How do i do binomial expansion

[hearty terrace]

hfomff f

[rugged grail]

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[bleak gorge]

What exactly are you interested in? How to calculate the binomial coefficients, or something else?

[hearty terrace]

i wanna know how to expand things such as (x+5)^5 using this method

i am learning in skl this but idk how it works

[bleak gorge]

Oh, easy enough. Like this.
Here C(n, k) = n!/(k! (n - k)!) are the binomial coefficients, which you can also see in Pascal's triangle.

For example:
(x + y)^3 = C(3, 0)x^3 y^0 + C(3, 1)x^2 y^1 + C(3, 2)x^1 y^2 + C(3, 3)x^0 y^3 = x^3 + 3x^2 y + 3xy^2 + y^3

[hearty terrace]

oh ok

i am confused by all the letters like the weird E, and C

is there a simpler explanation cuz i havent even finished skl yet!

[bleak gorge]

That's Σ - capital sigma, not E.

[hearty terrace]

ohh

is that sum of

[bleak gorge]

Yes, sum.

[hearty terrace]

oh ok

the umbers look like periodic table elments

[bleak gorge]

Well, binomial coefficients appear in many different places.

[hearty terrace]

yea

[hearty terrace]

at what age do u learn this thing

liek the formula

[bleak gorge]

Probably around grade 7-8 or so.

[hearty terrace]

how old is that in yrs cuz idk wat that is

[bleak gorge]

About 13-14, I believe.

[hearty terrace]

oh ok

in my slk i am not doing that yet

[bleak gorge]

That was quite a while ago for me, so I don't exactly remember.

[hearty terrace]

oh ok

i am currently learning differentiation

[bleak gorge]

Oh, that's good. It's a very important topic.

[hearty terrace]

yea

is there antyhing else i need to know apart from: when it is = 0 i can find turning point with it and i can find gradient with it

[bleak gorge]

Well, that's only the necessary condition for an extremum point, not a sufficient one.

[hearty terrace]

oh ok

what does sufficient mean

[bleak gorge]

"A is a necessary condition for B" means "if B is false, then A is false".
"A is a sufficient condition for B" means "if A is true, then B is true".

[hearty terrace]

o ok

dat is confusing

idk if i will learn that this year

[bleak gorge]

Well, probably.

[hearty terrace]

so when i differentiate and sub in x i can find gradient right

and when i make it = to 0 i can find where it turns around

is that how it works

[bleak gorge]

As I said, that's only a necessary condition.

[hearty terrace]

o yea

[bleak gorge]

For example, f(x) = x^3 has a critical point x = 0, but it's not an extremum point.

[hearty terrace]

ohh ok yea

does ur country have gcses

[bleak gorge]

What is that?

[hearty terrace]

it is like exams before 6th form

at the age of around 15-16

[bleak gorge]

Wait, grade 5 at 15-16 years old?

The grade system seems to also be different, then.

Here first grade is 7-8 years old, I believe.

Or something like that.

[hearty terrace]

yes

[hearty terrace]

ohh ok

is that in america

i need help in this too

[bleak gorge]

No, Russia.

[hearty terrace]

ohh

cool

[bleak gorge]

Ah, alright. First, we write the formulas for surface area and volume.
S = 2πr(r + h)
V = πr^2 h
Express h from the formula for S and substitute it into the formula for V.

[hearty terrace]

ok thanks

doi i need to use 600 pi yet>

[bleak gorge]

No need. We can solve this generally.

[hearty terrace]

ok thanks

so i rearrange for height first?

[bleak gorge]

Yes.

[hearty terrace]

ok will do

i got h= (S-2πr^2)/2πr

is that right

[bleak gorge]

Yes. In other words, h = S/(2πr) - r.

Now we need to substitute it into the expression for V.

[hearty terrace]

how do i get that

[hearty terrace]

o ok

[bleak gorge]

Divide the numerator by denominator term by term.

[hearty terrace]

oh

[hearty terrace]

so wat is the first thing u divide

[bleak gorge]

(S - 2πr^2)/(2πr) = S/(2πr) - 2πr^2/(2πr) = S/(2πr) - r

[hearty terrace]

ohhh

i did not know u could do that

[bleak gorge]

Well, why not? After all, (a + b)/c = a/c + b/c.

[hearty terrace]

i get it tho now

na i just didnt think to do that

i thought since the numerator has a + or - then u cant eliminate things so i cant simplify it

but now ik u can but it will have the - r at the end

[hearty terrace]

yh i didnt see it that way

[bleak gorge]

Alright. Now, what do you get when you substitute it into the expression for V?

[hearty terrace]

idk if its right

i got

(Sπr^2/2πr)-r^3

[bleak gorge]

You forgot to distribute π to the second term.

Also, the first term can be simplified.

[hearty terrace]

oh ok i will try

i will write down wat i am doing

im starting with πr(s-2πr^2)/2πr

cuz i didnt write down the siplified one u did

[bleak gorge]

You mean πr^2 (S - 2πr^2)/(2πr)?

[hearty terrace]

o yea

i forgot

πr^2(s-2πr^2)/2πr

[bleak gorge]

Right, So, what can you simplify first?

[hearty terrace]

hmm

i can expand ??

[bleak gorge]

No, that's not very rational.

[hearty terrace]

or i can divide pi?

[bleak gorge]

Try cancelling by πr first.

[hearty terrace]

can i eliminate pi and r

ok thanks

i forgot about that

now i have r(S-2πr^2)/2

after eliminating πr

[bleak gorge]

Yes. And now we should expand that, as we will need to differentiate it later.

[hearty terrace]

o ok

(rS-2πr^3)/2

[bleak gorge]

Yes. In other words, V = (S/2)r - πr^3.

Now you can see that we get the desired result in (a) for S = 600π.

[hearty terrace]

o yea

[bleak gorge]

Now, let's do (b).

We have V = (S/2)r - πr^3 and we need to find its maximum value. What do we need to do first?

[hearty terrace]

differentiate it i thini

[bleak gorge]

Yes, we need to find dV/dr.

What will it be?

[hearty terrace]

do i differntiate 300πr-πr^3??

[bleak gorge]

No, leave S for now.

Find dV/dr for V = (S/2)r - πr^3.

[hearty terrace]

oh ok

is r the thing i find the power and times the coefficient for?

[bleak gorge]

What do you mean?

[hearty terrace]

like

if a question is differentiate 5x^2 then the x is the thing im talking abt

[bleak gorge]

Oh, yes.

[hearty terrace]

the thing which u multiply its coefficient by power

ohh ok

is that what the d/dx means

the x is the thing im talking abt

[bleak gorge]

Yes, differentiation with respect to x.

[hearty terrace]

ohhh

amd wat abt the numerator of d/dx

[bleak gorge]

It's the function we are differentiating.

Let's get back to our exercise, though. It's 1:33 AM for me now, so it would be good if we finish it now and I can go to sleep.

[hearty terrace]

o ok

yea

i got S/2-3πr^2

after differentiating it once

do i need to differentiate twice now?

[bleak gorge]

No, not yet.

So, to find the critical points, we find the values of r where dV/dr = 0.

[hearty terrace]

o ok

[bleak gorge]

So, solve S/2 - 3πr^2 = 0.

[hearty terrace]

is critical point = turning point?

[hearty terrace]

ok

[bleak gorge]

Well, I don't really use the term "turning point".
Critical point of f(x) is where f'(x) = 0 or doesn't exist.
Extremum point of f(x) is where f(x) has a local maximum or minimum.

[hearty terrace]

is r = root S/6π

[hearty terrace]

ohh ok but in skl we say turning point

[bleak gorge]

Yes. Of course, there is a negative solution, but we don't need it since r is nonnegative.

[hearty terrace]

oh ok

[bleak gorge]

Now, we need to check that this solution corresponds to a maximum point of V.

To do that, find d^2 V/dr^2 first.

[hearty terrace]

yea cuz radius is in cm

[hearty terrace]

ok

so 2nd derivative?

[bleak gorge]

Yes.

[hearty terrace]

so do i differentiate rS/2-πr^3 now

waait i meant

S/2-3πr^2

[bleak gorge]

Yes.

[hearty terrace]

i got -6πr

[bleak gorge]

Yes.

[hearty terrace]

nice

[bleak gorge]

Now, this is, of course, always negative for relevant values of r.

So, whichever critical point we got before is a maximum point.

[hearty terrace]

aw man

so do i do -6πr=0

[bleak gorge]

Thus, r = √(S/(6π)) is a maximum point.
Now, to find the maximum value of V, just substitute that into its expression.

[bleak gorge]

No, we just need the sign.

[hearty terrace]

oh ok

how do we do that

like how do we know that it is going to be negative

[bleak gorge]

Well, just substitute r = √(S/(6π)) into V = r((S/2) - πr^2).

[bleak gorge]

Well, if r is positive, -6πr is obviously negative.

[hearty terrace]

ohhh

yea

cuz -6(3)(1)=-18

and it will stay negative

and it cant be negative cuz it is length

i get that now

[bleak gorge]

No, that's not the point.

[hearty terrace]

oh why not

[bleak gorge]

It's the second derivative, its sign shows whether the function is convex up or down.

[hearty terrace]

ohh ok

[bleak gorge]

If d^2 V/dr^2 < 0 at some point r = r0, that means V is convex up there. So, if that point was a critical point, that also means that it is a maximum point.

We did this just to check that it was really a maximum point.

So, what do get when we substitute r = √(S/(6π)) into V = r((S/2) - πr^2)?

[hearty terrace]

so when 2nd derivative is less than 0, it means its a max point

cuz the only direction the graphj goes after is down

[hearty terrace]

hmmm

let me try do that

[bleak gorge]

Note that I factored it because then there is an r^2 inside the second factor, which is nice as we have a square root in r.

[hearty terrace]

what is r((S/2)-πr^2) again

[bleak gorge]

It's V.

[hearty terrace]

ohh yh

wat happens if i subtitute r into 300πr-πr^3

[bleak gorge]

No, substitute S later.

[hearty terrace]

oh ok

isnt that longer

[bleak gorge]

It's more general, which is usually better.

We can substitute S at the end.

[hearty terrace]

ok

this question is super duper long

[bleak gorge]

Well, not really.

[hearty terrace]

dis wil take me the whoel exam

[bleak gorge]

We are just discussing it.

[hearty terrace]

thats true

[bleak gorge]

Usually it would take about 3-4 minutes.

[hearty terrace]

but if wer werent discussing it i wouldnt know what to do or how it works

[bleak gorge]

Well, you're learning.

Of course it might take a bit longer at first.

Anyway, what did you get?

[hearty terrace]

if i square a root can i jus remove the root sign

i think i can

[hearty terrace]

im almost done

woa i got some long answer

i havent factorised it but its
root(S/6π)x(S/2-π(S/6π))

[bleak gorge]

Yes, good so far. Now, simplify π(S/(6π)) first.

[hearty terrace]

it can be (-Sπ/6π)

[bleak gorge]

Well, you can cancel π.

[hearty terrace]

yes

is it negative

[bleak gorge]

Well, it was -π(S/6π), so we get -S/6.

[hearty terrace]

o ok

yea

[bleak gorge]

So far V = √(S/(6π))(S/2 - S/6).

Now you can simplify S/2 - S/6.

[hearty terrace]

S/2-S/6

can be 3S/6-S/6

and thats 2S/6

and thats S/3

[bleak gorge]

Yeah, good! So, V = (S/3)√(S/(6π)).

Now, we can leave it like that, but it's possible to simplify a bit more.

We have S in two places, which isn't too nice.

We can multiply and divide S/3 by 2π: S/3 = 2π(S/(6π)).

Then:
V = (S/3)√(S/(6π)) = V = 2π(S/(6π))√(S/(6π))

[hearty terrace]

oh y do we do that

[bleak gorge]

Then, as x√(x) = x^1 x^(1/2) = x^(1 + 1/2) = x^(3/2), we get:
V = 2π(S/(6π))^(3/2)

And now S is only in one place, which is nicer.

[hearty terrace]

i am super duper confused

i think i skipped a few pages

[bleak gorge]

I just used a couple of properties of powers:

1. √(x) = x^(1/2)
2. x^a x^b = x^(a + b)

[hearty terrace]

oh ok

but why do we multiply and divide by 2π

[bleak gorge]

So that S/3 turns into S/(6π).

Which is what's under the square root.

[hearty terrace]

ohh

[vestal dagger]

beekxy what year are u in?

[hearty terrace]

but shouldnt ie also be 2π(S)/(S/3)

[hearty terrace]

im year 11

which is 15

[bleak gorge]

Where did the extra S come from?

[hearty terrace]

year old

[hearty terrace]

oops

[vestal dagger]

are u doing further maths? why are u learning binomial excpansion already

[hearty terrace]

i meant 2π(s)/3(2π)

[hearty terrace]

yea

im doing that

[bleak gorge]

Yes, and that's 2π(S/(6π)).

[vestal dagger]

furhter maths in y11 is just year 12 maths beta access

[hearty terrace]

ohh ok yea

[hearty terrace]

i find it kinda hard

[bleak gorge]

Anyway, we got V = 2π(S/(6π))^(3/2).

[hearty terrace]

but some are not that hard like tending to infinity

[bleak gorge]

Now we can substitute S = 600π.

What do we get?

[hearty terrace]

V=2π(600π/(6π))^(3/2)

[bleak gorge]

Right, and how can we simplify?

[hearty terrace]

eliminate 6π

so 100π

[bleak gorge]

Yes.

Well, just 100, rather.

[hearty terrace]

o yea

2πx100=200π

[bleak gorge]

No, we can't do that.

[hearty terrace]

o

oh

[bleak gorge]

100 is raised to the power of 3/2, so we have to do that first.

[hearty terrace]

o yea

[bleak gorge]

What is 100^(3/2)?

[hearty terrace]

100x100x100

root

root 100^3

root(1000000)

[bleak gorge]

Well, probably easier to take the root first, then cube it.

But that also works, I guess.

[hearty terrace]

oh yea

10 cubed 1000

[bleak gorge]

Yes.

So, finally, we get that the maximum volume is V = 2000π.

[hearty terrace]

wow

[bleak gorge]

And that's it.

[hearty terrace]

thanks

[bleak gorge]

You're welcome!

[hearty terrace]

this waas very long but helpful

[bleak gorge]

Now I'm going to sleep, as it's 2 AM for me now 😅

[hearty terrace]

oh ok thanks for helping

should i close the post

[bleak gorge]

As you wish, it isn't really required.