help-33

[rocky scaffold]

C=a+7 does not mean

C²=a²+7

a²+c²-2ac

Does this seem familiar?

[sour flame]

it looks like the a-c=7 but theres a + instead and it has an exponent

[rocky scaffold]

Have u not seen this identity?

(A-b)²

If not we shall continue with what you were doing

[sour flame]

we didnt learn that i believe

[rocky scaffold]

Ok then

So a-c=7

So a=c+7

Converting to 1 variable will not help you

Also 2a-2c is not 7

It is 14

[sour flame]

wewahgt

but isnt 2a - 2c not the same distance?

[rocky scaffold]

2a-2c is not equal to a-c

Its 2(a-c)

[sour flame]

OHH

oopsies

so then a² + 14 + c² - 2ac

?

sory id ont know

qait buy

but

a = c - 7 too, no?

or not

[rocky scaffold]

No

A=c+7

[marsh citrus]

@sour flame Has your question been resolved?

[marsh citrus]

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[silent gulch]

im not sure what i’m doing wrong

first i find the scale factor, then work out the perpendicular height of the small cone, then of the frustum but something keeps going wrong and im not sure where

<@&286206848099549185>

[marsh citrus]

@silent gulch Has your question been resolved?

[slow cedar]

you probably calculated the height of the original cone, which is x+5

[silent gulch]

OMG THANKYOU YOURE THE BEST

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[marsh citrus]

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[modest shuttle]

In every living organism, the radioactive carbon isotope C14 can be detected in a certain constant ratio to normal carbon. If an organism dies, it no longer absorbs C14 and the existing C14 disintegrates with a half-life of 5730 years.
How do I calculate the growth factor?

[modest shuttle]

<@&286206848099549185>

[marsh citrus]

@modest shuttle Has your question been resolved?

[woeful cipher]

why do people call decay as growth factor lol

more accurate to be phrased as decay factor

use the formula

$N(t)=N_0*(\frac{1}{2})^\frac{t}{T_{1/2}}$

[elfin berry]

$Pure2$

[modest shuttle]

tbh I’ve never seen that formula in school, we never used something like that. Are there other solutions as well?

[modest shuttle]

.close

[marsh citrus]

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[tough abyss]

is there a way to write (1) as (2)?

[hidden plaza]

whats k1 k2

[iron meadow]

and what's with ( )

[main idol]

you can set them equal and solve for ( )

[tough abyss]

okay i will try to

i got it thank you

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[marsh blade]

just wanting to check my wording would achieve all 4 marks

[marsh citrus]

Please don't occupy multiple help channels.

[marsh blade]

Given a<0, a is a negative value. 2^-n = 1/2^n, as n increases the fraction 1/2^n expontetially tends towards 0. as a -ve fraction gets closer to 0 it is becoming more positive (hence increasing)

[marsh citrus]

@marsh blade Has your question been resolved?

[marsh blade]

<@&286206848099549185>

[marsh blade]

.close

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[lethal bridge]

How should I set up the trig equation for part a

[dry prawn]

The equation is already set up for you

[lethal bridge]

no because o need to find the days after 21 December

The length I mean

[dry prawn]

How many days after dec 21 is dec 21?

[lethal bridge]

idk

0?

[dry prawn]

In how many days from today will it be today?

0

[lethal bridge]

burt how do I do part a

[summer trench]

he just told u that i think

[dry prawn]

L(t) represents the length of daylight t days after dec 21

We've established that dec 21 is 0 days after dec 21

[lethal bridge]

So trig equation = 0

[dry prawn]

No

The trig equation models L, which is what we want to find

[lethal bridge]

So I don’t get what I’m supposed to do

[dry prawn]

If this is true, what is the value of t for dec 21?

[hollow sparrow]

remember that t represents the number of days after Dec 21st. We've already concluded that we have 0 days for part i. What does that make t?

[lethal bridge]

0?

[dry prawn]

0

So the length of sunlight (modelled by L(t)) on dec 21 (where t=0) can be represented by what?

[lethal bridge]

So L(0) = 12

[dry prawn]

L(0) I agree with, how did you get 12?

[summer trench]

cos 0 is 1 i suppose

[dry prawn]

I think you need to go look at how L is defined

water beam can answer my questions well enough without your input, thanks

[lethal bridge]

14

12 + 2

[dry prawn]

14 looks better

[lethal bridge]

How can I find out the shortest length of daylight

part ii)

[dry prawn]

How indeed

[lethal bridge]

[dry prawn]

The only part of L that changes over time is the cos() part, agree?

So you just need to show where the cos(...) function is at its lowest

[lethal bridge]

I guess? Sorta

Not sure what that means tho

[dry prawn]

You have to find the value of t that minimizes L(t)

[lethal bridge]

but how do I do that?

[dry prawn]

By finding this

[lethal bridge]

.. isn’t that the same question

I can’t do either

[dry prawn]

It's very similar but not exactly the same

For which value of x does cos(x) reach its smallest value? (Only looking at x in the interval [0, 2pi])

[lethal bridge]

just the regular cosx?

[dry prawn]

Yep

[lethal bridge]

I guess at pi

because it dips down

to y = -1

[dry prawn]

Well don't guess, I need a surefire answer

[lethal bridge]

well pi

Is what I think it is

[dry prawn]

Ok yes, it's pi

cos(x) is minimized when x = pi; similarly, cos(2pit/366) is minimized when 2pit/366 = pi

[lethal bridge]

Hm okay

So I’m solving for t or something?

[dry prawn]

Actually wait I completely misread part 2, oops

It's actually not this complicated

[lethal bridge]

[dry prawn]

If cos is minimized at -1, then what is the minimal value of 12+2cos(...)

[lethal bridge]

T = -1?

[dry prawn]

We're not solving for t

It's asking what the smallest value of L is, when I thought it asked where the smallest value of L occurs

[lethal bridge]

Smallest value of L uh

L(-1)?

[dry prawn]

The value of L, not the input

If we found the t that minimized cos(...), and plugged that t in, what would be L(t)

[lethal bridge]

t = -1?

[dry prawn]

No

cos(2pit/366) = -1, not t = -1

If cos(2pit/366) = -1, what is L(t)

[lethal bridge]

so I’m solving for t?

[dry prawn]

No

There is some value of t such that cos(2pit/366) = -1. At that value of t, what is the value of L(t)?

To be as explicit as possible, you are not solving for t. You are computing L(t)

[lethal bridge]

If we don’t do L(-1) and not solving for t and not doing t=-1

I don’t know what to do

[dry prawn]

If x = -1 what is 12 + 2x?

[lethal bridge]

12 + 2(-1)

10

[dry prawn]

If cos(2pit/366) = -1 what is 12 + 2cos(2pit/366)?

[lethal bridge]

12-2

= 10

[dry prawn]

Can you find the smallest value of L(t)?

[lethal bridge]

It’s 10

[dry prawn]

It's 10

[lethal bridge]

That was quite a confusing problem

ok thanks

[dry prawn]

The next one will probably feel even more confusing

[lethal bridge]

oh bruh I didn’t see that

Is that 11 = trig equation

If so I’m pretty cracked at trig equations ngl

[dry prawn]

Yes

[marsh citrus]

@lethal bridge Has your question been resolved?

[marsh citrus]

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[graceful turtle]

Hi ! I am watching a '3b1b lecture' on Complex Numbers. He is solving how to calculate cos(75°) using simple construction below. I do not however get why does he only multiply the real part and the imaginary part , which in the end will cancel out the imaginary part and only leave the real part { is this the reason , or is there a deeper mathematical meaning behind this?} Thanks!

[vapid sedge]

because cos theta is the real part of the complex number cos theta + i sin theta

$e^{i75}=e^{i45}.e^{i30}$ or cos(75)+i sin (75) = cos(45+i sin 45) (cos 30 + i sin 30) and he's solving only for cos 75 which is the real part

[elfin berry]

JS

[graceful turtle]

So in that case, he multiplies the imaginary parts because the multiplication results in a real number?

[vapid sedge]

yeah $i^2 = -1$ so if you multiply the two imaginaries together you'll get a real...

[elfin berry]

JS

[graceful turtle]

Thanks!

[marsh citrus]

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[glad vapor]

why does the u-sub match in the first question but not the second?

[fleet basalt]

ur still left with an x

like ur gonna have u^2 1/3 x du

[marsh citrus]

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[orchid breach]

Help pls

[safe badger]

,rotate

[orchid breach]

Do I make it xy^1/2=1
Then sqrtxy=sqrt1

Then make x the sub of the formula

[safe badger]

why are you sqrt the 1

[orchid breach]

Idk to remove it from the side of the equation?

Is that wrong😢

[marsh citrus]

@orchid breach Has your question been resolved?

[safe badger]

just raise it to the power of 2!

[marsh citrus]

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[vivid river]

I'd like help solving the first one. Then I'll do the second, and would appreciate a confirmation of my answer

[iron meadow]

$(4x+9)^{\frac{1}{2x}}=e^{\ln\left[(4x+9)^{\frac{1}{2x}}\right]}$

[elfin berry]

Soosh

[iron meadow]

you can start off with something like this, now you can use the exponent rule for logarithms to take the 1/2x in front of the ln

and your limit can be rewritten as e^ lim of...

$\frac{\ln(4x+9)}{2x}$

[elfin berry]

Soosh

[iron meadow]

now try lhopitals on that

[vivid river]

MM

Ok one sec

Im doing like

3 problems at once

[vivid river]

So its 1

Bc the limit ends up as 0

And e^0 is 1

Got it

Thank you, I'll do the other problem on my own

Could you confirm my work for this?

Not sure if I did it wrong

[vivid river]

If there is anything wrong, would you mind working me through this too :(

[frosty tundra]

try finding the derivative again

because that's wrong

[vivid river]

WHA

Oh oops

Its cos

I mean

[frosty tundra]

not just that

the -4/x^2 isn't subtracting

[vivid river]

Wdym

[frosty tundra]

it's multiply by (-4/x^2)

[vivid river]

Oh damn wait you're right

Ah

This ends up as another indeterminant form

Bc its 0*infinity

Ugh

[frosty tundra]

nuh uh

i think

yeah no

[vivid river]

Wdym

Itll be cos(4/x)*-4/x^2

[iron meadow]

you can do this a different way

[vivid river]

*10x^2

[frosty tundra]

$\frac{\cos\left(\frac{4}{x}\right)\left(-\frac{4}{x^{2}}\right)}{\frac{-1}{10x^{2}}}=\cos\left(\frac{4}{x}\right)\left(-\frac{4}{x^{2}}\right)\left(-10x^{2}\right)$

[vivid river]

cos(0) is 1

[elfin berry]

CH3

[vivid river]

Yeah

[iron meadow]

no need to use lhopitals at all

[vivid river]

1 times 0 times -infinity so 0*infinity

[frosty tundra]

$\cos\left(\frac{4}{x}\right)\left(-\frac{4}{x^{2}}\right)\left(-10x^{2}\right)=40\cos\left(\frac{4}{x}\right)$

[elfin berry]

CH3

[vivid river]

HUH??

[frosty tundra]

x^2 and 1/x^2 divide

[vivid river]

Ah

So its 40

I see

I mean that makes sense then

x^2 cancels

-4*-10

[iron meadow]

you can use use $\lim_{x\to0}\frac{\sin x}{x}=1$

[vivid river]

cos(4/infinity)=cos(0)=1

Oh I hate that

[iron meadow]

without applying lhopitals at all

[frosty tundra]

oh yeah that works too

[elfin berry]

Soosh

[vivid river]

I believe I have to use l'hop though

For this

[iron meadow]

well whatever works

[vivid river]

I mean if that were easier I'd do that

But given we're supposed to be applying what we've learned from the chapter

Im stuck with l'hop

How do I tackle this?

Am I supposed to do common denom

[frosty tundra]

$\frac{1}{2x}-\frac{1}{\sqrt{x}}=\frac{1}{2x}-\frac{2\sqrt{x}}{2x}$

[elfin berry]

CH3

[orchid breach]

I'm still lost😭

[frosty tundra]

which is also equal to $\frac{1-2\sqrt{x}}{2x}$

[elfin berry]

CH3

[vivid river]

So yeah

Common denom no?

[frosty tundra]

[vivid river]

Okokok

One sec

Let me see if I can do this

[safe badger]

make a new tickett

[vivid river]

This is just dne then right

[frosty tundra]

should be positive infinity

[vivid river]

Bc I mean you dont really have to L'Hop anything after this step

Since its 1-0/0

Or 1/0

Which is just pos infinity yeah

But that can also be written as dne no?

[frosty tundra]

don't think so

actually

err

[vivid river]

Lol

I mean the limit is said to not exist at infinity bc it isnt set in real numbers right

So you can write infinity as dne?

[frosty tundra]

well i don't think that's right so just write infinity
because dne would be something like 1/x as x->0, it would be negative infinity from the left and positive from the right
while in this case it's infinity

[vivid river]

Okokok

I hate these limits with a passion

[frosty tundra]

lol

[vivid river]

For this

And you're free ofc to not help if you're tired, etc

I ask many question I sorry

But for this

[frosty tundra]

it's ok lol

[vivid river]

I rewrite this as e^ln(5x^2+2^4/ln(x))

Right

[frosty tundra]

yeah

[vivid river]

e^ln(5x^2+2^4/ln(x))

Then I just

Do the limit of ln(5x^2+2^4/ln(x))

Yes

Itll be a product rule

[frosty tundra]

you wrote this right? $e^{\frac{4\ln\left(5x^{2}+2\right)}{\ln\left(x\right)}}$

[elfin berry]

CH3

[vivid river]

Basically

Yes

[frosty tundra]

yep

[vivid river]

After you power rule it

Honestly

That seems easier than product rule

[frosty tundra]

also i don't see a product rule

[vivid river]

Its bc

You combined the terms

But I have it as 4/ln(x)*ln(5x^2)+2

[frosty tundra]

wuh

[vivid river]

How do you type this out so fast

[frosty tundra]

desmos lol

[vivid river]

Wha

Ive never done it

[frosty tundra]

copy paste from desmos

[vivid river]

Ohh

[frosty tundra]

i don't see how this is the case

oh wait

i see what you mean

[vivid river]

When you power rule

yeah

You bring it down

Yadda yadda

[frosty tundra]

but that's wrong

[vivid river]

But

[frosty tundra]

remember l'hopital

[vivid river]

Yes

[vivid river]

I must

No ik

I was js

Oh wait

Yeah

It wouldnt have worked

Bc I have to have a quotient

Yes

Lol sorry

Im burning out and I still have like 6 word problems to do after this

Plus a practice exam

[frosty tundra]

i feel you lol

[vivid river]

Woo!

It be rough

[vivid river]

So it be this

1

[frosty tundra]

$\frac{\frac{4\left(10x\right)}{5x^{2}+2}}{\frac{1}{x}}=4\left(\frac{10x^{2}}{5x^{2}+2}\right)$

[elfin berry]

CH3

[vivid river]

O

Welp

Well

I mean

it still ends up as 40/30x

Which is still 0

No?

[frosty tundra]

nope if you l'hopital on the above it becomes 4(20x/10x) which is just 4(2)

[vivid river]

Hm

Im confused

Where does the 1/x go then

[frosty tundra]

?

[vivid river]

Oh wait

Nvm

x*10x

10x^2

[frosty tundra]

[vivid river]

Then l'hop that

I see

Thats where I messed up

Threw my problem way off

So it all becomes e^8

Got it

Now these I hate more

We know Area = l*w or in other words

A=x*y

The diagram for this would be something like

Right

So the equation is

A=2x*3y

[frosty tundra]

wouldn't x be y

[vivid river]

Is it?

But y is vertical

[frosty tundra]

like x would be in the place of y

not equal to y

[vivid river]

Oh wait

Yeah

It says that

[frosty tundra]

they said "x be the length of the shared side"

[vivid river]

You're right, I misread

Well

In that case is A=3x*2y

Right

[frosty tundra]

no

[vivid river]

Brain hurt

[frosty tundra]

it's the sum of those two areas

aka xy + xy

2xy

[vivid river]

Mm

So

I make 2xy set to 2000

[frosty tundra]

no

[vivid river]

But we have a total of 2000

[frosty tundra]

the perimeter is equal to 2000

[vivid river]

No

Ah

Yeah

Idk

Like the last problems I at least had an idea of what to do

But these are like being blind

[frosty tundra]

the perimeter is just the sum of all sides

[vivid river]

My worst problems

[frosty tundra]

so 4x + 3y

[vivid river]

Yes I get that

Hm

[frosty tundra]

4x + 3y = 2000

[vivid river]

So we just solve for y

[frosty tundra]

actually since we replaced x by y

[vivid river]

And x

Repectively?

Oh

[frosty tundra]

it would be 4y + 3x = 2000

[vivid river]

Right

[frosty tundra]

so find y in terms of x

[vivid river]

So we use perimeter to find y?

Then from that plug in y into

2xy?

[frosty tundra]

yep

[vivid river]

Hm

Okay

I mean

This seems weird

Bc for the first equation

y=500/3x

So when we plug in

We get 2x(500/3x)

From here

its 1000x/3x right

[frosty tundra]

wuh

[vivid river]

Idk

LMAO

[frosty tundra]

4y + 3x = 2000
4y=2000-3x
y = (2000-3x)/4

[vivid river]

Oh my

Yeah my brains actually suffering LMAO

[frosty tundra]

take a break

[vivid river]

Bamboozled by the sign

I cant

Have to finish this unfortunately

[frosty tundra]

🫡

[vivid river]

I sadly have 5 more of these problems

Want to

Flip off a very high place

Woo

[frosty tundra]

lol

[vivid river]

So

Using that equation

and plugging in

We're left with 2000x-3x^2/2

Do I like

Differentiate this

Then set it to 0?

[frosty tundra]

find the maximum yeah

[vivid river]

Then plug in for x

And thats the solution

[frosty tundra]

[vivid river]

OKAY BUT LIKE

I DID IT WRONG

AND STILL END UP WITH 1000/3

Lmaooo

I love that

Hmm

Idk

[frosty tundra]

lol but at first you found the area to be equal to 1000/3

but this time it's x = 1000/3

[vivid river]

But I have to plug this in to

2000x-3x^2/2 right

[frosty tundra]

yes

[vivid river]

Thats like

Tm work

[frosty tundra]

calculator

or

[vivid river]

2000(1000/3)*3(1000/3)^2/2

Im not allowed calculators on my exam later

So it has to be in my head

And that seems like

[frosty tundra]

(x(2000-3x))/2 at x = 1000/3 -> 1000/3 (2000-1000)/2 -> 500(1000)/3

[vivid river]

A big number

Thats basically 500000/3

Why does that feel so

Off

Lol

[frosty tundra]

do you have the answer?

[vivid river]

No

You mean like an answer key right

Bc no

[frosty tundra]

meh then it's probably right

99%

[vivid river]

Welp

More suffering!

[frosty tundra]

same thing

[vivid river]

Okokok

I got this

So

I solve for either variable then

?

So if we go with m

[frosty tundra]

are you asking what to do or are you saying what you're going to do

[vivid river]

Im saying this

m=40-15n/8

Then I differentiate this correct

[frosty tundra]

no

[vivid river]

Kill me

[frosty tundra]

you want to find the maximum for the product

mn

n(40-15n)/8

[vivid river]

Hm

Okay

Ah

Then I differentiate for n

Then plug in

Etc etc

My mom is joe

[frosty tundra]

[vivid river]

And my dad is cool

Got it

Okokok one sec

0kok

I get 10/3

But

Thats wrong

Oh

Nvm

Im right

More suffer!

[frosty tundra]

draw it, write out the coordinates and it should be easy

[vivid river]

Similar process to the other problems right

[frosty tundra]

somewhat yeah

[marsh citrus]

@vivid river Has your question been resolved?

[vivid river]

Alright

Le area is 50

Oh yeah

I appreciate it

[frosty tundra]

it is?

i got 6

[vivid river]

Well

No?

[frosty tundra]

the whole area of this triangle is 18 so it's impossible to be 50

[frosty tundra]

yeah no

[frosty tundra]

this is a new question...

[frosty tundra]

i'm talking about this one

[vivid river]

OH YEAH

LMFAO

OOPS

Theres 2 questions

That are similar MY BAD

LMAO

Oops

Sorry

[vivid river]

So Im uh

Having trouble with this

Ill be having so solve for 2 equations right

Instead of 2xy

I have to do

x*y of one equation

Plus x*y of the other equation

Right

[vivid river]

Is it not 9?

[frosty tundra]

woops yeah it is

[vivid river]

Oh cool

Well

Thank you for your help

[vivid river]

Have a great night or day

[frosty tundra]

you too

[vivid river]

.close

[marsh citrus]

Channel closed

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Use .reopen if this was a mistake.

[frosty tundra]

*buy me space engineer on steam *

[vivid river]

lol

[frosty tundra]

[vivid river]

if i had the money

[frosty tundra]

aww

[vivid river]

alas i do not

[marsh citrus]

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[ashen crown]

Is there some trick or easy way to break up a fraction? I assume the denominator is gonna be four but I cant figure out what to put for numerator.

[ashen crown]

Sorry 3 and 4

Cause then I can get a angle

[fervent rampart]

3 & 16

just need to find some powers of 3 and 4 that add to 19

[crisp imp]

Try that and tell us.

[ashen crown]

Ohhh

Okay

I was doing this backwards

Thank you

.close

[marsh citrus]

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[coarse bison]

I do not know how to start the question. My math teacher said I need to use differentials

[sweet pawn]

total derivative is the sum of the partials

[coarse bison]

so would i plug in 5% and 2% into r and h of the partials when im adding them?

[coarse bison]

sorry I dont know where to go from this point 😭

[sweet pawn]

dV = pi r^2 dh + 2pi r h dr

actually not enough info

you also need r and h

you are only given dr and dh

[coarse bison]

yeah thats where i was stuck

😦

thank you for your help

ill try bringing this up with my math teacher

[sweet pawn]

you cannot get a numerical answer here

because the percentage error gets arbitrarily large as r and h get arbitrarily small

because then you're just basically dividing by 0

[coarse bison]

oh dang

thank you so much for your help, im just gonna skip this one then 😭

.close

[marsh citrus]

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[marsh citrus]

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[agile sinew]

Is this good

[spark condor]

yes, its correct

[agile sinew]

is that the final solution?

nvm i found it.

i have another problem i want help with, but i am trying to solve it myself first. pls wait

does anyone know what the derivative of square root of x is?

[spark condor]

its
1/2(sqrt(x))

basically you are differentiating
x^(1/2)

you can use
y' = nx^(n-1)

[agile sinew]

this is the problem if you want to solve it. i need to solve g'(3.5) if g(x) = ...

[spark condor]

do you have doubt in this?

[agile sinew]

in what?

This is as far as I can go

[spark condor]

in this question

[spark condor]

you dont need to do this

[agile sinew]

wdym

[spark condor]

its just
x^(1/2) / x^4
= x ^ (1/2 - 4)
= x ^ (-7/2)

[spark condor]

ah

you know power gets subtracted when 2 same variables are in division?

[agile sinew]

never learned that

[spark condor]

um, its basic

wait maybe you thinking something diffrent

[agile sinew]

guess i forgot lol

i just dont see it in this problem

[spark condor]

do you know this?

[agile sinew]

you talking about x⁴ and x⁴?

[spark condor]

x to the power 1/2 and x to the power 4
yes

[agile sinew]

so the (x⁴)² will become x⁴? the denominator

Like this

[spark condor]

I cant see the whole thing you have written but if its
x to the (-7/2)
then its correct

now you can put down the power (-7/2) in front as its in log and just differentiate

[agile sinew]

i think i did it wrong actually

It’s supposed to be like this

[spark condor]

yes, you are going correct

now use this formula

to get -7/2 down

and differentiate

you have to find g'(x)

by the way you can tag me or reply to get my attention, I may not be looking here

[agile sinew]

What rule is that? And it says log instead of ln

[agile sinew]

this?

[spark condor]

ln just means log base e

[spark condor]

yes

[agile sinew]

i think i'm supposed to use derivative rules to solve this. the chapter is about derivatives. not really fair to simplify instead of using the chain rule or the product rule.

can we do both methods?

[spark condor]

if you want you can absolutely just differentiate as it is now,
but maths is about making connections with all the chapters, the thing I told you was just to make things easy, if you want you can do it other way too!

[spark condor]

chain rule ND division rule
yes you can use both

its just one can be easier than the other

so there are less chances for you to make mistakes

[agile sinew]

yeah

[agile sinew]

.

[marsh citrus]

@agile sinew Has your question been resolved?

[agile sinew]

@spark condor

[spark condor]

oops sorry I was eating

[spark condor]

no no,

chain rule is applied when the variable you are differentiating is inside a function which is inside another function,

here (-7/2) act as a constant number so you take it out and just differentiate the "ln x" part

@agile sinew you knwo this??

[agile sinew]

i am going with the quotient rule this time btw. i've come pretty far.

[spark condor]

um, okay

[spark condor]

you forgot to differentiate

ln (sqrt(x)/x^4) first

$\ln{\frac{\sqrt{x}}{x^4}}$

[elfin berry]

anshu

[agile sinew]

i'm doing the inside of the parantheses first. then i will do chain rule for ln and (x)

[spark condor]

okay

I will let you finish then send my method

[agile sinew]

ok

[agile sinew]

Here. Also brb

[spark condor]

I dont understand what you just did here

honestly you dont have to even put half the work in second method

here it is

I have elongated the step for you to understand easily

its waaaaaay simpler

[agile sinew]

its this here. the chain rule

[spark condor]

yes

[spark condor]

I will explain the chain rule too

but just tell me first do you understand this?

[agile sinew]

wouldn't -7/2 be 0 if you use derivative? derivative of a constant is 0

[spark condor]

yes, but its when its a constant term
like 1, 3, 4, 7, 2039, 12930, 1232134.41245

but if you got a x in the term you cant just write it as zero
like you cant write derivatives of these as zero
2x
4x
513x

heres the same thing in formula

you are saying the first thing

but you see here we have a function of x which is "ln x" here multiplied with 7/2

[agile sinew]

yeah

that was all i was confused about. i understand your solution now.

[spark condor]

great! so now about chain rule as I said

[spark condor]

you need to understand first whats g(x) and f(x) here

[spark condor]

you have taken $\frac{\sqrt{x}}{x^4}$ as g(x)

[elfin berry]

anshu

[spark condor]

thereforce f(x) which means an outer funtion becomes $\ln{x}$
as you assume $\frac{\sqrt{x}}{x^4}$ is x in the f(x)

[elfin berry]

anshu

[spark condor]

nah this will confuse you more

wait lemme think how to explain it to you

do you know differentiation by subsitution?

[agile sinew]

no

[spark condor]

um

how did you get to chain rule then?

chain rule is a short trick to do subsitutions faster

[agile sinew]

i used the quotient rule since $\frac{\sqrt{x}}{x^4}$ is division

[elfin berry]

Crawling Ham

[spark condor]

yes but you cannot forget that its inside a log function

ok lemme do it like this

LET t = $\frac{\sqrt{x}}{x^4}$

[elfin berry]

anshu

[spark condor]

so now you have $\ln{\frac{\sqrt{x}}{x^4}} = \ln{t}$

[elfin berry]

anshu

[spark condor]

you understand this?

[agile sinew]

hmm. yeah

[spark condor]

we use quotient rule to differentiate "t" here which is $\frac{\sqrt{x}}{x^4}$

[elfin berry]

anshu

[spark condor]

so $\frac{dt}{dx} = \frac{-7}{2x^4\sqrt{x}}$

[elfin berry]

anshu

[spark condor]

this was the answer you got

[agile sinew]

yeah

[spark condor]

now what we needed to do in the actual question was to differentiate $\ln{\frac{\sqrt{x}}{x^4}}$

[elfin berry]

anshu

[agile sinew]

and to do this i used the chain rule. thus $\ln{\frac{\sqrt{x}}{x^4}}$ becomes this

[elfin berry]

Crawling Ham

[spark condor]

becomes?

[agile sinew]

$ln\left(x\right):=:\frac{1}{\frac{7}{2x⁴\sqrt{x}}}\cdot :\frac{7}{2x⁴\sqrt{x}}$

[elfin berry]

Crawling Ham

[agile sinew]

ln(x) being the original thing

[spark condor]

isnt this equals to 1?

so your answer is 1?

but I got -1

[agile sinew]

didn't get that far. i'm stuck here lol

[spark condor]

stuck?

its the same thing divided and multiplied

it equals to one

[agile sinew]

oh

[spark condor]

its wrong basically

[spark condor]

did you follow upto here?

[agile sinew]

wdym

[spark condor]

I mean did you understand this statement so I can continue with this

I will tell you a interesting thing about this method in the last

[agile sinew]

i have understood

[spark condor]

so you basically want $\frac{d \ln{\frac{\sqrt{x}}{x^4}}}{dx}$

[elfin berry]

anshu

[spark condor]

just lemme know if you understand this by yes or no

[agile sinew]

no

i know that d/dx means derivative, but when it looks like this i dont understand

[spark condor]

ok so by question
$g (x) = \ln{\frac{\sqrt{x}}{x^4}}$
and you need $g'(x)$

[elfin berry]

anshu

[agile sinew]

yes

[spark condor]

so this is basically $\frac{d}{dx} g(x)$

[elfin berry]

anshu

[spark condor]

I just wrote ln function in place of g(x)

[agile sinew]

i get it now

[spark condor]

okay

so now we are going to connect all the dots

we will be using this you sent earlier

[agile sinew]

ok

[spark condor]

$\frac{d }{dx}\ln{\frac{\sqrt{x}}{x^4}} = \frac{d }{dt} \ln{t} * \frac{dt}{dx}$

[elfin berry]

anshu

[spark condor]

I used this and wrote it as ln t

and multiplied and divided by dt

$\frac{d}{dx} \ln{\frac{\sqrt{x}}{x^4}} = \frac{d}{dx} \ln{t} = \frac{d }{dt} \ln{t} * \frac{dt}{dx}$

[elfin berry]

anshu

[spark condor]

added a step

now you understand it?

[agile sinew]

yeah

[spark condor]

you know what $\frac{d }{dt} \ln{t}$ equals to?

[elfin berry]

anshu

[agile sinew]

$-\frac{7}{2x}$

[elfin berry]

Crawling Ham

[spark condor]

no

lemme put it this way

you know what $\frac{d }{dx} \ln{x}$ equals to?

[elfin berry]

anshu

[agile sinew]

1/x

[spark condor]

bingo

so $\frac{d }{dt} \ln{t} = 1/t$

[elfin berry]

anshu

[spark condor]

yes?

[agile sinew]

yes

[spark condor]

so here you got $\frac{d }{dt} \ln{t} * \frac{dt}{dx} = \frac{1}{t}* \frac{dt}{dx}$

[elfin berry]

anshu

[agile sinew]

so what i got wrong was $\left(\frac{1}{\frac{7}{2x⁴\sqrt{x}}}\right)$

[elfin berry]

Crawling Ham

[spark condor]

bingo!

yes

just lemme finish real quick

[spark condor]

do you understand this?

[agile sinew]

oh yeah true lol i used g'(x) instead of g(x)

[spark condor]

?

[agile sinew]

yeah

[spark condor]

we just then put in the value of t we assumed here : #help-33 message

and put in value of dt/dx we got here : #help-33 message

and you get your diffentiated function

[agile sinew]

$\frac{1}{\frac{\sqrt{x}}{x⁴}}\cdot \frac{7}{2x⁴\sqrt{x}}$

[elfin berry]

Crawling Ham

[spark condor]

and the interesting thing I told you about earlier was
this whole process is called differentiation by subsitution, where you subsituted t in place of $\frac{\sqrt{x}}{x^4}$
and the more interesting thing about this is what you took g(x) in your solution is the t here!!
so you see how chain rule is short method of subsitution?

[elfin berry]

anshu

[spark condor]

yes

[agile sinew]

yes

[spark condor]

I think you understand now how chain rule works and what you did wrong in your solution

[agile sinew]

yep. i realized back here

[spark condor]

by the way DONT FORGET YOUR MINUS!!

[agile sinew]

that too

[spark condor]

now you just solve this you get g'(x)
then just replace x with 3.5 to get g'(3.5)

just lemme know what your final answer came by this method and we can wrap up

[agile sinew]

$:-\frac{7}{2x}:=:-\frac{7}{2\cdot 3.5}:=:-\frac{7}{7}:=:-1$

[elfin berry]

Crawling Ham

[spark condor]

bullseye!

[agile sinew]

thanks

[spark condor]

your welcome :)

[agile sinew]

.close

[marsh citrus]

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[marsh citrus]

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[buoyant quest]

Someone can help me to understand the green part? (RSA Algorithm) ,this is what I have done for now.

[buoyant quest]

How can i get the a0, a1, b0 and b1 value

ecc...

and why i need a?

[marsh citrus]

@buoyant quest Has your question been resolved?

[buoyant quest]

.close

[marsh citrus]

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[restive sky]

my answer doesn't match the answer sheet

what went wrong

[dry prawn]

Everything up to you finding lambda looks ok

[restive sky]

only thing i can think of is to chuck all the negative signs

[dry prawn]

The step where you plug lambda into the right side of your initial equation seems to be the mistake

[restive sky]

what is lambda

[dry prawn]

$\lambda$

[elfin berry]

Steakanator

[restive sky]

whats the mistake when plugging it in

the negative signs should cancel right?

[dry prawn]

Your initial equation involves Pe^(-4lambda), but when you plug it in, you seem to be treating it as Pe^lambda

The negatives and the 1/4 should both be affected

[restive sky]

no b/c the initial equation was to find the half life that occured at 4 years

we don't know the time for quarter life

[dry prawn]

Ah right i misread that

Well the negatives should also cancel, hopefully that's the only issue

[restive sky]

yeah i just put my answer into symbolab

it still ended giving me an 8

so ig you can just rewrite it like that

[dry prawn]

log(1/x) = -log(x) so you can simplify your final answer still yes

[restive sky]

aii thanks

.close

[marsh citrus]

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[void igloo]

how to find surface area

[void igloo]

i still don’t get this

i looked at this but like

i just dont understand

[green badge]

find the area of all the sides of a shape

then add them together

[void igloo]

all of them is 2 m obviously right bc that’s what the porblem says?

then i do

oh wait

i do 2x3

2x2*

and then that’s the area?

[void igloo]

then i add the answerr from 2x2 3 times?

[marsh citrus]

@void igloo Has your question been resolved?

[marsh citrus]

@void igloo Has your question been resolved?

[marsh citrus]

@void igloo Has your question been resolved?

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[proper kite]

Anyone have a quick mnemonic of sorts to get down the main trig identities?
Also bonus points if theres one to learn the main integration rules

[marsh citrus]

@proper kite Has your question been resolved?

[proper kite]

<@&286206848099549185> Anyone there?
Oh and another question is, realistically how many trig identities should you know for calc 1

[red dagger]

i dont think a mnemonic is the right way to go about learning trig identities

[proper kite]

Hm

[red dagger]

id argue taking time out to understand them will help your recall and application of those identities better than some meaningless memorisation pattern

[proper kite]

Right
So mnemonics won’t particularly help that much

[red dagger]

they can help you pass the test but in a way that kinda defeats the point of the test

[proper kite]

Now realistically, how much should I get hang of till smth like calc 1?

[red dagger]

idk how deep calc 1 goes im european

[proper kite]

Fair enough, my teacher just told me to chisel to make it less of a strangle

[proper kite]

Iirc the American calc 1 is equivalent to gr12 calculus for Canada so yeah

[red dagger]

i know even less about the canadian system
any chance you could convert to british?

[proper kite]

Iirc tho my teachers did say they don’t particularly apply integrals with identities till first year uni calc

[proper kite]

When do you guys first learn calc? Year 10?

[red dagger]

year 12 usually

[proper kite]

Ah I see

[red dagger]

this is a surprise though

[proper kite]

What does it look like tho?

I think I can find an ss for a gr12 textbook

[red dagger]

differentiation, u sub, trig sub, ibp, parametric differentiation and integration, improper integrals, volumes of revolution, linear diffeqs, simple dynamics problems

oh also hyperbolic sub

[proper kite]

Huh I see, interesting
Wait so how much trig identities are you guys expected to know before like year 12?

Or within

[red dagger]

before year 12? sin^2+cos^2

after y12 p much all of them

[proper kite]

Huh I see, since here since Imm gr11, they only taught us the first main 3

[red dagger]

which are?

[proper kite]

The pythagorean one
The quotient one
And reciprocal one

I think the textbook said smth about 6 or smth

[red dagger]

ur gonna have to be more specific im afraid

[proper kite]

Like y’know csc=1/sin
For the reciprocals
Sin^2+Cos^2=1
For Pythagorean one
And ah what was it
Sin/Cos for example tan and Cos/Tan for cot
For the Quotient Identitt, apologies
Phone battery dropped

[red dagger]

sin/cos you can remember by looking at a triangle

sin^2+cos^2 you can remember by looking at the unit circle

then the reciprocal functions just have names you need to remember

[proper kite]

Ah I see, wait aren’t their like 30 smth of em used for functions?

[red dagger]

wdym

[proper kite]

Isn’t there like 36 different trig identities or so?

As in the main ones

[red dagger]

that depends entirely on your notion of "main"

and what exactly you mean by "identity" because some people wouldnt consider tanx=sin/cos to be one for example

[proper kite]

Well I mean in like sphere for general mathematics and calc.
Since ik you only need like 3-6 here for like highschool
And by identity I mean moreorless unique ones

[red dagger]

i dont think counting how many there are is ever useful

[proper kite]

Ah fair enough

[red dagger]

also this still depends entirely on what you consider to be unique

is sec^2=1+tan^2 unique given sin^2+cos^2=1?

complletely subjective

[proper kite]

Hm I see, fair enough
Wait so how are trig identities exactly applied when doing integration?

Since ik its just moreorless used as a way for simplification

[red dagger]

when they can simplify an integral you use them to simplify the integral

eg if i gave u the integral of 1/(1+x^2)

[proper kite]

Oh I see, is there any particular steps or rules when applying so?

[red dagger]

youd see that

[red dagger]

be reminded of the tan^2 +1identity

[proper kite]

Ah I see

[red dagger]

and then make a trig sub

[red dagger]

eh

just do the obvious things mostly

[proper kite]

Since ik if you apply them too early it can mess some things up

[red dagger]

and stuff like cos^2 is harder to integrate than cos(2x) so use double angle formulae to simplify that etc

[proper kite]

Right, I see

Trig identities, very handy
So @red dagger
How similar are arc identities and trig ones then?

Like smth of arctan of the sorts

[red dagger]

im not sure that ive ever used any
except obvious stuff like arcsin(sqrt(1-cos^2 x))=x

[proper kite]

Ah I see
Oh right, is there any particular trick to make it factoring and dealing with rational expressions easier or does just all come from practice and practice

[red dagger]

add 0 to the numerator

eg

x/(1+x)

=1-1/(1+x)

thats all of everything youll ever want to do with rational expressions

[proper kite]

Huh
Wait are you able to draw smth to make it a bit more easier to visualize

[red dagger]

?

[proper kite]

Are you able to like draw a quick visual of it
I mean of the equation or whatever

Idk, I’m not good with words

[red dagger]

theyre the same curve

[proper kite]

Interesting, wait are you able to like draw a quick rational expression equation thats harder than the one you gave, Using the 0 numerator method
(Sorry if I’m sounding like I’m asking for a bit much)

[red dagger]

$\frac{x}{1+x}=\frac{x+(1-1)}{1+x}=\frac{x+1}{x+1}-\frac{1}{1+x}=1-\frac{1}{1+x}$

[elfin berry]

bruhh

[proper kite]

I see, so what would it look like if you had an equation which required you to factor everything and applying the 0 numerator methodology

[red dagger]

example?

[proper kite]

Ah I’m not particularly good at examples, but lemme find some practice question

@red dagger So a rational expression like this, how would it to be solve applying the 0 numerator method if applicable

[red dagger]

oh thats a division not a partial fraction

[proper kite]

Huh, so you have to just factor and use pemdas?

Thats all?

[red dagger]

p much