# MATHS NERD HELP! (pleeease. don't make me beg)

A group of students are researching the rate at which ice thickens on a frozen pond. They have experimental evidence that when the air temperature is –T °C, the ice thickens at a rate T/14000x cm s^-1, where x cm is the thickness of the ice that has already formed.

On a particular winter day the air temperature is constant at –7 °C. At 12.00 noon the students note that the ice is 2 cm thick. Time t seconds later the thickness of the ice is x cm.

(a) Show that dx/dt = 1/2000x

(b) Solve the differential equation and hence find the time when the students predict the ice will be 3 cm thick.

Between me and my girlfriend we can't do it. And we're supposed to be clever.

On a particular winter day the air temperature is constant at –7 °C. At 12.00 noon the students note that the ice is 2 cm thick. Time t seconds later the thickness of the ice is x cm.

(a) Show that dx/dt = 1/2000x

(b) Solve the differential equation and hence find the time when the students predict the ice will be 3 cm thick.

Between me and my girlfriend we can't do it. And we're supposed to be clever.

Sloth/Mod

You and your girlfriend have homework and you are actually doing it!!!! What's the matter with the youth of today?? This time should be sexy time X)

Disclaimer - If any babies are born as a result of this comment i am not liable.

Original Poster

Later! Maths now, 'work' later...

Disclaimer - If any babies are born as a result of my actions tonight i am not liable.

Original Poster

Shh you.

Original Poster

On a particular winter day the air temperature is constant at –7 °C. At 12.00 noon the students note that the ice is 2 cm thick. Time t seconds later the thickness of the ice is x cm.

(a) Show that dx/dt = 1/2000x

(b) Solve the differential equation and hence find the time when the students predict the ice will be 3 cm thick

a/ Yes

b/

Tea time ...the last time i did anything remotely this hard was ... wait i've never done anything this hard ..

Original Poster

thanks.A group of students are researching the rate at which ice thickens on a frozen pond. They have experimental evidence that when the air temperature is –T °C, the ice thickens at a rate T/14000x cm s^-1, where x cm is the thickness of the ice that has already formed.

On a particular winter day the air temperature is constant at –7 °C. At 12.00 noon the students note that the ice is 2 cm thick. Time t seconds later the thickness of the ice is x cm.

(a) Show that dx/dt = 1/2000x

(b) Solve the differential equation and hence find the time when the students predict the ice will be 3 cm thick

(a) It says it right there

(b) I pick the lightest student to stand on it, when she stops falling through it is 3 cm

Done

Original Poster

I sincerely hope this isn't how the thread is going to go. I don't want to have a disappointed and upset girlfriend...

you want funny answers to serious equations ask us xxx lol

Sloth/Mod

Original Poster

That will come later when the sexy time starts X)

Thought you were doing that later?

this isn't going to end well ... i suggest you turn the pc off, solve it yourself then service your missus before she sees what you been writing

therefore dx/dt = 7/14000x = 1/2000x

b) you want dx to equal 1 cos 2 cm already exist.

therefore simply sub and solve for dt which gives 4000 seconds = 66.6 mins.

Therefore time is 13.06:67 (round to required accuracy)

Edited by:"amzmalhotra" 23rd May 2011Ok then a) Just substitute the temperature T(=7) in to the formula given as it relates to the rate of change of thickness.

b) 12:33 (Edit: Wrong! Dropped x when integrating)

Edited by:"james132" 23rd May 2011Original Poster

She's currently playing Limbo and is unaware I'm on HUKD...

And shh sancho and micoo ):

a) is putting the info into a mathematical form i.e. recognise that the ice thickens as a rate dx/dt is shorthand for that.

dx/dt= -T/1400x, subsitute for T = -7

b) answer is 5000seconds after. The proof is short but dificult to type on here!

b/ = 13:23

Original Poster

I'm off for now (:D), thank you once again

Screwed up my first time but this is what I get now too. Quick proof:

dx/dt = 1/2000x

2000xdx = dt

Integrating gives:

t = 1000x^2 + C

We know when t = 0 x = 2 therefore subbing in:

t = 1000*(2)^2 +C

C = - 4000

t = 1000x^2 - 4000

We want to know when x = 3 therefore

t = 1000*3^2 - 4000

= 5000 seconds (Approx 13:23)

thats cos i is cleva

It's easy when you see the answer though... as always with maths.

Edited by:"emhaslam" 23rd May 2011