help-13

[crimson sedge]

i know that the tanx will change to its quotient which is sinx/cosx but after wards i dont know whats the next step

hello?

somebody ?

[livid hound]

multiply numerator and denom by cos(x) to get rid of nested fractions

[crimson sedge]

@livid hound may I take a picture of my answer and check it if its right ?

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[dense wing]

Gave you help already

you evaluate g at a+h

and I told you what a and h were

and we already computed g(-1)

You also seem heart-set on using that form of slope, instead of the alternative I presented

I mean... it's the same

but frankly easier to use cause you're given 2 x values.

[cedar kiln]

Channel closed

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[dense wing]

K then.

K.

[cedar kiln]

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[keen jetty]

I need to calculate the volume of this equation:
y^2 <= z <= 4-x^2
To get a better visual understanding of the problem I have tried to plot it in geogebra, however I am not sure I have done this the right way

[keen jetty]

I wrote in the surfaces like this to try to plot them

Because if I put the equation as is (after separating it into two) I get two planes

[jaunty mural]

So you want to find the min/max of each variable (possibly in terms of other variables if inside)

and assuming you're integrating this in cartesian ig

I usually approach by trying to figure out which is the 'easiest' to put on the outside

[keen jetty]

Yeah the task is just to calculate the volume of the area enclosed within that equation

[jaunty mural]

But sometimes it doesn't matter which goes outside

Here, I 'feel' like dz should go on outside

[keen jetty]

This is what I get if I plot it like this and I have no clue why. Its not even on the z axis.

[keen jetty]

My question right now is to see if I plotted the equation correctly

[jaunty mural]

Like a visual understanding is important, but there is a lot you can do algebraically

uhhhh im not sure honestly

[keen jetty]

as in if it is the volume between those two planes I showed in the beginning

yeah because here I just have two overlapping planes on the xy plane which... has no volume you know

[jaunty mural]

$y^2 \le z \le 4-x^2$

[wraith dagger]

Shuri2060

[jaunty mural]

How do I put it

with this alone I think you can 'see' what bounds for z should be?

[keen jetty]

not really?

[jaunty mural]

When I say bounds for z

[keen jetty]

I'm missing something obvious aren't I

[jaunty mural]

we need all of the possible values z can reach

If we are having z as the outside integral

[keen jetty]

So like

$z: a\to b

nvm no clue how this tex bot works

[jaunty mural]

$$\int\int\int 1 \dd x\dd y \dd z$$

[wraith dagger]

Shuri2060

[jaunty mural]

Look, your integral will be the above

[keen jetty]

yeah That much I know

[jaunty mural]

but you are free to rearrange the dx dy dz

So first you should choose which one goes on outside

dz on outside makes most sense to me

[keen jetty]

If the first plot is correct then it should be z on the outside as the planes make a circle on the xy plane which makes the xy bit easy

[jaunty mural]

If dz is on the outside

[jaunty mural]

then the bounds for z

should be the minimal and maximal possible values of z

in the entire region

$y^2 \le z \le 4-x^2$

[wraith dagger]

Shuri2060

[jaunty mural]

This can be found via this inequality

[keen jetty]

0 to 4?

I found that graphically though so no clue how you would do this algebraically

[jaunty mural]

Exactly,

You are finding the min/max values of z algebraically

in that entire region

in other words, you are minimizing y^2

and maximizing 4-x^2

$0 \le z \le 4$

[wraith dagger]

Shuri2060

[keen jetty]

Yeah sorry I still dont quite get it

Im an engineering student for the record, we are as dumb as rocks

[jaunty mural]

dw about it, you're new to this.

Let's suppose for a sec

we have our integral ordered like this

$$\int^{z = u}{z = l}\int^{y = u(z)}{y =l(z)}\int^{x = u(y, z)}_{x = l(y, z)} 1 ,\dd x,\dd y,\dd z$$

[wraith dagger]

Shuri2060

[jaunty mural]

Note u and l are different functions in each

[keen jetty]

yes

[jaunty mural]

But I'm trying to say that the upper/lower bounds will be functions of various things depending on where its located

So for the outside bounds

They will just be constants

The have to represent the maximal and minimal value of the variable in the entire region

[keen jetty]

Right, that I get, but how do I get it

[jaunty mural]

Otherwise your integral won't cover the entire region

So here we have an inequality bounding z

This inequality tells us the maximal range of z

when the left side is minimized, and the right side is maximized

[keen jetty]

right this is the bit I dont get

probably a language thing

[jaunty mural]

This is actually 2 inequalities (we are technically treating them separately)

[keen jetty]

I study in Swedish

[keen jetty]

yeah that I realised

[jaunty mural]

$$y^2 \le z$$
$$z \le 4-x^2$$

[wraith dagger]

Shuri2060

[jaunty mural]

Right ok.

We only have these 2 inequalities which tell us everything about the region

We want to find the range of z --- every single possible value of z

[keen jetty]

Aye

[jaunty mural]

$$y^2\le z$$

[wraith dagger]

Shuri2060

[jaunty mural]

Since $y\in\bR$

[wraith dagger]

Shuri2060

[jaunty mural]

We know $$0\le y^2\le z$$

[wraith dagger]

Shuri2060

[jaunty mural]

Similarly $$z \le 4-x^2 \le 4$$ since $x\in\bR$

[wraith dagger]

Shuri2060

[jaunty mural]

$$0\le y^2\le z\le 4-x^2 \le 4$$

[wraith dagger]

Shuri2060

[jaunty mural]

So we have this

[keen jetty]

Right

[jaunty mural]

And those bounds can be achieved when y = 0 or x = 0

[keen jetty]

Im still not entirely sure what happened oh nevermind it just clicked hahaha

yep now it makes sense,
$$z: 0\to 4

ok texit

[jaunty mural]

$$\int^4_0\int\int1,\dd x,\dd y,\dd z$$

[wraith dagger]

Shuri2060

[jaunty mural]

So you have this (x and y might be swapped)

Then you need to repeat the same process for x or y

Find the smallest and largest possible value it can be depending on z

[keen jetty]

Right so how would I go about that for y then? Rearrange it somehow or just use y^2<=z

[jaunty mural]

Right so y...

[keen jetty]

Yep y would be 0 to 2

[jaunty mural]

not quite

[keen jetty]

nvm then

[jaunty mural]

you can square root that inequality

the left 3 things

[keen jetty]

right

[jaunty mural]

$$0\le |y|\le \sqrt{z}$$

[keen jetty]

0<=y<=sqrt{z}

[wraith dagger]

Shuri2060

[jaunty mural]

y can be negative, remember.

So can you see what the bounds for y should be?

$$\sqrt{a^2} = \abs{a}$$

[wraith dagger]

Shuri2060

[jaunty mural]

^just remember the sqrt of a^2 should be this in general

[keen jetty]

yep

Nah I still fell like it is between 0 and 2 since z can go up to 4

[jaunty mural]

The inside bounds need to depend on the outside bounds

The bounds will be a function of z

Like so.

[keen jetty]

I dont quite get that

[jaunty mural]

Imagine moving along the z axis

[keen jetty]

Yeah

[jaunty mural]

You are observing what the max y and min y

should be

depending where you are on the z axis

[keen jetty]

Right

and say z=1 gives me plus minus 1 due to the square root

Is this the bit I am missing or am I wrong again

[jaunty mural]

that is correct

if z = 1

-1 =< y =< 1

[keen jetty]

oOh

[jaunty mural]

This inequality tells you this, indeed

[keen jetty]

Right, I feel like I am getting on the right track

So y would go
-2 to 2

same for x really I think

[jaunty mural]

Its not

If z = 1

-1 =< y =< 1

The limits being -2 to 2 are incorrect

[keen jetty]

yeah I meant if I maximise z to 4

Right -1 to 1 if z = 1

Oh right in the integral Ill be stepping up the z axis

and getting different bounds depending on z

allright makes sense

So it goes from -sqrt(z) to +sqrt(z)

[jaunty mural]

exactly

And finally, you need to do the same thing for x

[keen jetty]

aight lemme try

https://cdn.discordapp.com/attachments/903430334681608232/942465104291704852/321445119171756033.png

[jaunty mural]

This kind of thing is where the geometrical intuition is useful

You imagine the difference slices along z
and need the max/min of y for each individual slice

[keen jetty]

z<=4-x^2<=4

[keen jetty]

yeah no this is great, I would have a much easier time visualising things if I was more confident on how things should look. By the looks of things the plot with two planes seems to be right, and so far the intervals are making sense

[jaunty mural]

👌

[keen jetty]

Right trying for x

sqrt(z)<=2-abs(x)<=2

Right no sqrt(4) should be plus minus 2 right

[jaunty mural]

uhhh

not sure you've done the right thing

[keen jetty]

yeah no it has to be a function of x and y

[jaunty mural]

$$z\le4-x^2\le4$$

[wraith dagger]

Shuri2060

[jaunty mural]

What did you do next?

[keen jetty]

The the square root of the entire thing

I mean I could have just done this instead

move 4 from the rightmost thing

to the middle

Taking it out

Nah that doesnt feel right at all

[jaunty mural]

You want to isolate x

in order to get an inequality with it

[keen jetty]

So get x alone

Actually hold on

You can rewrite 4-x^2 to (2-x)(2+x)

I dont think that helps nvm

[jaunty mural]

yh i dont think so either

[keen jetty]

Hm

Maybe I should separate them and see if I understand it any better

But if its the innermost inegral it should be a function of z and y as I will be stepping along those axis

No?

[jaunty mural]

It is a function of z and y

but actually, you should already see that x will turn out not to depend on y

[keen jetty]

Aye

[jaunty mural]

Because z gives the stronger bound

You need to take this and apply the same process as before

The process before was isolating x

$$0\le \abs{y}\le \sqrt{z}$$

[wraith dagger]

Shuri2060

[jaunty mural]

Before, we had this.

This actually isolates y

[keen jetty]

aye

[jaunty mural]

$$-\sqrt{z}\le y\le \sqrt{z}$$

cus it simplifies to this

[wraith dagger]

Shuri2060

[keen jetty]

yeah thats the bound

[keen jetty]

Right so we go 0 to 4 on the z axis, meaning for z=0 we get
0<=4-x^2<=4
which gives x: -2 to 2
and for z=4 we get
4<=4-x^2<=4
which gives x=0

[jaunty mural]

I'm not sure what you're doing

ok i guess I can see

You need to treat this inequality algebraically

[keen jetty]

me neither I will be honest, I am very sorry

[jaunty mural]

Like any other inequality/equation you are trying to solvefor x

You do this through algebraic manipulation

[keen jetty]

z-4<=-x^2<=0

times -1

4-z<=x^2<=0

[jaunty mural]

inequality signs flip

[keen jetty]

right

4-z>=x^2>=0

now square root of it all

sqrt(4-z)>=abs(x)>=0

meaning

-sqrt(4-z)<=x<=sqrt(4-z)

so for z = 0 I get -2<=x<=2, for 4 I get 0<=x<=0

[jaunty mural]

yh

[keen jetty]

Right so now I have my limits for the triple integral

Time to try it out

[jaunty mural]

👌

[keen jetty]

Eyoo it worked, thanks a lot @jaunty mural !!
Very sorry for being slow with understanding and thank you for being patient with that, its a bit funny because lagrange and jacobians and stuff with derivatives I get quite quickly while integrals have been slow for me.

I've had enough heruekas this session that I'll probably not forget this for a long while

[jaunty mural]

No worries 👌

[keen jetty]

Have a good day c:

.close

[cedar kiln]

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[cedar kiln]

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[real bone]

,w derivative 2sec(x)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by tiberiusrox#6304

[real bone]

,w derivative 2tan(x) sec(x)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by tiberiusrox#6304

[violet rapids]

Do you have a question?

[real bone]

yes

[violet rapids]

oh ok lol

[real bone]

i got 2 cos^3 x + 2 sin^3 x over cos ^4 x

does that equal 2secx (tan^2 x + sec^2 x) ?

[glad kestrel]

,w (2cos^3(x)+2sin^3(x))/(cos^4(x))=2sec(x)(tan^2(x)+sec^2(x))

[wraith dagger]

Results provided by WolframAlpha

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Requested by a disappointing son#0540

[glad kestrel]

no

[real bone]

;/

idk what I am doing wrong

[glad kestrel]

can i see your work

[real bone]

yea i can start with the first derivative

,w derivative 2sec(x)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by tiberiusrox#6304

[glad kestrel]

i don't mean see wolfie's answers

i mean see your paper

[real bone]

give me a sec

rather than take a picture of chicken scratch I will walk you through it

so i took 2tanxsecx

to mean

2sinx/cos^2x

if that is wrong then I messed up step 1

[glad kestrel]

why break it into a quotient instead of just using product rule

[real bone]

hmm

well i was trying to work only with sine and cosine

cuz at this point I only know derivatives of sine and cosine

also if I do (2sinx)(1/cosx)(1/cosx) idk how to use product rule for 3 factors

maybe ((2cosx)(1/cosx) + (1/-sinx)(2sinx))(1/cosx) ?

[glad kestrel]

if you only know sin and cos derivatives, that's alright

your first step is correct

[glad kestrel]

where did you go from here

[real bone]

my next derivative from here is

actually question

derivative of cos^2x = -sin^2x right?

just want to make sure I dont need to use power rule

[glad kestrel]

not quite

you do need to use the power rule along with the chain rule

[real bone]

-2sinx?

[glad kestrel]

no

do you know the chain rule

[real bone]

idk why they are having me do this in 2.4 we dont learn chain rule until chapter 3

[glad kestrel]

oh

hm

[real bone]

nope

[glad kestrel]

well yeah that's weird

cause you need the chain rule here to take that derivative

[real bone]

its ok I can skip it and come back

thanks

.close

[cedar kiln]

Channel closed

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Use .reopen if this was a mistake.

[glad kestrel]

👍

[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
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[dire geode]

Post the full question and show what you've tried so far

[crimson sedge]

just have to integrate it

ive tried U-sub

what would be the best way to do it

I can U-sub
make U be 49-x^2

i get du as -2x

[drifting matrix]

If you see something like

sqrt(constant - x²)

try substituting x = sqrt(constant) sin(u)

(or cos)

Due to sin² + cos² = 1 this will make the sqrt thing go away.

[crimson sedge]

wait what

[drifting matrix]

Just try it 🙂

[crimson sedge]

ok

but why would i do that

i probably cant use that either my teacher gonna be like where did you learn that from

[drifting matrix]

It's not an obvious thing if you have not seen it before.

Actually I think the substitution you started will be good too.

For me just sqrt(49 - x²) => Me immediately wanting to substitute sin/cos. xD

[crimson sedge]

is there anyother way besides u-sub

i want an optimal way

i can do the u sub but i feel their has to be another way

[drifting matrix]

The optimal way is to guess the right integral and to just check that it has the correct derivative.

I don't think there is another way.

[crimson sedge]

ok thanks man

i apprentice it

./close

.close

[cedar kiln]

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[turbid portal]

For each integer r in {0,1,2,3,4}, let A_r be the set of all integers which leave a remainder of r when divided by 5. (That is, x in A_r iff x=5q+r for some integer q.) Prove: {A_0, A_1, A_2, A_3, A_4} is a partition of Z.

[turbid portal]

I'm new to partitions and only saw examples, so I'm not entirely sure on how to prove that this is a partition

[dense wing]

do it with the checklist

need to verify they're all subsets, all non-empty, pairwise disjoint, and union to Z

[turbid portal]

I'm not sure how to tho

[dense wing]

I mean, 1st 2 are trivial

/ dont require much work.

[turbid portal]

I'm confused

[dense wing]

How specific

Could you explain your confusion...?

[turbid portal]

Do I use the x=5q+r? For what you're talking about?

[dense wing]

Yes, you'll need to use how A_r is defined.

Like, clearly A_r are subsets

$q,r\in\mathbb{Z}\implies 5q+r\in\mathbb{Z}$

[wraith dagger]

Mosh

[turbid portal]

That's how you'll start it off?

[dense wing]

First prove they are subsets.

[turbid portal]

Before I do so

I have a quick question about partitions

Like the definition

A partition is a family of non-empty subsets of A:

1. if any two classes, say A_i and A_j, have common element x (not disjoint), then A_i=A_j
2. every element x of A lies in one of the classes

Could you further explain?

[dense wing]

Further explain what?

1 and 2 say the same thing

[turbid portal]

(we did examples in class but not where we had to prove, nor did he say how)

[turbid portal]

Really? In my book, it says "and"

[dense wing]

I guess the 2nd leans on none of the partitioned sets can be empty

[turbid portal]

Oh

[dense wing]

But, as I said already, you just prove the 4 defining requirements

[turbid portal]

I see

I'm just super confused

[jaunty mural]

If I tell you that the evens and odds form a partition of the integers, would you understand why?

[turbid portal]

Not really no

[dense wing]

Check the requirements

[jaunty mural]

I think you need to go through some examples to get the concept yh 👀

[dense wing]

2Z and 2Z+1 are both subsets of Z
They're both non-empty
No integer is both even or odd, thus they're disjoint
Every integer is either even or odd, so they union to Z

[turbid portal]

We did, but they were a bit complicated to me

[jaunty mural]

$$\bigcup A_i = X$$
$$\bigcap A_i = \emptyset$$

Have you seen in this notation the conditions for a partition of X?

[dense wing]

\emptyset btw

[turbid portal]

I have not

[wraith dagger]

Shuri2060

[jaunty mural]

Perhaps this might be more intuitive than words? Well. In any case, you need to 'digest' what partition means in some way

[turbid portal]

Okay

[dense wing]

they're all subsets, all non-empty, pairwise disjoint, and union to Z

[turbid portal]

What does pairwise disjoint mean?

[jaunty mural]

Any pair of sets that form your partition have an empty intersection

[turbid portal]

What exactly does it mean to be an empty intersection?

[jaunty mural]

Their intersection is the empty set

ie. they share no elements

For example, the odds and evens have an empty intersection

Because no number is both odd and even

[turbid portal]

Ohh

[jaunty mural]

Next, the union of all of the sets making up the partition must be the original set

Another way to phrase is 'nothing is left out of your partition'

Intuitively, you are just chopping your original set up into a bunch of pieces. That's all.

[turbid portal]

Which is the A_r in this case?

[dense wing]

A_r would be the sets you're partitioning Z into.

[turbid portal]

Okay

Do we want A_i to = A_j? For some integers i and j?

(when I was stating the def)

[dense wing]

You want $A_i\cap A_j=\emptyset$ for $i\neq j$

[wraith dagger]

Mosh

[turbid portal]

Oh okay

So I'd have to start off by showing it's a subset?

[dense wing]

Yes.

You need to prove A_r is a subset of Z

As I've said many times, you prove the 4 requirements

[turbid portal]

Mhm

Im just verifying

When proving a subset, is it the same as subgroup?

[dense wing]

$x\in A_r\implies x\in\mathbb{Z}$

[wraith dagger]

Mosh

[dense wing]

$A_r\subseteq\mathbb{Z}$

[wraith dagger]

Mosh

[turbid portal]

Oh I show the iff part?

[jaunty mural]

not at all. There is no mention of groups here.

[turbid portal]

Okay

[dense wing]

there's no iff

[turbid portal]

Okay

No I know, I'm talking about the x=5q+r

[dense wing]

you don't show that

That's how A_r is defined

[turbid portal]

Oh okay

So if x is in A_r, then x is an integer?

[dense wing]

yes.

[turbid portal]

So it's a subset?

Of Z

[dense wing]

Yes.

$q,r\in\mathbb{Z}\implies x=5q+r\in\mathbb{Z}$

[wraith dagger]

Mosh

[turbid portal]

It's non-empty because we have more than one integer in this subset?

[dense wing]

You have at least 1 element in each set.

Namely $r\in A_r$ obviously

[wraith dagger]

Mosh

[turbid portal]

Oh okay

"At least" one, will remember that

Now to show pairwise disjoint?

[dense wing]

yes, once you've written the proof for subset and non-empty

[turbid portal]

Which were fairly short

How do I show that they're disjoint?

[dense wing]

Have you attempted to prove it or did you look at it and go "No clue"

[turbid portal]

I attempted

Then erased cuz I got confused

Can I start off by letting an x be an integer? Where x is in A_r and some other set, say A_s?

[dense wing]

yes, you do it by contradiction usually

[turbid portal]

Okay

I'm a bit stuck

How would this lead to contradiction?

We use the form of A_r? x=5q+r

x =5q +s

r=s?

[cedar kiln]

@turbid portal Has your question been resolved?

[jaunty mural]

You just did it

Suppose x in Ar and x in As

Then x = 5q + r = 5q + s

Then r = s

That's it. (btw no contradiction used here by me)

[turbid portal]

So it's not a contradiction?

[jaunty mural]

You could.

I could've started by saying 'suppose x in Ar and x in As and r =/= s'

Then I would get a contradiction.

But it isn't necessary.

[turbid portal]

Oh okay

Union to Z

[jaunty mural]

Sorry I wasn't paying attention

So you showed the intersection of any 2 is empty

Now you need to show union of all makes Z?

[turbid portal]

Could we start off that we know that any remainder, r, is an integer? Or is that irrelevant?

[jaunty mural]

Well there are a variety things you could probably do

[jaunty mural]

That is relevant, but too vague.

[turbid portal]

I see

[jaunty mural]

so what u wanna show is that any integer

belongs to one of those 5 classes

you can use contradiction if u want, up to u

[turbid portal]

Classes?

[jaunty mural]

sets

[turbid portal]

Like an example?

I'm confused

[jaunty mural]

which set is 11 in

[turbid portal]

1

A_1

[jaunty mural]

right.

now do the same for a general integer

maybe this is bad of me haha

it doesnt really show the way forward

You can just simply state that all integers leave remainder 0, 1, 2, 3, 4 upon division by 5

(maybe you're confused because you're overthinking things)

[jaunty mural]

If you don't think that's enough, you can try a contradiction via algebra or something

[turbid portal]

When am I never ovethinking things?

[turbid portal]

Alright

Thank you Shuri

.close

[cedar kiln]

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[turbid portal]

And Mosh

[cedar kiln]

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[burnt shard]

need help solving this question

[green belfry]

So

When it says slope your really looking for the gradient

[burnt shard]

right, so do i start of by plugging in root 3 in the equation ?

[green belfry]

Do u know differentiation

[jaunty mural]

what does this achieve?

[burnt shard]

u find y

[jaunty mural]

And what does that help with?

[green belfry]

But that’s just a point

Not the gradient of the tangent

[jaunty mural]

hint: it doesn't help, take omar's suggestion.

The derivative f'(x) says what the gradient of the curve is at each x.

[burnt shard]

not yet

[jaunty mural]

???

Then how on earth are you meant to do this question...

[green belfry]

How did they expect you to do this

What topic are you getting this question from

[burnt shard]

wait, could you define the word, maybe I have done it but it a diff word?

[jaunty mural]

Differentiation

Derivative

f'(x)

dy/dx

[burnt shard]

oh yes

derivative

yes

[jaunty mural]

All 4 of those things I just said roughly refer to the same concept ok?

...

[burnt shard]

Alright

[jaunty mural]

The derivative f'(x) says what the gradient of the curve is at each x.

They should have taught you this.

[burnt shard]

yes, we are just getting started with it

so first, do I find the eqaution of the slope

so 10x^4?

[green belfry]

Yes

That’s the tangent equation

[burnt shard]

and plug root 3 in after

[green belfry]

Or first derivative

Then plug in

Yup

[jaunty mural]

no

[green belfry]

I meant gradient function

[jaunty mural]

First derivative, yes.

[burnt shard]

for this one, I start by getting the derivative?

than plug in 0

[green belfry]

Yes

[jaunty mural]

no, read the question.

[green belfry]

No perhaps the opposite

[burnt shard]

what do you mean

[jaunty mural]

Last time, we wanted the tangent of the function at a certain value of x

This time they don't seem to want that.

Read the question.

[burnt shard]

we want x,y?

2 points

[jaunty mural]

???

what?

Read the question. Slowly.

Yes, we want a certain (x, y) since it asks for a point

But there is more to it.

[burnt shard]

The last part of the question kinda confuses me

"each parabola is zero"

[jaunty mural]

it says 'each parabola' because your question has multiple parts

[burnt shard]

multiple parts to what

[jaunty mural]

part i)

part ii)

iii) ..

I presume.

cut off from your screenshot

[jaunty mural]

'Determine the point at which the slope of the tangent to the parabola is 0'

Just read it like this for each part.

[burnt shard]

thers only 1 more part

[jaunty mural]

are you saying this entire question 4

only has a part i)

[burnt shard]

no

it has another part'ii

[jaunty mural]

alright

thats why the question says each

[burnt shard]

oh

so what am i missing

[jaunty mural]

'Determine the point at which the slope of the tangent to the parabola is 0'

[burnt shard]

ik the first part is 12x-3

[jaunty mural]

well yes, that is the derivative.

Do you understand what the question is asking for?

[burnt shard]

not really other than its a point on a line

[jaunty mural]

• Determine the point at which
• the slope of the (tangent to the parabola)
• is 0

Does that make the question clearer?

[burnt shard]

a point where slope is 0

?

[jaunty mural]

exactly

You are looking for where the slope is 0.

[burnt shard]

so how would i do that

[jaunty mural]

slope = 0

f'(x) = 0

understand?

[burnt shard]

does this has to do anything with 'b' in y=mx+b

[jaunty mural]

well no

do you know what f'(x) is

that is the derivative

f'(x) = 12x - 3 as you have already said

then you set this equal to 0 and solve.

[burnt shard]

what do i solve if its alr set to 0?

[jaunty mural]

0 = 12x - 3

??

solve for x?

you are trying to find (x, y)

[burnt shard]

ohhh

find x

and plug x in after to find y

wait no

the 0 is not y correct?

as a point

[burnt shard]

im getting (4, 45) is that right?

[jaunty mural]

f'(x) = 12x - 3

f'(4) = 48 - 3 = 45

that certainly isn't 0.

[burnt shard]

what am i doing wrong

[jaunty mural]

0 = 12x - 3

Solve for x.

How did you get 4

[burnt shard]

by solving for x

12/3

i mean the other way around

(1/4, 0)

[jaunty mural]

...

12x - 3 = 0

12x = 3

1/4, my bad

[jaunty mural]

Then you need to plug x = 1/4 back into the original equation

to find the y value at x = 1/4

This gives you the point (x, y) at which f'(x) = 0

[burnt shard]

for this one im getting (-4/3, 0)

[jaunty mural]

,calc (3/4)(-4/3)^2+2(-4/3)

[wraith dagger]

Result:

-1.3333333333333

[jaunty mural]

No, your y coordinate is wrong.

For both, you need to plug the x into the original equation

[burnt shard]

i did

oh wait

original meaning the non derivative one?

[jaunty mural]

yes

ofc

The derivative is just the slope equation

plugging in x tells you the slope

not the y coordinate.

[burnt shard]

so to find x

we use the derivative

and to find y

we use original

equations

[jaunty mural]

Thats for this very specific question yes...

[burnt shard]

so the answer for my first question is wrong

y is not 0

its 29/8

[jaunty mural]

its not 0, yes.

You can plot these equations in desmos to check your answers

or some similar software

[burnt shard]

I was wrong about there being no 'iii" part and the 'iii" part deals with desmos

[jaunty mural]

On new lines

(x, y)

^you can input just this

for the specific points you found

and it will display them.

[burnt shard]

wdym

vertexs

[cedar kiln]

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[coral creek]

Suppose that points F,G, and H are collinear, with FG = 7, FH = 2 and GH = 5.
a) explain carefully how ray GH is defined and decide if F is in ray GH
b) If R and S are on the same line containing F and G, and if RF = 12 and SG = 6, find all possible values for RS, assuming Ruler Postulate. What are all of the possibilities?

[coral creek]

For a, I started off with basically the def of ray GH => GH = {x | G-X-H or G-H-X or X=G or X=H}

then i said there can be a point in between g and h on the ray or there can be h in between g and that point x

F is on the ray GH since FH=2, FG = 7, GH =5. So if it's G-H-F where x is F, this shows 2+5 = 7

not sure if i was doing this right

but i have to state the other two parts as well

I was told that there are three parts to this

[cedar kiln]

@coral creek Has your question been resolved?

[coral creek]

<@&286206848099549185>

[cedar kiln]

@coral creek Has your question been resolved?

[coral creek]

Please help 🥺

[cedar kiln]

@coral creek Has your question been resolved?

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[crimson sedge]

Hiii please can someone help me on what I can cancel here ?

Please guide me in answering this question im still confused

[uncut hedge]

You can simplify by sin x

[crimson sedge]

which sin x ?

[uncut hedge]

all of them

[crimson sedge]

so the two sin x on the top will be canceled as well in the bottom ?

[uncut hedge]

(sin x - sin x / cos x) / sin2 x = (1 - 1 / cos x) / sin x = (cos x - 1) / (sin x cos x)

[crimson sedge]

like this ?

[uncut hedge]

Yes

[uncut hedge]

well don't forget the integral around it of course

[crimson sedge]

how did the sin^2x on the denominator become sinx ?

[crimson sedge]

i still dont understand it sorry...

[gentle lintel]

do you see how

(x + y)/z

is x/z + y/z

[uncut hedge]

sin x is a common factor of the numerator and denominator, hence you can simplify by it

[gentle lintel]

if you divide through sin x - tan x by sin²x, you obtain 2 integrals

both of them should be easy

[crimson sedge]

freak i cant visualize what you guys are saying sorry for being so dumb

[gentle lintel]

do you know what i mean

[crimson sedge]

its really hard to understand if I cant see how it was done im so sorryy....

im really bad at this

[gentle lintel]

for example

(2x + 2y)/2

= (2x)/2 + (2y)/2

= x + y

[glad kestrel]

$\frac{\sin(x)-\frac{\sin(x)}{\cos(x)}}{\sin^2(x)}=\frac{\sin(x)}{\sin(x)}\left(\frac{1-\frac{1}{\cos(x)}}{\sin(x)}\right)$

[gentle lintel]

here, we split it into two fractions

what

[wraith dagger]

a disappointing son

[crimson sedge]

hmmm

[crimson sedge]

so like we get the common factor and divide them and that will be the result ?

[gentle lintel]

in general $\frac{kx}{ky} = \frac{x}{y}$

[wraith dagger]

Chromium

[gentle lintel]

we cross out the k

[gentle lintel]

this is the detailed step

[crimson sedge]

are you guys available on a quick voice channel ?

[gentle lintel]

do you understand why this is

[crimson sedge]

i kinda get it now but here is the part where im still confused

that sinx/cosx on the numerator

how did it become 1/cosx?

[gentle lintel]

.

[crimson sedge]

like this ?

[gentle lintel]

do you know how to do something like

$\frac{x^2 - y^2}{x + y}$?

[wraith dagger]

Chromium

[crimson sedge]

what si the name of that topic, ill try to search it at youtube and try to learn it

[gentle lintel]

so you don't..?

[crimson sedge]

yes.

i think ill manage this one

[gentle lintel]

oh dear

[crimson sedge]

thank you for all the help

[gentle lintel]

this is one of the first things you learn in algebra

[crimson sedge]

appreciate it

[gentle lintel]

which is wayy before integrals

[crimson sedge]

math really isnt my thing thats why i hate this hahahaha

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

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[hallow folio]

I'm pretty sure I've already solved it but I wanna make sure what I did is valid.

So I figured if that represents one object, I can switch the r and s in both terms, which when rearranged does indeed switch the signs of both terms, same for exchanging i and j.

But what's tripping me up is that I basically just... moved the indices between the deltas, it works... but I don't know if that's allowed? I'm sort of confused.

I think it's allowed because all I was really doing was renaming the indices consistently.

.close

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[gentle lintel]

find y'' and y'

[covert island]

So this question is messing with me big time

for starters

can derivatives have derivatives?

like is that a thing

[gentle lintel]

wdym

second derivatives, third derivaitves etc exist

those are called higher order derivatives

[upper abyss]

y" is the derivative of y'

[covert island]

so the derivative of a derivative?

[upper abyss]

Which is just a single derivative of a function called y'

[gentle lintel]

yea

[covert island]

alright i kinda get it so far

so how would I find the y'' and y'

like where would I start?

[gentle lintel]

derivatives are just functions

if you know how to differentiate functions

you know how to differentiate derivatives (which themselves are functions)

it isn't really hard

[gentle lintel]

that i believe you know

[covert island]

??

[upper abyss]

y is given and is that quadratic.

[gentle lintel]

you don't know what second derivative means?

[covert island]

we just said the the second derivative is just the derivative of y'

right?

[gentle lintel]

yes

that's literally it lol

[covert island]

so if we have y then y' would be 2Ax+B

[gentle lintel]

yes

[covert island]

wait

then y'' would just be 2A

right
??

@gentle lintel would it be right then for y'' to just be 2A

[gentle lintel]

yes

[covert island]

the problem wants us to find the constants of A B and C

[gentle lintel]

yea

[covert island]

im not sure what the constants would be past this point?

[gentle lintel]

you've sovled for y' and y''

[covert island]

yes

[gentle lintel]

directly plug y, y' and y'' in the given equation

then solve

[covert island]

i got 2A + 2Ax + b - 2Ax^2 - 2Bx - 2C = x^2

right??

[gentle lintel]

yes

[covert island]

well thats a problem

cuz theres no like terms...

[gentle lintel]

you're solving for A, B, C such that this is true

you can do this by comparing coefficients

[covert island]

wait

is A B and C all = to 2

[gentle lintel]

wdym

$2A + 2Ax +B - 2Ax^2 - 2Bx - 2C = x^2$

[wraith dagger]

Chromium

[covert island]

yes

[gentle lintel]

$x^2 (-2A) + x (2A - 2B) + (2A + B - 2C) = x^2 (1) + x (0) + (0)$

solve for the corresponding A, B, C's

[covert island]

ok im sorry

im still really lost

as to this last part

[gentle lintel]

do you know comparing coefficients?

[wraith dagger]

Chromium

[gentle lintel]

maybe this looks better to you

[covert island]

i dont think so. maybe i know the concept but not the name

[gentle lintel]

you should obtain a system of equations in A, B, C

[covert island]

??

im sorry im so lost

[covert island]

what does this mean

[gentle lintel]

like this, for example

can you see how the system was set up?

[covert island]

kinda

they basically just split 1 into a b and c

right?

[gentle lintel]

wdym

[covert island]

they split the numerator (1)

into 3 variables

then split the OG equation

into 3 parts no?

[gentle lintel]

oh don’t care about the partial fraction part

[covert island]

nvm

we back to square one

im lost as hell

[gentle lintel]

you understand this, right?

[covert island]

ok yes to an extent

my only question is

what happens to the x

and x^2

[covert island]

ok after much review

i get this

goin back to the OG equation then

[gentle lintel]

so here

[covert island]

-2a would be X^2

[gentle lintel]

we can set up a system

[covert island]

or 1

-2a would be = to 1

2a-2b would be 0

and 3a+b-2c would also be = to 0

[gentle lintel]

2a*

yea

[covert island]

wait

isnt it just 2a

[covert island]

like here?

[gentle lintel]

$\begin{cases}
-2A = 1 \
2A - 2B = 0 \
2A + B - 2C = 0
\end{cases}$

like this

[wraith dagger]

Chromium

[covert island]

ok

A= -1/2

B = 1/2

C = 3/4

[gentle lintel]

yea it's just algebra

[covert island]

gotcha!

Thank you so much for your help!

I really appreciate it!

.close

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[eternal siren]

How do I find the surface area of a heart shape?

[violet rapids]

Is there a diagram of your shape

Cause if its 🫀 its gonna be hard

[eternal siren]

♥️ This shape

So the formula for it

[violet rapids]

There's not really a formula

But I would break it into two semicircles and one big triangle

Then add those areas together

[eternal siren]

Oh ok

Thanks

.close

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[tough osprey]

How does the statement quoted in red makes sense?

The proof is by contradiction

But where did the contradiction arise?

[toxic owl]

The sum of a rational number and a surd cannot be a surd

But from the inital assumption, we came to the conclusion that $m + \sqrt{b} = \sqrt{d}$ or, that the sum of a rational and a surd is a surd, which is mathematically impossible

Hence, the contradiction

[wraith dagger]

[TEB] darthlothins

[tough osprey]

Ok get it, thanks

.close

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[wanton glacier]

how is this a valid method of algebraic manipulation? Can't you just substitute c(1-b)?

[wanton glacier]

I thought you could only subtract or add between equations

when would you divide between equations?

[upper abyss]

Nothing wrong with that. That's like saying:
if a = x
And if b = y
Then a/b = x/y

Which should be obvious. Just take a/b, replace the a with x, and replace the b with y. You get x/y, so they're equal

Can do the same with other things, like a^b, provided a is positive.

[wanton glacier]

alright

[wanton glacier]

when would you do it?

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[gaunt trench]

this is on C

shouldnt this print decimals?

[fallen heath]

snake and lizard are both defined as integer variables

obviously C is going to execute an integer division

$3 \divisionsymbol 4 = 0$

[wraith dagger]

Ansh

[gaunt trench]

oh right

hey also

"register" is a valid variable name for C right

?

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[half trellis]

How to prove that F is derivable on R

[zenith sail]

Unless I'm missing something

Doesn't this just follow from the fundamental theorem of calculus?

[half trellis]

what is the primitive of the function inside the integral?

[zenith sail]

I don't believe there is an elementary primitive

but

So isn't it true that $F'(x) = e^{x-\frac{x^2}{2}}$?

[wraith dagger]

tatpoj

[half trellis]

lets call the primitive smthing other than F since the function is named F

but yes it is

so i don't have to write it as primitive then derivate to get this?

[zenith sail]

No, the FTC guarantees that this is true

[half trellis]

okey

thanks

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[indigo basin]

18 workers build a house in 15 days. In how many days would 30 workers build a house?

[indigo basin]

I know is easy but i just wanna know procces beacuse some are harder

And i forgot process

<@&286206848099549185>

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[cedar kiln]

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Remember:
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[dull anchor]

$\ln\left(\frac{k+3}{k}\right)-\ln\left(\frac{k+2}{k-1}\right)$

[wraith dagger]

Big xdddd

[dull anchor]

how can i show that this is for all k in N < 0?

first i did $$\ln(k+3)-\ln(k)-\ln(k+2)+\ln(k-1)$$

[wraith dagger]

Big xdddd

[dull anchor]

and then i did $$\ln\left(1+\frac{3}{k}\right) +\ln\left(\frac{k-1}{k+2}\right)$$

[wraith dagger]

Big xdddd

[dull anchor]

but idk how this is going to help me

[toxic owl]

Do you mean for N>0?

[dull anchor]

yes for all natural numbers

[toxic owl]

Did you try to prove by induction?

[vale pond]

You can do $$\ln\left(\frac{k^2+2k-3}{k^2+2k}\right)$$

Cmon

[dull anchor]

you forgot the }

[toxic owl]

This looks like a typical induction problem

[wraith dagger]

ku-sital

[dull anchor]

damn i can really do that

[vale pond]

Idk how this help

[toxic owl]

Idk how hard it will be to prove the induction hypothesis though

[dull anchor]

but this wouldnt be < 0 i guess because $$ \ln\left(1-\frac{3}{k^2+2k}\right) > 0, \forall n\in\mathbb{N}$$

[wraith dagger]

Big xdddd

[dull anchor]

or wait

it isnt

yea that works because i know that log(1) = 0

[dull anchor]

and this shit is log(1-sth) which is smaller 1

[dull anchor]

wait but how did you got that what did you do?

ok i got it

thank you!

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[cedar kiln]

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[viral tusk]

how would you calculate the area of a square, inside of a circle?

[viral tusk]

.close

[cedar kiln]

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