2000 question 10b (1 Viewer)

[Azreil]

How would you guys approach this question? I looked at it, went "Uhhhhh..." and wrote something that I'm pretty sure the markers described as 'nonsensical' in their comments.

 

[tommykins]

would help if you posted the question. or giev a link to it.

 

[lyounamu]

How would you guys approach this question? I looked at it, went "Uhhhhh..." and wrote something that I'm pretty sure the markers described as 'nonsensical' in their comments.

You really have to approach this question like you do in the 3 Unit rate of change question. Definitely a different type of question, but the way you approach is pretty similar.

First you outline all the given values and approach the question slowly.

Post the question I did that question before.

 

[lyounamu]

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

This is the question asked by OP: Q10 bi) and bii)

Here is my solution: I cannot find the solution that pwnage did and I don't know where that is.

v = A/h

Understand that v = dx/dt

In the question (in part i), we were to prove that

dx/dt = k/t but in the question they already provided that v = A/h which obviouls means that we need to rearrange this to fit what question asked.

GIVEN:

h = Kt (where K is a constant of proportionality)
v = A/h (v = dx/dt)

NOW,

v = A/h
i.e. dx/dt = A/(Kt)
Therefore, dx/dt = k/t (where A/k is another constant)

Part ii to come soon.
ii)

given:

dx/dt = k/t

In the question, 1km of snow was ploughed in two hours (i.e. from T = 6am to T +2 = 8am)

If we integrate dx/dt (T to T+2), the answer we yield should be 1.

i.e. Integral: T to T+2 (k/t dt) = 1
i.e. k (ln(t)) (T -> T+2) = 1 .....(1)

And another thing given was that, another kilometre was ploughed from T+2 to T+5.5 as it took further 3.5 hours to plough it off.

i.e. Integral: T to T+5.5 (k/t dt) = 1+1 (because another km was covered for further 3.5 hours)
i.e. k (ln(t)) (T+2 -> T+5.5) = 2 ....(2)

From (1), we get:

k ln((T+2)/T) = 1

From (2), we get:

k ln((T+5.5)/T) = 2

Double (1), so we get:

2k ln((T+2)/T) = 2 ...(3)

Equate both sides we get:

2k ln((T+2)/T) = k ln((T+5.5)/T)
2 ln((T+2)/T) = ln((T+5.5)/T)
i.e. ln(((T+2)/T)^2) = ln((T+5.5)/T)
Further equating:

((T+2)/T)^2 = (T+5.5)/T)
Further working out, you should get:

T = 8/3 hours = 2 hours and 40 minutes

Rest should be easily worked out too.

 

[Graceofgod]

Got part i straight away, Just tried part ii, and after 5 minutes, found that T = T + 3.5

Methinks time to try a different approach

 

[lyounamu]

Got part i straight away, Just tried part ii, and after 5 minutes, found that T = T + 3.5

Methinks time to try a different approach

That's a great work. It took me 10 minutes to work that out but funnily, it took me about same amount of time to work out the 2nd part...lol

 

[Graceofgod]

That's a great work. It took me 10 minutes to work that out but funnily, it took me about same amount of time to work out the 2nd part...lol

Yeah but part i isn't hard as such, just depends on how you think about it.

Just had dinner, so I will try part ii for the second time now. (still haven't looked at your answer)

 

[lyounamu]

Yeah but part i isn't hard as such, just depends on how you think about it.

Just had dinner, so I will try part ii for the second time now. (still haven't looked at your answer)

I know that part i is very easy. But it was actually quite taken aback because this type of question is very rare for 2 unit. In the end, I had to use my little brain to work that out.

 

[m00]

howd you get h = Kt? i dont get that lol
its like maths just pooped that concept out of its ass

 

[lyounamu]

howd you get h = Kt? i dont get that lol
its like maths just pooped that concept out of its ass

Here is an extract from the paper:

"The depth of the snow,​

h, increases at a constant rate through the night and the following day".

That means that the dh/dt = k i.e. h = kt where k is a constant of proportionality.

​

 

[m00]

oooooooooooooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

 

[Graceofgod]

Wow, I have no idea wtf i did. But I found an answer in about 5 lines for part ii. of T = -8/3

Here is what I did, and to be honest, I was just fooling around:

You have k/T, k/(T+2), k/(T+3.5)
This is simply substituting into dx/dt = k/t from part i. This can be done because x = 0, x = 1, x = 2 Are the three values of x each of these corresponds to.

I figured, since k is a constant, that this is a geo series.

Therefore (k/T)/[k/(T+2)] = the common ratio = (T+2)/T
Also, the devision of the second two, [k/(T+2)]/[k/(T+3)] = the common ratio = (T+3.5)/(T+2)

Since the common ratio is COMMON, and therefore stays the same for the whole series,

(T+2)/T = (T+3.5)/(T+2)

T = -8/3
EDIT I made an error, it is 8/3
Simple

(I too have no idea why it is negative. And it is quite possible I might be the luckiest person alive)

EDIT final answer: 3:20am

 

[Timothy.Siu]

lol i cudn't figure out the first part, but i'll look at it later, ..

 

[henry08]

I stil ldon't get it []

 

[lyounamu]

I stil ldon't get it []

Don't worry. I reckon this question would be close to being one of the hardest q10 from hsc mathematics. You can still get 98 or 97 without answering half of q 10 - but you will need to kill the rest of the exam though...

 

[trailblazer]

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

This is the question asked by OP: Q10 bi) and bii)

Here is my solution: I cannot find the solution that pwnage did and I don't know where that is.

v = A/h

Understand that v = dx/dt

In the question (in part i), we were to prove that

dx/dt = k/t but in the question they already provided that v = A/h which obviouls means that we need to rearrange this to fit what question asked.

GIVEN:

h = Kt (where K is a constant of proportionality)
v = A/h (v = dx/dt)

NOW,

v = A/h
i.e. dx/dt = A/(Kt)
Therefore, dx/dt = k/t (where A/k is another constant)

Part ii to come soon.
ii)

given:

dx/dt = k/t

In the question, 1km of snow was ploughed in two hours (i.e. from T = 6am to T +2 = 8am)

If we integrate dx/dt (T to T+2), the answer we yield should be 1.

i.e. Integral: T to T+2 (k/t dt) = 1
i.e. k (ln(t)) (T -> T+2) = 1 .....(1)

And another thing given was that, another kilometre was ploughed from T+2 to T+5.5 as it took further 3.5 hours to plough it off.

i.e. Integral: T to T+5.5 (k/t dt) = 1+1 (because another km was covered for further 3.5 hours)
i.e. k (ln(t)) (T+2 -> T+5.5) = 2 ....(2)

From (1), we get:

k ln((T+2)/T) = 1

From (2), we get:

k ln((T+5.5)/T) = 2

Double (1), so we get:

2k ln((T+2)/T) = 2 ...(3)

Equate both sides we get:

2k ln((T+2)/T) = k ln((T+5.5)/T)
2 ln((T+2)/T) = ln((T+5.5)/T)
i.e. ln(((T+2)/T)^2) = ln((T+5.5)/T)
Further equating:

((T+2)/T)^2 = (T+5.5)/T)
Further working out, you should get:

T = 8/3 hours = 2 hours and 40 minutes

Rest should be easily worked out too.

lol... you really want that state rank don't you. Well, you seem on the right track.

 

[lyounamu]

lol... you really want that state rank don't you. Well, you seem on the right track.

Yeah, I really do because there is a chance I may not even do 4 Unit Mathematics. If I have some great mark in 2 Unit Mathematics, I have something to fall back on.

 

[trailblazer]

Yeah, I really do because there is a chance I may not even do 4 Unit Mathematics. If I have some great mark in 2 Unit Mathematics, I have something to fall back on.

Nice, I know of someone from my school who got state ranks for both 3U(1st i think) and 2U(2nd 3rd?) and didn't do 4U because of it, his UAI was 99.8 wowers.

 

[lyounamu]

Nice, I know of someone from my school who got state ranks for both 3U(1st i think) and 2U(2nd 3rd?) and didn't do 4U because of it, his UAI was 99.8 wowers.

you from sydney boys? If that's the case, I think he did 4 unit too. His name is anthony morris and he definitely did 4 unit maths and got state rank for it - 3rd I think and his UAI was 99.75 - close enough, lol.

 

[5647382910]

The mother of all q 10's is the pulley one in 2003.....................I hate that q