wait
#help-42
spare void
they were saying (-∞, 0] U [2, 5] right
yes
spark stratus
yea then they mixed things up
do you want to see why this would lead to (-∞, 0] U [2, 5] (sort of)
spark stratus
this definition would mean "higher x-values must lead to higher y-values"
now this resolves the "increasing at a point": take a look at f(x) = -3/4 x^4 + 7x^3 - 15x^2
see that in the interval [2, 5], higher x-values do lead to higher y-values
spare void
Ah
I understand it
interesting
well thank you
for confirming my thoughts
i'll go by that definition for the rest of the assignment and hope it works
spark stratus
hope for the best, np
glad sinew
calm coralBOT
glad sinew
I got this as my eigenvector for -1+2i…
what am i doing wrong?
this is the question
pallid halo
how do you know your answer is wrong?
pallid halo
your answer is a scalar multiple of theirs, so they're both valid
glad sinew
hm alright
pallid halo
try multiplying their answer by (5+i)/2 to see why
glad sinew
ok
oh I see
perfect thank you
@upper sparrow hiii
also Im gonna stay in this channel for a bit
im gonna do some more differential equations questions, and then Im planning to do all of this worksheet
I might not finish everything but the idea is i try to😂
the worksheet^
calm coralBOT
@glad sinew Has your question been resolved?
calm coralBOT
@glad sinew Has your question been resolved?
glad sinew
for these half spaces, how do you decide which region to shade?
@upper sparrow
nvm i can just pick a point to test
oh no this is bad
I gotta review more
calm coralBOT
@glad sinew Has your question been resolved?
glad sinew
this is for question 1e)
why is the last two negative
I remembered the formula as this
the last part is n-1
in my image
calm coralBOT
limber kernel
nice gn!
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blazing coyote
Let $c\in \R$.Define $cA = {cA:a \in A}$ . If $c \geq 0$ show $\sup(cA)=. csup(A)$
\
By definition $a \leq \sup(A) ; \forall a \in A$ . Then $ca \leq c\sup(A), \forall a \in A$. Therefore $c\sup(A)$ is an upper bound.
Let the least upper bound be $d$. We then have $ ca\leqd \leq c\sup(A)$. Or $a \leq \frac{d}{c} \leq \sup(A)$. As the least upper bound of $A$ is $\sup(A)$. It follows $\frac{d}{c} = \sup(A)$. Therefore $d = c \sup(A)$ as desired.
potent lotusBOT
ƒ(Why am. I here)=I don't Know
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blazing coyote
**ƒ(Why am. I here)=I don't Know**
Compile Error! Click the <:errors:50426655605...
This is my proof
half sparrow
I think the better way is to deal with the case c=0 first
cA={0} so clearly sup cA=0
blazing coyote
Well, assuming $c \neq 0$ isn't my proof valid?
potent lotusBOT
ƒ(Why am. I here)=I don't Know
half sparrow
then we dont have to worry about dividing by 0
half sparrow
Well, assuming $c \neq 0$ isn't my proof valid?
yeah
bronze adder
Are you sure c is meant to be a real number
blazing coyote
a positive real
blazing coyote
Yeah, will do in my overleaf doc
I've been asked to postulate a similar statement for $sup(cA)$ when $c <0$
potent lotusBOT
ƒ(Why am. I here)=I don't Know
blazing coyote
I suspect $\sup(cA) $ need not necessarily exist. Like consider $(-\infty, 1]$. It has a supermum, but no infimum. Then let c=-1, the interval effectively becomes $[-1, \infty)$, which has no supermum, but now has an infimum
potent lotusBOT
ƒ(Why am. I here)=I don't Know
blazing coyote
Is that right?
thick sinew
Is that right?
yes
in some analysis modules there are extensions for those proofs with R∪{∞,-∞} for the supremum/infimum definition but I'd assume it doesn't apply to yours
calm coralBOT
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formal tulip
Julia has 100 light bulbs whose life times follow independent exponential distributions with mean 5 years. If the bulbs are used one at a time, with a failed bulb being replaced immediately by a new one, what is the probability that there is still a working bulb after 525 hours?
formal tulip
I'm not sure how to begin with this
What does it mean mathematically for a failed lightbulb to be replaced immediately
Define $T_i \sim \text{Exp} , (\lambda = 1/5)$ \
So I just need to find $P \left ( \sum_{k = 1}^{100} T_k \geq 525 \right )$?
potent lotusBOT
jewels!
calm coralBOT
finite oasis
Oh central limit theorem
the sum will follow a gamma distribution
finite oasis
I don't have a neat table for the gamma distrbution
Well what i am trying to say is that you should be thinking of approximating that gamma distribution to a normal one
when n is very large
which it is
formal tulip
what's wrong with just using the central limit theorem?
finite oasis
nothing, im just giving you a different viewpoint
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fluid swallow
Help me with verifying my solution for of the questions (a) (b) and (c) because I am really not good at math and want to get good grade but I absolutely have no clue if I do it right
fluid swallow
This picture is the question
And the 3 pictures are my attempts please point any flaw so that I can get a good grade on it
It’s time for math genius to show their genius! 🥰
<@&286206848099549185>
Guys please help me with my homework 😭😭😭
frozen talon
no
fluid swallow
Is it because of my wording or I have to do a proof for Df/dy?
frozen talon
Is it because of my wording or I have to do a proof for Df/dy?
tbh you dont, but just show it
since f(x,y)=f(y,x)
fluid swallow
Okay I will change that immediately! 🥰🥰 is b and c fine if I may ask😭😭
frozen talon
b) looks sloppy tbh
fluid swallow
Then I shall redo that one 🥰
frozen talon
you just need to show the limit $$\lim_{{(x,y)}\to{(0,0)}} \frac{ \sqrt{(x^2+y^2)}}{\sin\frac{1}{\sqrt{x^2+y^2}}}$$ exists
potent lotusBOT
benadryl
frozen talon
i hope thats correct
fluid swallow
you just need to show the limit $$\lim_{{(x,y)}\to{(0,0)}} \frac{ \sqrt{(x^2+y^2...
Okay I will fix this part! But c is correct right it has most points
You’re really good at this so please help 😭😭
frozen talon
Okay I will fix this part! But c is correct right it has most points
mhm just let $\sqrt{x^2+y^2}=k$ then use the epsilon-delta definition or just do sin(h)/h=1 i guess
potent lotusBOT
benadryl
fluid swallow
mhm just let $\sqrt{x^2+y^2}=k$ then use the epsilon-delta definition or just do...
For c, can’t I do sequential tests?
frozen talon
For c, can’t I do sequential tests?
that was for b
c looks correct
again you might want to do it for y as well, or just say the function is symmetric
instead of saying WLOG
i guess
fluid swallow
that was for b
Perfect! Omg you’re mighty!!!!!
I have been so scared of my homework can’t get good grades
And have been refining it for 4 days
frozen talon
do u get grades for this
fluid swallow
Yes from 1 to 10
And it’s actually reflecting on my final too
If c is perfectly correctly then I will be have at least 5 though
fluid swallow
Like 30%
fluid swallow
This piece is graded all together for another 3 questions, account for 25 points
And we have 4 piece of homeworks which accounts to 100 that is 30% reflected on final unless final exam s grade is strictly higher then the final is only accounted for final exam
frozen talon
And we have 4 piece of homeworks which accounts to 100 that is 30% reflected on ...
oh
do you guys have midsems?
fluid swallow
Yes but this is usually not accounted into grade it’s usually replica final exam for only first 6 chapters
frozen talon
Yes but this is usually not accounted into grade it’s usually replica final exam...
wtf whats the point of the exam then
fluid swallow
Because our final is like hell
The last final for cal 2 is strictly to hell and linear algebra is even worse with a staggering failure rate of 40% and the professors are making the exam excruciatingly hard since student cheats with ChatGPT for good homework grade which cause the higher difficulty
frozen talon
The last final for cal 2 is strictly to hell and linear algebra is even worse wi...
oh
good luck lol
vocal leaf
find the integral of 1/(x^2 + a^2)^1/2 dx using the substitution x =atantheta
remote mural
torn gorge
vocal leaf
uhh
unkempt sphinx
you dont need to find the diff of atantheta?
just plug it in for x and then simplify
$\tan^2(\theta)+1=\sec^2(\theta)$ may be useful
potent lotusBOT
lpieleanu
vocal leaf
oh wait
potent lotusBOT
remote mural
but we have atan $\theta$ over here instead of tan $\theta$ but this still applies becuase a is a constant so you can just bring it outside
potent lotusBOT
remote mural
:D
calm coralBOT
@vocal leaf Has your question been resolved?
blazing coyote
(c)
calm coralBOT
@blazing coyote Has your question been resolved?
blazing coyote
<@&286206848099549185>
ykw, I'll do these (c and d) tomorrow, been stuck on them for 3 hours
.close
calm coralBOT
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pseudo wedge
If $ax^2-bx+c=0$ has two distinct roots lying in $(0,1)$ where $a,b,c\in\bN$ then we need to comment on what range $abc$ can be in
potent lotusBOT
kheerii
pseudo wedge
so the conditions I made were $b^2-4ac>0,\frac{b}{2a}\in (0,1), f(0)>0$ and $f(1)>0$
potent lotusBOT
kheerii
marble pendant
f(0) * f(1) > 0?
pseudo wedge
yes, but a is also positive
pseudo wedge
get some conditions
i wrote the conditions I got
marble pendant
f(0) > 0 is not always true
marble pendant
have you done location of roots?
pseudo wedge
**kheerii**
the first thing gives me $b>2\sqrt{ac}$ the second thing gives me $b<2a$ and $f(1)>0$ gives me $a+c>b$
kind axle
where V is the vertex
potent lotusBOT
kheerii
pseudo wedge
but then I get $a+c>2\sqrt{ac}$
potent lotusBOT
kheerii
pseudo wedge
wait shit
I thought it was the other way around
i goofed
this just tells me a != c
i thought this was giving me a contradiction lol
so the only two relevant conditions I have are $2\sqrt{ac}<b<2a$ and $b<a+c$
potent lotusBOT
kheerii
lyric ravine
You should get a nice bounded region if you graph all the inequalities
pseudo wedge
in 3d?
lyric ravine
Yeah
pseudo wedge
😭
lyric ravine
And then you just need to check points with natural number coordinates
But i suppose thats not what youre looking for
pseudo wedge
i mean i can't really do a 3d plot on paper
I feel like I haven't done anything with the roots
let alpha and beta be the roots
then I have $\alpha+\beta=\frac{b}{a}\in (0,2)$ and $\alpha\beta=\frac{c}{a}\in (0,1)$
potent lotusBOT
kheerii
pseudo wedge
okay so this tells me a>c
which is good
so abc is given by $a^3\alpha\beta(\alpha+\beta)$
potent lotusBOT
kheerii
calm coralBOT
@pseudo wedge Has your question been resolved?
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distant creek
How do I find the vertex?
calm coralBOT
How do I find the vertex?
blazing coyote
do you know calculus ?
distant creek
no
blazing coyote
What about completing the square ?
distant creek
I know how to find (h,k) but for some reason I am having an issue solving it in fraction form
Yes, I sort of know how to complete the square
I usually just use the -b/2a and find x, then solve for y
pseudo socket
blazing coyote
Well, what did you get for -b/2a
distant creek
I did the math and it is correct, sorry I was mistaken
tame forge
,w substitutte x = -3/2 in x^2 + 3x + 4
potent lotusBOT
wintry echo
calm coralBOT
wintry echo
why is that dotted line r2-r1 like this?
torn gorge
you just need t travel first r1 and then -r2
wintry echo
And why is that?
potent lotusBOT
BananabrainJoe
wintry echo
?
wintry echo
But doesn't it overlap first?
atomic quarry
To calculate the difference r₂ - r₁ we do like this r₂ - r₁ = (x₂-x₁ , y\₂ - y₁)
wintry echo
rotate r1 anti clockwise until it superimposes itself on r2
Do you mean until it forms 90° with r2?
atomic quarry
this difference gives you a new vector that point from the position of the charge +q (end of r₁) to the charge -q (end of r₂)
Ok so far?
atomic quarry
To draw it you have to start form the position of +q (head of r1). Draw a line that arrive to the position of -q (head of r2)
the arrow you get is r2 - r1
wintry echo
Can you make a diagram pls
atomic quarry
The direction and the lenght of the vector r2 - r1 show : the distance between +q and -q
and the direction from the +q to -q
winged inlet
Its differerent from the original
Imagine a clock. There's a point in time when second and minute hand overlap
so you've to basically rotate r1 like that so that it overlaps with r2
calm coralBOT
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atomic quarry
@wintry echo Hello ... ?
wintry echo
Hello
atomic quarry
Its clear now ?
wintry echo
No
atomic quarry
<@&286206848099549185>
agile rapids
yes
agile rapids
add what
agile rapids
shut up bruv
wintry echo
Hello
wintry echo
Hi
agile rapids
what do you need for math?
wintry echo
Do you know what a clock is?
Yes
atomic quarry
@agile rapids how its going on roblox
winged inlet
Yes
okay, now do you know that there comes a point in time when second and minute hand overlap? let's assume second hand to be longer than the minute hand we can calculate difference in their length by
length of second hand(say r2)-length of minute hand(say r1)
agile rapids
im a grade 12 student
winged inlet
<@1105483091952275519> right ?
I think so
agile rapids
anyone knows this?
wintry echo
!occupied
calm coralBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
atomic quarry
anyone knows this?
Yes
wintry echo
okay, now do you know that there comes a point in time when second and minute ha...
Ok
agile rapids
@atomic quarry k
agile rapids
who me?.
atomic quarry
!occupied
wintry echo
Have you got your answer now
No
calm coralBOT
Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).
atomic quarry
@agile rapids
atomic quarry
You cant ask for help here
agile rapids
Why
agile rapids
huh
wintry echo
anyone knows this?
What
agile rapids
@moderaters
dull wagon
anyone knows this?
This channel is occupied. If you seek help, follow the instructions to claim your own personal channel. #❓how-to-get-help
ebon pagoda
I HATE YOU MODS
just send a message here with your question it's not that hard
nothing against you
atomic quarry
@wintry echo So its clear ?
atomic quarry
wintry echo
Bye
calm coralBOT
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cobalt escarp
hey. So I have a trainning homework from robotics class. Work is accumulating and also just signed a student job contract in the lab, so time is very short and I want to impress the professor. ( I am a very hands on guy and very bad at doing maths) Can someone help me with it ?
cobalt escarp
I'm completely lost. and this was not very well explained by the professor. When he talks about O to A. Is it form that O y x frame above link 0 ? Why is there no rc ?
nvm no rc because I'm supposed to use l2 and sin and cos right ?
This could also be because I did all of my studies in Portuguese and now doing maths in english has been way more chanlenging than I expected. I thought There would be 0 difference but there is. it is hard to understand what is asked some times.
calm coralBOT
@cobalt escarp Has your question been resolved?
frank shoal
I'm completely lost. and this was not very well explained by the professor. When...
yes, Rc should be a necessary calculation but its not asked in the question i guess..
so when it talks from O to A, yes its talking about the origin, (where the xy plane starts) and then A is the node above where it says "link 0"
for the questions where you find the relative position, of let's say B to A, (im assuming the arm and body rotates at a constant speed), you can express B to A in polar form then convert it to rectangular
actually using polar might make you confuse the variables, so i suggest parameterizing the x and y coordinates instead
calm coralBOT
cobalt escarp
.reopen
calm coralBOT
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cobalt escarp
but where is 0 . is 0 that 0 in the picture ?
frank shoal
but where is 0 . is 0 that 0 in the picture ?
yeah, 0 is O, where the y and x axes are
to the up left of A
cobalt escarp
ok. so for 1 its rd = rb + [ l1 cos (teta 1 ) // l1 sin (teta 1) ] + [l2 cos ( teta 1 + teta2 ) // sin ( teta1 + teta 2 ) ]
?
I thin I got 1 and 2 done. any ide where for 3 ?
frank shoal
ok. so for 1 its rd = rb + [ l1 cos (teta 1 ) // l1 sin (teta 1) ] + [l2 cos ( t...
yeah thats right
frank shoal
I thin I got 1 and 2 done. any ide where for 3 ?
linear velocity is equal to r* d(theta)/dt
so you just have to find d(theta)/dt for each function
calm coralBOT
@cobalt escarp Has your question been resolved?
wise ore
How was the green line for V/U calculated when V/U = {v + U | v element of V}?
From that image it makes sense that dim(V/U) = 1, but I don't see how it makes sense from the definition
lyric ravine
How was the green line for V/U calculated when V/U = {v + U | v element of V}?
F...
well can you find a basis for V/W
hint : ||write R^2 as a direct sum of W and W perp||
calm coralBOT
@wise ore Has your question been resolved?
wise ore
well can you find a basis for V/W
From this definition: V/U = {v + U | v element of V}, I don't know how to find a basis. In my mind if we adding any vector from v to the set U, we should get R2
wise ore
hint : ||write R^2 as a direct sum of W and W perp||
As for this, how do you know for sure it is perpendicular?
lyric ravine
for any subspace W and vector space V you have V = W direct sum W perp
wise ore
for any subspace W and vector space V you have V = W direct sum W perp
From this are you saying that V/U is the "reverse" of a direct sum?
lyric ravine
im basically saying, take a basis of W and extend it to a basis of R^2 orthogonally
wise ore
im basically saying, take a basis of W and extend it to a basis of R^2 orthogona...
I understand that's what you saying, I just don't understand how one sees that from the definition "V/U = {v + U | v element of V}"
lyric ravine
thats because the component of v along U gets "absorbed into" U when you do v + U
the remaining bit is the component perpendicular to U
wise ore
I see why that would be true for spaces that are perpendicular, but it is more the definition that has me confused
lyric ravine
I see why that would be true for spaces that are perpendicular, but it is more t...
what about the definition is confusing
wise ore
what about the definition is confusing
In my mind, the set {v + U | v element of V} is V. Like if you add every vector from V to every vector in U, you get V
lyric ravine
ah okay i see the confusion
V/U is a vector space
the elements of V/U are sets here
so just by that you can see V/U cant be R^2
since R^2 consists of points, and V/U consists of sets
wise ore
Fair enough
wise ore
How was the green line for V/U calculated when V/U = {v + U | v element of V}?
F...
So labeling that line as V/U makes no sense then?
wise ore
I understand that now, but now what confuses me is the formula: dim(V/U) = dim(V) - dim(U)
wise ore
Well if V/U has sets as elements, what does that dimension even mean
lyric ravine
Well if V/U has sets as elements, what does that dimension even mean
V/U is a vector space, as mentioned above (in this case finite dimensional)
if it is a vector space, it has a dimension
the dimension of V/W here = the dimension of the green line (=1)
calm coralBOT
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sonic pawn
The surface of the cardboard has the shape of a rectangle measuring 8 cm and 15 cm. In the four corners of this cardboard, squares of 2.5 cm side were cut out. The remaining part was used to make a box.
Calculate the volume of this box. Record the calculation.
sonic pawn
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calm coralBOT
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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
potent lotusBOT
Alex Waldh
sonic pawn
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west horizon
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calm coralBOT
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west horizon
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potent lotusBOT
Kenzo
west horizon
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potent lotusBOT
Kenzo
west horizon
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potent lotusBOT
Kenzo
potent igloo
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dx = 0.01 
west horizon
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west horizon
dx = 0.01 <:bleakkekw:913344117981999125>
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river pine
if you need an equation solve for y' = ----
west horizon
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river pine
$y' = \frac{dy}{dx}$ and you can just treat it as a fraction and multiply both sides by dx
potent lotusBOT
Adum
river pine
dy = y' dx
west horizon
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potent lotusBOT
Kenzo
west horizon
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potent lotusBOT
Kenzo
west horizon
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hollow perch
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you could've just rearranged for y and differentiated normally...
river pine
factor out the dy/dx, and solve for it
west horizon
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potent lotusBOT
Kenzo
west horizon
you could've just rearranged for y and differentiated normally...
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west horizon
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hollow perch
rearranging?
river pine
y=0, which is implied from the given equation in the question and x=0 which is given
west horizon
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river pine
you include it, i was responding to the other person
west horizon
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river pine
dy/dx can be in terms of y, meaning you can move it to the right side of the equation
you subtract the y first, then divide by (1-x) then multiply by dx in order to follow the order of operations
west horizon
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potent lotusBOT
Kenzo
river pine
$dy = \frac{1-y}{1-x} dx$
potent lotusBOT
Adum
west horizon
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calm coralBOT
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Available help channel!
eternal shard
Let $G, H$ and $K$ be groups.$\$
Show that [ G \cong H \Rightarrow G \times K \cong H \times K. ]
$\textbf{Proof.}\$
Since $G \cong H$ there exists an isomorphism $\Phi : G \to H$ s.t. [ \forall a,b \in G : \Phi(ab) = \Phi(a) \Phi(b).]
Define $\Psi : G \times K \to H \times K$ s.t. $(g,k) \mapsto (h,k) = (\Phi(g),k).\$
Then for $g_1,g_2 \in G \times K$ and $h_1,h_2 \in H \times K$ holds
\begin{align*} \Psi((g_1,h_1)(g_2,h_1)) = \Psi((g_1g_2, h_1h_2)) = (\Phi(g_1g_2),h_1h_2) = (\Phi(g_1)\Phi(g_2),h_1h_2) = (\Phi(g_1),h_1)(\Phi(g_2),h_2) = \Psi(g_1,h_1)\Psi(g_2,h_2). \end{align*}
Therefore $\Psi$ is a homomorphism between $G \times K$ and $H \times K.\$
$\ \Psi$ is also bijective because for some $(h,k) \in H \times K$ exists one $(g,k) \in G \times K$ s.t.
[ \Phi^{-1}(h) = g \text{ and } I(k) = k. ]
The identity and $\Phi$ are bijective therefore there exists exactly one ordered pair.$\$
By that $\Psi$ is an isomorphism between $G \times K$ and $H \times K.\$
potent lotusBOT
𝔸dωn𝓲²s
eternal shard
@glass heart I haven't forgotten it
eternal shard
I thought it was a one liner
Also I was a bit unsure about the bijectivity part
if it was sufficient
glass heart
well ok oneliner was a bit exaggerated
but everything except for the line in the middle is just bookkeeping
I assumed you had maybe proven before that if you combine bijective functions that way then you get a bijective function back
eternal shard
no
glass heart
well not a big deal
calm coralBOT
tiny robin
can anyone help me in my channe
eternal shard
.unsolved
calm coralBOT
✅
eternal shard
well not a big deal
I think I actually realized that it's wrong
big time, I am supposed to show for two pairs a,b of G x K it holds Psi(ab) = Psi(a)Psi(b)
unique jackal
I think all you need to change is g_1, g_2 in G, h_1, h_2 in K
the computation should be the same
eternal shard
It's supposed to be$\$
Then for $(g_1,h_1), (g_2,h_2) \in G \times K$ where $g_1,g_2 \in G$ and $k_1,k_2 \in K$ holds
\begin{align*} \Psi((g_1,k_1)(g_2,k_1)) = \Psi((g_1g_2, k_1k_2)) = (\Phi(g_1g_2),k_1k_2) = (\Phi(g_1)\Phi(g_2),k_1k_2) = (\Phi(g_1),k_1)(\Phi(g_2),k_2) = \Psi(g_1,k_1)\Psi(g_2,k_2). \end{align*}
potent lotusBOT
𝔸dωn𝓲²s
unique jackal
glass heart
whoops didnt even notice. just skipped past the bookkeeping
eternal shard
Also I just wanted to make sure
Is the argument correct, that because of phi being bijective and k basically mapping to itself making it the identity and it being also bijective
that there exists exactly one ordered pair jeweils
glass heart
let psi(a,b)=psi(c,d)=(e,f). then phi(a)=phi(c)=e but phi is injective, hence a=c
eternal shard
that shows just injectivity right?
tiny robin
can anyone help me in my channel
eternal shard
let psi(a,b)=psi(c,d)=(e,f). then phi(a)=phi(c)=e but phi is injective, hence a=...
So I know that there is always at least one ordered pair, I think, and that part of yours would make it bijective I think
calm coralBOT
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drifting mist
how to do this?
calm coralBOT
how to do this?
supple needle
Know what tangent is in terms of sine?
drifting mist
no it has not
drifting mist
Know what tangent is in terms of sine?
Tangent is opp/adjancet
leaden thunder
no it has not
Do the bot command
calm coralBOT
Available help channel!
drifting mist
how does that apply to the question
calm coralBOT
how does that apply to the question
drifting mist
how to do this?
leaden thunder
cot = 1/tan
leaden thunder
Tangent is opp/adjancet
Plug this into cot
drifting mist
so 1/(opp/adjancet)?
drifting mist
but how does that pply to this
Write the first trigonometric function in terms of the second for 𝜃 in the given quadrant.
cot(𝜃), sin(𝜃); 𝜃 in Quadrant II
leaden thunder
so 1/(opp/adjancet)?
Do you know what opp and adj are in terms of sin and cos?
leaden thunder
,tex .sohcahtoa
drifting mist
and cos is adjancet/hyp
potent lotusBOT
riemann
leaden thunder
<@&268886789983436800> uhhhh
shy spindle
<@&268886789983436800> uhhhh
Stop abusing the ping?
slate field
Stop abusing the ping?
that's not abusing the ping. please don't voice your complaints about this here.
shy spindle
how to do this?
Cot(0) = 1/tan
leaden thunder
and cos is adjancet/hyp
Just plug these into tan
shy spindle
Cot(0) = 1/tan
!done
calm coralBOT
Cot(0) = 1/tan
If you are done with this channel, please mark your problem as solved by typing .close
leaden thunder
Tangent is opp/adjancet
tan = sin/cos from here
drifting mist
leaden thunder
**riemann**
Also from here
drifting mist
so idk
leaden thunder
Which part ?
drifting mist
would i type in 1/tan theta?
leaden thunder
tan = sin/cos?
drifting mist
so it would be cot0 =sin0/cos0?
leaden thunder
No
leaden thunder
cot = 1/tan
Go back to here
drifting mist
but it says in terms of sin
leaden thunder
so 1/(opp/adjancet)?
But instead of opp/adj, use cos, sin
dreamy lance
Cosine can be written in terms of sine using pythagorean identity
leaden thunder
but how do i type that in to it
You're not done solving
You have like 3 more steps
drifting mist
hOOWWW
dreamy lance
sin^2 (x) + cos^2(x) =1, solve for cosine
This is after you expand tangent in terms of sine and cosine fwiw
drifting mist
i just need the answer for this its the last question on my hw and it counts as a grade
dreamy lance
We can't just give you the answer
drifting mist
i dont understand the concept
leaden thunder
i just need the answer for this its the last question on my hw and it counts as ...
Then do the work
dreamy lance
but how do i type that in to it
Here, next you see tan(t)=sin(t)/cos(t)?
shy spindle
I think b is trolling…
dreamy lance
i don't think so
drifting mist
sec(t)/csc(t)
dreamy lance
It's simpler than that
drifting mist
I think b is trolling…
how is this really that easy because i dont understand
dreamy lance
cot=1/tan=1/(sin/cos)=cos/sin see?
drifting mist
wrong answer
dreamy lance
Don't recommend that
shy spindle
wrong answer
Not always…
drifting mist
cot=1/tan=1/(sin/cos)=cos/sin see?
not really
dreamy lance
cot=1/tan=1/(sin/cos)=cos/sin see?
This isn't the final answer either btw
dreamy lance
not really
You see cot=1/tan and tan=sin/cos?
drifting mist
i understand the first cot=1/tan but not tan=/sin/cos
but if thats what i need to know i got it
dreamy lance
That part is just soh cah toa
drifting mist
soh cah toa is sides though
how does tan equal other trigonometric functions
so your saying
opp/add =(opp/hyp)/(adj/hyp)
ohhhh ok yes
i got that part now
they have the same denominator
dreamy lance
Yeah that works
So now we almost have everything in terms of sine
Except the cosine so we need to write cosine in terms of sine
Pythagorean theorem should tell you sin^2(t) + cos^2(t) = 1 and if you solve this for cosine you'll get cosine in terms of sine and 1
drifting mist
ut cosine isint in the equationm
dreamy lance
so your saying
opp/add =(opp/hyp)/(adj/hyp)
This one?
drifting mist
oh ok
dreamy lance
Sine is opp/hyp and cosine is adj/hyp so it is in that equation
You have both
But you only want the sine so you have to find a way to rewrite the cosine in terms of sine
drifting mist
-1?
dreamy lance
Nah
drifting mist
would we do th einverse os cos
dreamy lance
Pythagorean theorem should tell you sin^2(t) + cos^2(t) = 1 and if you solve thi...
You have to use pythagorean theorem for this part
dreamy lance
Yeah but if you divide the whole thing by c^2 it reduces to soh cah toa stuff again
dreamy lance
Pythagorean theorem should tell you sin^2(t) + cos^2(t) = 1 and if you solve thi...
Which means you can reduce it to the equation I'm stating here
And this equation lets you solve for either sine or cosine to rewrite one in terms of the other whenever you want.
dreamy lance
Almost
dreamy lance
sin^2(t)+cos^2(t) = 1 is what we have and we want to isolate sine
So subtract sin^2(t) from both sides
You get cos^2(t)=1-sin^2(t)
But we still have an ugly square on the cosine
drifting mist
so square root
dreamy lance
So take the plus/minus root on both sides next
We haven't figured out the sign yet. When you square root both sides you have to do plus or minus until you know whether it's the square root or the negative square root
drifting mist
this is torture how do people like math
dreamy lance
So doing that gives us cos(t)=plus or minus sqrt(1-sin^2(t))
You get really used to all this sort of stuff and the questions get more interesting over time
drifting mist
idk if i can do another 3 years of this
dreamy lance
Well if you keep going you'll just get used to the parts of this that you dislike and it will eventually feel really smooth and easy
Learning it at the time can be fairly tedious and annoying though
dreamy lance
So doing that gives us cos(t)=plus or minus sqrt(1-sin^2(t))
So we have this, when t is in quadrant 2 what is the sign of cosine?
(Is cosine positive or negative in quadrant 2?)
drifting mist
negative
dreamy lance
So doing that gives us cos(t)=plus or minus sqrt(1-sin^2(t))
Yeah so then is it plus sqrt blah bkah here or minus sqrt blah blah here?
drifting mist
mius
dreamy lance
Yep
dreamy lance
Which thing are you claiming that is?
drifting mist
cot(theta)
dreamy lance
cot=1/tan=cos/sin and you subbed in the - sqrt blah on the top for that right?
dreamy lance
Yep that's how this goes
drifting mist
i also need help with another problem wher eim positive my logic is right but its marking wrong
dreamy lance
Yeah you can just post it here or close this help channel and open a fresh one if you want.
drifting mist
A pilot flies in a straight path for 1 h 30 min. She then makes a course correction, heading 14° to the right of her original course, and flies 2 h in the new direction. If she maintains a constant speed of 650 mi/h, how far is she from her starting position? (Round your answer to the nearest mile.)
So i got c^2=975^2+1300^2-(2x975x1300xcos(14))
which gives that answer
dreamy lance
which gives that answer
This is what you think the answer is?
dreamy lance
I think maybe your angle is wrong
drifting mist
would it be 180-14?
drifting mist
why though
dreamy lance
She heads straight along the vertical path I drew and turns right by 14 degrees like in my pic
But the angle we want for law of cosines is the one behind it that you mentioned is 180-14=166
dreamy lance
You basically got the whole question right except a small angle issue
drifting mist
no not this
Ik the trig part but its 3 parts
its the logarithm stuff, trig part, and one about finding domain range and axis of stuff
i got a 31 58 and 71 on respective test but he will drop the lowest grade and take the hw grade as test grade
which i got a 90 on i think
So I need a 51 or more
dreamy lance
Feels like you just need more practice is all
calm coralBOT
@drifting mist Has your question been resolved?
echo apex
calm coralBOT
echo apex
Find H
what i got so far is fiinding the angle thats vertically oppsite in the ED triangle
but i dont know where to go from here
nvm got it
.close
void viper
calm coralBOT
void viper
never, always, sometimes, always, always, sometimes
Can anybody let me know if I made an error
inner zealot
never, always, sometimes, always, always, sometimes
The last one is incorrect.
void viper
.close
formal thicket
calm coralBOT
formal thicket
i get pi/12 but how did they get 5pi/12
drifting seal
$2\theta = \frac{1}{2} \implies 2\theta = \frac{\pi}{6} \quad 2\theta = \frac{5\pi}{6}$
potent lotusBOT
knief
drifting seal
5pi/6 gives 1/2 but because we have frequency 2 we get 5pi/12 which is in Q1 still
formal thicket
ohhh ok i get it, thanks
drifting seal
you’re welcome
calm coralBOT
@formal thicket Has your question been resolved?
calm coralBOT
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small crag
for part d, why is the answer not 5?
calm coralBOT
for part d, why is the answer not 5?
fringe reef
for part d, why is the answer not 5?
because f(5-5) = f(0) which is not zero
median marsh
for part d, why is the answer not 5?
you are plugging x-5 for c everywhere in f
fringe reef
f(0) = 15 in fact
small crag
because f(5-5) = f(0) which is not zero
but it is 0? ... im a bit confused
small crag
how?
small crag
i just substituted 0 into f(x)
okok i understand that but then i dont understand what to do next?
fringe reef
okok i understand that but then i dont understand what to do next?
ok what did you get for c?
small crag
5
fringe reef
that 5 is a root?
small crag
oh no, like the y intercept is 5 from part c
bc the quadratic i found had no real roots
small crag
yup
fringe reef
for part d, why is the answer not 5?
but then for part a
from the fact that (x+3) is a factor
small crag
-3 is a root
fringe reef
x=??? is a root
small crag
yes
fringe reef
so for d)
you want to have f(-3) = 0
f(? - 5) = 0
wwhat should go instead of '?'
small crag
2... ohhh
fringe reef
glad to help :)
small crag
.close
shadow trout
can someone please help with iii)
shadow trout
<@&286206848099549185>
unkempt sphinx
which part(s) do you need help on @shadow trout
shadow trout
iii)
hollow perch
- do you know how to find length with complex numbers
- lengths RT and SU are equal => RT/SU=1
shadow trout
You mean |RT|/|SU|=1 ?
shadow trout
- do you know how to find length with complex numbers
- lengths RT and SU are eq...
yeah sqrt(a^2+b^2)
hollow perch
i meant like the length between two complex numbers, like r and t for example
shadow trout
i don't think so?
hollow perch
if you imagine the complex numbers as just normal points on a plane, it turns out as just RT = |r-t|,
if you had their coordinates, it'd be sqrt( (a1-a2)^2 + (b1-b2)^2 )
basically minus one from the other and find that length
shadow trout
if you imagine the complex numbers as just normal points on a plane, it turns ou...
so its just vector theory, that RT = OT - OR
ok
calm coralBOT
@shadow trout Has your question been resolved?
calm coralBOT
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blazing coyote
,, \lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}}
calm coralBOT
,, \lim_{(x,y) \to (0,0)} \frac{xy}{\sqrt{x^2 + y^2}}
potent lotusBOT
ƒ(Why am. I here)=I don't Know
blazing coyote
I feel the limit exists and is 0
remote mural
Finally a question that I know and wai doesn't
blazing coyote
Lol, there will be a lot more of these next sem as I cover multi var and vector calc
remote mural
I feel the limit exists and is 0
What made you come to that conclusion?
blazing coyote
I approached it along a few paths, which all gave 0 as the limit
though I should probably try epsilon-delta to be safe
remote mural
Have you tried the path y=mx?
unique jackal
! 
blazing coyote
oh
hollow totem
I feel like it works, no?
blazing coyote
that gives $\frac{m}{\sqrt{1+m^2}}$ as the limit
potent lotusBOT
ƒ(Why am. I here)=I don't Know
unique jackal
mm, you've only shown the limis is 0 along one kind of path
hollow totem
,w limit as x approaches 0 of (x*mx)/sqrt(x^2+(mx)^2)
potent lotusBOT
unique jackal
that doesn't mean it's 0 in general
hollow totem
I think he thought it wasn't 0
remote mural
:p
unique jackal
oh, but it is 
unique jackal
you may try 
swift laurel
i think polar coordinates would be a good choice
unique jackal
you can do without them too
hollow totem
Oh there's a nice solution
I kinda don't want to give it though and ruin the problem solving process
Can I just spoiler it and ur not allowed 
potent igloo
you can do without them too
Polar is always just secretly epsilon delta 
unique jackal
Polar is always just secretly epsilon delta <:thinkies:827642567775748163>
no eps-delta needed either
blazing coyote
,w lim_{x \to 0} x^4/\sqrt{x^2+x^6}
bronze adder
i think polar coordinates would be a good choice
start using polar coordinates after you're comfortable with finding δ and/or disproving the limit exists otherwise
potent lotusBOT
unique jackal
this one, you can bash with squeeze theorem if you do it right
hollow totem
||squeeze theorem with AMGM to get (x^2+y^2)/2sqrt(x^2+y^2) which is r/2||
blazing coyote
oh right the squeeze therom works here
potent igloo
||graph it|| 
blazing coyote
I squeeze it between $\frac{x}{\sqrt{x^2+y^2}}$ and $\frac{-x}{\sqrt{x^2+y^2}}$
potent lotusBOT
ƒ(Why am. I here)=I don't Know
blazing coyote
to get 0 as the limit
hollow totem
Try approaching that along y=0
obtuse briar
personally I'm team polar + squeeze that limit in polar
blazing coyote
Try approaching that along y=0
That gives 0
hollow totem
,w limit of x/sqrt(x^2+0^2) as x goes to 0
potent lotusBOT
hollow totem
np
blazing coyote
personally I'm team polar + squeeze that limit in polar
so x=rcos(t), y=rsin(t)?
that gives rcos(t)sin(t) as the limit
obtuse briar
it's a bit vague how you have it, before you had (x,y) -> (0,0) what's happening to r and t now?
obtuse briar
closer, but actually t could do anything, you could think of it as an arbitrary function of r if you wanted here
obtuse briar
cool, yeah yw
blazing coyote
.close
tall moon
find all positive integers (a,b) such that $a^2+2016=3^b$
calm coralBOT
find all positive integers (a,b) such that $a^2+2016=3^b$
potent lotusBOT
Skissue ping4response
tall moon
one thing i noticed is that a^2 mod 3=0 => a=3x
dusky flax
a is a 3 multiple then?
tall moon
so we can get $9x^2+2016=3^{y+2}$
potent lotusBOT
Skissue ping4response
tall moon
$$x^2+224=3^y$$
potent lotusBOT
Skissue ping4response
dusky flax
and x is odd
tall moon
thats basically where my ideas run out
blazing coyote
Cool, this can be solved by inspection, I suppose
dusky flax
by mod 2
tall moon
oh yea
dusky flax
maybe you can do two cases fro mod 4 (even and odd y, giving 1 and -1 remainder)
i have to go now
potent lotusBOT
Skissue ping4response
calm coralBOT
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Available help channel!
blazing coyote
Can someone walk me though this
calm coralBOT
Can someone walk me though this
blazing coyote
So I have to use the eps-\delta defn here
Basically $0 < \sqrt{x^n-a^n}< \delta \implies ||x|- |a||< \varepsilon$
potent lotusBOT
ƒ(Why am. I here)=I don't Know
calm coralBOT
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Available help channel!
unborn stone
calm coralBOT
calm coralBOT
@unborn stone Has your question been resolved?
silver tapir
how are the coordinates (0,2a sintheta ) and (2costheta , 0) via parametric form
leaden thunder
Look for the right triangle
silver tapir
Look for the right triangle
ik it can be solved via trigonometry
my teacher said to solve via parametric form
idk how
leaden thunder
Sounds like not enough information given then
silver tapir
oh
maybe ill work on the question then
thanks for ur help
also when i derived length of tangent from a point to a circle i got it as S1 denoting the circle eqn when u put x1,y1 in it
but the actual formula is √S1?
wait mb
i forgot the squared part
.close
calm coralBOT
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unborn stone
calm coralBOT
unborn stone
I did u=tan instead and got a different answer?
Why doesn’t tan work
My answer was 1/3 times u^3
calm coralBOT
@unborn stone Has your question been resolved?
leaden thunder
I did u=tan instead and got a different answer?
What's the derivative of tan
unborn stone
What's the derivative of tan
Yeah I realized before after I posted
Can u help with this tho
How am I suppose to put this into an equation
Idk how to complete the square for x and y
leaden thunder
Idk how to complete the square for x and y
,tex .cts
potent lotusBOT
riemann
unborn stone
,tex \.cts
So x^2-y +(z+1)(z+1)=0?
calm coralBOT
So x^2-y +(z+1)(z+1)=0?
unborn stone
Yes that's equivalent to the answer here
I need to get it into one of these forms
unborn stone
So x^2-y +(z+1)(z+1)=0?
How I get this to them
leaden thunder
It's bottom left but shifted
unborn stone
It's bottom left but shifted
I forgot any of the variables can be on the left
Am I suppose to keep z as (z+1)(z+1)?
X^2 +(z+1)^2?= y?
I thought the z would need to be by itself