modelling question

[magic hazel]

where would i start with this?

[little flax]

so first of all yuck

well so what do you wanna find

[magic hazel]

time

i think

[little flax]

yeah so you wanna find x(t) wont you

aka you wanna somehow solve this

so what kinda ways do you know how to solve these kinds of equations

[magic hazel]

i was thinking of integrating but cos its with respect to t, you cant do it so idk

or sub in the values

[little flax]

yeah so one method for these is called using seperability

seperation of variables

is that familiar?

[magic hazel]

no, i dont think so

[little flax]

maybe ill do an example

would you be able to solve something like

dy/dx = y

[magic hazel]

solve for what??

im confused

[little flax]

finding y

as a function of x

ok lets go thru this then

so im gonna be using this dy/dx = y as a practise

[magic hazel]

okay

[little flax]

so what were gonna do is not super precise but it works

so what you can do is imagine dy/dx as a fraction

and so dy/dx = y would be the same as

dy = y dx

ill then divide by y to get 1/y dy = dx

if thats ok?

[magic hazel]

yep

im following so far lol

[little flax]

and now we can just integrate both sides

its not super precise but

you then have integral of 1/y dy = integral of 1 dx

ok?

and then do you know how to integrate both of those

[magic hazel]

1/y goes to ln(y) right??

[little flax]

yep

[magic hazel]

dx = x

??

im not sure

[little flax]

so integrating 1 you can think of

as like 1 * x^0

would you know how to integrate x^0?

[magic hazel]

add 1 to the power, divide by the new power, so just 1x^1 = x

[little flax]

yep so

integrating 1 dx is just x yep

we do have to remember a +c tho so

we get ln(y) = x + c

is that ok?

[magic hazel]

yes

[little flax]

ok so

were gonna try use the same idea

except your equation is much worse

just reposting

[magic hazel]

lovely

this looks awful

but ill try

[little flax]

yep ok

[magic hazel]

so first, do you multiply by dt

[little flax]

yep

so dx = 2(x-6)(x-3) dt

now cuz you wanna integrate

you really want all the same variables on the same side as the d(..)

yep

divide by the brackets so

1/(x-6)(x-3) dx = 2 dt

ok?

[magic hazel]

yes

[little flax]

and now you just

integrate both side

which is where itll be not good

so lets do the right first because its easier

integrating 2 dt

[magic hazel]

2t +c

[little flax]

perfect

now lets deal with the other side

what do you think about integrating 1/(x-6)(x-3) dx

[magic hazel]

split up the fraction first maybe

then deal with 2 separate ones

or is that making it worse

[little flax]

no that sounds good

how are you gonna split it tho

theres a special name for this method

[magic hazel]

partial fractions?

[little flax]

perfect yep

so do you know how to do partial fractions?

[magic hazel]

yep

[little flax]

ok awesome so

you do that lmk when you work it out

we can compare answers

[magic hazel]

okay

im done

[little flax]

so am i

what did you get

[magic hazel]

1/3(x-6) - 1/3(x-3)

[little flax]

ok awesome

now integrating both of those

what do you think

[magic hazel]

natural logs

[little flax]

yep

so what will it integrate to

[magic hazel]

1/3 ln(3x-18) - 1/3 ln(3x-9)

[little flax]

i dont think i agree with that

so what would just 1/(x-6) integrate to

[magic hazel]

ln (x-6)

[little flax]

yep and so

we can think of 1/(3 (x-6)) as

1/3 times 1/(x-6)

right?

[magic hazel]

yes

[little flax]

and so integrating that

itll just be 1/3 ln(x-6) right

[magic hazel]

okayy, i expanded the bracket on the denominator and did it that way but ig you cant do that

[little flax]

nop youre safer just factoring out consatnts

leaving them outside

ok so

integrating the whole thing will then be

1/3 ln(x-6) - 1/3 ln(x-3)

ok?

[magic hazel]

yep

[little flax]

awesome and so

putting these back into the equation before

we have 1/3 ln(x-6) - 1/3 ln(x-3) = 2t + c

[magic hazel]

what do you do with the c though

[little flax]

ah great question

so we need to use some other stuff we got in the question to find c

so it says that

assume x=0 initially

so what do you think we can use that to say

[magic hazel]

t=0

[little flax]

exactly

when t=0, x=0

[magic hazel]

sub those in to find c then

[little flax]

yep

i have just realised that

your ln's should have absolute values round them

so it shoudl really be 1/3 ln|x-6| - 1/3 ln |x-3| = 2t + c

[magic hazel]

yep

[little flax]

but yep so

try find c

[magic hazel]

1/3 ln 4

[little flax]

i dont agree with the ln(4)

i get a different number inside

[magic hazel]

wait nvm ive realised what ive done wrong i think, let me try again

1/3 ln 2

[little flax]

perfect i agree

so then our like

final equation will be

1/3 ln|x-6| - 1/3 ln|x-3| = 2t + 1/3 ln(2)

cool?

[magic hazel]

yep

[little flax]

ok awesome so

now we can actually

answer the question

how long does it take to make 2 grams

[magic hazel]

7 minutes

[little flax]

youll need to show working 😦

how might you show its 7 minutes

[magic hazel]

do we just sub in x=2 and solve for t

[little flax]

exactly perfect

[magic hazel]

im doing something wrong, i get 0.115524... as the value for t

[little flax]

ah so

be careful

what units are t in in the question

[magic hazel]

minutes

[little flax]

2nd line in the question

[magic hazel]

hours

[little flax]

mmmm

so your 0.1155 is actually that many hours

and so how many minutes is it

[magic hazel]

i assume 7 ??

[little flax]

its like 6.93 ish yea

which is about 7

[magic hazel]

wait how do you work out how many minutes it is? ik theres a button on the calculator but idk which one it is

nvm i found it

[little flax]

just

times by 60

[magic hazel]

oh yeah

[little flax]

1 hour has 60 mins

😭

ok anyway

so part a) done

🥳

[magic hazel]

that was not as bad as i thought it would be though

[little flax]

for 10 marks tbf

like it was gross but yeah i agree

[lofty osprey]

go and help my one

[magic hazel]

for part b, is it just because if you sub in x=3 into the equation we worked out, it doesnt have an answer

[little flax]

so i think we should maybe like

look at the original dx/dt thing

youre right tho plugging x in everything dies

so in our dx/dt thing

when x is 0, what is dx/dt

[magic hazel]

change in g of the compound with respect to time

[little flax]

yep and

when x = 0, that value is like

36 right

[magic hazel]

yeah

[little flax]

but if you pick a aslightly larger value for x so when you have slightly more compound made

say x=1

dx/dt is now gonna be like 20

aka smaller than when x was 0 right

so having more compound has slowed down how fast we make it

and as you get closer and closer to 3

that dx/dt

itll get smaller and smaller right

[magic hazel]

yep

[little flax]

and in fact like

at x=3 you have dx/dt is 0

so it takes longer and longer and it just

takes infintely long effectively

the amount of extra compound goes to 0 as the amount we have gets close to 3

does that kinda make sense?

[magic hazel]

yes

[little flax]

awesome!

[magic hazel]

i always struggle with the wording of these questions though, i never know what i should write to get the marks

[little flax]

something along the lines of

as x gets close to 3, dx/dt gets close to 0

and so you never actually reach x=3

[magic hazel]

thank you so much honestly

[little flax]

its no worries!

now you just gotta remember it lol

[magic hazel]

ik, its a lot but at least ill know where to start next time