help! 2U Maths question (1 Viewer)

[danal353]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics23U.pdf

hey guys, I was wondering if someone had the time to explain to me how to do this question

It's question 10b) and I can't do any of it!

 

[raniaaa]

http://community.boredofstudies.org/12/mathematics/225547/heardest-hsc-paper/3.html2nd last post

 

[lyounamu]

http://www.boardofstudies.nsw.edu.au/hsc_exams/hsc2000exams/hsc00_maths/00mathematics23U.pdf

hey guys, I was wondering if someone had the time to explain to me how to do this question

It's question 10b) and I can't do any of it!

LOL
I knew that someone would ask this question eventually.

Since I have done enough 4 unit study for the day, I will reward myself with 2 unit maths xD

i) The depth of snow increases at a constant rate i.e. dh/dt = C (constant)
so h = Ct +c1 = Ct (c1 = 0 because there is no snow when t = 0)

Then we have v = A/h
i.e. dx/dt = A/h = A/Ct = k/t where k = A/C

EDIT: raniaaa!!! damn...hang on he didn't finish the Q. I will continnue then LOL

Part 2 (this is the tricky part so bear with me):


T = time when it started to snow
T + 2 = time when it covered 1 km of snow

Integration (terminal t = T and t = T+2)

After integration, dx/dt = k/t becomes
1 = k (ln (T+2) - ln T)
1 = k ln ((T+2)/T) ...i

Now T+5.5 = time when we covered 2km of snow

Integration (terminal t = T and t = T+ 5.5)

After inegration dx/dt = k/t becomes

2 = k (ln (T+5.5) - lnT)
2 = k ln ((T+5.5)/5)...ii

NOW EQUATE part i and ii

2k ln ((T+2)/T) = k ln((T+5.5)/T)
2 ln ((T+2/T) = ln ((T+5.5/T)
ln ((T+2)/T)^2) = ln ((T+5.5)/T)
((T+2)/T)^2 = (T+5.5)/T
find for T
T = 8/3

So 8/3 hours before 6am...i.e. whatever

 

[danal353]

i also don't understand part ii) =(

 

[raniaaa]

LOL
I knew that someone would ask this question eventually.

Since I have done enough 4 unit study for the day, I will reward myself with 2 unit maths xD

i) The depth of snow increases at a constant rate i.e. dh/dt = C (constant)
so h = Ct +c1 = Ct (c1 = 0 because there is no snow when t = 0)

Then we have v = A/h
i.e. dx/dt = A/h = A/Ct = k/t where k = A/C

EDIT: raniaaa!!! damn...hang on he didn't finish the Q. I will continnue then LOL

haha sorry didn't mean to steal your thunder

 

[lyounamu]

haha sorry didn't mean to steal your thunder

lol it's alright

 

[danal353]

LOL
I knew that someone would ask this question eventually.

Since I have done enough 4 unit study for the day, I will reward myself with 2 unit maths xD

i) The depth of snow increases at a constant rate i.e. dh/dt = C (constant)
so h = Ct +c1 = Ct (c1 = 0 because there is no snow when t = 0)

Then we have v = A/h
i.e. dx/dt = A/h = A/Ct = k/t where k = A/C

EDIT: raniaaa!!! damn...hang on he didn't finish the Q. I will continnue then LOL

Part 2 (this is the tricky part so bear with me):


T = time when it started to snow
T + 2 = time when it covered 1 km of snow

Integration (terminal t = T and t = T+2)

After integration, dx/dt = k/t becomes
1 = k (ln (T+2) - ln T)
1 = k ln ((T+2)/T) ...i

Now T+5.5 = time when we covered 2km of snow

Integration (terminal t = T and t = T+ 5.5)

After inegration dx/dt = k/t becomes

2 = k (ln (T+5.5) - lnT)
2 = k ln ((T+5.5)/5)...ii

NOW EQUATE part i and ii

2k ln ((T+2)/T) = k ln((T+5.5)/T)
2 ln ((T+2/T) = ln ((T+5.5/T)
ln ((T+2)/T)^2) = ln ((T+5.5)/T)
((T+2)/T)^2 = (T+5.5)/T
find for T
T = 8/3

So 8/3 hours before 6am...i.e. whatever

thanks! it took me a while to understand but thanks a lot - was the bottom bit explaining the top bit? since T = 8/3 both times?

tried to give rep for all the help but it says I can't =(

thanks again

 

[lyounamu]

thanks! it took me a while to understand but thanks a lot - was the bottom bit explaining the top bit? since T = 8/3 both times?

tried to give rep for all the help but it says I can't =(

thanks again

Sorry for not being clear enough. T simply meant when it started to snow but T =1 means 1 hour before 6am.

6am is when the machine started to take snow. Before, that it already started snowing. T simply refers to amount of time it took to snow to accumulate before 6am.

EDIT: I just realised how improper my explanation was. T is the time we started ploughing. But t = 0 is the time when the snow started. Since we got T = 8/3, t = 0 must be 6am - 8/3 i.e. 3:20 am

 

[sheizaminelli]

Namu ur doing the four unit test tomorrow right???
just wondering.

 

[lyounamu]

Namu ur doing the four unit test tomorrow right???
just wondering.

Unfortunately...yes....

 

[sheizaminelli]

oh shoot i just read what u wrote
sorry
good luck tomorrow though

but that means ur 100 in 2 unit wont count :O

 

[lyounamu]

oh shoot i just read what u wrote
sorry
good luck tomorrow though

but that means ur 100 in 2 unit wont count :O

Nope. FML
Instead I am going to take up some random mark that is hardly close to 100 scaled mark

 

[sheizaminelli]

are u repeating three unit or keeping ur mark???

 

[lyounamu]

are u repeating three unit or keeping ur mark???

I somehow chose to repeat that...FML again...

 

[Cloesd]

LOL
I knew that someone would ask this question eventually.

Since I have done enough 4 unit study for the day, I will reward myself with 2 unit maths xD

i) The depth of snow increases at a constant rate i.e. dh/dt = C (constant)
so h = Ct +c1 = Ct (c1 = 0 because there is no snow when t = 0)

Then we have v = A/h
i.e. dx/dt = A/h = A/Ct = k/t where k = A/C

EDIT: raniaaa!!! damn...hang on he didn't finish the Q. I will continnue then LOL

Part 2 (this is the tricky part so bear with me):


T = time when it started to snow
T + 2 = time when it covered 1 km of snow

Integration (terminal t = T and t = T+2)

After integration, dx/dt = k/t becomes
1 = k (ln (T+2) - ln T)
1 = k ln ((T+2)/T) ...i

Now T+5.5 = time when we covered 2km of snow

Integration (terminal t = T and t = T+ 5.5)

After inegration dx/dt = k/t becomes

2 = k (ln (T+5.5) - lnT)
2 = k ln ((T+5.5)/5)...ii

NOW EQUATE part i and ii

2k ln ((T+2)/T) = k ln((T+5.5)/T)
2 ln ((T+2/T) = ln ((T+5.5/T)
ln ((T+2)/T)^2) = ln ((T+5.5)/T)
((T+2)/T)^2 = (T+5.5)/T
find for T
T = 8/3

So 8/3 hours before 6am...i.e. whatever

O lawd, when i did part i), i simply said that dx/dt = speed. Speed is inversly propoportional to depth of snow. depth of snow is proportional to time. THus speed is inversly proportional to time.

Hmm...

Would you lose marks for not explaining it.. "mathamatically"

 

[danal353]

Sorry for not being clear enough. T simply meant when it started to snow but T =1 means 1 hour before 6am.

6am is when the machine started to take snow. Before, that it already started snowing. T simply refers to amount of time it took to snow to accumulate before 6am.

EDIT: I just realised how improper my explanation was. T is the time we started ploughing. But t = 0 is the time when the snow started. Since we got T = 8/3, t = 0 must be 6am - 8/3 i.e. 3:20 am

ah no no, my inability to understand is not because you didn't explain properly - I just simply lack the brain capacity...