help-10

[knotty geyser]

Yo

[obtuse pebble]

Please don't occupy multiple help channels.

[knotty geyser]

Help

[tiny seal]

not enough context

about?

[knotty geyser]

Surface area

[tiny seal]

of the whole cylinder?

[knotty geyser]

or a cylinder

yes

[tiny seal]

its a rectangle and two circles

[knotty geyser]

Yes

[tiny seal]

what is the width of the rectangle

[knotty geyser]

5mm

I mean 6

I mean 7

[tiny seal]

no

the width of the rectangle if we like, "unwrap" it from the cylinder

basically, the circumference

[knotty geyser]

Oh ok

[tiny seal]

what is the circumference

[knotty geyser]

How would I get it log

Length times height*

[tiny seal]

right

[knotty geyser]

But I don’t know the height

[tiny seal]

of the rectangle?

[knotty geyser]

Yes

[tiny seal]

open a channel for it

[frosty anvil]

wrong channel

sry

[tiny seal]

yg

[knotty geyser]

Don’t worry

So

When I unfold it

I will have 2 circles and a rectangle

Length times height

What would the length and the height be

[tiny seal]

we are given the height

[knotty geyser]

Which is 7

But the length is what?

2

?

[tiny seal]

the length is the perimeter of the circle

[knotty geyser]

Oh

Ok

Perimeter

Is

[tiny seal]

circumference

do you remember the formula?

[knotty geyser]

2 pie r

[tiny seal]

right

whats r

that will give you the width of the rectangle

[obtuse pebble]

@knotty geyser Has your question been resolved?

[obtuse pebble]

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[runic thunder]

Would it be at t = 5?

[viral blade]

assuming up means moving right then yes

after that point it starts moving left

[runic thunder]

Alright thought so

.close

[obtuse pebble]

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[ruby elm]

Problem 5.

[ruby elm]

I do not know how do I approach this one.

[timber island]

Well you know that 0.5^|x| is a decreasing function

[ruby elm]

Yes.

[timber island]

And x^2 - a can only be decreasing when?

[ruby elm]

x^2-a is always increasing right?

[timber island]

Yeah not always

If x^2 < a, then it must be decreasing no?

[ruby elm]

I understand, when x is less than 0

[timber island]

Or 0

[ruby elm]

Yes.

[timber island]

Wait oh shit

0.5^x can never be negative as long as x is real right

[ruby elm]

I did it, using graph makes it easier.

.close

[obtuse pebble]

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[fading cedar]

Can someone explain the exponent rule?

of this?

[red knoll]

it becomes this so

[fading cedar]

ohhhhhh

lmfao

thnak you

[obtuse pebble]

@fading cedar Has your question been resolved?

[obtuse pebble]

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[shadow lava]

@dark stirrup https://www.reddit.com/r/askmath/comments/12rhf3y/comment/jguhpoe/?utm_source=share&utm_medium=web2x&context=3

[shadow lava]

(FYI)

[wild swallow]

lol

people dont understand that implications dont go the other way

sure, f'(x) > 0 => f(x) is increasing

but you dont have the other way

[shadow lava]

Organic Chemistry Tutor, my university professor, Reddit, they are all wrong?

Everyone else is telling me rounded brackets

So i'm super confused

[wild swallow]

realise that the concept of a function increasing is not restricted to differentiable functions

so it makes no sense to define it in terms of derivatives

[shadow lava]

this is what was given to me

I'm just passing on the message

(don't shoot the messenger)

just trying to get to the bottom of this once and for all

[tardy epoch]

That's a theorem, not a definition

[wild swallow]

also the reddit comment saying that a function being constant at a single point is a joke

[wild swallow]

you could say any function is constant at a single point

its a nonsensical statement

[shadow lava]

lmao alright..

so that's 3/3 now

[rocky goblet]

yeah a function isn't increasing/decreasing/constant at a point, it does that over an interval

[shadow lava]

but why does everybody else have it wrong? @nocturne minnow included

[rocky goblet]

well if f is increasing on [a, b] then it is also increasing on (a, b), because that's a slightly smaller interval

[tardy epoch]

Are you sure you're not just misreading again

[shadow lava]

https://tenor.com/view/blinking-eyes-white-guy-blinking-meme-what-huh-gif-26334322

[rocky goblet]

if someone is specifically claiming that this theorem is false in general, then either they're wrong or the people claiming this is correct are wrong

[shadow lava]

() vs []

that is the question

[tardy epoch]

That's not a coherent question

[shadow lava]

for interval notation: i don't know how i would be misreading, it's fairly a simple question

when it comes to increasing/decreasing

[viral blade]

both can be true

[] is just stronger

[shadow lava]

everybody is telling me don't include f'(0)

[viral blade]

derivative positive on () implies increasing on []

hmm

[rocky goblet]

...f'(0) is completely irrelevant

[viral blade]

that's not actually true is it

Oh

[shadow lava]

actually it's very relevant

[viral blade]

But if f is continuous it should be

See there's your problem

f isn't continuous

[wild swallow]

how can f not be continuous if its differentiable

[shadow lava]

lmao

[viral blade]

f needs to be continuous on [a,b]

not just (a,b)

continuity on [a,b] is not implied by differentiability on (a,b)

[wild swallow]

classic MVT

[viral blade]

so the problem is that f isn't continuous on the closed interval

that's why the theorem doesn't apply

[wild swallow]

its a polynomial

(3/2) x^2 not 3/(2x^2)

[viral blade]

oh I misread I thought it was a rational

[rocky goblet]

yeah f is definitely continuous everywhere

and also even if it wasn't the point 0 doesn't matter

[shadow lava]

https://tenor.com/view/ross-chewing-smug-stare-friends-gif-15826364

[rocky goblet]

the boundaries of the intervals are -4 and 3

[viral blade]

ok then yeah this is a certified calc 1 moment

To make things easier on students sometimes increasing is just described as positive derivative

and vice versa

[wild swallow]

"1/x is discontinuous at x=0"

[shadow lava]

i'm just gonna use regular parenthesis for this question (). take me to jail. do not pass go. do not collect $200. but in the end, i win Monopoly

[gilded needle]

ha what, this again?

i thought in the last channel, the consensus was, follow whatever bullshit convention your instructor is using, but recognize that it's not strictly correct when you get to more pedantic mathematics like real analysis

[shadow lava]

https://tenor.com/view/simpsons-games-hard-gamenight-gif-7531474

[gilded needle]

in all seriousness i think it's just a matter of convention.. and maybe of how you interpret "the interval on which the function is increasing".. that could mean one of two slightly different things: (1) "if x and y are in the interval and x < y, then f(x) < f(y)" [in this case you'd include the endpoints in your example], or (2) "the set of points x where f is increasing at x" [here you might say that f is not actually increasing at a critical point if "increasing at a point" is interpreted to mean "increasing on an open neighborhood of the point"]

[obtuse pebble]

@shadow lava Has your question been resolved?

[fathom flicker]

typical Bungo W moment

[gilded needle]

hey now

[fathom flicker]

🇼

[obtuse pebble]

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[celest hazel]

Can someone help with this?

[obtuse pebble]

@celest hazel Has your question been resolved?

[celest hazel]

<@&286206848099549185>

Pls can anyone help

I’ve tried doing a system for the first parabola but idk how to proceed

[celest hazel]

Pls

.close

[obtuse pebble]

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[celest hazel]

.reopen

[obtuse pebble]

✅

[celest hazel]

<@&286206848099549185>

Nvm the other question someone helped me alr

How do I write the equation of a parabole knowing the coordinates of the tangent point to a line of which we know the equation?

[obtuse pebble]

@celest hazel Has your question been resolved?

[celest hazel]

Dude it’s been 3 hours

[celest hazel]

@shrewd sphinx

The tangent point is A(0,1)

The equation of the line is y=4x+1

And we also know the parabole passes through b(2,5

[shrewd sphinx]

So first you have the general equation of a parabole p(x) = ax^2 + bx + c . I would start differentiate the equation p'(x) = 2ax + b

Then you know the gradient from the tangent equation equals 4 at the point A(0,1)

[celest hazel]

Ye

@shrewd sphinx so I replaced a and b with the coordinates of b

Idk what to do next

[shrewd sphinx]

no don't replace them

with the equation p'(x) = 2ax + b you can now put in p'(x) = 4

so it would be 4 = 2ax + b

because p'(x) is the gradient

and then you put in x 0 . So it would be 4 = 2 * a * 0 + b

and if you calculate it the result is b = 4

[celest hazel]

Damn

Thanks man

@shrewd sphinx what if I have the equation of the parabole and I have to find the equation of the lines tangent to the parabole (I know the lines are parallel to y=x)

[shrewd sphinx]

first you differentiate the parabole equation, second you put in the derivative function f'(x) = 2ax + b the gradient 1

f'(x) = 1 = 2ax +b

and then you transform the equation to x

x = (1-b) / (2*a)

[celest hazel]

Alr thanks so much

.close

[obtuse pebble]

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[obtuse pebble]

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[quiet flume]

how to prove the derivative of sinx is cosx

[sage geode]

Tried using the definition of a derivative?

[quiet flume]

i am not understanding how to use it

[sage geode]

Here you simply need to evaluate the limit $\lim_{x\to{a}}\frac{\sin{x}-\sin{a}}{x-a}$

[warm shale]

A Lonely Bean

[sage geode]

Hm, it's better to use the other definition though, wait

$\lim_{h\to0}\frac{\sin{(x+h)}-\sin{x}}h$

[warm shale]

A Lonely Bean

[sage geode]

Any ideas on how to simplify this?

[quiet flume]

i dont know how to simplify

[royal basin]

do you know angle sum identities for sin and cos

[sage geode]

Are you familiar with sum of angles identity for sine?

[royal basin]

sniped

[quiet flume]

no, i am weak in trigo

[sage geode]

Well that's needed here

[royal basin]

then you need to refresh yourself in trigonometry

[quiet flume]

i just need the solution

[sage geode]

$\sin{(x+h)} = \sin{x}\cos{h}+\sin{h}\cos{x}$

[warm shale]

A Lonely Bean

[sage geode]

Consider the limit knowing this

[quiet flume]

ok

.close

[obtuse pebble]

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[sage geode]

Oh and btw may want to keep the double angle identity for cosine in mind as well

[obtuse pebble]

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[quiet flume]

how to differentiate non separable equation in 2 variables

[royal basin]

what

can you maybe show us the problem you're looking at

[obtuse pebble]

@quiet flume Has your question been resolved?

[obtuse pebble]

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@wooden blade Has your question been resolved?

[obtuse pebble]

@wooden blade Has your question been resolved?

[obtuse pebble]

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[timid silo]

Cube ABCD.A'B'C'D'. Vecto AB equals vecto a, vecto AD equals vecto b, vecto AA' equals vecto c. Demonstrate vecto AC, vecto B'D', vecto BD', vecto DB', vecto BC' and vecto AD' according to vecto a, vecto b and vecto c.

[obtuse pebble]

Please don't occupy multiple help channels.

[timid silo]

can anyone help me

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[quartz hollow]

the derivative of a definite integral is so confusing
i have no idea

[tiny seal]

the fundamental theorem of calculus

$\frac{d}{dx}(\int_{a}^{x} f(t)dt) = f(x)$

[warm shale]

Cycadellic

[stone spoke]

it looks like a fund thrm at first but wouldn't you just take the cos(x) out of the integral?

[tiny seal]

well, we want G'

that will undo the integral, then it will set t=x

cause x is the top bound

[stone spoke]

oh I only looked at part a

yeah that's my bad

[tiny seal]

yeah, i should specify

FTOC for both

[quartz hollow]

theres x and t at the same time which should i deal first

[tiny seal]

t=x just for b

[tiny seal]

look at ftoc

its saying we can remove the integral

then we can plug in t=upper bound

so plug in t=x

for problem b

[quartz hollow]

the issue is in part a there is both x and t within the cos

[tiny seal]

yeah

[quartz hollow]

isnt it a function of both x and t

[tiny seal]

ignore the x

just plug in t=x

(youll see how we do tx)

soo

given t=x

what is xt

[quartz hollow]

t^2

[tiny seal]

right but lets do x

x^2

[quartz hollow]

so the derivative is just cos x^2

[tiny seal]

right

and since its a bounded integral, the C canceled so we ignore it

even if we had a C, the derivative would take care of it and make it 0

so theres your answer

[quartz hollow]

part b is much harder

G’(x) is just cos x if we use the same method in (a)?

and then sub x = 1

[trail cloak]

That's only true if your bounds are 0 to x

$\int_{\frac 1x}^{x} \cos(\sqrt{xt}) \dd t$

[quartz hollow]

part a goes from 0 to 1

not to x

[warm shale]

VulcanOne

[trail cloak]

Oh ye

Well

[quartz hollow]

then what happens

[trail cloak]

Well when you integrate, you will place the bounds

You will get something like

$\int_{\frac 1x}^x f(t) \dd t= F(x) - F\left(\frac 1x\right)$

[warm shale]

VulcanOne

[trail cloak]

Now when you differentiate, the left side will have the derivative of this integral

The right side will have

$\dv{x} \left( F(x) - F\left( \frac 1x\right)\right) = f(x) \cdot 1 + f\left( \frac 1x\right ) \cdot \frac{-1}{x^2}$

[warm shale]

VulcanOne

[quartz hollow]

so we replace the “t” inside the square root first

with x

[quartz hollow]

and then plug into this thing

[trail cloak]

Kinda

$\dv{x}\int_{\frac 1x}^x \cos(\sqrt{xt}) \dd t = \cos(\sqrt{x^2}) \cdot 1 - \cos(\sqrt{\frac xx}) \cdot \frac{-1}{x^2}$

[warm shale]

VulcanOne

[trail cloak]

We took away the t from the original function and we replaced it with the bounds

There is one more part missing from that

Are you familiar with something called Leibniz differentiation rule?

[quartz hollow]

no

[trail cloak]

Okay

Oki so

When you differentiate an integral that has 2 variables

Like the one we have

We expect an additional term to appear

It's the integral of the partial derivative of the function

So like

Something like your question has both x and t

So we do the same thing we did but we add on to it the integral of the partial derivative of cos(sqrt(xt)) with respect to x

$\int_{\frac 1x}^{x} \pdv{x} \cos(\sqrt{xt}) \dd t$

[warm shale]

VulcanOne

[quartz hollow]

what is partial derivative

[trail cloak]

Like normal derivative

But you ignore everything except for what you are differentiating for

Like here you are differentiating with respect to x

So you treat everything else as a constant

While when you find any x, you take it's derivative like normal

[quartz hollow]

understanding this problem is just way too hard for me

let alone solving it

[trail cloak]

[quartz hollow]

i can do normal integrals just fine

but not integrals with differentiation

[trail cloak]

It needs some getting used to

And a bit of practice

[quartz hollow]

is it possible to just evaluate the indefinite integral first then apply derivative

[trail cloak]

It's quite difficult to evaluate the integral

[quartz hollow]

i would honestly evaluate integral then do derivatives seperately in part a

[trail cloak]

Part a can work

Because it is simpler

[quartz hollow]

instead of using fundamental theorem of calculus

[trail cloak]

Part b needs Leibniz integral differentiation rule

[trail cloak]

FTC came from this concept

Evaluating the integral and applying the bounds

Actually

$\int f(x) \dd x = \int_c^x f(t) \dd t = F(x) + C$

Where c is a constant

And we consider F(c) = constant

[warm shale]

VulcanOne

[trail cloak]

Now when you take the derivative of this

You get back your f(x)

However in the case where you have an integral with an integrating variable and variable bounds, it differs

$\int_{a(x)}^{b(x)} f(x,t) \dd t = F(x, b(x)) - F(x, a(x))$

[warm shale]

VulcanOne

[quartz hollow]

the existence of “t” in the problem is the biggest issue

it confuses everyone

[trail cloak]

Yeah

Or rather

The x

In the cos

You will have to consider an additional step

$\pdv{x} \cos(\sqrt{x t})$

You'll have to take this derivative

Treat the t as a constant

[warm shale]

VulcanOne

[quartz hollow]

it already feels so weird differentating this

[trail cloak]

Why?

Chain rule

[quartz hollow]

oh ye

dont know how to type so i have to write

[trail cloak]

Correct

Missing one thing

t

Or rather the derivative of xt

[quartz hollow]

t + x (dt/dx) by product rule

[trail cloak]

No you're treating t as a constant in this case

[quartz hollow]

so just t

got used to implicit differentation

[trail cloak]

Mood

Lol

[quartz hollow]

like this

[trail cloak]

Yep

Now so far

$\dv{x} \int_{\frac 1x}^x \cos(\sqrt{xt}) \dd t = \cos(\sqrt{x^2}) +\frac{\cos(\sqrt 1)}{x^2} \ \+ \int_{\frac 1x}^x \frac{t}{2\sqrt{xt}} \cos(\sqrt{xt}) \dd t$

[warm shale]

VulcanOne

[quartz hollow]

hope ill never see problems this hard in exam

[trail cloak]

I hope things get easy

[quartz hollow]

the hardest example in our notes is much easier than this

luckily this is just homework

might as well skip this question

[trail cloak]

Wait

https://integral-calculator.com

[quartz hollow]

time to quit wolfram alpha

bye

.close

[obtuse pebble]

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[obtuse pebble]

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[steel goblet]

so for this problem, I did (10C2)(10C3)(10C5)

is that correct?

[haughty coyote]

it's not 10 spots left everytime

[steel goblet]

okay so the first 2 digits would be 10C2

now there are 8 places left

so for the three 1's, would it be 8C3?

[haughty coyote]

yes

[steel goblet]

okay so now I have (10C2)(8C3)

do I multiply that by 5!

[haughty coyote]

the rest of the digits are forced to be 2 already

it's 5C5 = 1

[steel goblet]

so the final answer is just (10C2)(8C3)

[haughty coyote]

yes

[steel goblet]

oh okay, thank you

one more question:

i know i can easily use the stars and bars theorem to obtain 1140, but im trying to understand it visually

i drew this:

the 3 dashes to the left represent possible integers that are 0

and the 17 circles represent the number 17. So then I will draw 3 "separators" to divide 17 into 4 integers

So for the first separator, there are 19 possibilities, right?

.close

[obtuse pebble]

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[solar crown]

this is one idea i came up with, what do you think and do you have other ideas?

rotate the bottom layer of the cube once to the left, then rotate the right layer of the cube down once, then rotate the bottom layer of the cube right once, then rotate the right side of the cube up once

rotate: ±
bottom layer: !
bottom: @
layer: #
left: $
right: %
down: ^
up: &

±@#$
±%^
±@%
±%&

with left, right, up, down: blrdbrru

[obtuse pebble]

@solar crown Has your question been resolved?

[obtuse pebble]

@solar crown Has your question been resolved?

[obtuse pebble]

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[runic plume]

people say that x/0 is undefined but I think its infinite as,
z/0.0000...1 is a possible expression which will have a large ans so as you decrease the denominator of an expression the larger the answer is.
Eg: x = 1
1/0.1 = 10 and 1/0.01 = 100 hence it proves my point
is this true?

[candid sigil]

what about z/-0.000001

-inf

[timber fox]

no

[runic plume]

why

[candid sigil]

usually people say that there is no limit

[solar trellis]

0.000..1 is not very precise

[drifting wraith]

i don't know why people say no

if you already believe infinity is a number, then yes

[solar trellis]

0.9999... is precise because it means the infinite sum 9/10 + 9/10^2 + ...

And this converges

[drifting wraith]

i guess people see infinity as some limit thing

[runic plume]

dang I am confused this is way too much for my 13 year old brain

[candid sigil]

i saw a video that disscussed something called wheel algebra

[drifting wraith]

when you say infinity math peple think of something complicated

[runic plume]

can you send a link of that video pls

idk

[candid sigil]

i cant find it hold on

[dark stirrup]

Yeah there's wheel theory, and complex analysis will sometimes consider "the point at infinity"

All approaches have tradeoffs

[candid sigil]

https://www.youtube.com/watch?v=5WbK9O9JNDk

[runic plume]

thanks

should I leave this open or should I close it

[tardy epoch]

do .close

[timber fox]

.coose

[runic plume]

.close

[obtuse pebble]

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[obtuse pebble]

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[vivid creek]

Why is this true I do t get it

[hard citrus]

look at a unit circle

[earnest tulip]

or a graph

[timid silo]

sine is an odd function

while cosine is an even one

[vivid creek]

Oh ok thanks

.close

[obtuse pebble]

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[carmine surge]

need help with the question all the way at the bottom

[obtuse pebble]

@carmine surge Has your question been resolved?

[obtuse pebble]

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[low topaz]

is this correct?

[low topaz]

$1^2 + (r-0.6)^2=r^2$

[warm shale]

Badekarret

[native inlet]

I'm confused what are you asking here? @low topaz

[obtuse pebble]

@low topaz Has your question been resolved?

[low topaz]

is my equation correct?

[native inlet]

In what regard?

I still don't understand what you're trying to do

[obtuse pebble]

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[mighty agate]

Calculate the area of ​​the triangle determined by the lines: y=2x+6, y=-x+6 and the 0x axis

[mighty agate]

I already calculated all the three coordinates: (0,6), (-3,0) and (6,0)

And I know the formula for an area of a triangle is A=ab/2

but how am I gonna calculate that using these lol

[leaden ginkgo]

use the coordinate geometry formula

[mighty agate]

Right. Which one though? 😅

[obtuse pebble]

@mighty agate Has your question been resolved?

[mighty agate]

<@&286206848099549185>

[night plaza]

@mighty agate

what are the lengths of the base and height?

[mighty agate]

I don't know... I would have to calculate, I don't know how.

[night plaza]

so the points comprising the base are (-3,0) and (6,0)
the horizontal distance between them is x2-x1

sorry (6,0)

[mighty agate]

Yes

[night plaza]

6-(-3)

and the height is 6

[mighty agate]

Waiiiit. Do I have to draw it to get it? I didn't draw it, I just calculated these 3 coordinates and that's it.

I can clearly see that the height is 6 on the graph xD

[night plaza]

i think the drawing helps visualize yeah

got it?

[mighty agate]

One sec

It's correct!

thank you!

[night plaza]

cool have a agreat night

[mighty agate]

.close

[obtuse pebble]

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[obtuse pebble]

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[fathom flicker]

Let p(D) be the operator whose unit impulse response is given by $w(t)=e^{-t}-e^{-3t}$

[warm shale]

AustinU

[fathom flicker]

I am supposed to use convolution to find the unit step response of this operator

[warm shale]

AustinU

[fathom flicker]

and I'm not sure where to go from there or how convolution plays into this

Can't I just take the laplace transform of w(t) to get 1/p(s) where s is the characteristic polynomial of the system?

[obtuse pebble]

@fathom flicker Has your question been resolved?

[obtuse pebble]

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[white quest]

Could someone walk me though through how to do this problem? Thanks!

[obtuse pebble]

Please don't occupy multiple help channels.

[fierce lagoon]

#help-14 you have this channel open

I'll close it for you

[white quest]

Mb, sorry about that.

[fierce lagoon]

What have you tried

[white quest]

I know how to get the magnitude and find the angle of a vector, but I have no idea how to go backwards

[fierce lagoon]

Are you familiar with polar coordinates?

You can apply a similar concept

[white quest]

Yea

[fierce lagoon]

in this case, r is |v|

And you know |v| = 14

so $v = \vecb{|v| \cos(\theta), |v| \sin(\theta)}$

[warm shale]

Umbraleviathan

[white quest]

so, (14cos(295),14sin(295) ?

[fierce lagoon]

Yeah

[white quest]

Would v = 5.92i - 12.69j be the correct answer? if you round to the nearest hundredth?

[fierce lagoon]

Seems legit yeah

[white quest]

Okay, thank you!

Appreciate it

[fierce lagoon]

Remember to close the channel

[white quest]

.close

[obtuse pebble]

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[peak barn]

!status

[obtuse pebble]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[jolly talon]

what do i do ?

[peak barn]

Using the table, do you know what 1 yard equals in inches?

[jolly talon]

36

[peak barn]

yes!

so you are given the depth of the crater in yards

[jolly talon]

ooooohhhhhh

i get it

[peak barn]

$$ 1 \text{ yd} \ra 36 \text{ in} $$
$$ 536.451 \text{ yd} \ra \text{ ?} $$

[warm shale]

Talent Unlimited

[obtuse pebble]

@jolly talon Has your question been resolved?

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[brave moon]

is this sets cardinality 2 or 3 {1,2,2}

[warm canopy]

2

[late hawk]

i need help

[warm canopy]

{1,2,2} is the same as the set {1,2}

[brave moon]

ok cool thank you<3

[brave moon]

do you know what m and b mean

[late hawk]

m means slope

[brave moon]

yes

[late hawk]

b means the y intercept

[brave moon]

yes

do you know how to get slope

[obtuse pebble]

Please read #❓how-to-get-help

[late hawk]

rise over run

but i got 1

[brave moon]

yes

[late hawk]

1/1

[brave moon]

so it would be y = x + y intersept

[peak barn]

yes

[nocturne minnow]

This is not your channel, please read #❓how-to-get-help and open your own

[peak barn]

what is the y intercept?

[late hawk]

1

[brave moon]

yes good

so what would be the equation

[late hawk]

it just be 1

[peak barn]

wouldn't it be 4?

the y-intercept

[nocturne minnow]

If you are done with your channel, please close it

[brave moon]

yeah its 4

[late hawk]

oh

[obtuse pebble]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[brave moon]

mb i didnt look at the graph

[peak barn]

lol

[brave moon]

so the equation is y= slope multiplied by x plus 4

[peak barn]

and also the slope isn't 1

[brave moon]

how would you write that in y=mx+b

lmao bro let me look at the graph

[late hawk]

y=1x+4

[peak barn]

@late hawk can you show how you got the slope?

[late hawk]

slow is 1/1

[brave moon]

if its 1x you dont neef to wright the 1

[late hawk]

run over rise

[brave moon]

just do x

[late hawk]

y=x+4?

[peak barn]

it's -1, the rise is negative

[brave moon]

yes but -x

[late hawk]

u right

slope going down

[peak barn]

yup

,w graph y=-x+4

[warm shale]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by Talent Unlimited#7292

[dark stirrup]

y'all know this ain't addax's channel right?

[late hawk]

what channel then

[brave moon]

mb

.close

[obtuse pebble]

Channel closed

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Please read #❓how-to-get-help

[dark stirrup]

!help

[obtuse pebble]

Please read #❓how-to-get-help

[obtuse pebble]

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[short warren]

how do i solve this help

[short warren]

i cant prove identities

i tried taking the left handside and simplfy it

but i cant get the answer

any tips on how to solve pls?

[obtuse pebble]

@short warren Has your question been resolved?

[short warren]

.close

[obtuse pebble]

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[timid silo]

how do i do this using calculus

[uneven palm]

related rates

you may want to draw a diagram

[timid silo]

i tried

idk how to do it

[blissful solstice]

can you show me ur diagram if not i'll draw one real fast

this looks kinda bad but whatever

[timid silo]

yeah basically what i drew

then i took the derivative of x^2 + y^2 = c^2

when x = 80 and y = 20

and then (dx/dt) = 40 and (dy/dt) = 30

[blissful solstice]

then , did you stop or ?

[timid silo]

i subbed it all in but i got the wrong answer

[blissful solstice]

whats the right answer

or if you dont know its fine

[timid silo]

130/root 17

[blissful solstice]

hmm umm let me sub it in too

[timid silo]

alr

[blissful solstice]

wait did you substitute before or after differentiating

[timid silo]

after differtiating

thats what ur meant to do right?

[blissful solstice]

yes

[timid silo]

idk why its not working

did you try it?

[blissful solstice]

uhh yea did not work for me

Imma probably redo it

like restart from scratch

[timid silo]

alr

ty

[timid silo]

i got it

its all good

the 30 is meant to be negative

[blissful solstice]

oh

welp

[timid silo]

since its going back to the reference point

ya lmao

mb

[blissful solstice]

I was so confused

why it wasnt working

[timid silo]

yeah lol

can u help with one more?

[blissful solstice]

okay let me see im not that good tho

[timid silo]

alr

[blissful solstice]

U see I don't remember how ramps effect the speed so, yea I can't unless u remind me how that works

[timid silo]

idk either lol

all good

[blissful solstice]

I mean I'll google it

managed to literally find the question so , like

https://www.wyzant.com/resources/answers/67771/a_waterskier_skis_over_the_ramp_with_a_base_of_5_m_and_a_height_of_1_m_at_a_speed_of_12_m_s_how_fast_is_she_rising_as_she_leaves_the_ramp

yea

now

I dont have to look up physics yay

[timid silo]

alr ty

.close

[obtuse pebble]

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[timid silo]

What's 3y²+4y²

[timid silo]

I need help with the exponent part

[gilded needle]

just pretend y^2 is a fixed number

3 times some number + 4 times the same number = what?

[timid silo]

Ok

Idk

[gilded needle]

would you know how to simplify 3y + 4y ?

[timid silo]

Adding them

[gilded needle]

sure, that's what the + indicates

the result is what?

[timid silo]

7y

[gilded needle]

yep

it works exactly the same way with y^2 instead of y

[timid silo]

So 7y^4?

[gilded needle]

no

7y^2

[timid silo]

So I don't add exponents

[gilded needle]

right

you would do that if you were multiplying

(y^2)(y^2) = y^4

y^2 + y^2 = 2y^2

[timid silo]

So If I multiply exponents it is adding?

[gilded needle]

if you are multiplying things of the form y^m and y^n, then you add the exponents:

$(y^m)(y^n) = y^{m+n}$

[warm shale]

Bungo

[timid silo]

So say y⁶+y⁵ would = y¹¹ instead of 30

[gilded needle]

no, you can't combine y^6 and y^5 by adding

or rather you can't simplify their sum

it's just y^6 + y^5

you can only simplify addition when the exponents are the same

y^6 + y^6 = 2y^6

[timid silo]

What about 3y^6+4y^5

[gilded needle]

that's as simple as you can make it

[timid silo]

Ok

So if the exponent are the same u simplify

[gilded needle]

yep

for addition

[timid silo]

.close

[obtuse pebble]

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[bleak eagle]

is the SVD of a real matrix unique? im finding conflicting answers in my research

[gilded needle]

almost unique

the columns of the orthogonal matrices can be multiplied by -1 and it's still valid

[bleak eagle]

ohhh that makes sense!

[gilded needle]

also the diagonal elements can be reordered, but the standard representation is to put them in descending order

[bleak eagle]

thats exactly what i was going to ask next

if they were not ordered conventionally in descending order, would that make the columns of the orthogonal matrices be in different order?

[gilded needle]

yes

oh also one more thing

when you have repeated singular values (same value appears multiple times)

[bleak eagle]

okay!

yes

[gilded needle]

then the columns of the U and V matrices can be shuffled around (permuted)

[bleak eagle]

just the columns that correspond to the repeated singular values though, right?

[gilded needle]

yes

exactly

think about the identity matrix

[bleak eagle]

okay!!

[gilded needle]

it can be represented in the form U^T U for any orthogonal matrix

[bleak eagle]

yes, totally

for sure! thank you for the clarification! thats essentially what my research was suggesting but i wanted to make sure

[gilded needle]

sure, gl

svd is the coolest matrix decomposition

according to me anyway

[bleak eagle]

no i think so too!

[gilded needle]

btw, if you don't know about it...

[bleak eagle]

ive been looking into its real-world applications, and I just had no idea something like it even existed

[gilded needle]

the book by trefethen and bau has a very nice discussion of the SVD in the first few sections

[bleak eagle]

ill remember that! yeah the textbook in my class is just getting into it now in the final chapter

[gilded needle]

it does a good job demystifying it and explaining it geometrically

to the point where it's almost obvious

[bleak eagle]

for sure! thank you for all the help!

[gilded needle]

cheers, enjoy!

[bleak eagle]

.close

[obtuse pebble]

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[timid silo]

How do I do this?

[tacit gate]

@timid silo what have you tried

[timid silo]

Tried making right triangles with the side with 3 units

[tacit gate]

so yes both triangles are right triangles since they are tangent with the circle

[timid silo]

Wait what

[tacit gate]

oh you made it yourself

[timid silo]

They were already right triangles to begin with?

[tacit gate]

basically a line that passes through 1 point of a circle is a tangent

and tangent is 90 degrees with radius

don't add extra work

[timid silo]

I'm so confused cause I can't see it

[tacit gate]

point B

a line goes through point B

it's an axiom

this line going through B is called a tangent and tangents are 90 degrees with radius at that point @timid silo

also 2 tangent lines that connect together outside a circle have their angle halved by the line coming from center of circle

so each triangle has 15° angle

[timid silo]

Like that?

[tacit gate]

yes

[timid silo]

I see

[tacit gate]

so now you have a right triangle with 3 given angles and 1 given side

[timid silo]

Yeah

[tacit gate]

you can use the sin rule to find your answr

[timid silo]

I got it now

Yeah

Just couldn't see it before and didn't realize

Ty

.close

[obtuse pebble]

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[thorny hound]

this is just screaming to me that it's wrong but I'm not sure where I messed up

[teal turret]

What do u see

[thorny hound]

oh ig it is right then lol

ty

[drifting wraith]

you somehow miscalculated, you got sin(2pi) − 2pi = 2pi

oh unless you meant the x

sure then it's right

[thorny hound]

.close

[obtuse pebble]

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[thorny hound]

oh that's a good point I didn't realize the abs max/min is the actual value and not where it's at

ty

[obtuse pebble]

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[pearl charm]

I have a precal question

[pearl charm]

<@&286206848099549185>

[obtuse pebble]

@pearl charm Has your question been resolved?

[pearl charm]

.close

[obtuse pebble]

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[tardy epoch]

it's the same as BFC

[obtuse pebble]

@timid silo Has your question been resolved?

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[uncut dirge]

I just need help with part A the "Do" section. I always get confused with it. What formula do I use?

[obtuse pebble]

@uncut dirge Has your question been resolved?

[ancient crystal]

Which formulas do you have to choose from?

[uncut dirge]

I don't know if this is all of them, but this is most of the ones I use

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[fathom wyvern]

is anyone able to explain what fF means in this context? i thought it was the scalar field applied to the vector field but now it seems like product rule is being applied?

[obtuse pebble]

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[safe galleon]

hi

[safe galleon]

what do i do

i just need to know

do i find the anti derivative

[ruby path]

A few of them need substitutions but yes

[obtuse pebble]

@safe galleon Has your question been resolved?

[safe galleon]

@ruby path hi sorry

didnt see this

but

for number 30

id find the anti deribvative then what

[ruby path]

I'd recommend first substituting x^2 = u

[safe galleon]

hmm

then what do i do

[ruby path]

Then finding an antiderivative becomes easy

[safe galleon]

and thats the answer?

[safe galleon]

thanks

but yeah is that it?

[obtuse pebble]

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[molten cosmos]

Is P(X<=Y) the same as P(0 <= X <= Y, 0 <= y <= 1)

for this questions?

[obtuse pebble]

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[molten cosmos]

<@&286206848099549185>

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[buoyant pewter]

I genuinely just don't know

[unique spear]

The answer is so obvious

[buoyant pewter]

I'm not good at math 😭😭

[unique spear]

dude

it’s simple math

you can express your answer in fraction form

[timid silo]

then help him

he doesn't understand the math so help him

[unique spear]

I don’t think I need to explain further as expressing your answers in terms of fraction is common

correct me if I’m wrong

@buoyant pewter when you put x as 3, what will y look like

from -20/x

if you change the x to 3

What will it become

[buoyant pewter]

-20/3?

[unique spear]

yes

so that’s your answer

[buoyant pewter]

Oh

[unique spear]

that’s it

[buoyant pewter]

Makes sense

💀

Thank you

[unique spear]

@buoyant pewter type .close

[buoyant pewter]

.close

[obtuse pebble]

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[jagged kernel]

https://www.youtube.com/watch?v=3pXALn2ovIE&ab_channel=TheOrganicChemistryTutor

minute 14:19

in the video he uses sin x as the u value

while my method I used cos x as the u

[royal basin]

both are ok here

[jagged kernel]

I plugged x = 10 into my answer and the video's answer

it's not the same

[royal basin]

hold up

what's the video's answer?

[jagged kernel]

can you go through it real quick and see where I mess up?

[royal basin]

don't wanna open it rn

[royal basin]

aight let's see

,w simplify 1/6 sin^6(x) - 1/8 sin^8(x) - (-cos^4(x)/4 + 2 cos^5(x)/5 - cos^6(x)/6)

[warm shale]

Wolfram Alpha doesn't understand your query!
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Requested by Ann#0413

[royal basin]

tf

,w 1/6 sin^6(x) - 1/8 sin^8(x) - (-cos^4(x)/4 + 2 cos^5(x)/5 - cos^6(x)/6)

[warm shale]

Results provided by WolframAlpha

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[royal basin]

okay so you did get it wrong by the looks of it

let me try to look through your steps...

ah.

you expanded (1-u^2)^2 as 1-2u+u^2 and not as the 1-2u^2+u^4 it should've been.

[jagged kernel]

oh crap

aight thank you!

.close

[obtuse pebble]

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[craggy sundial]

I missed a class and the notes aren't up but was emailed a question to solve.

Find lineralization of f(x,y,z) = xy +2yz - 3xz @ P(1,0,0)

Help.

[craggy sundial]

Are these the correct steps to follow? Based off my understanding on the textbook.

Substitute P(1,0,0) to
F(x,y,z)
F(x,y,z)'x
F(x,y,z)'y
F(x,y,z)'z

Then substitute those answers into L(x,y,z) = f(x,y,z)+f'x(x-1)+f'y(y-0)+f'z(z-0) and simplify?

[obtuse pebble]

@craggy sundial Has your question been resolved?

[craggy sundial]

<@&286206848099549185>

[wary vigil]

isn't it f(1, 0, 0)

not f(x, y, z)

[craggy sundial]

Yes that is what i meant.

[craggy sundial]

besides that, is that the correct steps?

[craggy sundial]

What i meant when I said this

Substitute P(1,0,0) to
F(x,y,z)
F(x,y,z)'x
F(x,y,z)'y
F(x,y,z)'z

was

F(1,0,0)
F(1,0,0)'x
F(1,0,0)'y
F(1,0,0)'z

[wary vigil]

yeah so a multivariable taylor expansion around a point

but in 3d not 2d

[craggy sundial]

okay good now i have a different question

this is for chain rules
H(x,y,z)=x^3y^2z+e^2z
x=p(lnq)r
y=pq^(4)+1
z=qe^(r)-2

Find dH/dq @ p=2, q=1, r=0.

Since the formula for chain rule states
dH/dx × dx/dq + dH/dxy× dy/dq + dH/dz × dz/dq

I found the derivatives of dH/dxyz and dxyz/dx (all seperatly).

What do i do now?

[wary vigil]

you gotta find all the terms

and then just use the formula

multiply and add them

[craggy sundial]

I did get all the terms

[wary vigil]

to get the total derivative to q

then fill in P q and r

[craggy sundial]

so i have

3x^(2)y^(2)z + 2x^3yz + x^(3)y^(2)+e^(2z)2

and since i have given values for P Q and R, i substitue it in right?

[wary vigil]

you substitute them into the definitions for x y and z

so that you find the value of those

and then also into the equation

which will allow you to find the answer

[craggy sundial]

im sorry, im not following. could you show me?

[wary vigil]

well x = p r ln(q)

and q = 1

so x = 0

because ln(1) = 0

x = 0

y = 2 * 1^4 + 1

y = 3

etc

and then fill those into the equation and also p q and r

[craggy sundial]

okay what do i do with those x y and z values?

ohhh

[craggy sundial]

plug into this equation?

the x y z p q and r

[wary vigil]

yes

[craggy sundial]

and what does that give though? I just get a number?

what does the number mean

[wary vigil]

it is the value of dH/dq at the point p, q, r = 2, 1, 0