#help-42
1 messages · Page 115 of 1
this because we can do this
hold on
$x^2+y^2 = 400-2xy \geq 2xy \implies 400\geq 4xy$
knief
so 100 >= xy
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wdym
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Kenzo
$f(x,y) = x^2+y^2 \quad g(x,y) = x+y$
knief
$\nabla f = \langle 2x, 2y \rangle \quad \nabla g = \langle 1, 1\rangle$
knief
$\langle 2x,2y \rangle = \lambda \langle 1, 1\rangle$
knief
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lagrange optimization
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$\lambda$
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knief
you won’t unless you take calc 3
partial derivatives
i think i might’ve shown you those
like when we did implicit
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remember my implicit differentiation trick
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$\frac{dy}{dx} = \frac{-\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}$
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knief
where $F(x,y) = c$
knief
i definitely showed you
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partial derivative
nah
implicit function theorem
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$\frac{\partial F}{\partial x} dx + \frac{\partial F}{\partial y} dy = 0$
knief
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black pen red pen
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they really did do a good job on the ap calc khan course
like the most effort they ever put into it
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the differential equations course on the other hand….
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not the differential equations in ap calc
like
further differential equations
more advanced
wasn’t so good
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needs a lot of work
multivariable was good on khan academy actually
because 3b1b did it
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you might’ve seen his essence of calc series
he’s probably the most popular math youtuber
makes all those sick animations
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Closed by @drifting seal
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b right?
yep
i think its a simply cuz its heavier
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Help pls
which part are you exactly confused with?
@obsidian swan Has your question been resolved?
For a, what does it mean about y intercept equals to 8 and for b, it says through point 0,-3 what does it mean and how should I apply in the equation?
y intercept means that the function intersects y-axis at -8
How should I apply it in the equation
remember that if a function passes through that point it is a solution of that equation
Alright
so, if a function passes through (0, -3) then it has a solution x=0 and f(x)=-3
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I have no idea what to do here
what are you even trying to do
Completing squares with fractions trip me out
Convert to vertex form so I’m tryna complete da square
How’d you get that?
I know that’s correct, but how do we reach dat conclusion bc the fraction within the parameters is tripping me out
this is just going in endless circles
???
what parameters
yes but no
correct as in the number is correct
but wrong as in the wrong step to take from the previous line
I should have left it as (3/2)^2?
yes
cuz what are you gonna do with this
Okay
So I leave it at 3/2 squared, and then what do I do from that?
I just hug it with brackets n make it (x +3/2)^2 n we good?
yes
Ok so
I understand
Now
Is this something I apply with all completing the square procedures that involve fractions?
That is, getting the (b/2)^2 n hugging dat + squaring it n eliminating da real b variable?
Ik how to complete squares yfm, it's just when they fractions it tweaks me out a bit
yes
treat a fraction like a regular number
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@sinful oak Has your question been resolved?
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4.- Let $B = {(0,1,1), (2,0,1), v}$ and $B' = {(-3,1,0), v, (0,0,1)}$ be bases of $\mathbb{R}^3$. Find, if possible, $v$ and $w$ in $\mathbb{R}^3$ such that the coordinates of $w$ in the basis $B$ are $(w)B = (3,-4,1)$ and the coordinates of $3w$ in the basis $B'$ are $(3w){B'} = (7,2,-4)$.
xbz
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✅
good
,av @onyx berry
9(0,1,1)-12(2,0,1) +3v = 7(-3,1,0)+2v-4(0,0,1)
3v - 2v = 7(-3,1,0) -4(0,0,1) + 12(2,0,1) - 9(0,1,1)
,w 7(-3,1,0) -4(0,0,1) + 12(2,0,1) - 9(0,1,1)
v = (3,-2,-1)
w = 3(0,1,1) -4(2,0,1) + v
w = 3(0,1,1) -4(2,0,1) + (3,-2,-1)
,w 3(0,1,1) -4(2,0,1) + (3,-2,-1)
good that I could help
,av @onyx berry
.solved
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I need to solve for x
try to write left side as (x-1)^smth
well you can also write 8/4(x-1) as 2/(x-1)
(2/(x -1))^(1/x) looks promising
Did that
assuming it’s xth root and not multiplied by x
okay so where are you getting stuck?
And finally get y^y^2 = 2
can i see all your work
holding
I was solving it with someone here but I didnt really like the solve
sorry i got class rn someone else gon help
i would usually just plug into desmos or smth to find it quickly
but an exact answer can’t give right now
Its pre differentiation
And I can't use a calculator
Also idk what desmos is
Anyways
It's alright
Good luck
graphing calculator
ty
Npnp
@wheat owl Has your question been resolved?
Try let a be x-1
Look
Do u see?
You only get one real solution x =0
It's the xth root of the mentioned thing
Not X . The square root
So what you did is wrong
Ifk then
@wheat owl Has your question been resolved?
Ima head to bed now
I'd appreciate if someone solves this by the time I wake up
Good night!
@wheat owl Has your question been resolved?
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Evaluate the function ƒ(x) = −x^2 − 1 and simplify at the indicated value: ƒ(−a) = ?
I got f(-a) = a^2 - 1 but it says its wrong
be careful about order of operations
-x^2 means first compute x^2, then multiply the result by -1
so lets say x = 1 -(1)^2 you do 1 * 1 = 1 then mulitply it by a negative to get -1?
ok that makes sense ty
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got 4.5 but it was wrong :p
yes
and the midpoint of walnut street is 0.9 miles
yes
isnt it 0.9
its the midsegment so its half of spruce street 
triangle midsegment theorem "the segment connecting the midpoints of two sides of a triangle is paralle to the third side and is half as long as that side"
none of these are right triangles right

yep we got it 
woo hoo
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hi i need help on a stats question
im just confused on what the
7% is used for
actually idek what formula to use
i think its the
n = z^2 ( p(p-1)) / e^2
@fleet cairn Has your question been resolved?
<@&286206848099549185>
ok so
i found the solution for it
but
im still confused
they said e = .07
so WHYYY in the bottom right is it .03?
.close
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formula
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cool cool
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Kenzo
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uhm
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uhm
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Kenzo
you see the problem is we don’t know r’
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uhhh
yes i know
water level
which is height no?
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so i’m thinking more like
uhh
i wanna say plug in r before differentiating but it isn’t constant
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yea
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i believe is it asking what is dh/dt when h = 5
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Kenzo
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or else you have 2 variables to work with which doesn’t work
this one is really easy man
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i don’t have a pencil
one second
yoy agree that the radius and the height wit change in proportion to one another
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so r = 45h/6
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plug thag back in
THEN differentiate with respect to h
and you win
gg ez
(i am also doing related rates in calc rn 😜)
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i’m a sophomore in our calc 1 advanced clas
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anyways u got the rest of the problem or any more questions
Kenzo
nonono
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h not h’
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Kenzo
combine terms???
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225h^2/4 * h/3
75h^3 * pi =V
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now
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WOAH
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im on my phone watching arcane eating famous amos cookies
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oki
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2
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V = 75h^3*pi
then differentiate
wait
oop
i forgot
V = 75/4 * h^3 * pi
not differentiate
now
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…
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i combined???
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okay
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do we understand
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Kenzo
???
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Okay
if you product rule
and you differentiate anything but h^3
you’ll have 0
so like
yes
do i needa restate
what’s the derivative of a•x^n
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yea
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Closed by @west horizon
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Kenzo
Looks good so far
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Kenzo
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Kenzo
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Think hard on this answer
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correct
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But +12 (don't forget your units) is still the correct answer to (a), btw
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Kenzo
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Wait, how did you get 12 for part a? 
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This is incorrect, but y=5 is correct. I presume you mean minus here
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Kenzo
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I was doing dy/dx derp
Kenzo
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yeah 12 ft/s is the rate at which the ladder falls
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naw area is $A=\frac12xy$
SWR
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and $\dot{A}=\frac12(x\dot{y}+y\dot{x})$
SWR
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,calc (12*(-12)+5*5)/2
Result:
-59.5
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discord hid the asterisks
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you multiplied by 2
Result:
-59.5
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Closed by @west horizon
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fr tho you had it all along. Just calculator mistakes
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.close
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how does it go from line 1 to line 2, more specifically the derivatives?
@odd heart Has your question been resolved?
<@&286206848099549185>
if you try simplifying the second line you get the derivative of $kp^k$ or $k^2 p^{k - 1}$, then multiply by $q$
south
so yeah you can interchange the derivative and the summation like that
also for the second summation just swap p and q everywhere that they appear
so instead of d/dp, you do d/dq and so on
i still don't get it
what keeps it equal?
oh nvm
yeah it's just weird
so it's because of this right $k^2p^{k-1}=k\frac{\mathrm{d}}{\mathrm{d}p}p^k$?
jacky
yeah
ok makes sense thanks
you just have to do the process again to fully get rid of the k right
yeah so that's $\frac{d}{dp} \left(p \frac{d}{dp} p^k \right)$
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south
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if I were to look at the left branch of this graph, so when x < 4, what's an intuitive way to see if the slope is negative & increasing or negative & decreasing at, when glancing at it.
My first assumption was negative & increasing but turns out it's negative & decreasing
Which side ?
the best way i'd like take a look is graphs go from left to right
as i go from negative infinity to approach 4 i am rapidly exponentially decreasing and becomign negative
^
When you got from left to right in x axis you should get increase in y for a increasing function
I guess my assumption was based on the fact that it's getting steeper, so the slope was "Increasing" but It's decreasing in value at a rapid rate, so the slope is "decreasing at a rapid/increasing rate"
That's if you were going from right to left, however graphs as you can see go left to right,
The tangent line compared to the exponential is rapidly at a decreasing rate
The best way is to get two points and see the slope
if you were to say x > 4 if you were to analyze that then it would be increasing, positive as the slope never goes below 0 and just flattens out
Its slope is negative because it's going down. And it's decreasing because it's going more and more down (becoming larger negative values)
always direction is left to right.
hmm well the way my class analyzed the the right side was that It's also negative slope, negative m value, except it's increasing
ahh I see but for the right branch, when x>4, the slope is negative as well, but couldn't you say it's decreasing since the y-values are getting closer to 0, when they initially started out bigger while being closer to the asymptote
A negative number decreasing means that the magnitude is increasing
So -3 decreases to -4 and so on
On the other hand a negative number increasing means its magnitude is decreasing, -3 increases to -2
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I’m confused, shouldn’t it be n-3 as the power not n-2?
what's the derivative of sec theta?
secxtanx
yep...
sure, yw
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Okay, so solving part a) is pretty much just plugging in number into the formula. 2pi * integral (g(y)+2) * \sqrt(1+(5y^4)^2)dy. Probably typed something in wrong (not sure how to do latex on discord).
But for part b, rotating around the x-axis, I'm not sure what I need to do.
The formula is basically the same, but do I need to change my limits of integration? Do I still use g(y) in my equation?
To be clear, this is an old exam. I'm going back over it in preparation for finals and this is one that I wasn't sure how to do.
You gotta integrate x in terms of y with the limits
So just [y^6/6]
Sorry, I think I'm being dense and missing what you're saying. I can see how integrating x would give us y^6/6, but I'm not seeing why that's the solution to rotating around the x-axis.
'I'm not doubting you, just trying to understand.
Ok so there is 2 ways to understand this
1 way : they gave limits for y (which for lower cases means you gotta integrate in terms of y)
If you gonna ask can't you get x from y , you can't here
The 2nd is trying it out
See when the rotation axis is x axis
And we gotta find an area between the two x axis line (y=a)
Which is only possible if we integrate the y axis
Okay, this is helpful. I feel like I need to draw the diagrams myself and work on some more problems. This was useful. Thank you.
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how to do these types of questions? i never understand how they just know
do you know binomial expansion
yeah to some extent
i can do normal binomial expansion but doing that wastes too much time in an exam
so they do the binomial expansion with combinatorics and algebra
Yes you shouldnt actually expand it
andi dont get the shortcutted combinatorics version
need help with this
use this formula Tr+1 = nCr x^n-r y^r
Basically, for a binomial of the form $(a+b)^n$ the $k$-th general term of its expansion is given by [
T_k = \binom nk \cdot a^{n-k} b^{k}
]
Thus, for your binomial, the $k$-th general term can be represented as[
T_k = \binom 6k \cdot \left(\frac{3}{x^2}\right)^{6-k} \cdot \left(-\frac{x}{2}\right)^k
]
Your objective is to simplify the above expression, and find under what conditions is the exponent of $x$ equal to 0
#❓how-to-get-help check here
Aero
oh, i see so from here it kind of is like guess and check to see what makes x's power = 0
You needn't guess
it is algebra
Try to simplify first
distribute the powers, group like terms, etc.
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In high school, does yall teacher make quizzes?
why do you ask
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In need of dire help, the answer sheet (very unreliable) tells that it is option (B).
It should be 1500Pa right?
it should be 0 because there's only atmospheric pressure in those points
like yeah, it will flow to the right, so there's a pressure difference, i'm just really confused
yeah if i pretend the tube is 1m² in section there's 150 kg difference
so 1500 Pa
@remote mural Has your question been resolved?
<@&286206848099549185>
yes
ok are u here now?
pls ping me when u r
anyway
you agree we can just cut of the length b from both of the parralel sides, withouth changing the distance we want to find?
yes
so now we have a traingle
with the top angle 90*
and the bottom length a-b
and we want to find the median length
can you show the diagram?
yes
now you contract CD around the center by length AB
can you do that?
just make the midline first
do you get this?
@remote mural
Oh so AE = DF?
yes
and you know that the median length of a rightangle triangle is half the hypotenuse
since
if one side is a, and the other sidelngth is b
I know this
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welcome
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If $F(x)=\int^x_1 e^x+e^t dt$, then $F'(x)=2e^x$ is this correct?
Volas
yes.
Closed by @prisma wyvern
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wait
the way you wrote it
$F(x) = (x-1)e^x + e^x - 1$
rafilou is not not born in 2003
so $F'(x) = (x+1)e^x$
rafilou is not not born in 2003
you can only use FTC if you have $F(x) = \int_a^x f(t)dt$
rafilou is not not born in 2003
✅
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so i split this into 2 integrals
and when im solving the left integral
im using x = 2tan(theta)
but when im using photomath to double check its saying the answer is ln something
how to do
@fleet cairn Has your question been resolved?
this should work
the answer is gonna be a hyperbolic function, which can be also written through lns
btw there is a simpler approach
of what "degree" is the first fraction? Of what degree is the second one?
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Let V be a finite set with at least two elements. Does the structure of a Q-vector space exist on V?
how would I go about showing or disproving this?
@weary cape Has your question been resolved?
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<@&286206848099549185>
if by absurd you could create such a structure
call v1 the element that would correspond to the 0 vector
and v2 a different element
Can you explain why there must exist different p,q in Q such that p * v2 = q * v2?
then conclude as to the existence of such a structure
$\mathbb{Q}$ is infinite right?
aName
now I am honestly already confused at what it means for a Q-Vectorspace to exist on a set.
because if I take any element/vector in V and use scalar multiplication to multiply it with a rational number then it might not exist in V anymore
well it literally should
if you have a Q-vectorspace structure
it means that you have
an addition operation
meaning for v,w in V
v+w in V
so (V,+) is a commutative group
and a scalar multiplication operation
meaning that for q in Q, v in V
q * v is in V
and it satisfies (pq) * v = p * (q * v), 1 * v = v and distributivity
so yes, Q is infinite
hm if v2 is the only other vector existing in V aside from the 0 vector then multiplying it can't do anything to it right?
cuz if p*v2 ≠ v2 it can only be v1
well for example
that would only be possible when p = 0
in a vector space, if p * v = the zero vector, then p = 0 or v = the zero vector
yes but there are no other vectors besides v1 (zero vector) and v2 so scalar multiplication wouldn't work with any elements of Q besides 0
and therefore it cannot exist?
?
you haven't told me why yet
i could very much have p*v2 = v2 when p is not 0
why doesn't it work?
the neutral element wouldn't be unique?
i remember a proof we had where if there are two elements having the property of e * v = v = e' * v then e = e'
but then for all q in Q q*v2 = v2
well try to apply that
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hi
how can i solve this?\
given an angle and radius, what is the formula for the arc length, and the formula for the area of the sector?
$\frac{\theta}{2 \pi} \cdot \pi r^2 = 6\pi$
oops
the area of the sector can be described like this
now you are mixing degree and radians
tenshi
$\frac{\theta}{2 \pi} \cdot 2 \pi r = 2\pi$
tenshi
hm i got r = 6?
I got the same radius
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wait how do you do this again 💀
just supstitue t to be 60?
i think?
not that simple lmao
