx haven
Product rule:
$$
\frac{d}{dx}(f(x) \cdot g(x)) = \frac{d}{dx}(f(x))\cdot g(x) + f(x)\cdot \frac{d}{dx}(g(x))
$$
flat frigateBOT
abfish
onyx haven
in these we have to just use the addition substraction or UV or chain rule right...
Let
f(x) = x\cdot sin(x) + cos(x)
g(x)=x\cdot cos(x) - sin(x)
Apply the product rule
rocky hemlock
the uv rule?
onyx haven
does it matter if we do product rule first or addition rule first
no you can apply them whenever you have those forms of functions.
Two functions added together? -> Can apply addition rule
Two functions multiplied together? -> Can apply multiplication rule.
They are quite powerful
Doesn't matter which order
onyx haven
Apply the product rule
$$
\frac{d}{dx}(f(x) \cdot g(x)) = \frac{d}{dx}(f(x))\cdot g(x) + f(x)\cdot \frac{d}{dx}(g(x))
$$
flat frigateBOT
abfish
onyx haven
If
f(x) = x\cdot sin(x) + cos(x)
g(x)=x\cdot cos(x) - sin(x)
Then
$$
\frac{d}{dx} (f(x) \cdot g(x)) = \frac{d}{dx}(x\cdot sin(x) + cos(x))\cdot (x\cdot cos(x) - sin(x)) + (x\cdot sin(x)
$$
$$
- cos(x))\cdot \frac{d}{dx}(x\cdot cos(x) - sin(x))
$$
flat frigateBOT
abfish
rocky hemlock
Then
$$
\frac{d}{dx} (f(x) \cdot g(x)) = \frac{d}{dx}(x\cdot sin(x) + cos(x))\cd...
but it is y = g(x) +f(x)
in the question
onyx haven
In plain english:
You apply the "derivative" operator to the first function and multiply it by the second, and apply the derivative operator to the second function and multiply it by the first, and add them together.
rocky hemlock
In plain english:
You apply the "derivative" operator to the first function and ...
ik the rules
onyx haven
which one are you doing?
rocky hemlock
which one are you doing?
10th
rocky hemlock
oh my bad. i meant 11)
So is the answer I got correct
onyx haven
it doesn't seem like you have reduced the function quite yet. you need to apply the derivatives until* there is no d/dx'es
rocky hemlock
it doesn't seem like you have reduced the function quite yet. you need to apply ...
Like this
?
rocky hemlock
This was the final answer I was getting and I had already sent it once before
onyx haven
$$
(x^5e^x)' = 5x^4\cdot e^x + x^5\cdot e^x
$$
Indeed
But you other half, you done something not quite right it seems.
flat frigateBOT
abfish
onyx haven
For the other half
$$
(x^6 \cdot log(x))' = 6x^5\cdot log(x) + x^6\cdot 1/x
=6x^5log(x)+x^5
$$
flat frigateBOT
abfish
onyx haven
You add them together, you get:
$$
(x^5e^x + x^6\cdot log(x))'= 5x^4\cdot e^x + x^5\cdot e^x
$$
$$
- 6x^5\cdot log(x) + x^6\cdot 1/x
$$
flat frigateBOT
abfish
rocky hemlock
Yesss I got it
I got confused on what to do so even after getting the right answer I again took d/dc
Dx
I got the answer on step 3 but still managed to fuck it up 😭😭
onyx haven
To be clear, you have the function of the form:
f(x)g(x) + h(x) i(x)
You want to differentiate it. So you apply the addition rule. Think of F(x) = f(x)g(x). G(x) = h(x) i(x)
Applying the addition rule:
(addition rule)
(F(x) + G(x))' ~> F'(x) + G'(x)
And
F'(x) + G'(x) = (f(x)g(x))' + (h(x) i(x))'
Now apply the product rule for each part..
Think of functions as some sort of object which can you split apart and combine
So you can fit them to the addition rule and multiplication rule
(That's the intution I have. May be hard to understand, because intuition is different from person to person)
rocky hemlock
To be clear, you have the function of the form:
f(x)g(x) + h(x) i(x)
You want ...
How is F(x) = f(x)g(x)G(x)
Oh it's a fullstop
rocky hemlock
I should kms
onyx haven
I should kms
anyways good luck. you seem to be doing it correctly (most of the time). If you want to check your work you can use a math tool to do i
rocky hemlock
Yeah on phone it's like .G(x)= at one line and h(x)i(x) on next line
rocky hemlock
anyways good luck. you seem to be doing it correctly (most of the time). If you ...
Which tool
onyx haven
you can use this one:
https://www.wolframalpha.com/
safe radishBOT
onyx haven
rocky hemlock
Oops
onyx haven
anyways you can use that to do it
rocky hemlock
anyways you can use that to do it
It has everything dam
onyx haven
It will not tell you the steps. You will have to figure out that yourself. 😄
But it will tell you the correct answers. If your teacher requires the steps, it is a good way to correct yourself..
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foggy mural
how do i write down that a number is prime chat
safe radishBOT
how do i write down that a number is prime chat
@foggy mural Has your question been resolved?
foggy mural
.reopen
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✅
foggy mural
.close
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safe parrot
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flat frigateBOT
Merineth
safe parrot
would result in a circle with a radius of sqrt(33)
but could it also be drawn by saying that highest x can be is x = 5 when y = 0
and y = 5 when x = 0?
tardy mango
(sqrt 33, 0) and (0, sqrt 33):
safe parrot
$x^2 + 0 < 33 \implies x < \sqrt{33} \implies x < \pm \sqrt{33}$
and the same thing for y?
flat frigateBOT
Merineth
tardy mango
**Merineth**
junior smelt
In fact-
tardy mango
$x^2<33 \implies |x|<\sqrt{33} \implies -\sqrt{33}<x<\sqrt{33}$
flat frigateBOT
Civil Service Pigeon
tardy mango
$\sqrt{x^2}=|x|$
flat frigateBOT
Civil Service Pigeon
safe parrot
$x^2<33 \implies \sqrt{x^2} < \sqrt{33}$
flat frigateBOT
Merineth 🇸🇪
safe parrot
Is this not what we are doing?
safe parrot
**Civil Service Pigeon**
Never seen this before afaik
safe parrot
**Merineth 🇸🇪**
$\sqrt{x^2} < \sqrt{33} \implies |x| < \sqrt{33}$
flat frigateBOT
Merineth 🇸🇪
craggy sedge
Never seen this before afaik
that is used in many cases
safe parrot
where $\sqrt{33} = + \sqrt{33}$ and $-\sqrt{33}$
flat frigateBOT
Merineth 🇸🇪
craggy sedge
**Merineth 🇸🇪**
i would suggest you dont write this
safe parrot
Well it's the correct answer
all it's meant to do is help me sketch the Df of a function
I need to learn lineintegrals, harmonisk integral, tripple integrals, flow integrals, double integrals , integral circulation
until tomoorw (:
craggy sedge
I need to learn lineintegrals, harmonisk integral, tripple integrals, flow integ...
thats a lot all the best
safe parrot
if i were to kms, now would be a prime opportunity lol
I unironically can't remember anything of mult calc
craggy sedge
if i were to kms, now would be a prime opportunity lol
yeah dont tho
craggy sedge
oh multi then off i go kek
safe parrot
(:
Iirc double integrals are calculated by going outwards inwards where we take the primitive function wrt y first and then wrt x
but first i need a definition area for D
My guess would be
$\int_0^{\pi /2} \int_0^{\pi /2} = D$
flat frigateBOT
Merineth 🇸🇪
junior smelt
Careful, that’s a rectangle 
junior smelt
oh multi then off i go kek
Also hiii 
craggy sedge
Also hiii <:Hiii:983878629987614752>
hello chartbit 
(im doing RA now 
)
junior smelt
and y goes from 0 to pi/2
Be a bit careful, while both of these statements are “technically correct”, x and y are “restricted” by each other
junior smelt
maybe 0 to pi/4?
For example, when x = pi/4, then y does run from 0 to pi/4, but not in general-
safe parrot
Not 100% sure i remember
junior smelt
See like this here 
If you fix a given choice of x in that triangle, then y runs between 0 and x
(as the red line’s equation is y = x)
safe parrot
I can't remember at all
i imagined x was the black
and red was the y
so one of the integrals goes from (0,0) to (pi/2, 0)
and the other goes from (pi/2, 0) to (pi/2,pi/2)
junior smelt
If you fix a given choice of x in that triangle, then y runs between 0 and x
If you choose to integrate wrt y first, then y runs from 0 to x, and then the x integral will run between 0 and pi/2
If you decided to do it wrt x first instead, you’d want to consider the values x can take on, which would be looking at it like-
safe parrot
junior smelt
It’s a bit pain to get around 
(But that’s why drawing out the region you’re considering helps out quite a bit 
)
junior smelt
Uhhh good question 
lemme see if I can find a good one
safe parrot
I only find indian ones on YT :(
Does determining the D for the double integral revolve somehow around finding the bounds?
As in
$0\le x \le \frac{pi}{2}$
flat frigateBOT
Merineth 🇸🇪
flat frigateBOT
Merineth 🇸🇪
flat frigateBOT
Merineth 🇸🇪
junior smelt
Kinda like that 
if you integrate wrt y first, that’s exactly what it is!
junior smelt
(And in fact, integrating wrt y first is what we really want to do for this integral!)
safe parrot
$\int_0^x \int_0^{pi/2} \frac{cosx}{x}$dxdy
flat frigateBOT
Merineth 🇸🇪
safe parrot
no sorry it has to be
other way around
since we gotta calc pi/2 first?
$\int_0^{\pi /2} \int_0^x \frac{cosx}{x}$dydx
junior smelt
Also \pi 
flat frigateBOT
Merineth 🇸🇪
safe parrot
Shouldn't this be right?
junior smelt
That’s it
safe parrot
**Merineth 🇸🇪**
And that's because this is the bounds for x
hence why we want to integrate wrt dx first?
junior smelt
Well, we’re really integrating wrt $y$ first, we’re working out the integral $\int_0^x \frac{\cos(x)}x \dd y$ first
flat frigateBOT
@junior smelt
safe parrot
hm?
junior smelt
Then, when we find what that works out as, you integrate that result between 0 and pi/2 wrt x
flat frigateBOT
Merineth 🇸🇪
junior smelt
**Merineth 🇸🇪**
This is really kinda saying
[
\int_0^{\pi/2} \qty( \int_0^x \frac{\cos(x)}x \dd y ) \dd x
]
safe parrot
Wasn't the rule for double and tripple integrals to integrate outwards inward?
flat frigateBOT
@junior smelt
junior smelt
(Also note that the inner integral depends on x as well)
flat frigateBOT
Merineth 🇸🇪
junior smelt
That’s for derivatives, the “equivalent” is by parts, but we won’t need either here-
Notice how the inside doesn’t depend on y at all, so cos(x)/x is effectively a constant with regards to y
safe parrot
oh
junior smelt
Would you know how to integrate $\int_0^x 1 \dd y$ at least?
flat frigateBOT
@junior smelt
flat frigateBOT
Merineth 🇸🇪
junior smelt
Would you know how to integrate $\int_0^x 1 \dd y$ at least?
This one works out as $y \eval_0^x$
flat frigateBOT
@junior smelt
junior smelt
Well, almost, remember how that cos(x)/x was basically a constant wrt y?
So really you’re working out $\frac{\cos(x)}x\cdot y\eval_0^x$
junior smelt
(Or at least you can )
🔫
junior smelt
Other way around, remember it’s “top limit - bottom limit”
junior smelt
Yep 
safe parrot
$\int_0^{\pi/2} \cos(x) dx$
flat frigateBOT
Merineth 🇸🇪
safe parrot
i unironically can't remember by heart
what cos becomes
i think it was
-sin
or sin
no wait
sin = cos
cos = -sin
in derived
so sin?
sin(pi/2)
which is 1?
That was all the questions which i had a vague idea how to solve
the rest i haven't even touched
Calculating line integrals ._.
safe parrot
not hard for uuuuu
since you are smart
and i'm dumb ||and ugly sad lonelyhatednotlovedunwantedwasteofair||
i'll look up a yt video and see if i can make any sense of it
junior smelt
since you are smart
I’m really not
junior smelt
i'll look up a yt video and see if i can make any sense of it
Awwww 
obsidian oracle
Yo merineth been a while
obsidian oracle
and i'm dumb ||and ugly sad lonelyhatednotlovedunwantedwasteofair||
Just came by to say you're good fam
safe radishBOT
@safe parrot Has your question been resolved?
safe parrot
hello
and noo i'm not good at all :I
Unironically been my worst Fall ever
and now i got to go back to doing some math 
chartbit ;-;
i don't think i'll pass tomorrow
but can you help me with this one? Q.Q
junior smelt
i don't think i'll pass tomorrow
Awwww
junior smelt
but can you help me with this one? Q.Q
Sure, I’ll need to go in a few mins though 
junior smelt
safe parrot
junior smelt
I don't think they're expecting this to be done the long way tbh 
junior smelt
Yea, that e.g. if you have some $f : \bR^n \to \bR$ such that $\mathbf{F} = \nabla f$ and some curve $\gamma$ in $\bR^n$ that starts at $\mathbf{p}$ and ends at $\mathbf{q}$, that $\int_\gamma F \cdot d\mathbf{r} = f(\mathbf{q}) - f(\mathbf{p})$
flat frigateBOT
@junior smelt
safe parrot
what the fuckk
junior smelt
You might have it stated a bit differently maybe?
What do you have for line integrals?
junior smelt
This is the one I'm referring to
safe parrot
Yeah that's under Curve integrals of vector fields
the problem here is that i haven't ever touched this before
so i' have no idea what i'm doing
Or rather no knowledge about it
The other 3 questions were easy becuase they only refered to integrals
and Df
I'm not sure how to apply your method since i don't know it :(
junior smelt
Awwww 
junior smelt
This is the one I'm referring to
Well do you at least get the statement they're making here?
(I have to go now so dead, have a meeting 
)
wind frigate
My teacher didnt explain them, i would like an assistance
safe radishBOT
My teacher didnt explain them, i would like an assistance
balmy goblet
My teacher didnt explain them, i would like an assistance
explain what
wind frigate
Could someone explain how to solve these typa questions
wind frigate
explain what
Sent!
wind frigate
Could someone explain how to solve these typa questions
.
wind frigate
safe radishBOT
wind frigate
Idk how to complete question 3,
But here is whats wanted
Determine if f is continuous . If possible, extend f to a new function that is continuous on a larger domain
abstract jacinth
can some help me with radical in maths im in the 8 grde but pls in call
wind frigate
.close
safe radishBOT
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bitter shadow
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bitter shadow
can i have an example
obsidian ginkgo
I mean any function involving the absolute function works
kindred gust
Range is -4,inf domain is R
obsidian ginkgo
$f(x) = |x + 1| - 4$
flat frigateBOT
Adarsh
obsidian ginkgo
Do you need help finding the range and domain?
bitter shadow
yeah
obsidian ginkgo
Do you agree the minimum value of the absolute function is 0?
bitter shadow
yes
balmy goblet
the range
obsidian ginkgo
And the maximum value is +inf?
bitter shadow
yup
obsidian ginkgo
So what would the maximum and minimum value of $|x| + 1$ be?
flat frigateBOT
Adarsh
bitter shadow
max 1 min inf?
obsidian ginkgo
Inf is not less than 1
bitter shadow
oh nvm other away around
bitter shadow
is that all
i think i get it now
but say for the original example i picked
is the answer like
domain (-inf,inf) range is [-4.inf)
obsidian ginkgo
Yup
safe radishBOT
@bitter shadow Has your question been resolved?
unique canopy
how to express the integral of ln(u) over u^n du
safe radishBOT
how to express the integral of ln(u) over u^n du
unique canopy
$\int{\frac{ln(t)}{t^n}dt}$
by using substitution
i've been told we should use v=ln(u)
digital sparrow
Let t = u 😂
unique canopy
Let t = u 😂
sure wait a sec
flat frigateBOT
Osmium
unique canopy
there you go
digital sparrow
I was jk
unique canopy
okay but how to solve it
wraith swift
just use by parts
wraith swift
take the function to differentiate as the log
unique canopy
oh ok
so i determine 1/u^n's integral
alright wait a sec
alright so it goes like
$\int{\frac{ln(u)}{u^n}du}=[\frac{u^{1-n}}{1-n}\timesln(u)]-\int{\frac{\frac{u^{1-n}}{1-n}}{u}du$
flat frigateBOT
Osmium
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
wraith swift
yea
1-n constant take it out from the right integrale
u^1-n/u is simply u^-n
which is 1/u^n
unique canopy
$\int{\frac{ln(u)}{u^n}du}=[\frac{u^{1-n}}{1-n}\times\ln(u)]-\frac{1}{1-n}\int{{\frac{u^{1-n}}{u}du$
wraith swift
nah u made a mistake first simply the fraction
flat frigateBOT
Osmium
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
unique canopy
like that?
unique canopy
💀 my bad
$\int{\frac{ln(u)}{u^n}du}=\left[\frac{u^{1-n}}{1-n}\times\ln(u)\right]-\frac{1}{1-n}\int{u^{-n}du}$
like that?
wraith swift
nahhh
u simplified the u wrong
it would simplify to 1/u^n
thats the most u can do
that should be the ans here
unique canopy
oh yeah i might be dumb
wraith swift
that is the ans yes
flat frigateBOT
Osmium
unique canopy
wat
oh wait
$\int{\frac{ln(u)}{u^n}du}=\left[\frac{u^{1-n}}{1-n}\times\ln(u)\right]-\frac{1}{1-n}\times\frac{u^{1-n}}{1-n}$
flat frigateBOT
Osmium
unique canopy
$\int{\frac{ln(u)}{u^n}du}=\left[\frac{u^{1-n}}{1-n}\times\ln(u)\right]-\frac{u^{1-n}}{(1-n)^2}$
flat frigateBOT
Osmium
unique canopy
what then?
$\int{\frac{ln(u)}{u^n}du}=\left[\frac{u^{1-n}\times\ln(u)}{1-n}\right]-\frac{u^{1-n}}{(1-n)^2}$
flat frigateBOT
Osmium
unique canopy
$\int{\frac{ln(u)}{u^n}du}=\frac{u^{1-n}\times\ln(u)}{1-n}-\frac{u^{1-n}}{(1-n)^2}$
flat frigateBOT
Osmium
unique canopy
$\int{\frac{ln(u)}{u^n}du}=\frac{u^{1-n}\times\ln(u)(1-n)}{(1-n)^2}-\frac{u^{1-n}}{(1-n)^2}$
flat frigateBOT
Osmium
odd crest
put n=1 to see if ur answer matches
odd crest
no it doesnt
unique canopy
thanks btw
odd crest
ur integral answer isnt even defined at n =1
unique canopy
$\int{\frac{ln(u)}{u^n}du}=-\frac{u^{1-n}\left(\ln(u)(n-1)+1\right)}{(n-1)^2}$
flat frigateBOT
Osmium
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bitter matrix
So I have a triangle with points (0,0,0) (0,3,0) and (2,1,0) and I have the plane z=2x+7y+8
safe radishBOT
So I have a triangle with points (0,0,0) (0,3,0) and (2,1,0) and I have the plan...
bitter matrix
And by extending the triangle in the z direction I can make a prism that is cut off by the plane
Now how do I find this area of the prism using volume integration?
Cause from what I know I need a vector function or a regular function to use the volume integral
So imagine this but as a volume
safe radishBOT
@bitter matrix Has your question been resolved?
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@bitter matrix Has your question been resolved?
safe radishBOT
@bitter matrix Has your question been resolved?
safe radishBOT
@bitter matrix Has your question been resolved?
lean otter
Have a question
lean otter
plucky elk
if you're done in your other channel, please close it
neat cypress
use ".close"
lean otter
K i closed it
So i need help how do u cross in math
The equation supposed to be crossed
@plucky elk @plucky elk @neat cypress
safe radishBOT
@lean otter Has your question been resolved?
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@lean otter Has your question been resolved?
true jolt
How can I speed up my solving time?
safe radishBOT
How can I speed up my solving time?
thin bridge
practice
true jolt
Can I get a more concrete answer?
Because yeah, sometimes I practice but then sometimes I do mistakes when doing the exercise fast.
thin bridge
do them slow / carefully / right first
worry about speed later
speed comes naturally the more you do it
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inner spear
hi i just need help cuz google won’t answer my question and it’s too late for me to think
no homework just wondering what X is
4.25% of X = 700
safe radishBOT
hi i just need help cuz google won’t answer my question and it’s too late for me...
split kayak
safe radishBOT
@inner spear Has your question been resolved?
mystic vale
if i have a function f(x,y) and x and y depends on t. the function f depends on one variable only, so it's graph will be with the t at the bottom, as independent variable, correct? if i were to graph the function in 3d, it will be a plane in z that shifts based on t.
is this correct?
safe radishBOT
if i have a function f(x,y) and x and y depends on t. the function f depends on ...
mystic vale
like f(x,y) = x^2+y^2
x = t
y = t^2
then f=t^2+t^4
gloomy scaffold
if i have a function f(x,y) and x and y depends on t. the function f depends on ...
Why do you want to plot this in 3D 
mystic vale
in another problem they were asking to find contour lines. i am confused
median vigil
please send the full context
mystic vale
gloomy scaffold
What is a contour line?
mystic vale
level couves sorry
median vigil
so you should draw the level curves on the xy plane
each curve is given by setting T = constant
so for example the level curve for T = 0 would be the curve given by 0 = 100e^-(x^2 + y^2)
you pick a few different values for t, and label each curve by the value of T it corresponds to
mystic vale
if i plug the x and y in the equation, i end up with 100e^-t^2
can i do
0 = 100e^-t^2 and try different t values?
median vigil
you should not plug in x and y to that equation
mystic vale
this is what i don't understand
this is the same as f(g(h)) right?
oh. i am plotting just f in this case, regardless of what inputs are
T is not the combination of 100e^-(x^2 + y^2) and the other functions
median vigil
r(t) and T are unrelated here
mystic vale
it's just 100e^-(x^2 + y^2)
ok then it makes sense ok ok
jeeeeesus
thanks!!
.close
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half shadow
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sleek sentinel
can you do the first bit? Draw AB with |AB| = 7cm?
can you determine its midpoint?
after that just draw another line CD where |CD|= 1/2 |AB|
wherever i guess
its just an unrelated line
uhh not mm
just .5
yeah that's kina the entire question
just drawing two lines
safe radishBOT
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dense sage
yall how do i do this
safe radishBOT
yall how do i do this
dense sage
pls tag me when someone gets to this
left gyro
yall how do i do this
do you remember chain rule?
dense sage
do i differentiate it?
left gyro
you need h'(x) so yes
dense sage
does diffing it make it h'(x) = g(x) f(g(x))
left gyro
youre forgetting the 's
left gyro
yea and h'(-k) = f'(g(-k)) g'(-k)
dense sage
oh oka
left gyro
now what you do here is simplify f'(g(-k)) g'(-k)
since h'(k) = m, that means f'(g(k)) g'(k) would be m too
dense sage
mhm
left gyro
now since g is even, what would f'(g(-k)) simplify to
left gyro
...it would simplify to f'(g(k))
dense sage
i dont know :c
dense sage
ohhh oka
left gyro
yea that is what simplifying means
dense sage
mhm
left gyro
in general derivatives dont share much in common with the original function beyond broad strokes
dense sage
mhm oka
left gyro
if g is even, then the negative side is a mirror copy of the positive side
dense sage
oka
left gyro
that also means positive slopes get mirrored to be negative slopes
and negative slopes get mirrored to be positive slopes
so if g is even, then g' is odd
left gyro
I didnt see it coming either
dense sage
ASDIHASDKJH
left gyro
I think I had to manually think through it to find that out
left gyro
we're simplifying f'(g(-k)) g'(-k)
the left term got simplified, so f'(g(k)) g'(-k)
whats left is g'(-k), so we gotta know if g' is even or odd or not
dense sage
why is the diff of the even g(x) odd
dense sage
we're simplifying f'(g(-k)) g'(-k)
the left term got simplified, so f'(g(k)) g'(...
mhm
wait wha
sobb
dense sage
um
left gyro
do you remember what we were talking about
dense sage
nah
left gyro
we need the equation of the tangent to y = h(x) at x = -k
dense sage
let me pull up the q again
oka so finding eq of tangent -> find m of tangent diff equation
h'(x) = g'(x) f'(g(x))
then h'(k) and h'(-k)
left gyro
its also given that h(x)'s tangent line at x=k is y = mx + c
left gyro
that would mean h'(k) = m
dense sage
yessir
dense sage
so g'(k) f'(g(k)) = m too
yes that makes sense
left gyro
yea
left gyro
with g'(-k) f'(g(-k))
dense sage
ya
left gyro
also, since g(x) is an even function,
the left side is a mirror copy of the right side
left gyro
thats not used here
g(k) = g(-k)
so f(g(k)) = f(g(-k))
g(k) and g(-k) are the same value
theyre equal
left gyro
the - sign disappeared
dense sage
so f(g(k)) = f(g(-k))
oh oka
left gyro
also f'(g(k)) = f'(g(-k))
dense sage
wait what doesnt that mean
left gyro
since g(k) and g(-k) are still the same number
dense sage
h'(k) = h'(-k) then
dense sage
wait g' is odd because g is even
left gyro
yes
left gyro
yes
dense sage
h'(k) = g'(k) f'(g(k))= m
h'(-k) = g'(-k) f'(g(-k))
g'(-k) odd so = -g'(k)
f'(g(k)) even so = f'(g(-k))
h'(k) = -g'(k) f'(g(k)) = -m
is that right
left gyro
yes it is
dense sage
MWAH
left gyro
consider if h(x) is even or odd
left gyro
yea
left gyro
you gotta at least wait for a response before you collapse
dense sage
wha
left gyro
at x=k theres a tangent line on h(x)
now since h(x) is even, this tangent line would be mirrored when at x=-k
dense sage
now since h(x) is even, this tangent line would be mirrored when at x=-k
does this not apply if h(x) is odd?
left gyro
if h(x) is odd, the tangent line would be rotated to the other side when at x=-k
left gyro
do you need a refresher on odd and even functions
dense sage
yessir
left gyro
why didnt you ask for one
dense sage
i didnt know it to this extent
my syllabus only defines odd and even functions as odd -f(x) = f(-x) and even as f(x) = f(-x)
left gyro
oh you didnt see the visualizations
left gyro
oh also uh the whole thing we did about f(g(x)) is still necessary, but visualizing the even h(x) is a shortcut
dense sage
oh okaa
left gyro
we'll have to prove it
dense sage
okaa
left gyro
algebra says a tangent line for h(x) at x=k is y = h'(k) (x - k) + h(k)
left gyro
is the tangent line equation not in the syllabus either
left gyro
thats point-slope form, not a tangent line
left gyro
thats you using point-slope form to find the tangent line
dense sage
they give me a point then i diff the original eq its tangenting off of
left gyro
the tangent line equation skips the initial "using point-slope form" part