is it that?
#help-38
limpid dawn
rustic nest
yeah
rustic nest
yes hah
rustic nest
$\frac{y^2-4y+4}{4y}$?
solid kilnBOT
wakamole
limpid dawn
no
rustic nest
ugh
rustic nest
why isn't it y^2 - 4y + 4?
hollow tangle
make them one fraction
Hey bacc, why are you making it into a single term?
Like just break it into all the terms and then integrate
limpid dawn
why isn't it y^2 - 4y + 4?
,w expand (y-1)^2
rustic nest
just for the ((y-1)/(ysqrt(y))^2
solid kilnBOT
solid kilnBOT
bacc the sigma😔🤞
rustic nest
get rid of the root now and split the fraction
wait split the fraction?
rustic nest
is it $\int_1^9\frac{3y + 1}{2} dy$ ?
solid kilnBOT
wakamole
limpid dawn
is it $\int_1^9\frac{3y + 1}{2} dy$ ?
🙏🏻
solid kilnBOT
bacc the sigma😔🤞
limpid dawn
Now split
rustic nest
oh i kept the power ^ 2
limpid dawn
And a sudden 3
rustic nest
Now split
sorry i dont understand
$(y+1)y^{-1/2} = \frac{y^{1/2} + y^{-1/2}}{2} = \frac{1}{2}$
?
solid kilnBOT
wakamole
rustic nest
And a sudden 3
i had expanded it then also lef the denominator as 4y so weird simplyfication
limpid dawn
still struggling...
rustic nest
but idk how to get the actual integral of that
it;s 1/3sqrt(y)(y+3) but how to get that idk
solid kilnBOT
bacc the sigma😔🤞
rustic nest
oh
limpid dawn
split
solid kilnBOT
bacc the sigma😔🤞
rustic nest
wolfram said the integral was $\frac{1}{3}\sqrt{y}(y-3)$
solid kilnBOT
wakamole
limpid dawn
ok
rustic nest
oh
solid kilnBOT
bacc the sigma😔🤞
limpid dawn
Do the rest
limpid dawn
Enough handholding
hollow tangle
It's been what an hour?
rustic nest
**bacc the sigma😔🤞**
i got $2y^{\frac{3}{2}}+{2y^{\frac{1}{2}}$ howw is that the same thing as $\frac{1}{3}\sqrt{y}(y+3)$
rustic nest
Enough handholding
?
limpid dawn
It's been what an hour?
Since you wanted ro sleep
solid kilnBOT
wakamole
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limpid dawn
Oh dear
limpid dawn
And divide by 2/3 too
solid kilnBOT
wakamole
rustic nest
i did that part correctly
i sitll need to multiply it by 1/
1/2
so it will be over 6 instead
limpid dawn
,w integrate 1/2 * ( y^(1/2)+y^(-1/2) )
solid kilnBOT
solid kilnBOT
bacc the sigma😔🤞
limpid dawn
applying power rule
limpid dawn
,, \frac{1}{2} \left [ \frac{y^{\frac{3}{2}}}{\frac{3}{2}} + \frac{y^{\frac{1}{2}}}{\frac{1}{2}} \right ]_1^9
solid kilnBOT
bacc the sigma😔🤞
rustic nest
mhm
limpid dawn
Your 2s cancel
solid kilnBOT
bacc the sigma😔🤞
limpid dawn
factorize sqrt(y)/3
solid kilnBOT
bacc the sigma😔🤞
limpid dawn
now sqrt(y) too
solid kilnBOT
bacc the sigma😔🤞
limpid dawn
i did integrate ok then
lies
rustic nest
**bacc the sigma😔🤞**
can you just explain what you did with the exponent here and how it leads to the next step
limpid dawn
damn what
rustic nest
i finished it but i would not have been able to do all that
limpid dawn
you need to review immediately calc 1
and algebra
anyway
so
2/2 + 1/2 = 3/2
i wrote it like that
rustic nest
i know that
rustic nest
yea
solid kilnBOT
bacc the sigma😔🤞
limpid dawn
yes
rustic nest
that's cool
limpid dawn
**bacc the sigma😔🤞**
FTC
solid kilnBOT
bacc the sigma😔🤞
rustic nest
i got 32/3
limpid dawn
,w 12-4/3
solid kilnBOT
limpid dawn
bet
rustic nest
.close
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unique harness
im confused with my answer, I was taught that the antiderivative of 1/x would ln |x| so the same thing should apply here, right? I wrote 2ln(|X|) but the answer key just shows 2ln(x)
unique harness
i just want some calcification
i know that my -2/5x^5 is wrong but the |x| would be right, right?
zinc ginkgo
your answer's right
unique harness
wait
zinc ginkgo
,w int 3x^4 + 2/x - 2/x^6 + 3sqrt(x)
solid kilnBOT
zinc ginkgo
even the -?
which -?
unique harness
-2/5x^5 is wrong
austere cedar
Technically it should use ||. However, most sources are lazy and use (). It often doesn't matter
solid kilnBOT
riemann
unique harness
i know what that one is, i messed up the signs then
just wondreded about the ||
cus the teacher said that they were imporatnt
but the answer key did not have it
.solved
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tulip quest
I have this optimization problem
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I have this optimization problem
tulip quest
My critical number is negative which doesn’t make sense
Wait
I see my mistake
I still get a weird nevstive
Negative
I get this instead
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@tulip quest Has your question been resolved?
polar dragon
yo
lean pike
use tests
whole coral
Remember that you're meant to try and use either limit or direct comparison for these, and if you can't use either, that's when you put n/a (so if you don't fit the conditions)
polar dragon
bro can u do one of them
whole coral
How about stating or showing what each of those says?
polar dragon
it should give me atleast one point
at bottom it says 0 percent
it should say at least 5percent if i got one right
and i refuse to believe i missed every single one
whole coral
it does say that you need all answers right to get credit though?
polar dragon
ah hell nah boy she said get em all right
i didnt even read it
i just do problems
you put na when the an>bn right
instead of doing the limits of an/bn i ignore that
whole coral
Well, worth stating what you mean by an and bn (i.e. the statements of both the direct comparison and limit comparison tests)
You especially may want to check the first two answers...
whole coral
(1) is wrong, as is (2)
whole coral
Remember, if you cannot apply either comparison tests, it's an n/a, even if you know convergence by some other method
polar dragon
i dont use AST then?
whole coral
Nope, you're not allowed to use it
polar dragon
- NA then let me check 2
for 2 i got 2 by taking the limits
and converges right
thats what i put the first time
whole coral
Got 2 for what, and what limits?
polar dragon
i did comparison with 1/n^2
then i did an/bn for lim to infinity
and limit equaled 2
but do i not use that?
is that using a different way like AST
polar dragon
is everything else good
i believe 1.s the only one i used
AST on
i used p series for rest
whole coral
Yep, happy with the rest 
polar dragon
holy fuck ur the goat
i kept on changing my answers
cuz i thought it was like others
i didnt know i had to get all right
i didnt read the problem i just answered the questions
@whole coral when i do an/bn to find the limit can i set the bn to anything or does it have to be the one that i compared to
whole coral
Well, you can choose your bn to be anything "legal" such that you get a finite limit, if you can find it, such that it gives you a series whose convergence you know
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urban copper
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urban copper
Let $A = (2, -2, 1)$, $L: \mathbf{X} = \lambda (0, 1, -1) + (-1, 1, 0)$, and $\Pi$ be the plane that passes through the points $(2, 1, 1)$, $(1, 3, 1)$, and $(2, 3, 0)$.
Find $B \in L$ such that the line passing through $A$ and $B$ does not intersect the plane $\Pi$.
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
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night patio
Let $A = (2, -2, 1)$, $L: \mathbf{X} = \lambda (0, 1, -1) + (-1, 1, 0)$, and $\P...
!status
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Let $A = (2, -2, 1)$, $L: \mathbf{X} = \lambda (0, 1, -1) + (-1, 1, 0)$, and $\P...
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
1
night patio
1
ok lets break it down so its easier to start
at least i find that helpful whenever im not sure what to do
rose heath
Let $A = (2, -2, 1)$, $L: \mathbf{X} = \lambda (0, 1, -1) + (-1, 1, 0)$, and $\P...
First, Parametrize the line L.
night patio
to find B, we have some conditions to follow
a) the vector joining A and B must be parallel to the plane to not intersect
b) We also know B is on the line L
Now lets try to find a parametric vector expression for the plane $\Pi$
solid kilnBOT
Dootud
night patio
Let $A = (2, -2, 1)$, $L: \mathbf{X} = \lambda (0, 1, -1) + (-1, 1, 0)$, and $\P...
do you know how the plane can be written in parametric vector
urban copper
We dont have the equation of the plane
Is the the points enough for finding the plane pi
night patio
Is the the points enough for finding the plane pi
it generally is as long as we can check the points are not collinear
urban copper
rose heath
Trying but how
Identify the direction vector, Identify a point on the line and then write the parametric equation.
night patio
so if u want to convince yourself or draw a diagram, we know that a plane can be formed from the following
- a base point (random point on plane)
- and 2 linearly independent direction vectors
urban copper
direction of
line L is (0,1,-1)
night patio
the diagram here is convenience, if u look at the red arrows, they represent 2 different direction vectors in which the plane spans
having 2 linearly independent direction vectors is sufficient to form a plane if we also have a basis point that is on the plane
night patio
ok so lets try with the 3 points we have here
lets first check that they aren't collinear
rose heath
direction of
line L is (0,1,-1)
Exactly. Now, Identify a point on the line.
urban copper
(-1,1,0)
urban copper
lets first check that they aren't collinear
how
night patio
Identify the direction vector, Identify a point on the line and then write the p...
there already is given a parametric form of the line L:x given
night patio
how
so if 3 points are colinear
then the vectors formed by joining 2 of the points must be a scalar of the other 2 points
since they have the same direction
rose heath
and here, λ = to ?
@urban copper No, not 1.
night patio
so if we can show $(2,1,1) - (1,3,1) \neq k[(2,1,1) - (2,3,0)], k \in \mathbb{R}$
solid kilnBOT
Dootud
night patio
then we prove that these 3 points are not clinear
urban copper
night patio
ok so now that we know these 3 points are not cllinear, it gives us some information
night patio
- we know that all 3 points are on the plane
- they do not lie on the same line,
and also i would encourage u to draw it out in a diagram just to visualise it
we can generate 2 different vectors
namely direction vectors for the plane
for example, if we pick (2,1,1) as a point of reference
the point (1,3,1) and (2,3,) from the point (2,1,1) are going towards 2 non parallel directions
then we know that actually, the vectors formed from $(1,3,1)-(2,1,1)$ and $(2,3,0) - (2,1,1)$ become the direction vectors of your plane spanning out from (2,1,1)
solid kilnBOT
Dootud
night patio
so we now have 2 direction vectors for our plane, namely $(-1,2,0)$ and $(0,2,-1)$ spanning out from the point (2,1,1)
solid kilnBOT
Dootud
night patio
if it makes it easier to understand, a plane is essentially formed from 2 lines that intersect at a point
which is why u have 2 direction vectors
we can then write this in parametric form as $(2,1,1) + \alpha (-1,2,0) + \beta (0,2,-1)$
solid kilnBOT
Dootud
night patio
where alpha and beta are arbtirary real numbers that enables us to span an entire plane
urban copper
i don't rlly get it
night patio
i don't rlly get it
I think it might be better to watch a video to understand because I lack the knowledge to generate a visual animation or diagram to show you how it works on discord
I think it’s quite a visually heavy topic
urban copper
just give me a minute
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@urban copper Has your question been resolved?
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@urban copper Has your question been resolved?
slender shard
i don't rlly get it
May I show one way to approach this?
urban copper
yes
slender shard
This isn't the "correct" correct way as I'm still learning this myself but it still works to find the answer.
Do you know how to find the plane of three non-collinear points?
urban copper
using the parametric equation for the plane
slender shard
What you're actually interested in is finding the Normal vector to the plane.
You want to find that vector.
slender shard
using the parametric equation for the plane
Yes/no?
urban copper
why the normal vector
slender shard
Because you will use that to make an equation for the plane that goes through the points.
urban copper
geogebra?
urban copper
scalar equation of the plane
slender shard
Yes.
The partial derivatives are the components of the Normal vector to the plane.
So if you know the Normal vector and a point on the plane, you can make an equation for the plane.
limpid dawn
cursed notation
slender shard
cursed notation
Still learning. Partial fraction better?
urban copper
how do i find the normal vefctor
slender shard
slender shard
f in this case will be a three-variable function in standard form that is equal to some arbitrary constant.
urban copper
perfect
slender shard
For your purpose, f(x,y,z) can equal any constant. It doesn't matter because the Normal will be the same regardless of what f(x,y,z) equals.
N = <a, b, c>
Does that make sense?
slender shard
Yes.
urban copper
okay sweet
Let $A = (2, -2, 1)$, $L: \mathbf{X} = \lambda (0, 1, -1) + (-1, 1, 0)$, and $\Pi$ be the plane that passes through the points $(2, 1, 1)$, $(1, 3, 1)$, and $(2, 3, 0)$.
Find $B \in L$ such that the line passing through $A$ and $B$ does not intersect the plane $\Pi$.
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
slender shard
(2,1,1), (1,3,1), and (2,3,0).
Probably a poor choice of names since A is already given and you are looking for B.
urban copper
A' = (2,1,1)
B' = (1,3,1)
C' = (2,3,0)
slender shard
That works.
urban copper
B' - A' = A'B' = (1,3,1) - (2,1,1) = (1-2, 3 - 1, 1- 1) = (-1,2,0)
C' - B' = B'C' = (2,3,0) - (1,3,1) = (2-1, 3-3, 0-1) = (1,0,-1)
,w (-1,2,0) x (1,0,-1)
solid kilnBOT
urban copper
thats a vector that is perpendicular to A'B' and perpendicular to B'C'
slender shard
Good. Now you can use the aforementioned plane formula.
urban copper
urban copper
-2(x-x1) -1(y-y1) -2(z-z1) = 0
slender shard
Good. Now, because you are looking for a point B that forms a vector with point A that is parallel to the plane that goes through A', B', and C'; any vector on the plane that goes through point A with the same Normal vector will also be parallel.
The light shade of blue has the same Normal vector but it goes through point A.
And both planes will be parallel.
Make sense?
urban copper
yes
slender shard
Now you want to find a point B that intersects that second plane that goes through point A.
There are two sets of coordinates in that equation; (x,y,z) and (x_1, y_1, z_1).
You know one set of coordinates, that is given by point A.
What would the other set of points, point B, be? (x_1, y_1, z_1)?
urban copper
You know one set of coordinates, that is given by point **A**.
is that given?
oh you mean (2,1,1), (1,3,1), and (2,3,0).
those points are contained in the plane
slender shard
A is given by the problem; A = (2, -2, 1).
B is given by the parametric equation L.
urban copper
B=(-1,1,0)
slender shard
We don't know what B is yet, you have to determine that.
B will be some point on the grey line in the image above.
urban copper
the normal vector is perpendicular to all the vectors contained in the plane
i am not sure but how do we find B then?
slender shard
The first line is given by L in the problem. You can convert that to a parametric equation.
slender shard
urban copper
okay perfect
slender shard
That will be your second second points for the plane equation.
Plug in both points and you can solve for t.
A and B.
Once you find that, I will show you the correct way to find B. It's quicker but I wanted to show you a way to verify your answer.
urban copper
slender shard
And plug in each of those points.
x=2, y=-2, z=-1, x_1 = -1, y_1 = 1+t, z_1 = -t
solid kilnBOT
slender shard
Plug that into the parametric point B and you will get your answer. Now I will show you the correct way to solve this problem.
It is quicker.
You can make a vector BA using B - A.
urban copper
-2(x-(-1))-1(y-1-t)-2(z+t)=0
urban copper
You can make a vector **BA** using **B** - **A**.
ok
slender shard
Here you will use a property of the Dot Product. Two orthogonal vectors will have a dot product that equals zero.
You know the parametric form of B, all you need to do is subtract point A from B to make a vector for AB.
urban copper
yes
slender shard
Set the dot product of AB · N equal to zero and you can solve for t that way as well.
solid kilnBOT
938c2cc0dcc05f2b68c4287040cfcf71
urban copper
Set the dot product of **AB** · **N** equal to zero and you can solve for **t** ...
how do I grt the normal of the plane
using the three points?
urban copper
true
slender shard
So the Normal vector from point A will be the same as the Normal vector from the plane that goes through the three given points.
Those two Normal vectors are equal even though they start at different points.
urban copper
yeah let this three points be named C D E
slender shard
All you are concerned about now though are these two vectors.
The dot product of those two vectors will equal zero.
BA · N = 0
slender shard
You want B - A.
urban copper
B-A=(-1-2,1+t+2,-t-1)
slender shard
Good so far, now simplify that.
urban copper
(-3,3+t,-t-1).N=0
slender shard
And what does N equal?
urban copper
is the
slender shard
You found it earlier.
urban copper
B-A x D-C
slender shard
You used it to make the equation of the plane earlier.
slender shard
No. Two parallel planes will have the same Normal vector.
urban copper
ehat is the normal vector of the other plane
slender shard
N = <-2, -1, -2>
urban copper
,w (-3,3+t,-t-1).(-2,-1,-2)=0
solid kilnBOT
slender shard
You get the same value for t as before.
slender shard
It's always a good idea to solve difficult homework problems from scratch again. This time write out your thought process as you do so.
Start with some basic principles of what you should understand is true.
- Two parallel planes have the same Normal vector.
- The dot product of perpendicular vectors is equal to zero.
These two planes are parallel and therefore have the same Normal vector.
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agile gull
can someone confirm this is correct
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can someone confirm this is correct
agile gull
you can probably do the math in your head (assume the fractions are right lol) but i just wanna know if im doing the Q right
ionic pendant
the process is fine but arithmetic errors make the fractions wrong
,calc 3-6*(-11)
solid kilnBOT
Result:
69
isnt it 3 - (6*-11) ? so -63
solid kilnBOT
Result:
69
why would that be the case?
agile gull
oh wait im wrong again
so 1,3 should be 69 and from there on i just need to do it correctly again
agile gull
why would that be the case?
agile gull
make 25/3, 26/3
ionic pendant
make 25/3, 26/3
it's correct other than that
ionic pendant
the result is correct
agile gull
yeah but i got there wrong so i gotta redo it
agile gull
the result is correct
eugh i already see a mistake
time to redo again
nah im gunna end it
agile gull
the result is correct
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wet echo
Question regarding vector analysis:
How to properly differentiate the function exp(|s|²) where s is a vector r - a (both in x,y,z) and |r|² is the dot product
Should I first open the dot product to get r² + a² + r a cos(theta)?
and how to proceed while differentiating this exp(r² + a² + r a cos(theta) )
zinc ginkgo
Yes expand the dot product
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wet echo
okay, but then I would have ∇exp(r² + a² + r a cos(theta) ) = (r² + a² + r a cos(theta) ) * exp(r² + a² + r a cos(theta) ) * ∇(r² + a² + r a cos(theta) ) right?
Them how to proceed with ∇(r² + a² + r a cos(theta) ) ? (btw, the ∇ acts on r. I forgot to mention it)
Can I just treat it the same as one variable calculus and do:
∇(r² + a² + r a cos(theta) ) = 2r + 0 + a cos(theta) ??
I ask it because r² is r · r and I think there may be some special properties in diferentiating this
I mean, is ∇(r·r) = 2r?
zinc ginkgo
Depends what your coordinates are
Is r spherical or cylindrical
https://profoundphysics.com/gradient-in-different-coordinates/
This has the formula for gradient for both
wet echo
Is r spherical or cylindrical
Sorry.
It is spherical coordinates
'll take a look at the links. Thanks for the help
.close.
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cold haven
Someone give me the answer, straight up. The question is asking [What is the smallest solution of x strictly bigger than pi] please give me the process on how you found the answer
viscid flower
say $\cos y = \frac 12$. what do we know about $y$?
solid kilnBOT
jan Niku
cold haven
? Don’t know
cold haven
??
solid kilnBOT
jan Niku
viscid flower
where is cos(theta) = 1/2
cold haven
5pi/3
solid kilnBOT
jan Niku
viscid flower
seems like the first place this could happen is at $y = \frac{5\pi}{3}$
solid kilnBOT
jan Niku
viscid flower
is that enough info to finish out the problem?
cold haven
No
They are saying it has an infinite solutions. What is the smallest x strictly greater than pi
viscid flower
well, you found this
cold haven
In the answer machine it’s wrong
Which means it’s not the correct answer
I’m not sure how to proceed
viscid flower
if $\cos x = \frac 12$, and $x$ is the first solution that's bigger than $\pi$, then $x = \frac{5 \pi }{3}$
solid kilnBOT
jan Niku
solid kilnBOT
jan Niku
solid kilnBOT
jan Niku
cold haven
Because, there’s another one 5pi/3
viscid flower
why doesnt $5x = \frac \pi 3$ work
solid kilnBOT
jan Niku
cold haven
I’m not sure
viscid flower
what would this make x
cold haven
X = pi/15
viscid flower
so whats wrong with this solution
I'm not saying its good, I'm just asking since you're saying you're having trouble to know what to do
cold haven
Don’t know
viscid flower
if $x = \frac{\pi}{15}$
solid kilnBOT
jan Niku
viscid flower
then $\cos (5x) = \frac 12$
solid kilnBOT
jan Niku
viscid flower
but this x is not the answer you want, why?
cold haven
Probably because it isn’t bigger than pi
viscid flower
Probably because it isn’t bigger than pi
yea
so, lets look at the next one up
$5x = \frac{5\pi}{3}$
solid kilnBOT
jan Niku
viscid flower
how about this one?
cold haven
X = 5pi/15
viscid flower
any good?
viscid flower
you need to try the next biggest root
cold haven
?
viscid flower
i mean, heres a thought
you know that x has to be bigger than pi
then $5 x > 5 \pi$, right?
solid kilnBOT
jan Niku
cold haven
Ye
viscid flower
wheres the first place we find a root that big
how many times do you have to go around the circle
cold haven
Twice?
viscid flower
more than that, but yea, at least twice
2pi is once
4pi is twice
5 pi is like
twice and a half
so lets say we went around twice, and another half
were now sitting at 5pi, at this red dot
wheres the next solution?
it will definitely make 5x bigger than 5pi
and it should be the first time this happens, too
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steel hound
Why wouldn't the next one be at 5x=5π+2/3π, so x = 17π/15?
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final reef
ermm i dont rlly understand this qn
languid trench
bru
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latent notch
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latent notch
so i got the magnitude of the resultant
but idk what angle he wants
like
BAR ?
cause that wouldnt be over 90
anyone know ?
prime crystal
latent notch
doesnt he want the angle of the resultant
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lapis cosmos
"Find a point in the first quarter where tangent line to "2(x^2 + y^2)^2 = 25(x^2-y^2) is horizontal"
lapis cosmos
please help \o/ ive been seriously lost with this one
dapper swift
"Find a point in the first quarter where tangent line to "2(x^2 + y^2)^2 = 25(x^...
have you tried implicitly differentiating both sides
then solving dy/dx = 0
marsh forum
Well, if a line is horizontal, what does that mean about the derivative?
lapis cosmos
when I say dy/dx = 0, I get another equation
and thats where I get lost
from plugging equations in desmos, apparently I need to make my derivative equal to my original equation to get the point I want
but I dont understand why
dapper swift
from plugging equations in desmos, apparently I need to make my derivative equal...
did you get $x\left(-4x^{2}-4y^{2}+25\right)=0$?
solid kilnBOT
higher's secret twin brother
lapis cosmos
exactly
carmine shuttle
need help
lapis cosmos
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dapper swift
exactly
okay, you have to note that x can't be 0
subbing into the original equation, if x = 0, y = 0
but then you get dy/dx = 0/0
lapis cosmos
I see
dapper swift
that implies $-4x^2 - 4y^2 + 25 = 0 \implies x^2 + y^2 = 25/4$
solid kilnBOT
higher's secret twin brother
dapper swift
sub in to get $2(25/4)^2 = 25(x^2 - y^2)$
solid kilnBOT
higher's secret twin brother
dapper swift
you get x^2 + y^2 = 25/4 and x^2 - y^2 = something
you can eliminate to find both x^2 and y^2
lapis cosmos
so I need to make the derivative equal 0, and then use that new equation to find my point
dapper swift
so I need to make the derivative equal 0, and then use that new equation to find...
well yes
lapis cosmos
my confusion comes from, shouldnt the derivative only give me points that exist in my equation?
so I shouldnt have to search for them, I dont know if that makes sense
dapper swift
my confusion comes from, shouldnt the derivative only give me points that exist ...
that's not true actually
it just gives you some other relation g(x, y) = 0
where all the points on your original relation f(x, y) = 0 and on g(x, y) = 0 satisfy dy/dx = 0
lapis cosmos
so I have dy/dx = f(x,y)
if i set dy/dx equal to any value, I get a new equation of all the points in my original equation where thats the derivative?
dapper swift
if i set dy/dx equal to any value, I get a new equation of all the points in my ...
yes
but the points where the derivative is 0 is a subset of the points on your dy/dx = f(x, y)
doesn't mean that all the points on your f(x, y) will lie on the original curve
lapis cosmos
hmm that makes a lot of sense
I thought I was doing something wrong
what if in f(x,y) = dy/dx, I set my x or y to a specific value
would I get all my possible derivatives when x or y equal that on my original function?
dapper swift
would I get all my possible derivatives when x or y equal that on my original fu...
no, say if x = a, you'd get dy/dx only for points on x = a
only for points on the curve that lie on that vertical line
sonic moss
Can you see where I went wrong?
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sonic moss
<@&286206848099549185>
nova spire
<@&286206848099549185>
uh
sonic moss
howdy
sonic moss
yeah that part was a little jank
nova spire
only works when a,b,c,d are real numbers
sonic moss
oh... 😦
sonic moss
only works when a,b,c,d are real numbers
I mean there is only.... 1 non-real...... yeah good point
sonic moss
so w = .../(2-i)
ah got it thanks!
a beautiful 1
very good helpering!
/close
.close
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marsh forum
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marsh forum
so here I have to take 2 vectors at a time and prove their product is 0?
wooden plover
yea
marsh forum
Ah, that's boring. Any novel way of doing it?
wooden plover
you'll survive computing 3 dot products
dull temple
haha
wooden plover
not rlly
dull temple
can't you row reduce it and find that it's full rank
like that'll probably be slower
marsh forum
you'll survive computing 3 dot products
You wish. I've lost 2.5 marks on a quiz because i got one entry in a matix product wrong
wooden plover
you could have fun computing A^TA (where A has your vectors in rows or cols doesn't matter here) and check that it's diagonal, but it's essentially the same thing and oops you have a matrix there
wooden plover
can't you row reduce it and find that it's full rank
how would that tell you orthogonality tho ?
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limpid belfry
Let ABC be a triangle where AB =AC, BAC = 80 degrees, if D is a point inside the circle such that DAB = DBA = 10 degrees, then find ADC
hidden dew
Angle ABC and ACB are equal
wind cloak
I am not sure if you'll find answer or not
but you could make the use of law of sines
wind cloak
I think assuming the angle you want is a better idea
rather revolving around some other angle
by this I mean 60+w = x thing
wind cloak
angle ACD, ADC, ADB and ABD are some angles that might be useful
you'll be able to make an equation in x but I don't think it is solvable easily
limpid belfry
well that sucks
ivory vault
I am not sure if you'll find answer or not
but you could make the use of law of...
i think this is a good idea
wind cloak
you'll be able to make an equation in x but I don't think it is solvable easily
but this is the problem
ivory vault
ah
wind cloak
well that sucks
I think you might be able to solve it
limpid belfry
what😭
wind cloak
angle ACD, ADC, ADB and ABD are some angles that might be useful
have you gotten something with law of sines for these?
wind cloak
probably has something to do with sin160 though
ofcourse it does
limpid belfry
at least try dude
im horrible at geometry man😭😭
wind cloak
angle ACD, ADC, ADB and ABD are some angles that might be useful
I don't think there is much geometry left after I told you the angles of interest
(geometry sucks anyway)
limpid belfry
i agree
ivory vault
dang i just watched a vid with a guy who loves geometry
and he said everyone else hates it
limpid belfry
uhhh i got sin(110 - ADC)/sinADC = sin(110)/sin(10)
wind cloak
uhhh i got sin(110 - ADC)/sinADC = sin(110)/sin(10)
what
hidden dew
I think sin(160)/AB would be enough
wind cloak
how did you got right hand side
wind cloak
I think sin(160)/AB would be enough
how you'll deal with AB?
limpid belfry
how did you got right hand side
sin(10)/AD = sin(160)/AB so AD = ABsin10/sin160
hidden dew
how you'll deal with AB?
oh wait … ngl my logic is off
wind cloak
sin(10)/AD = sin(160)/AB so AD = ABsin10/sin160
yeah
wind cloak
uhhh i got sin(110 - ADC)/sinADC = sin(110)/sin(10)
but that's not sin10/sin160
limpid belfry
uhhh i got sin(110 - ADC)/sinADC = sin(110)/sin(10)
so it should be sin(160)/sin(10)
wind cloak
whatever, I recommend expanding sin(110-ADC)
limpid belfry
alright
wind cloak
so it should be sin(160)/sin(10)
sin10/sin160*
wind cloak
after that it is just awful simplification nothing else
(like converting bigger angles to 10 and 20 degrees)
(turning cos(ADC) into sine)
and then forming an awful quadratic equation in sinADC
better sub sinADC=u
hidden dew
and that’s why nobody loves geometry
limpid belfry
and im supposed to spend 2 minutes per question what
wind cloak
and im supposed to spend 2 minutes per question what
I don't think you have to do every question
some questions are meant to be un-touched
limpid belfry
oh yeah for this question theres choices
wind cloak
oh yeah for this question theres choices
lmfao
limpid belfry
a) 20, b) 23, c) 26, d) 29
limpid belfry
nope
limpid belfry
are entrance exam questions always this hard😭
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vagrant prism
kinda clueless here
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vagrant prism
i'm supposed to use induction here
base case is fine:\
$(1-\frac{1}{\sqrt{2}}) < \frac{2}{2^2} = \frac{1}{2}$
pulsar dust
base case is fine:\\
$(1-\frac{1}{\sqrt{2}}) < \frac{2}{2^2} = \frac{1}{2}$
\frac{1}{2} not 2....
nova spire
wut
vagrant prism
oh sorry
solid kilnBOT
artemetra
vagrant prism
yes
checks out now
$1-\frac{1}{\sqrt{2}} = \frac{2-\sqrt{2}}{2} < \frac{1}{2}$ because $\sqrt{2} > 1$
solid kilnBOT
artemetra
pulsar dust
alright go on
vagrant prism
induction hypothesis is
$$\left(1-\frac{1}{\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{3}}\right)\cdots\left(1-\frac{1}{\sqrt{n}}\right) < \frac{2}{n^2}$$
solid kilnBOT
artemetra
vagrant prism
multiplying both sides by (1-1/(sqrt(n+1)))
$$\left(1-\frac{1}{\sqrt{2}}\right)\left(1-\frac{1}{\sqrt{3}}\right)\cdots\left(1-\frac{1}{\sqrt{n}}\right)\left(1-\frac{1}{\sqrt{n+1}}\right) < \frac{2}{n^2} \left(1-\frac{1}{\sqrt{n+1}}\right)$$
solid kilnBOT
artemetra
vagrant prism
so my task is to show that $$\frac{2}{n^2} \left(1-\frac{1}{\sqrt{n+1}}\right) < \frac{2}{(n+1)^2}$$
solid kilnBOT
artemetra
vagrant prism
wait is that necessarily true
vagrant prism
yes!
thanks
so my task is to show that $$\frac{2}{n^2} \left(1-\frac{1}{\sqrt{n+1}}\right) \leq \frac{2}{(n+1)^2}$$
solid kilnBOT
artemetra
nova spire
yep
pulsar dust
correct
vagrant prism
ookay
okay i am sorta stuck
hmm
$$\frac{2}{n^2} \left(1-\frac{1}{\sqrt{n+1}}\right) = \frac{2}{n^2} \left(\frac{\sqrt{n+1}-1}{\sqrt{n+1}}\right) = \frac{2}{n^2} \left(\frac{n+1-\sqrt{n+1}}{n+1}\right)$$
solid kilnBOT
artemetra
umbral dragon
okay i am sorta stuck
I think you can try to move the n stuff from the denominator into the enumerator
vagrant prism
**artemetra**
oo i have an idea
$= \frac{2}{n^2} \left(\frac{(n+1)^2-(\sqrt{n+1})(n+1)}{(n+1)^2}\right)$
solid kilnBOT
artemetra
vagrant prism
$= \frac{2}{(n+1)^2} \left(\frac{(n+1)^2-(\sqrt{n+1})(n+1)}{n^2}\right)$
solid kilnBOT
artemetra
vagrant prism
and since the goal is < 2/(n+1)^2, i need to show that the thing in the brackets is less than 1
nova spire
(or equal)
solid kilnBOT
artemetra
solid kilnBOT
artemetra
vagrant prism
$n^2+2n+1-(\sqrt{n+1})(n+1) \leq n^2$
solid kilnBOT
artemetra
vagrant prism
$2n+1-(\sqrt{n+1})(n+1) \leq 0$
solid kilnBOT
artemetra
vagrant prism
$2n+1\leq (\sqrt{n+1})(n+1)$
solid kilnBOT
artemetra
umbral dragon
If you can show this is true for all n >= 2 then I think so
vagrant prism
$\iff 2n+1\leq \sqrt{n+1}n+\sqrt{n+1}$
solid kilnBOT
artemetra
vagrant prism
actually i am not quite sure here
i would say this is definitely true for all $n \geq 3$ because $\sqrt{n+1} \geq 2$ and $\sqrt{n+1} \geq 1$
solid kilnBOT
artemetra
vagrant prism
and for $n=2$ i can check separately
nova spire
**artemetra**
you only need to square both sides
solid kilnBOT
artemetra
vagrant prism
you only need to square both sides
fuck
lmao
$\implies (2n+1)^2 \leq (\sqrt{n+1}n+\sqrt{n+1})^2$\
$\iff 4n^2 + 4n + 1 \leq (n+1)n^2 + 2(n+1)n + (n+1)$\
$\iff 4n^2 + 4n + 1 \leq n^3 + n^2 + 2n^2 + 2n + n+1$\
$\iff 4n^2 + 4n + 1 \leq n^3 + 3n^2 + 3n+1$\
okay
solid kilnBOT
artemetra
vagrant prism
hm
nova spire
everything on one side and factor
vagrant prism
$\iff n^3 + 3n^2 + 3n+1 \geq 4n^2 + 4n + 1$\
$\iff n^3 - n^2 - n \geq 0$\
$\iff n^2 - n - 1 \geq 0$ (because $n\neq 0$)
nova spire
pretty easy
umbral dragon
Tbh I find a combination of both nicer
solid kilnBOT
artemetra
umbral dragon
Tbh I find a combination of both nicer
Normal reasoning for n>=3 and square for n=2
But both works in the end
vagrant prism
hi!
i finally got to algebra lol
show
really not sure what is going on here
okay
this isn't true tho
take n=3
yeah the reciprocals always trip me up
i think i am done with mine
pretty much what was written here
but i'll rewrite it nicely now
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vagrant prism
okay i rewrote it nicely
here
.
i saw a typo and had to fix it 😆
thank you!
i think i'll close this one
thanks rafilou and convergence
.close
eternal vapor
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eternal vapor
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hollow tangle
!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
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hollow tangle
and which question?
eternal vapor
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hollow tangle
for 12
first check the condition
then factorize all the numerators and denominators
eternal vapor
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hollow tangle
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it can be further done
hollow tangle
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denominator must be non-zero in this case
eternal vapor
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hollow tangle
it can be further done
read the question it has a hint in it
eternal vapor
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dapper swift
yeah
eternal vapor
yeah
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hollow tangle
first factorize them
dapper swift
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you should factor first
hollow tangle
it will be easier
eternal vapor
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hollow tangle
split the middle term
eternal vapor
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hollow tangle
yes
eternal vapor
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dapper swift
yeah it's pretty clear that it factors as (5x ....)(x ....) cause 5 is prime
hollow tangle
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no
eternal vapor
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hollow tangle
for 5x^2 - 7x - 6 right?
eternal vapor
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hollow tangle
nope
eternal vapor
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eternal vapor
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eternal vapor
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eternal vapor
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hollow tangle
for 5x^2 - 7x - 6 right?
.
eternal vapor
for 5x^2 - 7x - 6 right?
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eternal vapor
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hollow tangle
yess
eternal vapor
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eternal vapor
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eternal vapor
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