if it's real it would be much more convenient for me rlly
#help-38
wraith hinge
@amber python come here be the middleman and translate the chinese to english
okay sorry ill continue writing my solution
give me a bit
amber python
wraith hinge
,align
W_{\t{tot.}} &=W_{sp.} + W_f + W_w = \Delta KE\
W_{\t{tot.}} &= W_f - \Delta U_{el} - \Delta U_w = 0 \
W_{\t{tot.}} &= -\mu_k w \m\cos\theta - \f12 kx^2_{\t{comp.}} - mg(y_2 -y_1) = 0
solid kilnBOT
wraith hinge
(i am putting my origin where the spring is uncompressed from now)
@kindred rapids do you agree with the above for now?
kindred rapids
basically yes, i just dont know what y2 and y1 is
wraith hinge
y_1 = height of the box at the initial point (before it got released)
y_2 = height of the box at the final point (maximum compression)
kindred rapids
yes
wraith hinge
okay time to figure those out one second
actually hm
how do i do this actually
how do i recover y_2 and y_1
kindred rapids
i usually saw 53.1 degrees when the question says sin53.1 ≈ 0.8 tho😂
kindred rapids
how do i recover y_2 and y_1
hmmmm, my suggestion is better not bring it into calculation
or more specifically, replace y2-y1 as something related with x_comp
kindred rapids
kindred rapids
y_1 = height of the box at the initial point (before it got released)
y_2 = heig...
as u just said
y1 obviously bigger than y2
wraith hinge
yeah
wraith hinge
ahhhhh okay i understand now ! this becomes clearer now
lemme try
so lets see
our equation becomes
,align
W_{\t{tot.}} &=W_{sp.} + W_f + W_w = \Delta KE\
W_{\t{tot.}} &= W_f - \Delta U_{el} - \Delta U_w = 0 \
W_{\t{tot.}} &= -\mu_k w \m\cos\theta - \f12 kx^2_{\t{comp.}} - mg(y_2 -y_1) = 0 \
W_{\t{tot.}} &= -\mu_k w \m\cos\theta - \f12 kx^2_{\t{comp.}} + mg\2\m\sin\theta\2(x_c+D) = 0
solid kilnBOT
wraith hinge
fingers crossed i didnt mess up something lol
rearranging,
[
W_{\t{tot.}} = -\f12kx^2_c + mgsin(\theta)x_c+ mgsin(\theta)D -\mu_k w\m\cos\theta = 0
]
solid kilnBOT
kindred rapids
kindred rapids
but isnt that contained in w
what?
solid kilnBOT
wraith hinge
oh lmoao
ok lets see
,w -1/2 1.2010^2 x^2 + 29.81sin(53.1 degrees)x + 29.81sin(53.1 degrees)4 -0.229.81*cos(53.1 degrees) = 0
there is something slightly off
kindred rapids
yes...
wraith hinge
huh should it now
solid kilnBOT
wraith hinge
oh wdym
wraith hinge
ah i see i forgot to multiply by the displacement
[
W_{\t{tot.}} = -\f12kx^2_c + mg\sin(\theta)x_c+ mg\sin(\theta)D -\mu_k w\m\cos\theta(x_c + D) = 0
]
?maybe this then
kindred rapids
yes
solid kilnBOT
wraith hinge
,w -1/2 1.2010^2 x^2 + 29.81sin(53.1 degrees)x + 29.81sin(53.1 degrees)4 -0.229.81*cos(53.1 degrees)(x +4) = 0
solid kilnBOT
wraith hinge
AYEEE WE GOT IT :D
wraith hinge
because u split x_comp and D out
😭
okay i can think i can work out the rest on my own probably
but i wanted to ask you two questions if u may
kindred rapids
just ask
wraith hinge
just ask
1- is the total mechanical energy ever conserved during all of this?
2- what differs exactly if we had placed our origin somewhere else
kindred rapids
1- no, because friction can create heat
2- if D is small enough, it wouldn't rebound out of the spring
kindred rapids
if we placed x = 0, at say, the inital point
x = 0?
kindred rapids
u mean u just created a coordinate system?
wraith hinge
i think my confusion is a bit weird so its okay
i think otherwise its fine
thank you again !
i will close this
wraith hinge
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lucid perch
me
wraith hinge
part 2 please
what?
steep bridge
!status
wraith hinge
what's the context
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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
in the pic
jagged wharf
I'm trying to find the Laguerre series of $e^{x/3}$, i.e. determine the coefficients $c_n$ in the formula $$e^{x/3}\sim\sum_{n=0}^\infty c_nL_n(x),\quad x\geq0.$$
jagged wharf
But how can I proceed with this last integral?
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subtle lava
Second thought. That derivative should have a general form
Find the derivative for different value of n
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main flower
it didnt show up for me either intially, might have to open another help channel
sick onyx
Yessir
royal orbit
fr?
royal orbit
yes
royal orbit
It’s reverse
sick onyx
yeah
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pale summit
can someone help me
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can someone help me
pale summit
I know that f(x) is decreasing thus it explains why f(n+1) <= f(n) but I dont seem to understand the intuition for why integral of f(x)dx for x= n and x=n+1 is bonded by f(n+1) and f(n)
question 8)c)
can some one help me figure the logic behind it
kindred pier
!15m
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west basin
Let $A = M_2(\mathbb{Z})$ be the set of $2 \times 2$ square matrices with integer coefficients. $A$ is a non-commutative subring of the ring $M_2(\mathbb{R})$. Show that its invertible elements are the matrices in $A$ whose determinant is $1$ or $-1$.
solid kilnBOT
lilisworld.
west basin
can someone xplain me why is it like this? Isn't it always true when det is different from 0?
i missed something
whole coral
The matrix may be invertible as an element of M2(R), but the inverse may not have integer coefficients entries
west basin
The matrix may be invertible as an element of M2(R), but the inverse may not hav...
oh yes thank you
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fierce hatch
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fierce hatch
how on earth
does tis equal this
following a^2 + 2ab + b^2
this
you get 2 - 1 - 2h + h^2
aka 1 - 2h + h^2
split chasm
missing ()
you should be subtracting the whole of the expansion,
not just what term happens to appear firat
$2 -\red{(} 1 - 2h + h^2\red{)}$
solid kilnBOT
ℝαμΩℕωⅤ
fierce hatch
oh i see
math is weird
anyways thanks for helpin, sorry it was just a simple algebra error on my end
take care
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glossy geyser
I'm having a tough time understanding this problem.:
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I'm having a tough time understanding this problem.:
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nova spire
It's not the probability that you get to 2 on the next step that is asked, It's the probability that one day, eventually, you get to 2
(Starting from 1)
glossy geyser
That makes sense, but I know how how to calculate that. It's not said explicitly in my book
glossy geyser
<@&286206848099549185>
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glossy geyser
For (a), p_{1,2} = 1, because the only way to the recurrent state 3 is through state 2.
Now for part (b), I'm still confused on what to do
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<@&286206848099549185>
deft sail
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deft sail
I am unsure of how to find x, y, and z.
tribal jetty
are 31z and 25z the whole segment or just the tiny parts
tribal jetty
does this mean angles draw the same way are the same? or no
also for the line drew**8t09awujo0ibfgjoawko6 with a tiny / in between QN, LS
deft sail
yes they are congruent
qn is congruent to ls
adn the angles with the same number of arcs are congruent to each other
deft sail
yes
deft sail
wait waht
tribal jetty
deft sail
oh yes
yes
um so i was gonna do pythagorean theorem but i'm not sure how to do it with letters
like (y^2)+(31z^2)=c^2
tribal jetty
ok
now what
nevermind there are many ways to find those sides
see how far you can go
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edgy obsidian
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edgy obsidian
i missed the first classes cuz i was sick so i have no idea what im doing
proper crest
you prolly need to know the formulars first
pearl tendon
just look at what the first numbers are doing.. find the pattern
proper crest
both arithmetic as well
faint patrol
I see
pearl tendon
@edgy obsidian what is the pattern here:
1, 3, 5, 7, 9, 11, 13...
yes, odd numbers, but what are you doing to each number in terms of arithmetic
see meme's pics
faint patrol
I wouldn’t say crazy high, but it will give a higher number because of the fact it’s multiplying instead of adding in the sequence
proper crest
yeah
edgy obsidian
so how do i solve this exactly
i mutiple whatever 5^11 is with 3?
and what next
or do i use a log function
proper crest
thats it
faint patrol
so how do i solve this exactly
i mutiple whatever 5^11 is with 3?
You do exactly that and you should get your answer
proper crest
its all formulas in this stuff
faint patrol
Mhm!
edgy obsidian
yeah but what about a12
cuz now i have a12 = 146484375
i just divide both sides right
faint patrol
It just indicates what part of the sequence it is, so it is good from there
edgy obsidian
or is that number my asnwer
faint patrol
The number is your final answer
proper crest
its a subscript
faint patrol
A12 just means its the 12th number in the sequence is all
edgy obsidian
feels wrong idk why
proper crest
so it just says the number in your pattern
edgy obsidian
but thanks tho
proper crest
feels wrong idk why
well it should be right
edgy obsidian
atleast im getting somwhere
faint patrol
Dont think of the A as a variable if that helps
solid kilnBOT
whole coral
,calc 189/16
solid kilnBOT
Result:
11.8125
and no problem 
proud crypt
whole coral
Well at least you know you're better than the book 
solid kilnBOT
whole coral
They’re really trying to make you doubt yourself huh 
pine mortar
I am so lost
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faint patrol
You’re trying to find one of the angles in a triangle technically, so each of the angles added together would equal 180.
Add angle 1, 2 and 3 together to total up to 180
(3x+1) + (7x-11) + 90 = 180
Solve from there
faint patrol
Alright, you too
pine mortar
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brave raven
I have this problem.
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I have this problem.
brave raven
!status
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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
I found out what a1 equals to just like in this solution
But I can't figure out how did they manage to make (1-cosx-sqrt(cos2x))/x^2 that way and just leave the 1 aside
Because I assume this is what makes the problem harder
I am referring to the last line, by the way.
That a1 just teleported its way there, haha.
wintry stag
I have no idea what you are asking because nowhere is there a 1-cosx-sqrt(2x).
brave raven
Ohh my bad.
Sorry...this problem really made my head spin, I meant 1-cosx-sqrt(cos2x)
wintry stag
...
brave raven
Sorry...this problem really made my head spin, I meant 1-cosx-sqrt(cos2x)
Yeah, I meant this thing for a_2 by the way.
vernal mica
Sorry...this problem really made my head spin, I meant 1-cosx-sqrt(cos2x)
$1 - cos(x)[1 + \sqrt{cos(2x)} - 1] = 1 - cos(x) + \sqrt{cos(2x)} - 1 + cos(x) = \sqrt{cos(2x)}$
solid kilnBOT
Bishop
brave raven
So we added and subtracted 1 and multiplied it term by term?
Ohhh
No, no, I forgot the PEMDAS
1 - (cos x)(1+sqrt(cos2x)-1 basically
vernal mica
$lim(x + y) = lim(x) + lim(y)$
solid kilnBOT
Bishop
brave raven
Hmm, I start to see a bit of it...
solid kilnBOT
Bishop
brave raven
Okay, but how did we get here? I see cosx multipling with whatever is in the other paratheses
But how did we get 4 terms out of 3 multiplications?
vernal mica
what
brave raven
I don't seem to understand some things regarding the multiplication above which is a limit sumation
solid kilnBOT
Bishop
brave raven
I mean, yes
brave raven
I can try and understand handwriting, but I would get 1 - cosx - cosx(sqrt(cos2x)) + cosx and cosines would cancel, that's what I saw
Ohhh, 1 - cosx is grouped?
vernal mica
yes in paranthesis
i'm just a little bit diseased so my organism is veryyyyy bad with dealing with computations and i do not think about stuff like that at all.
brave raven
I'm shocked we can even do that, I thought that 1 stayed here and I'd have -cosx multiplied with the members on the paratheses
vernal mica
wait we can't?
brave raven
Uhh...
vernal mica
oh nah
brave raven
Hm?
vernal mica
brave raven
Mhm...I did note that, now...we just divide this limit into 2?
vernal mica
what
brave raven
I got into the steps you've shown, now what?
Like, (1-cosx)/x^2 *sqrt(cos2x) ?
The first one would be a1 it self, and the second is left to go
vernal mica
$\lim_{x \to a} (f + g) = \lim_{x \to a}(f) + \lim_{x \to a}(g)$ , therefore we wish that
$$\lim_{x \to 0} \frac{1 - cos(x)[1 + \sqrt{cos(2x)} - 1]}{x^2} = a_1 + \lim_{x \to 0} ... = \lim_{x \to 0} \frac{1 - cos(x)}{x^2} + \lim_{x \to 0} ...$$ which
$$ = \lim_{x \to 0} \frac{1 - cos(x)}{x^2} + ...$$
solid kilnBOT
Bishop
vernal mica
Therefore if we choose $... = \lim_{x \to 0} \frac{-cos(x)[\sqrt{cos(2x)} - 1]}{x^2}$ we realize that:
$$\lim_{x \to 0} \frac{1 - cos(x) - cos(x)[\sqrt{cos(2x)} - 1]}{x^2}$$
solid kilnBOT
Bishop
vernal mica
$\lim_{x \to 0} \frac{1 - cos(x) - cos(x)[\sqrt{cos(2x)} - 1]}{x^2} = \lim_{x \to 0} \frac{1 - cos(x) - cos(x)\sqrt{cos(2x)} + cos(x)}{x^2} = \lim_{x \to 0} \frac{1 - cos(x) \sqrt{cos(2x)}}{x^2}$
solid kilnBOT
Bishop
brave raven
I understand the idea, make what you wrote last there be a limit sum, or a1 + another limit
I did that and I got what you got as well
Now, how do we make this 1 - cosx(sqrt(cos2x)) be a1 + another limit?
vernal mica
that explains how they got a_1 + lim...
vernal mica
https://cdn.discordapp.com/attachments/839950625269088276/1205552956682928188/42...
you do this
so you realize that a_2 equals to THAT in the middle
vernal mica
**Bishop**
next you go back all the way to the first step and you write the limit as a _1 + lim(x --> 0) [ something that you dont know yet]
you use limit property to show that you will need something (...) such that $\frac{1 - cos(x)}{x^2} + (...) = \frac{1 - cos(x)[1 + \sqrt{cos(2x)} - 1]}{x^2}$
solid kilnBOT
Bishop
vernal mica
$\lim_{x \to a} (f + g) = \lim_{x \to a}(f) + \lim_{x \to a}(g)$ , therefore we ...
just as i did here
vernal mica
**Bishop**
next you choose $(...) = \frac{-cos(x)[ \sqrt{cos(2x)} - 1]}{x^2}$ which satisfies the property
solid kilnBOT
Bishop
vernal mica
you then do some simple algebra to show that adding these two fractions results in $\frac{1 - cos(x) \sqrt{cos(2x)}}{x^2}$
solid kilnBOT
Bishop
vernal mica
which is what you got all the way in the beginning from manipulating $\frac{1 - cos(x)[1 + \sqrt{cos(2x)} - 1]}{x^2}$ and thus you proved the equality and thus know everything about all the steps now
solid kilnBOT
Bishop
brave raven
**Bishop**
Well, how do I choose this?
brave raven
**Bishop**
I just...subtract the RHS with what we know in the LHS?
I've got this far, actually
Trying to find that (...) and justify my answer
Seems wrong because I do have some trouble at the end
vernal mica
Well, how do I *choose* this?
you don't need to answer that question
It's trivial
Regardless of whenever they gave you the answer or not its assumed to be trivial due to algebra
brave raven
I see.
brave raven
Well, I found it with algebra (correct photo)
And now we just plug it in and say a2 = a1 + that monstrosity
I can see they made some conjugate multiplication
Did they multiply only the negative at the cos to get cosx(1-sqrt(cos2x)) ?
vernal mica
what
brave raven
Asking about the next steps
vernal mica
Did they multiply only the negative at the cos to get cosx(1-sqrt(cos2x)) ?
where
i don't see it
i gtg
brave raven
Okay
But I meant this
How did they actually turn the numerator this way
I understand the conjugation
But how did the cos x dissapear?
Ohhhhhh
cos 0 is 1
Yeah!
Yup, found about the rest.
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wraith hinge
(Asking again to elaborate on a different method of doing this)
Hello, so for a), I will first establish my coordinate system. I made the positive x axis be along the incline towards the sky, with the positive y axis pointing perpendicularly to the sky as well. I made the origin be the initial point of release.
So setting up the total work formula of
W_tot = W_f + W_sp. + W_w = ∆KE
Gets me the following after doing the necessary substitutions:
wraith hinge
,w 1/21.2010^2x^2+0.29.812cos(53.1 degrees)(-x-4)+9.81*2sin(53.1 degrees)(x+4) = 0
solid kilnBOT
wraith hinge
this is almost right, albeit for the sign of the first term representing the work of the spring
it in fact should be:
,w -1/21.2010^2x^2+0.29.812cos(53.1 degrees)(-x-4)+9.81*2sin(53.1 degrees)(x+4) = 0
solid kilnBOT
wraith hinge
with the positive solution being the correct answer
My question now being: I tried to reason out why that negative sign is there geometrically but i can't quite retrieve that?
tepid rock
Which negative sign?
ionic pendant
the difference is the sign of the work term for the spring, right? the spring is resisting being compressed, so it slows the block down
wraith hinge
my integral was: [
W_{sp.} = \int_{D}^{D + x_c} k(x-D) \dd x
]
solid kilnBOT
wraith hinge
which once evaluated, gets you 1/2 kx^2_c
wraith hinge
Which negative sign?
that of the first term in the WA input. 1/2kx^2.
tepid rock
That's because work is F.dx
The direction of F and the displacement are opposite.
Dot product.
wraith hinge
The direction of F and the displacement are opposite.
oh hmmm i guess you are right so in reality it should have been
[
W_{sp.} = \int_{x_1}^{x_2} \vec{\vb* F} \vd \dd \vec{\vb* s}
]
solid kilnBOT
wraith hinge
okay so it pretty much translates to like
,, W_{sp.} = \int_{x_1}^{x_2} \vec{\vb* F} \vd \dd \vec{\vb* s} =- \int_{x_1}^{x_2} F \dd s
solid kilnBOT
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rugged sundial
linear algebra, i just need a run down on what to do here
zinc ginkgo
Quite confusing indeed
proper kernel
it is what it is
sleek canopy
are you a troll?
rugged sundial
is that a serious question
zinc ginkgo
linear algebra, i just need a run down on what to do here
Should just pretend the last equation isn't there and then find the matrix representation of T and see if the matrix is invertible
sleek canopy
is that a serious question
go away if you want some easy trolling
rugged sundial
you need to chill... i miswrote 2 numbers
rugged sundial
Should just pretend the last equation isn't there and then find the matrix repre...
isn't there also something like the rank determines if it's either injective or surjective?
can i prove it's bijective if it's surjective and injective?
zinc ginkgo
isn't there also something like the rank determines if it's either injective or ...
Surjective and injective is the definition of bijective
rugged sundial
yeah
ok so what i got was that L_A is not bijective, but is injective and not surjective because the rank of the standard matrix A is 2 ?
raw magnet
are you a troll?
dude calm down literally just made a mistake
we all make mistakes
this seems like a legit q
sleek canopy
sure my bad
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rugged sundial
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molten vale
<@&286206848099549185>
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molten vale
kindred pier
!15m
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kindred pier
!noans
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kindred pier
@molten vale I'm willing to help. You're not willing to try. Until you accept that, nothing will change.
kindred pier
Tell me how you got your answer and I will verify it for you
split kettle
Crazy
kindred pier
But I am not going to tell you what the answers are
molten vale
i did sin 55 x 30
kindred pier
for question 1?
molten vale
yes
kindred pier
Okay. where did 55 and 30 come from? I don't even see them on the triangle.
molten vale
i dont know how
kindred pier
$\frac{a}{\sin{A}}=\frac{b}{\sin{B}}$
solid kilnBOT
SWR
kindred pier
Looks good to me
kindred pier
40/ sin 110 =sin -1 (x)
Not quite. It's law of sines still
molten vale
ohhh
kindred pier
It seems like all the problems are using law of sines
kindred pier
16, not 15, but otherwise correct
kindred pier
This is known as an ambiguous case. It could be either 56 or 124, because sin(56)=sin(124). But you're given that the angle must be obtuse, so choose the obtuse angle
btw, the sines are equal because 180-124=56 and sin(x)=sin(180-x)
kindred pier
yup
molten vale
okok thanks
last one i did
sorry 4*
i did
same thing as other and got 38
right?
kindred pier
38 looks good
kindred pier
how'd you get 82.74?
molten vale
i did
i dont know i did whatever
its messy
let me restart it
so what is the equation for this?
kindred pier
You want to get the total length of all three sides
You know the length of one side, so you need to get the lengths of the other two sides
Each length can be found using law of sines still
kindred pier
which sines and which angles?
molten vale
sin24 x 23
then i got 9. something and then i did that x sin 30 and got other side
- something and added together
right?
@kindred pier
u there
<@&286206848099549185>
@kindred pier plss
kindred pier
sin24 x 23
You're forgetting another term again
molten vale
i am?
kindred pier
same question, show me how you got it
kindred pier
same thing but i added 126 sin x 4
126 sin x 4?
molten vale
yes?
kindred pier
what is sin x 4?
molten vale
sin 126 x 4*
kindred pier
Where'd you get 4 from?
molten vale
4. something and added together
..
oops
its the 4 from earlier
i got
pls is it right
kindred pier
what did you get?
molten vale
83
kindred pier
and how did you get it?
molten vale
i did sin 24 x 23 to get 9 x sin 126
kindred pier
you're multiplying by sin of some angle twice. One of them needs to be divided
molten vale
how do we do it?
kindred pier
**SWR**
You still need to be using this. You're just solving for the length incorrectly. As an example, try solving for a in this equation up here, and see what you did different
molten vale
ohhhhhhhhhh
23 / sin 24
56
then
so
yeah i was right it is 83
right?
pls just tell me if im right or wrong im wasting time now
sorry
kindred pier
23 / sin 24
You're forgetting terms again
molten vale
howwww
kindred pier
23/sin24 is equal to (length) / (sin of opposite angle)
molten vale
that gives me 2
kindred pier
gives you 2 how?
molten vale
okay pls send me the full equation so i know how and ill ask mhs to regenerate another question like that one
cause i need to know how
@kindred pier
plss
kindred pier
**SWR**
This is the full equation. a and b are your lengths and A and B are your angles. If you know at least three of them, you can solve for the fourth one
molten vale
pls tell me the answer with the numbers i will generate a new one dw i just need to know how
kindred pier
If I had this triangle
split kettle
Rip
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SWR
solid kilnBOT
SWR
solid kilnBOT
SWR
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@molten vale Has your question been resolved?
kindred pier
you can still learn the material even tho the assignment ended. It can help you in the future
wraith hinge
Can someone help me understand: Solve $3*\sin(\frac{\pi}{5}x)=2$ for the four smallest positive solutions
wraith hinge
I got a basic idea, but i'm still confused on it
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BlazeStorm81
wraith hinge
I'm mostly confused about the period of the graph, and how it relates to solving for x. Usually I'd subtract 2pi - (answer of the arcsin) but I'm not sure if I can do that here because of the period change
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tribal fractal
what should my u function be here, for u substitution?
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what should my u function be here, for u substitution?
ionic pendant
if you have a function inside another function, that's a good one to try
solid fulcrum
it could yeah
but it really depends
just look for a function who's derivative appears in the integral
tribal fractal
if you have a function inside another function, that's a good one to try
oh so arctan(2t)
tribal fractal
if you have a function inside another function, that's a good one to try
so i got it to
$\int^{arctan(2)}_{arctan(5)}{\frac12 \cos(u)\mathrm{d}u}$
solid kilnBOT
talk_less
tribal fractal
which turns into -(1/2)sin(x) and then i use FTC
ionic pendant
be careful with signs there
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short silo
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short silo
hi can someone help me simplify the top expression to a single trig ratio
i got it to sinx /cos^2 x but idk what to do from there
sharp heart
That's about as simple as you'll get it.
lethal anvil
I don't know if that just has a single expression
can you post the original problem and the prompt?
charred tiger
Maybe sin(x) / (1-sin^2(x))?
lament reef
what a terrible question
charred tiger
what a terrible question
That's what I was thinkin
short silo
Maybe sin(x) / (1-sin^2(x))?
ill try it
did i put it in right ? cause it was also wrong
lethal anvil
I suppose you could flip them...
Try sec^2(x)/csc(x)
charred tiger
Idk how to help u I'm sorry gurl<\3
charred tiger
I suppose you could flip them...
Try sec^2(x)/csc(x)
bahaha
lethal anvil
yeah I'm sorry I'm not sure what they're smoking here
short silo
I suppose you could flip them...
Try sec^2(x)/csc(x)
nope 😭😭😭
ok i have one more question
lethal anvil
yeah then I'm sorry 😦
I have no idea what they want. Maybe screenshot each attempt and send it to your teacher and see if it's an error on their end
lethal anvil
yeah it's just like the other one lmao
short silo
yeah then I'm sorry 😦
I have no idea what they want. Maybe screenshot each at...
yeah no problem !
lethal anvil
similarly try cot(b)/sin(b)
charred tiger
r they just trying to break ur ego or something damn this is evil 😭😭
digital bolt
tan/cos?
short silo
yeah this website has a lot of wrong answers for questions idk why my teacher still uses it
lethal anvil
I have literally no idea what it wants for an answer then lol
charred tiger
yeah this website has a lot of wrong answers for questions idk why my teacher st...
I'd complain
short silo
OK i have one more question
which i think has an answer to it i just don't know how to get it
lament reef
which i think has an answer to it i just don't know how to get it
hint: start by multiplying the fraction on the left by cos on the numerator and denominator
short silo
should i switch the csc and sec into cosine and sine terms ?
short silo
for the numerator would it be like cosx / sinx ?
lament reef
I don't like these problems a lot bc you need to be incredibly familiar with trig to know how to start them, otherwise it's a whole mess
but it is what it is
short silo
yeaaah me too 😭
lament reef
for the numerator would it be like cosx / sinx ?
of which fraction?
that's cotx if that's what you're asking
short silo
the left one numerator
digital bolt
I don't like these problems a lot bc you need to be incredibly familiar with tri...
eh
short silo
i did 1/sinx multiplied by cosx/1
digital bolt
if all else fails you can just turn everything into sin and cos and then simplify a ton
short silo
so isn't cscx times cos x = cosx/sinx
lament reef
yes
lament reef
if you're just talking about the left one
short silo
yes
short silo
right
lament reef
so what happens when you multiply that WHOLE THING by cosx?
keep in mind the only thing that's a fraction is 1/cosx
that 1 + is not a fraction
lament reef
cosx(1 + 1/cosx)
short silo
right
short silo
i did that so isn't 1/cosx times cosx/1 just one ?
lament reef
it is, but what happened to the 1 that's added in the parentheses?
short silo
OH cosx +1?
lament reef
there you gooooo
lament reef
no, just the left
digital bolt
it should be -, not +
short silo
oh right
short silo
it should be -, not +
so wouldn't the numerator just cancel out
short silo
so i'm left with 1+cosx?
short silo
oooh
digital bolt
like 1/2 - 1/2 = (1-1)/2 = 0/2 = 0
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sudden stream
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sudden stream
,rotate 270
solid kilnBOT
sudden stream
So this is just concerning my u substitution
Idk why I forgot what to do when the problem gets to this situation
Because wouldn't -(1/32)x du = dx be bad?
Because I couldnt evaluate that
blissful bison
sudden stream
Ohhhhhhh
is that because
16x^2 is like saying 4x*4x
So I can u substitute just 1 of those
blissful bison
So I can u substitute just 1 of those
sudden stream
Ohhhh
lament reef
if you're familiar with trig sub, now's the time to use it :)
Unless you're given the formula for that integral already
sudden stream
Yeah that's sec right?
blissful bison
Yeah that's sec right?
there is no need to use trigonometric functions, use what I wrote above
that is the fastest way in your case
sudden stream
oh yeah no I got it from here now that I know the u sub technique I think
working on it rn tho gimme a sec
alright I got it thank you very much!
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lofty stone
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lofty stone
Can someone help with this?
pearl tendon
Where are you lost?
lofty stone
I j dont know what to do for problem 1.1
digital bolt
integrate the absolute value of the velocity
lofty stone
Do i input the t=0 and -8 into 16t+64
pearl tendon
yes
that's your velocity function
you're just going from the indefinite integral to the definite integral
lofty stone
I kinda just got lost what to do with those two numbers once i inputted them
lofty stone
So i use the first equation given of s(t)?
pearl tendon
Do you know how to take a definite integral?
lofty stone
Nope
pearl tendon
So.. where did you get your indefinite integral?
lofty stone
erm do you mean the derivative ?
pearl tendon
ha! I do
lofty stone
Sorry im not familiar with the terms
pearl tendon
you're in a calculus based physics class, no?
lofty stone
I still dont know what to do LMAO so i input the -8 and 0 into the equation and then what do i do with that?
pearl tendon
Ah, ok.. rereading the problem..
Read part 1 again..
you've got an equation that gives you the robot's position at whatever time (t) you plug into it
pearl tendon
No.. you're looking for distance, not speed, right?
lofty stone
Oh so i add the two values goven?
pearl tendon
not quite
you have to find the distance between the two points
so if I gave you two points... let's say 4 and 37
wait
-4 and +37
What's the distance between them?
lofty stone
41
pearl tendon
assume everything is on the x axis
There you go
same idea.. you're going to get two points, just find the distance
lofty stone
Soo when i input -8 into s(t) i get 6 and when i input 0 i get 6 as well
pearl tendon
Walk me through your process for -8
lofty stone
8(-8)^2+64(-8)+6 is 6 right
lofty stone
uh its 0?
lofty stone
It still says the answer is wrong when i put 0 😞
pearl tendon
=/
lofty stone
Idk what it wants
pearl tendon
one sec.. looking at numbers
Ahh.. ok
So.. I assumed the robot's total delta x was in one direction
it's not.. it moves one direction, then reverses and ends up back where it started
so the start and stop points are the same, but because of the way the numbers roll out... you get the idea
lofty stone
Yess so what would the answer be then?
pearl tendon
have you been given any delta x formulas to use with problems like this?
lofty stone
Nope
pearl tendon
Ok.. so stop and look at the equation for a moment...
8x^2, etc....
what does that look like? Just looking at the formula, what shape do you think that would look like on a graph?
pearl tendon
Yep.. do you know how to find the minimum and maximum of a parabola?
lofty stone
not rlly i forgot
pearl tendon
ok.. if I said that the derivative of f(x) gives you the slope, does that sound familiar?
lofty stone
Sort of yeah
pearl tendon
Great! So.. at the minimum or maximum of a parabola, what is the slope?
lofty stone
Idk 😭
lofty stone
2?
lofty stone
Oh oops
pearl tendon
There's no rise.. so it's 0 over... any other point
lofty stone
Idk if this pertains to the question im trying to work out though
pearl tendon
It does, because you recognize 8x^2 as a parabola
And since the start and stop points are the same, we know the position formula localizes the function to the same sort of parabola shape in my example
lofty stone
so i put 0 though and it still says incorrect
pearl tendon
... the derivative gives you the slope
You want a slope of 0
Set the derivative = 0 and solve for x
that will give you a value of t equal to the point where the robot stops and turns around
You want the total distance... going to that halfway point where the robot stops and turns around is half the distance
Remember that the start and stop points are not 0.. they're 6
lofty stone
So i set the v(t) to 0 and i got -4
pearl tendon
Find the distance between s(-8) and s(-4)
or s(-4) and s(0)
s(-8) and s(0) are both 6
so what's the difference between either of those points and the point at s(-4)
pearl tendon
I'm going to assume the program you're in only wants the absolute value
so what's the distance from 6 to -122
lofty stone
128?
lofty stone
So 256
pearl tendon
Try that
pearl tendon
Before you go
lofty stone
Oh shit it asked something else no 😭
pearl tendon
ROFL
lofty stone
could you help me with it too…
pearl tendon
it wants a? thinks
That... should be just the 2nd derivative of your position function
pearl tendon
What's that telling you?
lofty stone
oh my fuck it asked another question
lofty stone
Because thats what v(t) equaled when finding the t
pearl tendon
and what was its velocity at that time?
lofty stone
0?
pearl tendon
Yep
When you come to a red light, you're slowing down.. then you stop... then what happens?
lofty stone
You’re at 0 velocity?
pearl tendon
I mean what happens when the light turns green
pearl tendon
you speed up from velocity = 0 to whatever, right?
lofty stone
Yes
pearl tendon
so if it's asking when the robot is speeding up, and you've already got -4 correct in the problem as it's velocity = 0 position...
then the other side of that interval is...
.. if your interval was -8 to 0
pearl tendon
Try that and then try 0 and see if you can think of why that might be
lofty stone
I tried both n wrong 😭
pearl tendon
If that's the answer it wants, it's looking for the entire interval, not just the 8 second slice it's had you working on