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

tomwatts Avatar
5y, 6m agoPosted 5 years, 6 months ago
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. :S
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tomwatts Avatar
5y, 6m agoPosted 5 years, 6 months ago
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#1
Stop trying to cheat in your maths exam :0)
[mod] 3 Likes #2
Between me and my girlfriend we can't do it. And we're supposed to be clever. :S


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. :D
3 Likes #3
Syzable
Between me and my girlfriend we can't do it. And we're supposed to be clever. :S


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. :D


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

Disclaimer - If any babies are born as a result of my actions tonight i am not liable. :D
#4
Not sure the in-laws would agree with that!!
#5
sancho1983
Not sure the in-laws would agree with that!!


Shh you.
#6
MATHS THREAD NOW. NOT A SEXY THREAD, K?
1 Like #7
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/ Yes

b/ Tea time ...

the last time i did anything remotely this hard was ... wait i've never done anything this hard .. :p
#8
arcangel111
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/ Yes

b/ Tea time ...

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


thanks.
#9
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
#10
sancho1983
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


I sincerely hope this isn't how the thread is going to go. I don't want to have a disappointed and upset girlfriend...
#11
lol - you want sincere - check the Charlie Scene thread :)
you want funny answers to serious equations ask us :):p xxx lol
[mod] 1 Like #12
When i was a lad we got a Gold Star if we knew the 7 times table. Good luck. lol
#13
F**k
3 Likes #14
tomwatts
sancho1983
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


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


That will come later when the sexy time starts X)
1 Like #15
tomwatts
sancho1983
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


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


Thought you were doing that later?
1 Like #16
tomwatts
F**k

Syzable
When i was a lad we got a Gold Star if we knew the 7 times table. Good luck. lol

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 :)
#17
A) T is constant at 7

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 on May 23, 2011 20:55
#18
I'm enjoying the banter a bit much to help now....

Ok 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 on May 23, 2011 21:25: Wrong!
#19
arcangel111
tomwatts
F**k


Syzable
When i was a lad we got a Gold Star if we knew the 7 times table. Good luck. lol


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 :)


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

And shh sancho and micoo ):
#20
Hubby says that part:
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!
#21
a/ = 7
b/ = 13:23
#22
Thank you all!!! Still having fun with it though as we keep getting different answers for the last bit, but we're going to settle on 13:23 :)

I'm off for now (:D), thank you once again :)
1 Like #23
chesso
Hubby says that part:
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!


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)
#24
tomwatts
Thank you all!!! Still having fun with it though as we keep getting different answers for the last bit, but we're going to settle on 13:23:)

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

thats cos i is cleva :)
#25
I'm not going to lie. I couldn't do it, only a year after doing my A-Level and still doing some maths at uni.

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

Edited By: emhaslam on May 23, 2011 23:10

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