Hey mathematicians

[Teh]

I need a 2nd opinion on a fucking maths question that is doing my head in. I've got a solution but i wouldn't mind a 2nd opinion since i'm pretty positive i'm cocking up somewhere.

So yeah anyway, i know it's basic as rates of change get but still:

A spherical balloon is inflated at a constant rate of 0.5 m^3/s. Calculate the rate of change of its surface area at the instant when the radius is 0.75m

Oh and just in case:
( A=4[pi]r^2 )
( V=4[pi]r^3/3)

i'm starting with dA/dt = dA/dr . dr/dt and going from there.

Danke in advance

[b i o n i c]

like 2, duh!

[Teh]

Well i get 1.33333333333333333 which seems a bit odd to me.

[b i o n i c]

well if you wanna get technical an everything... but, i mean, come on!

[zorra_chiflada]

hey, i almost understand what's going on there. almost

[Teh]

Well its actually a question from last years university coursework, but i'm revising differentiation before we do laplace transforms.

[zorra_chiflada]

i thought you wrote lapdance there for a second

[Teh]

my maths skills are verbally based panty removers, after all

[zorra_chiflada]

right

[Teh]

Right anyway, ill post my solution for now:

dA/dt = dA/dr . dr/dt

dA/dr = 8(pi)r

dA/dt = 8(pi)r . dr/dt

dr/dt = dV/dt / dV/dr <- i think im going wrong here cos i end up with an expression for dr/dt that doesn't actually have 't' as a variable in it

dr/dt = 0.5 / 4(pi)r^2 <- see?

dA/dt = 8(pi)r . (0.5/4[pi]r^2)

dA/dt = 1/r = 1.3333

[zorra_chiflada]

are you drunk?
when people are drunk they like to show off their "skills"

[Teh]

I'm not really skilled in this bit of maths, but its so key to all the maths and shit i do this semester and im curious to know if i've gotten this right or not since nobody else i know personally can (be bothered to) get a solution.

plus i thought if i posted my solution y'all coulds correct me without having to attempt the whole bastard question yo

[icy manipulator]

the rate of change is the derivative of something. so the derivative of the surface area of a sphere is the derivative of 4 pi r squared. which is 8 pi R. so if the radius is 0.75cm than it shall be 6 pi. that make sense yeah? i've gotta say you worded that question strangly but from what i understood of the way you said that question that is the right answer 6 pi sq cm.

[zorra_chiflada]

none of you guys will ever touch a boob

[b i o n i c]

lol boob

[yooooo]

1+1= 2 ?

[icy manipulator]

hmm, the last pair the had total access to were 14F's. i nearly suffocated in them :( seriously. sex wasn't enjoyable when she was on top. unfortunately unintentional auto erotic asphixiation isn't something i get off on :(

[yooooo]

hmhm yes

[Teh]

the rate of change is the derivative of something. so the derivative of the surface area of a sphere is the derivative of 4 pi r squared. which is 8 pi R. so if the radius is 0.75cm than it shall be 6 pi. that make sense yeah? i've gotta say you worded that question strangly but from what i understood of the way you said that question that is the right answer 6 pi sq cm.

It can't be, but i'm having difficulting explaining why. In this question it's asking for the rate of change of the surface area, but w.r.t time, t. The balloon is expanding as time goes on, ie, dV/dt = 0.5 m^3/s

plus its worth 20 marks. sorry for the wording, i copied it straight from the exam paper.

[Planetary]

0+0=1

[Teh]

I just remembered how much i want your avatar as a tattoo planetary.

[icy manipulator]

what do you need this question for? one of my degrees at uni is maths and i can tell you that the rate of change of what you're telling me is the derivative of the surface area of a sphere. 1.3333 makes no sense. Pi must be part of the equation, thats for sure.

[Teh]

what do you need this question for? one of my degrees at uni is maths and i can tell you that the rate of change of what you're telling me is the derivative of the surface area of a sphere. 1.3333 makes no sense. Pi must be part of the equation, thats for sure.

Did you see my solution? i posted it kind of midway down. Like i said, i'm pretty positive it's the rate of change with time, at the instant where r = 0.75m

all the questions i've done that've been similar have been w.r.t time, anyway. i just cant explain why


1.3 m^2/s is what i got having subbed in 0.75m to get the rate of change at the instant where r = 0.75m

oh and im studying civil engineering since i think thats what you just asked?

[zorra_chiflada]

oh for fuck's sake

[Teh]

oh for fuck's sake

?

[Rancid_Beasties]

I used to be really good at maths and now it all looks like gobledy gook to me.

[jabumbo]

its been so long since i've had to do that shit, and now i know why i was happy when it was over with

[Teh]

hey in fact, time is the independant variable because its a dynamic system! every quantity is changing with time.

sorry for the confusion, i should've said that before! you were right though icy (duh)

you encounter rates of change much these days jabumbo?

[jabumbo]

unless its consolidating clay particles, no

[Teh]

laplace?

[jabumbo]

fuck laplace


terzaghi is where its at

[Teh]

:D

So anyway, if anybody can still be arsed getting a solution to this question then it'd be appreciated, cos like i said, i'm not sure if my answer for dA/dt is correct or not.

[HAL 9000]

I need a 2nd opinion on a fucking maths question that is doing my head in. I've got a solution but i wouldn't mind a 2nd opinion since i'm pretty positive i'm cocking up somewhere.

So yeah anyway, i know it's basic as rates of change get but still:

A spherical balloon is inflated at a constant rate of 0.5 m^3/s. Calculate the rate of change of its surface area at the instant when the radius is 0.75m

Oh and just in case:
( A=4[pi]r^2 )
( V=4[pi]r^3/3)

i'm starting with dA/dt = dA/dr . dr/dt and going from there.

Danke in advance

Im pretty drunk ripht now so forgive me if this is wrong.... but this is my attempt

V = 4/3(pi)r^3

dv/dt = 0.5 = 4/3(pi)(3r^2) (dr/dt) (where r = 0.75)
= 4(pi)(r^2) (dr/dt) (where r = 0.75)
so

dr/dt = 0.5/(4(pi)0.75^2) = ((pi)0.75^2)/8 = (pi *0.5625)/8 = 0.22m/s

now

da/dt = da/dr*dr/dt

If a = 4(pi)r^2
da/dr = 8(pi)r and if r = 0.75 then da/dr = 8(pi)0.75 = 18.85

and from earlier dr/dt = 0.22

so Da/dt =0.22*18.85 = 4.15m/s^2

the arithmatic may be wrong but i think the method is right. hope tht helps

[kate]

i haven't done this in 5 years, my head hurts :( i think hal is right though.

[Dorothy Wood]

lol @ calling it "maths"


yeah, and I don't know how to do that fancy kind of math. I mainly stick to simplifying rational expressions.

[Teh]

Im pretty drunk ripht now so forgive me if this is wrong.... but this is my attempt

V = 4/3(pi)r^3

dv/dt = 0.5 = 4/3(pi)(3r^2) (dr/dt) (where r = 0.75)
= 4(pi)(r^2) (dr/dt) (where r = 0.75)so

dr/dt = 0.5/(4(pi)0.75^2) = ((pi)0.75^2)/8 = (pi *0.5625)/8 = 0.22m/s

now

da/dt = da/dr*dr/dt

If a = 4(pi)r^2
da/dr = 8(pi)r and if r = 0.75 then da/dr = 8(pi)0.75 = 18.85

and from earlier dr/dt = 0.22

so Da/dt =0.22*18.85 = 4.15m/s^2

the arithmatic may be wrong but i think the method is right. hope tht helps

Yeah mate, that's my solution to the letter! (posted it on the last page). The bit in bold isn't really needed though, that's like a proof of checking you got dr/dt right... anyway, your arithmetic is wrong (where it's 0.22 i got 0.07 or something) but it's basically the same solution. If your aritmetic was correct you'd have gotten 1.3, same as me.

Thanks for the elaboration, it's much appreciated!

Ps - why lol dorothy?

Thanks for the elaboration.

[HAL 9000]

Right anyway, ill post my solution for now:

dA/dt = dA/dr . dr/dt

dA/dr = 8(pi)r

dA/dt = 8(pi)r . dr/dt

dr/dt = dV/dt / dV/dr <- i think im going wrong here cos i end up with an expression for dr/dt that doesn't actually have 't' as a variable in it

dr/dt = 0.5 / 4(pi)r^2 <- see?

dA/dt = 8(pi)r . (0.5/4[pi]r^2)

dA/dt = 1/r = 1.3333

**Warning Nerd Post Coming Up**

Your solution is very elegant the way it simplifies to 1/r, its nice when the figures do that, I should not have plugged the figures in so early!

Your right too, I should have got 0.07 for dr/dt - I have not used a scientific calculator in about 5 years so I have some excuse!

[Lyman Zerga]

now stop this

it's getting too entertaining and i cant take the thrill anymore

[Schmeltz]

This is exactly how I feel watching two people speak Chinese or something.