help-10

[sleek valley]

which pass through 0,0? you mean you want me to find the corresponding y value?

solve:

,tex $0=2(0) + 2$ which is never true?

[warm shale]

Hairo

[pseudo swift]

I want the tangent to pass through (0,0)

Not the curve itself

[sleek valley]

ohh

[pseudo swift]

I wasn't clear enough oops

[pseudo swift]

I mean pretty much like with your ellipse stuff

[sleek valley]

it has to have absolute offset (+b) equal to 0

[pseudo swift]

Just without the implicit diff part

[sleek valley]

my solution is that you would need to shift it down by 2 otherwise the point (0,0) is not being crossed by the tangent line

[pseudo swift]

I checked and there are tangents passing through (0,0)

It's not a trick question

[sleek valley]

of y = x^2 + 2?

[pseudo swift]

Yes

[sleek valley]

ohh, I am so lost

wait

[pseudo swift]

When I mean the tangent has to pass through (0,0)

It doesn't mean the point of tangency and (0,0) have to be the same

[sleek valley]

you mean at what point would the tangent of y=x^2 + 2 pass through 0,0

solution:

anypoint (a,b) on the graph of y=x^2 + 2, which has a slope of b/a

which means

,tex$2x=x/y$ ?

[warm shale]

Hairo

[pseudo swift]

I mean you know a point on the curve will be of the form (a, a^2+2)

From the equation of the curve

[sleek valley]

agreed

[pseudo swift]

So the tangent at a of our curve will pass through (a, a^2+2) and its slope will be 2a

[sleek valley]

,tex$2x = \frac{x}{x^2+2}$

[pseudo swift]

(2a being the derivative of the curve at x=a)

[sleek valley]

yes

[warm shale]

Hairo

[sleek valley]

any x that satisfies the equation is such point, right?

[pseudo swift]

You should be able to find the equation of the tangent at a from that

I'm not sure what your equation is supposed to mean here

[sleek valley]

if you try to imagine y=x^2 + 2, now think about what points' tangent line will pass through (0,0), you will realize that the slope of the tangent line at that point has to equal the y/x in order to cross 0,0

so any point that satisfies the equation above (2x =...) is a valid point for your question, right?

[sleek valley]

sorry, if I'm talking nosense. But I understood what you said, I think

[pseudo swift]

Uh yeah actually your thing makes sense

You just mixed up x^2+2 and x actually

[sleek valley]

whoops, sry

[pseudo swift]

You should have $$2x = \frac{x^2+2}{x}$$

[warm shale]

aPlatypus

[sleek valley]

rise over run 😄 not run over rise xD

[pseudo swift]

Indeed indeed

[sleek valley]

so with that new found knowledge, I now have to find the correlation?

[pseudo swift]

You should be able to find the suitable x from there

Should be +/- sqrt(2) from what I gathered

[sleek valley]

correct!

[pseudo swift]

But yeah I can show you the more "academic" way I thought of

[pseudo swift]

Using these pieces of info you can get the equation of the tangent at a

Which should be $$y=2a(x-a) + (a^2+2)$$

[warm shale]

aPlatypus

[pseudo swift]

Which exactly fits the description I gave of it above

And then you can plug in the point (0,0) and solve for a from there

[pseudo swift]

Honestly I like the way you intuited the thing here

[sleek valley]

ty 😄

[pseudo swift]

But yeah for more complicated stuff it might be more difficult to intuit

So yeah the ellipse stuff

[pseudo swift]

So it's nice to have a kinda standard way to approach it sometimes

Anyway

What did you get for the implicit derivative again?

[sleek valley]

-x/4y

[pseudo swift]

Okay

[pseudo swift]

Here the idea of taking a point (a,b) on the curve is a good one

Cause it's not like we have a direct relation between our two variables here

[sleek valley]

agreed, you mean because of the y^2 ?

[pseudo swift]

Yeah

So we take a point on the curve (a,b) (I.e. a^2 + 4b^2 = 5 )

What's the slope of our tangent at that point ?

[sleek valley]

the slope? -a/4b 😄

right?

or am I missing something

[pseudo swift]

Yup nice correction

[sleek valley]

you mean equation right?

[pseudo swift]

Nah as in swapping the x y for a b

[sleek valley]

ohh, xD

[pseudo swift]

So we got ourselves a tangent of slope -a/4b

Which passes through (a,b)

Can you get the equation of the tangent from there ?

[sleek valley]

1 sec, lemme think

[pseudo swift]

(You can inspire yourself from the "academic" method I showed earlier)

[sleek valley]

question: you don't want to hop into a voice channel, do you?

[pseudo swift]

I guess I should explain a little bit more my method if you're lost

[pseudo swift]

I can't right now

[sleek valley]

alright

so what I am suck with is the following:

$$y=2a(x-a) + (a^2+2)$$

[warm shale]

Hairo

[sleek valley]

the meaning of a,b,x and y

more like a,x

[pseudo swift]

(x,y) are the points on the tangent

[pseudo swift]

And a is the place on the curve where you're taking the tangent

[sleek valley]

oh, yeah

like this?

[pseudo swift]

Exactly

Well with a circle (a,f(a)) is difficult

But that's the idea yes

[sleek valley]

and how did we derive your formular again?

does it come from $$y=m(x2 - x1) + y2$$

[warm shale]

Hairo

[pseudo swift]

That's the idea yes

If you have a line of slope m passing through (a,b)

The points (x,y) on this line satisfy $$m=\frac{y-b}{x-a}$$

[warm shale]

aPlatypus

[sleek valley]

got it

[pseudo swift]

Rearranging that gives you pretty much my equation

[sleek valley]

okay let's see

so

we know that slope (m) = -a/4b, right (if (a,b) is the point were we are taking the tagent line)

[pseudo swift]

Yeah indeed

[sleek valley]

okay so

$$\frac{-a}{4b} (x-a) + b = y$$

[warm shale]

Hairo

[pseudo swift]

Yeah ok

[sleek valley]

know we know that (-5,0) is one the line and plug it in

so

$$\frac{-a}{4b}(x-(-5)) + 0 = y$$

[warm shale]

Hairo

[sleek valley]

$$\frac{-ax-5a}{4b} = 0$$

[pseudo swift]

Well you had 50% chance of getting it right :/

[sleek valley]

:d

😄

[pseudo swift]

The idea here is that we want to find the (a,b) such that the tangent passes through (-5,0)

[sleek valley]

= 0

[warm shale]

Hairo

[sleek valley]

nvm

[pseudo swift]

Yeah you want to fill in the x y not the a b

It's the (x,y) that represent all the points of your tangent remember

[sleek valley]

wait what the hell did I do?

I plugged in -5 and 0 into only one a 😄

not x and y

I can't focus 😄 I'll go again

$$\frac{-a}{4b}(-5-a)+b = 0$$

[warm shale]

Hairo

[sleek valley]

$$\frac{5a+a^2}{4b} + b = 0$$

like this?

[warm shale]

Hairo

[pseudo swift]

Yup indeed

[sleek valley]

$$5a+a^2 = -0.25$$

[warm shale]

Hairo

[sleek valley]

should I use the quadratic formula to solve it?

[pseudo swift]

uh

[sleek valley]

😄

[pseudo swift]

what did you do to b

you killed it

[sleek valley]

cancled

I did -b then *4b

[pseudo swift]

yeah

-b*4b is not 0 as far as I know

[sleek valley]

but it's -0.25

I can just +0.25 and use the quadratic formular

or not?

[pseudo swift]

nah

-b*4b is -4b²

not -1/4

[sleek valley]

ohh what the hell am I doing

sorry, I really can't focus rn

$$5a+a^2 = -4b^2$$

[warm shale]

Hairo

[sleek valley]

now we just plug in any b or a? I guess

[pseudo swift]

yeah indeed

now you might be wondering

"wtf there's two unknowns and one equation"

[sleek valley]

nope

that's fine

it's a linear equation

for any x there is a f(x)

[pseudo swift]

[sleek valley]

for any a -> (5a + a^2) there is some b -> (-4b^2)

xD

ohh wait yeah, I am wandering

just ingnore me 😄

[pseudo swift]

but yeah actually

we have two equations

cause it's not like we took a random (a,b)

[sleek valley]

a,b are on the circle

[pseudo swift]

our (a,b) still has to be on the curve

yeah

[sleek valley]

so their ratio is well defined

[pseudo swift]

so we have $$\begin{cases} a²+4b²&=5 \ a²+5a+4b²&=0\end{cases}$$

oops

[warm shale]

aPlatypus

[pseudo swift]

we'd think we're fucked at first looking at this quadratic systems of equations

but actually we're alright

[sleek valley]

5a = 5 ?

[pseudo swift]

well 5a+5=0

almost

[sleek valley]

5a = -5

$$a^2 + 4b^2 -5 = 0$$

[warm shale]

Hairo

[sleek valley]

which is equal to $$a^2+5a+4b^2$$

[warm shale]

Hairo

[sleek valley]

doing (-a^2 -4b^2) in both equation gives 5a = -5

[pseudo swift]

we went a bit full circle pun not intended didn't we ?

I mean from the second equation you get a=-1

[sleek valley]

yes, and that's what we need

so know we know that the tangent line crosses (a, + or - f(a))

(where f(a) is the original equation solved for y)

and we can find the exact slope and rise

we know that the lamp is on x=3 and we had our point on x=-5

which gives us delta x of abs(-5) + abs(3) = 8

so delta x = 8

and now we can find delta y

[pseudo swift]

ok

do we have the slope though ?

[sleek valley]

yes.

$$\frac{-(-1)}{4\sqrt{\frac{5-(-1^2)}{4}}}$$

which is 1/2 or 0.5

that's the slope

what you are seeing above is the equation rearanged and solved for y

wrong

[pseudo swift]

the slope is -a/4b right

[sleek valley]

there is a mistake

yes

that's what I wanted to say

[warm shale]

Hairo

[sleek valley]

so the slope is 2

[pseudo swift]

[sleek valley]

0.5 * 4 = 1 😄

[pseudo swift]

,w sqrt((5-(-1)²)/4)

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by aPlatypus#3790

[pseudo swift]

algebra

[sleek valley]

haahhaa

I can't

[pseudo swift]

the calculus student killer

[sleek valley]

I calculated 4 over 4 and said it's 0.5

I can't

I swear, I can't focus 😄

really 😄

hahaha

so know we know that the slope is 4 right? we got everything we need

[pseudo swift]

it's time to learn how to do algebra full drunk mode

[pseudo swift]

**+**1/4

[sleek valley]

ahahah

I can't

I can't

yes

delta x * 1/4 = delta y

delta y = solution to entire problem

[pseudo swift]

yeah

I guess we won now

almost

[sleek valley]

😦

why almost :/

[pseudo swift]

we still have to multiply 1/4 by 8

to get the answer

[sleek valley]

yeah, obv 😄 that's what I mean by delta x

xD

so the solution is 2.1

,w (1.025/4) * 8

[warm shale]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Hairo#0567

[sleek valley]

nice finally

[pseudo swift]

1/4 is just 1/4, not 1.025/4

[sleek valley]

it was joke xD

[pseudo swift]

yeah the answer is 2.00000000000000000000000000001

[sleek valley]

xD

do you think the question was supposed to be solved like that?

yeah, fair enough, that's actually an easy problem

[pseudo swift]

- find the derivative at a point (a,b) on the curve
- finding the equation of the tangent of the curve at (a,b)
- solve for (a,b) which fit the conditions given in the question
- find the final point of the lamp

it's actually not that hard conceptually what we did

it's always longer when you gotta explain the thing though

+algebra

[sleek valley]

algebra and 10h of mental work is my death 😄

[pseudo swift]

10h!

what did you do for 10h?

[sleek valley]

programming

work

[pseudo swift]

ah yeah, so you have a job or you're just on your own learning ?

[sleek valley]

just got my first job xD

so work

[pseudo swift]

ah noice

ok

[sleek valley]

btw, one question:

I'm self studying calc and want to study calc I, II, and III and linear algebra. I have a book with about way more than 1000 pages, where 99% of it are problems. Should I do all problems or should I skip most of them. How did you study calc?

[pseudo swift]

don't do all problems that's for sure

in books with lots of problems there's redundancy in the problems

so if you feel like a problem is easy for you just skip it

(i.e. if you have a clear idea of how the solution should look like just skip the problem)

[sleek valley]

merci

can I add you on discord so I can ask for some help, when I'm not completly dying inside?

[pseudo swift]

and also for calc I mostly just did the exercises the profs provided us

in lin alg I found myself a book

[pseudo swift]

sure

what book are you using btw ? @sleek valley

[sleek valley]

Stewart, James

calculus 7e

[pseudo swift]

ah stewart ok

just wondering

[sleek valley]

I'm using a digital copy, why? would you recommend something different?

[pseudo swift]

most of the books I have are pdfs lol

if you want a book to change yourself from programming all day you could

your call I guess

let's go on DM I guess

[sleek valley]

.close

[obtuse pebble]

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[hazy rose]

Ok so I seem to be doing something wrong as finding the anti derivative of x^2(x-6)^7 is much harder than what I am supposed to do

[novel knoll]

You just need to do ibp many times

And you still aren’t writing dx as you should

[hazy rose]

I see what I did

[charred plume]

Why not try u=x and dv=(x-6)^8 dx?

[hazy rose]

It would be simple to have u as x

[charred plume]

Hah, sorry to steal your thunder there.

[hazy rose]

Thanks for all the help : ) I’m sorry I don’t see these things as quickly as I should

No worries, I’m just glad ppl like you two are there to help ; )

.close

[obtuse pebble]

Channel closed

Closed by @hazy rose

Use .reopen if this was a mistake.

Please take a minute to participate in [our survey](#changelog message) if you haven't already!

[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[somber sierra]

How is the second options equal to 0?

[somber sierra]

It’s -0.5

[idle shoal]

Are you referring to x=2.2 and x=8?

If so, you can see that they have the same value

[somber sierra]

Yes

[idle shoal]

So their difference is zero

[somber sierra]

explain?

[idle shoal]

f(2.2) and f(8) have the same value, 2.9

on the graph

And you are looking at the value of the expression

(f(8) - f(2.2))/(8-2.2)

But the numerator is zero from what was mentioned above

Does that make sense?

[somber sierra]

No, what mention?

This is equal to -0.5

[cinder sparrow]

I have the feeling it's referring to the derivative of the function

[idle shoal]

f(8) is 2.9, not 0

[cinder sparrow]

f(8)=0 in the graph

[somber sierra]

Confused

[idle shoal]

Ok I misread something my bad

[somber sierra]

how is f(8) = to 2.9?

Oh kk

[cinder sparrow]

Do you now a bit about derivatives?

[somber sierra]

No

[cinder sparrow]

Ever heard of those?

[somber sierra]

No

[cinder sparrow]

Hmm

That's weird

[somber sierra]

?

[cinder sparrow]

Cause that's what the question is referring to I would suppose

I think the answer on that website aren't correct

[somber sierra]

It’s asking what is equal to zero right?

The first and last equal zero

The middle equal 0.5

So I get how that is wrong, but than the one it said is correct is equal to -0.5 if I did math correct

[cinder sparrow]

The thing all the correct answers have in common is that the x-values all have a responding value of zero for the derivative of the graphed function (but that doesn't mean anything for you, since you've never heard of those).

But that's not the question, the textbook asks, weird

So yes, the first and last one

[somber sierra]

I probably have heard of derivatives but can’t remember exactly what off the top of my head. Can you explain to me what derivatives are?

I am in UNI and taking a summer alg/pre Calc course

[cinder sparrow]

I'll give you a quick summary

[somber sierra]

Kk

[cinder sparrow]

A derivative says something about the slope of a function at any given point

[somber sierra]

example?

I do best when given a example paired with a explanation

[cinder sparrow]

So let's say I were to graph the position of a car. With on the x-axis the time in seconds and on the y-axis the meters it has traveled.

Then the derivative of that function would graph the speed of the car at x seconds, so it would basically graph what the speedometer shows at any point in time

[somber sierra]

Should I interpret ‘graph’ as the position of the car on the x and y axis?

[cinder sparrow]

X axis is time, y is position in meters traveled

So imagine the car is driving in a straight line

If the car would move at a constant speed, you'd get a linear graph like this

[cinder sparrow]

Exactly

[somber sierra]

Okay

[cinder sparrow]

In this example, the position of the car is plotted in red and the speed (so the derivative of the red plot) in blue

[somber sierra]

I’m confused because I see two cars now. How can one car have different values for time and distance

[cinder sparrow]

This is quite the concept to grasp, so don't worry if it doesn't really click directly

[somber sierra]

Mhm

So both lines rep. The one car?

[cinder sparrow]

It's the same car, both plots are just different properties of the car

[somber sierra]

But the only properties are distance and time

How can speed be involved

[cinder sparrow]

Wait, let me think how to explain this

[cinder sparrow]

But shouldn't you be able to calculate how fast the car is moving, just looking at the graph of distance and time?

[somber sierra]

If their is a upward slope than you could presume that the car is moving faster

[cinder sparrow]

Precisely

[somber sierra]

Is the derivative a context concepts or hard math?

[cinder sparrow]

It is doable

[somber sierra]

Is the derivative the slope

[cinder sparrow]

You just have to get it intuitively

[cinder sparrow]

Yes

[somber sierra]

Ohhh

So why is it two separate things tho

[cinder sparrow]

Let's take this example again

[somber sierra]

So is this true, m = f’(x)

[cinder sparrow]

Could you tel me it's slope at any given point?

[somber sierra]

Yes because it’s constant

[cinder sparrow]

(lets assume the function describing this graoh is something like 3x=y)

[cinder sparrow]

Yup

So if i were to plot on the x axis time, and the slope on the y axis

What would that look like?

[somber sierra]

Idk

[cinder sparrow]

Remember what you said

The slope is constant

It's always the same, regardless of time

[somber sierra]

Wdym?

[cinder sparrow]

You know how we plotted position on the y axis and time on the x axis?

[somber sierra]

Plotting a dot on the graph

Coordinate

[cinder sparrow]

Now we don't plot its position on the y axis, but it's speed (it's slope)

So, you know how a car has a speedometer

[somber sierra]

Yes

[cinder sparrow]

Well, imagine if id plot the value this speedometer shows on the y axis

[cinder sparrow]

Kind of, you have to plot something for evey value of t,

[somber sierra]

Any derivative of a constant is 0?

[cinder sparrow]

Yes

That is most certainly true

[somber sierra]

What classifies as a constant?

is 3.2 a constant?

[cinder sparrow]

Yes

Just a number

Something that is not dependent of an input, like x

So the graph y=3.2 would show a constant

[somber sierra]

Y is dependent of an input?

[cinder sparrow]

This would be the graph of y=3.2

Just a flat line

[somber sierra]

Why

[cinder sparrow]

Cause for every value of x, the value you plot on the y axis is 3.2

If you have the function y=2x

[somber sierra]

Okay

[cinder sparrow]

Then for the input x=1, you get y=2*1=2

[somber sierra]

y=2x is not a constant?

[cinder sparrow]

If the input is x=3, you get y=2*3=6

But now, if I take the function y=3.2

Then for the input x=1, y=3.2

And for the input x=3, y=3.2

You see where I'm going with this?

[somber sierra]

Okay so the derivative of y=3.2 is constant

[cinder sparrow]

Yes, the derivative of this function is constant

Could you tell me what constant?

[somber sierra]

Wdym?

what constant?

[cinder sparrow]

Which number?

[somber sierra]

3.2

[cinder sparrow]

Hmmm

Look at the graph

What is the slope of this graph at any point?

[somber sierra]

Oh 0

[cinder sparrow]

Exactly!

[somber sierra]

There is no slope

Technically

[cinder sparrow]

So the derivative of the function is y=0

Do you agree?

[somber sierra]

Is the the same as saying “The slope of the graph is y=0”

[cinder sparrow]

If this were a plot of the position of a car, the car would just be standing still

[cinder sparrow]

Yup

[somber sierra]

So derivative is just the output?

[cinder sparrow]

And if this were to describe y=3.2x, what would be it's derivative?

[somber sierra]

0

Oh wait 3.2x

Idk

[cinder sparrow]

Look at the slope of the graph again

[somber sierra]

It’s constant upward

[cinder sparrow]

Yeah, so it's slope is constant, it's the same at any point in time

Agree?

[somber sierra]

Yeah

[cinder sparrow]

So would you then also say that the derivative of that function would look something like this?

[somber sierra]

Yes but it’s tilted

So it’s increasing instead of being straight

[cinder sparrow]

Is it increasing though?

The position of the car is increasing

But is the slope of the graph as well?

[somber sierra]

The slope is still “3.2” but multiplied by x every increment

Is y = 3.2x the same as y= 3.2?

[cinder sparrow]

No

If I would have x=2, what would be the y values for both functions?

[somber sierra]

Y = 3.2(2) = 6.4; Y = 3.2

[cinder sparrow]

Yeh

[somber sierra]

The slope is changing than

?

[cinder sparrow]

Wait a minute

Were gonna take a small step back

[somber sierra]

Kk

[cinder sparrow]

Let's take our function again, 3.2x=y

It looks like this

[somber sierra]

Yes

[cinder sparrow]

What slope does it have at the point x=2?

[somber sierra]

6.4

[cinder sparrow]

Think again

I'm not asking for it's position, but it's slope

[somber sierra]

3.2?

[cinder sparrow]

Remember how you said that this guys slope was always zero?

[cinder sparrow]

Yeah

Indeed it is

Do you see why?

Give it some thought, you shouldn't have to get it immediately

[somber sierra]

But than how is y = f(c) slope?

[cinder sparrow]

Wdym?

[somber sierra]

3.2 is the slope

Than what is the multiplied by x part?

[cinder sparrow]

It just describes the graph

So 3.2 multiplied by the x value gives you the y value

Now how fast is that y value increasing? That is it's slope

[somber sierra]

it is decreasing at a rate 3.2x

[cinder sparrow]

If you take one step along the x axis, you take 3.2 steps along the y axis

[somber sierra]

Or 3.2 per ‘step’

[cinder sparrow]

So it's slope is 3.2

[somber sierra]

okay or other words it’s derivative is 3.2?

[cinder sparrow]

Jup

So the derivative of y=3.2x is y=3.2

Agree?

[somber sierra]

Yeah

[cinder sparrow]

Nice!

[somber sierra]

^

The derivative is the whole graph or just each part

[cinder sparrow]

Okay, yeah, now let's tie this newly acquired knowledge onto the original question

[cinder sparrow]

We don't know the function that describes this graph, but we can figure out a few things just by looking at it

What would you say is the derivative of this function at the point x=2.2?

Or in other words, what is it's slope at that point?

[somber sierra]

0

It’s flat at 2.2?

[cinder sparrow]

Yup

Exactly right

[somber sierra]

5.2 isn’t 0?

It’s on a slope?

[cinder sparrow]

At the point 5.2, the derivative indeed isnt zero

[somber sierra]

so when a plot is at a height or point where it’s like at the top of the hill the slope is zero?

[cinder sparrow]

Can you find more points on the graph where the derivative is zero?

[cinder sparrow]

Exactly right

[cinder sparrow]

Yes!

Precisely

[somber sierra]

I see wym now that the only common trait it the derivative or slope is zero

[cinder sparrow]

Now, let me restate the original question: "for which of these pairs of x-values is the derivative equal to 0?"

[somber sierra]

Oh

[cinder sparrow]

Yes

[somber sierra]

The formula “deltaF(x) over deltaX” means slope

“For which of these pairs of x-Values is the slope equal to 0”

[cinder sparrow]

Yes

Well

Sorta

Actually, the most used notation for this is dF(x) over dx

That's why I got so confused

[somber sierra]

That’s like derivative notation?

I’ve seen that before

[cinder sparrow]

Yes

[somber sierra]

But here it’s using point slope notation?

[cinder sparrow]

Yeah, the question is wrong

So it's not your fault you got the question wrong

[somber sierra]

Should I email my prof abt?

[cinder sparrow]

Yeah, do that

[somber sierra]

Idk if they are in control of the question

How should I word it?

[cinder sparrow]

Still, couldn't hurt

[somber sierra]

“I noticed the question is using point slope notation but the answer relies on derivative notation…”?

[cinder sparrow]

Something along those lines, if you refer to the exact question and the test it came from

And maybe send a picture of the question in the appendix

[somber sierra]

Wait wtf

[cinder sparrow]

What is it?

[somber sierra]

Didn’t it reverse

[cinder sparrow]

It changed?

[cinder sparrow]

Those are different

The hell

[somber sierra]

It changes to what I put prior right?

[cinder sparrow]

Yes it did

[somber sierra]

I did get email that smth was wrong with the HW so it was extend to Sunday

But the email didn’t clarify what was wrong

[cinder sparrow]

That is really weird

[somber sierra]

Bro Yk what stupid

[cinder sparrow]

Well, you got it right in the end

And now you know what derivatives are

[somber sierra]

I had to pay $69 to access this HW portal and textbook

[somber sierra]

True

The $69 lead me to here

Should be paying you $69

[cinder sparrow]

Your poor money

[cinder sparrow]

No need

[somber sierra]

UNI is harsh man

Going in for comp sci 🤓💀

[cinder sparrow]

Nice

Planning on doing that as well

[somber sierra]

Sick, I’m thinking of data science as a career but I don’t want to be a code monkey because I like talking to ppl and doing business. My goal is to be a head of a team or manager

[cinder sparrow]

I hope you get there

You have it in you

[somber sierra]

Now that I made it to UNI I’m actually trying to apply myself and do smth. HS I just did whatever and passed

So I’m learning and catching up hence the ALG and precalc

[cinder sparrow]

Well, we secretly discussed a bit of calc already

[somber sierra]

Yes it is algebra infused precalc!

Anyways thank you very much. I’ll close the channel now

[cinder sparrow]

Very well

It's been nice helping you

Good luck!

[somber sierra]

.close

[obtuse pebble]

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[thorny zephyr]

am i supposed to use discriminant formula? if so, how because i only know it for quadratic equations?

[low ermine]

U just need to look at the power of x

All the x^k is k even

x^even number is always positive

Then, +1 means graph is above X axis 1 unity

So the function doesn't have any root because doesn't cross the axis

[thorny zephyr]

i see now, it makes. i thought of that before but wasn’t sure if the +1 would affect it

thank you!!

[low ermine]

Yeah, if it didnt have the +1 the 0 would be a root

Uw

[thorny zephyr]

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[raven plume]

Lemma 1.2.3. The set $B(a ; \delta)$ is open in $\mathbf{R}^{n}$, for every $a \in \mathbf{R}^{n}$ and $\delta \geq 0$.
Proof. For arbitrary $b \in B(a ; \delta)$ set $\beta=|b-a|$, then $\delta-\beta>0$. Hence $B(b ; \delta-\beta) \subset B(a ; \delta)$, because for every $x \in B(b ; \delta-\beta)$
$$
|x-a| \leq|x-b|+|b-a|<(\delta-\beta)+\beta=\delta .
$$

[raven plume]

can someone explain why $B(b ; \delta-\beta) \subset B(a ; \delta)$ implies the set $B(a, \delta)$ is open?

[raven plume]

.close

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[vague mural]

hey can yo add/subtract items in parentheses to the other side of the equation
ex: (t+3)(t-357)=0
can i just do
(t+3)(t)=357

[idle shoal]

You can't do that

[vague mural]

ok

would u solve this type of equation by factoring

[idle shoal]

Well remember that if a*b=0, then a=0 or b=0

So with (t+3)(t-357)=0

Either t+3=0 or t-357=0

[vague mural]

oh my b its asking for the sum of the solutions to the equation

[idle shoal]

Ok but you can get the solutions from above and add them up

[vague mural]

add what up?

[idle shoal]

Those two solutions

You rearrange t+3=0 and t-357=0 to find the values of t that solve it

Then you add those two numbers together because they are the only solutions

[vague mural]

i dont understand could u show me how to do that

it looks like it would be t=3

and t=357

[idle shoal]

Almost you have t=-3

[vague mural]

oh ye

[idle shoal]

Because you subtract 3 from each side

[vague mural]

ye

[idle shoal]

but t=357 is correct

So those are the two solutions

Now you add them up

[vague mural]

and then u subtract/add

354

[idle shoal]

Yes

Well done!

[vague mural]

oh dope

alr cool ty homie

good explination

[idle shoal]

You're welcome

Thanks

[vague mural]

.close

[obtuse pebble]

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[obtuse pebble]

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[sly sinew]

What would u and du be in this case?

[dreamy basin]

14x as u, as you want to remove the x when differentiating

[sly sinew]

then what is du?

[dreamy basin]

du/dx = 14

du=14 dx

[sly sinew]

ahh as the e term is its own int/deriv so i dont worry about it?

[dreamy basin]

yes assuming this is int. by parts

[sly sinew]

ok thank you ill try it

hmm didnt work

[brave bramble]

You're going to have trouble integrating e^(-x²)

Are you sure you want integration by parts?

[sly sinew]

ill just show full problem

[brave bramble]

Okay so you want u-sub

A good hint in general is if you have a trouble function, try setting whatever's in the function to u

In this case, e^(-x²) is making life hard. You'd try u = -x²

This works nicely because the derivative of x² is right there.

[sly sinew]

that kinda makes sense, would du be 14x dx then?

[brave bramble]

You tell me

[sly sinew]

😆 i have no clue

[brave bramble]

How do you normally find du, given a u?

[sly sinew]

derivative of it, sorry very new to thiis

[brave bramble]

Very nice!
Since I'm suggesting u = -x²
Then du = -2x dx

[sly sinew]

🤦‍♂️ ok I overcomplicated that horribly

thank you! I spend more time on that then I should have, that cleared it up

@brave bramble

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[elder violet]

I'm struggling with this

[devout solar]

why

[elder violet]

Im not sure how to prove it algebraically

[idle shoal]

What are you supposed to prove?

[elder violet]

read from 8a down

from a*

[obtuse pebble]

@elder violet Has your question been resolved?

[elder violet]

<@&286206848099549185>

[obtuse pebble]

@elder violet Has your question been resolved?

[slim cove]

yo

so you can think of 84 as 8*10 + 4

and 48 as 4*10 + 8

therefore the way to generalize this is that you want to compute (a * 10 + b) - (b * 10 + a)

does that make sense?

[elder violet]

yesyes thankyou vedi much

.close

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[steady knot]

Currently trying to solve quadratic equations using the quadratic formula

Im not sure how to get from the top equation to the bottom 2

When i got stuck i turned to using this calculator online

here is the full thing

[royal basin]

calculator.

you're not expected to evaluate sqrt(65) by hand.

[steady knot]

even when i put it in a calculator it doesnt give me the same answer

[royal basin]

so you are putting -15/8 + sqrt(65)/8 into your calculator, and getting an answer that isn't -0.867218 or thereabouts?

is that what you are saying?

[steady knot]

let me just double check

never mind lol

.close

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[fallen oar]

did i do this right
dl^2 = dx^2 + dy^2
introduce polar coordinates.
x = r cos(θ)
y = r sin(θ)
take the differentiate
dx = cos(θ)dr - r sin(θ)dθ
dy = sin(θ)dr + r cos(θ)dθ
add their squares
dx^2 + dy^2 = (cos(θ)dr - r sin(θ)dθ)^2 + (sin(θ)dr + r cos(θ)dθ)^2 = (cos(θ)^2 dr^2 + r^2 sin(θ)^2 dθ^2 - 2r sin(θ)cos(θ)drdθ) + (sin(θ)^2 dr^2 + r^2 cos(θ)^2 dθ^2 + 2r sin(θ)cos(θ)drdθ) = dr^2 + r^2 dθ^2
So:
dl^2 = dr^2 + r^2 dθ^2
Now factor out dθ^2.
dl^2 = ((dr/dθ)^2 + r^2)dθ^2
then take the square root?

[fallen oar]

<@&286206848099549185> hello can I please get some help

.close

[obtuse pebble]

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[frigid jasper]

did you get it?

[obtuse pebble]

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[hazy rose]

The blue line is f(x) the green is g(x)

[hazy rose]

The intersection is at 1/6

Im stuck at taking the integrals

[solar igloo]

that's simple as long it's linear

yo do know the integral of ln( x )?

[hazy rose]

Well I’ve been told it’s (ln(x)-1)x

I think, you can just solve it with integration by parts

[solar igloo]

pardon g2g

[hazy rose]

Oh

Ok, have a good day

<@&286206848099549185>

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[wooden cipher]

This feels like a contest problem, wheres it from?

[gilded topaz]

For this equation to be true, one of the expressions in parentheses must be 2 and the other two must be 1, right?

[gilded topaz]

Copyrighted AwesomeMath

[wooden cipher]

Its impossible to make one of them 1

[gilded topaz]

not a contest, a program

[wooden cipher]

Ah

[gilded topaz]

exactly, thats how I came up with no solutions

[wooden cipher]

Mm not quite

Heres a start, we can use a wlog right?

[gilded topaz]

what is that?

[wooden cipher]

Without loss of generality
It is used for symmetric equations so that we can narrow down a solution withiut having to deal with permutations

[gilded topaz]

Yes, we can use that because there is no difference between x, y, z and their conditions, etc.

[wooden cipher]

For example, in this scenario we can say
$WLOG x \leq y \leq z$

[warm shale]

GarlicBredFries

[gilded topaz]

Agreed.

[timid silo]

If one term will be 1 then u will have to make 1/x = 0 which seems impossible

[gilded topaz]

Yep.

[wooden cipher]

And since we know x has to be greater than 1 we can bound it
$WLOG 2\leq x \leq y \leq z$

[warm shale]

GarlicBredFries

[gilded topaz]

I'm confused -- why does x having to be greater than 1 make the inequality above?

[wooden cipher]

x must be at least 2 now

[gilded topaz]

Oh right, my bad.

[unkempt cargo]

I believe this was a BMO problem

[gilded topaz]

Can you give a link to the source?

[gilded topaz]

So how do I use this?

[unkempt cargo]

BMO 1995 Problem 1

[gilded topaz]

Lol, variables changed

[unkempt cargo]

It’s a relatively easy problem, bounding is correct I’m pretty sure. When I solved it I think I expanded it

The LHS

[gilded topaz]

Oh, I see, that would make the problem easier than I thought

Thanks!

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[unkempt cargo]

Wish I could make AwesomeMath : ( my teachers would never give me a letter of rec

[obtuse pebble]

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[gilded topaz]

What grade are you in?

[gilded topaz]

Oops, go to DMs

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[timid silo]

Did u get the answer?

[gilded topaz]

Yes

[timid silo]

How did u do it then?

[gilded topaz]

I expanded and factored then used casework

(I found 5 sols)

[timid silo]

Ouh alright thanks!

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[fluid snow]

Determine if $\sqrt[3]{0.8}$ is irrational or rational.

[warm shale]

Kepe

[fluid snow]

I don't really know how to continue though

[solid sun]

I mean before looking at what you have, you know 0.8 = 8/10 right?

[fluid snow]

yes

[solid sun]

and 8 = 2^3?

[fluid snow]

yes

$\sqrt[3]{\frac{8}{10}} = \frac{2}{\sqrt[3]{10}}$

[warm shale]

Kepe

[haughty echo]

Huh

Yea

Just prove cbrt 10 irrational

[solid sun]

just try showing $\sqrt[3]{10}$ is rational or not.

[haughty echo]

[warm shale]

BenJr

[fluid snow]

Why not just show if $\sqrt[3]{0.8}$ is rational or not right away?

[warm shale]

Kepe

[fluid snow]

Is showing that easier?

[royal basin]

it may be harder that way.

[haughty echo]

Hm yea

It will be easier just by proving cbrt 10 irrational

[fluid snow]

alr

[solid sun]

after you have done that, it fun to prove if p is a rational number not equal to 0 and q is irrational then pq is also irrational.

[haughty echo]

Why are you still carrying "2" here

[warm shale]

GG・Goof

[fluid snow]

After you get $\frac{2}{\sqrt[3]{10}} = \frac{a}{b}$ you right away want to prove that, right?

[warm shale]

Kepe

[haughty echo]

Hm yea

[fluid snow]

alr

[fluid snow]

Is this correct?

[haughty echo]

Yes

[fluid snow]

thx!

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[glossy wharf]

can someone explain why it's n-1 choose r-1 instead of n choose r?

[obtuse pebble]

@glossy wharf Has your question been resolved?

[glossy wharf]

<@&286206848099549185>

[drifting wraith]

the implication is that last trial is fixed to be successful

that's how variable number of trials comes up, you stop when you have your r

[glossy wharf]

so basically there are n-1 choose r-1 ways to have exactly r-1 failures in n-1 trials, and since we are taking our assumption that trial n will be another failure we don't account for it in the binomial? and since we have a total of r failures we use (1-p)^r and a total of n-r sucesses so we have p^(n-r). am i thinking about this correctly?

[drifting wraith]

yes, my bad, fixed to not succeed

[glossy wharf]

ah wasn't correcting you just making sure i understood correctly 🙂

thank you!

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[willow ravine]

Complex variables:

This is what I got so far, but don’t make sense because I don’t think it should be zero.

@slim cove

[obtuse pebble]

@willow ravine Has your question been resolved?

[willow ravine]

<@&286206848099549185>

[slim cove]

yo

LOL that picture

looks good! it's surprising to me that it's zero though lol

I wonder if there was some easier way to see that everything cancels

[willow ravine]

If it is zero that would mean the function is analytic on the entire contour

[slim cove]

not quite

[willow ravine]

Which shouldn’t be true because it has a singularity in the contour

[slim cove]

you're thinking of cauchy's theorem, if a function is analytic in the whole region then the contour integral is zero

I don't think the converse is also true

[willow ravine]

Ahh

But it all checks out and is supposed to be zero?

[slim cove]

I don't see any problems with your work hmmm

lemme try doing it with the residue theorem and see if I get the same answer

[willow ravine]

Ok

I’m allowed to use residue theorem I’m just not good at understanding it and how to work through it.

Are you sure I’m supposed to take the derivative ?

[slim cove]

yeah

hmm do you think it'd be a good idea to also do it using the residue theorem then to see if you can do it

we can go through it together

[willow ravine]

We can try.

[slim cove]

okay! so recall that the residue theorem tells us that the result of this integral should be the sums of the residues inside the circle

there is only one residue inside the circle, that at z = 2i

therefore the answer to this integral is just the residue of $\frac{z}{(z^2+4)^2}$ at $z=2i$

[warm shale]

Eric Tao (he/him) 🌈

[slim cove]

make sense?

[willow ravine]

Would you want to hop on a voice chat?

[slim cove]

can't sorry :(

[willow ravine]

Okay.

That makes sense.

[slim cove]

awesome, so next notice that we can factor the function as $\frac{1}{(z-2i)^2}\frac{z}{(z+2i)^2}$

[warm shale]

Eric Tao (he/him) 🌈

[willow ravine]

Yes,

I see this

[slim cove]

okay awesome so hmm notice that it has a double pole at z=2i because of the (z-2i)^2 term

[willow ravine]

That makes sense.

Can you just say it is a pole of order 2?

[slim cove]

yep

the residue is the coefficient of the 1/(z-2i) term, so if we find the taylor series of z/(z+2i)^2 at z=2i, we should be able to figure out the residue

namely, the residue will be the coefficient of the first-power term of the series for z/(z+2i)^2 term right

[willow ravine]

Okay. That makes sense. Basically find a Taylor series and equate coefficients to like power terms.

[slim cove]

yep!

but now notice the coefficient of the first-power term of a taylor series is just the derivative of z/(z+2i)^2 at z=2i

which is exactly what you found using the previous method

so, zero again

hopefully that should give you some reassurance that the answer is zero 🙂

[willow ravine]

Okay. It does.

[slim cove]

awesome!

[willow ravine]

I’m going to move on to the next question in the homework set. You rock!

.close

[obtuse pebble]

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[dawn yew]

I have several linearly independent vectors $v_i$ in many-dimensional euclidean space. I want to find the (I believe unique up to a scalar) vector $V$ such that $V$ is in the span of the $v_i$s and for any $v_i$, $v_j$, $V \cdot (v_i - v_j) = 0$. For the case of 3 vectors in 3-space this would be easily solved with a cross product, but I don't know how to expand that beyond 3-space.

[warm shale]

TheZachMan

[dawn yew]

specifically, the number of vectors I have is potentially lower than the dimension of the space

[kind hawk]

maybe get an orthonormal basis for span(v_i) using gram schmidt and then take (1...1) in that basis?

[dawn yew]

I'm not sure that works, if I'm understanding you right?

if my vectors are [1 0] and [1 1], I get the standard basis as my orthonormal basis, but the correct value would be V = [1 0]

[kind hawk]

you need to write the standard basis in your basis consisting of v_i

I think

you have to do a few basis switches but I think it works. just make sure you always know with respect to which basis you are currently doing stuff

[solar trellis]

Perhaps I'm misunderstanding something but for 3d I don't think this is possible. Eg. when v_1 = (1,0,0) and v_2 = (0,1,0), what would V be?

[kind hawk]

(1,1,0) or am I doing a stupid mistake

=1v_1+1v_2=(1,1) in basis (v_1, v_2)

[solar trellis]

Ah, true. There's no use of the cross product tho

[kind hawk]

why do something that's specific to R^3 when you don't have to

[solar trellis]

I was thinking it had to be (0,0,1) or something but that's not necessary

[kind hawk]

(0,0,1) is not in span(v_i)

[solar trellis]

Yep

[dawn yew]

I guess my concern with the change of basis thing is

I don't really know what that gets me?

[kind hawk]

well you want something about orthogonality

[dawn yew]

like, converting to an orthonormal basis sounds plausible

[kind hawk]

but?

[dawn yew]

but I don't see what the actual procedure would be

[kind hawk]

get an orthonormal basis u_i from v_i using gram schmidt. then take the vector u_1+u_2+...+u_k

no wait

uhm

I just realized I made a mistake

let me rethink

shit I can't think anymore, it's too late

wait can't you just calculate the kernel of the matrix consisting of the rows v_i-v_j

[dawn yew]

that sounds plausible

and I'm guaranteed it's 1d because the v_is are independent

hm

well, no it isn't

[kind hawk]

I'm actually not sure such a vector V exists if you have too many v_i

[dawn yew]

that gets me the space of vectors orthogonal to all the v_i-v_j s, but it doesn't impose the constraint that V in span[v_i]

[dawn yew]

I'm pretty sure it does?

[kind hawk]

you could add rows for each vector in a basis of the orthogonal complement of span(v_i)

god gets even more complicated

forget that

[dawn yew]

oh wait, hang on

maybe the orthonormal thing does work

because I get v_1 = u_1, v_2 = a u_1 + u_2, v_3 = b u_1 + c u_2 + u_3

and then I take the kernel of the difference matrix in that space

[slim cove]

just wanted to add a fun fact, the cross product is only defined in 2, 3, and 7 dimensions

(that I know of)

[dawn yew]

that is a fun fact

[haughty coyote]

Do you have a link to that ? Kind of interested

[slim cove]

https://en.wikipedia.org/wiki/Seven-dimensional_cross_product

technically, it's only 0, 1, 3, or 7 dimensions

but I count there to be a cross product in 2 dimensions even though it's not really a cross product (it outputs a scalar instead of another vector)

[haughty coyote]

That gives the determinant of the 2 vector ?

[slim cove]

yeah, in 2 dimensions you can say the "cross product" is the determinant of the two vectors put together, but that's kinda hand-wavy since it's not really a cross product

[obtuse pebble]

@dawn yew Has your question been resolved?

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[timid silo]

the way i did it was separate and integrate both sides giving me y=tan(x^2/2)

and sub x^2/2 for x in taylor series for tan(x)

is that right?

[obtuse pebble]

@timid silo Has your question been resolved?

[slim cove]

yeah, sounds right what was the taylor series you got?

[timid silo]

1/2x^2 + 1/24x^6 +...

[slim cove]

seems good

[timid silo]

hence find y(0.5), do i just sub that into tan(x^2/2)

@slim cove u the goat man u helped me so much these couple days

[slim cove]

no problem!

[slim cove]

yes

or well

is the question to use the taylor series to find it

or just to find y(0.5)

[timid silo]

i had the same question

i emailed my teacher

[slim cove]

oh hm

[timid silo]

she says 'it shows it clearly on the question'

-.-

[slim cove]

bruh

????

it says "this" I assume this means the taylor series

so I guess just go with plugging it into the first two terms of the taylor series?

what a mean teacher

[timid silo]

okay i think i did it wrong cos its nowhere near the right answer

[obtuse pebble]

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[timid silo]

.reopen

[obtuse pebble]

✅

[slim cove]

what did you get using each method?

[timid silo]

tan(0.5^2/2) = 0.00218

taylor 0.125ish