3 Unit Maths HSC Exam Revision (1 Viewer)

[random-1005]

Another past 2 unit question, this time sydney boys.

A petrol tank is design by the rotation of the curve y= x/5 ( x-40) about the x axis between x=0 and x=40. If the units are in centrimetres how many litres will the tank hold??

Do you guys know how many cubic cm in cubic metre and how many cubic metres to a litre etc, these things will pop up in harder 2 unit and 3 unit ( orbetter yet in questions like this it would be easier to change your limits, ie integrate from 0 to 0.4, and im pretty sure 1 m^3 is a litre, correct me if im wrong )

 

[undalay]

I think you may have forgotten to use this part of the question

"A stone is thrown so that it will hit a bird at the top of a pole."

That is, at t=T1 the stone must be at (W,L). This together with the other info, puts restrictions on the whole scenario leading to set of unique solutions, including those for W and L.


So overall, its saying all 6 unknowns including W and L must be of specific values or else the stone cannot possibly satisfy all the required conditions at the same time, and in particular these two: 1.being at (W,L) and 2.hitting the bird. There wont be a set of unique solutions if the condition to be satisfied was just to hit the bird, which is what you are thinking i believe.

There are infinite solutions.
Take case 1 for example.
Imagine L to be 10m.
Take W to be 0.

What does Vx has to be to hit bird ? 10m/s

Take w to be 1. Will Vx = 10 hit bird? No.

"That is, at t=T1 the stone must be at (W,L). This together with the other info, puts restrictions on the whole scenario leading to set of unique solutions, including those for W and L. "
No. For any specified W and L, there will be one Vx and Vy which causes the stone to hit the bird. Given Vx > 10.

If you still need more proof, i'll try to make a more rigour proof at the end of my exams.

 

[undalay]

1. A given height of pole, H specifies the time of impact. (As 2H is max height).
That is TOI = f(h)


2. X distance between bird and stone is.

W + 10t - Vx*t = distance between bird and stone.

Take impact( x distance between bird and stone = 0)

Vx*t = W + 10t
Vx = W/t + 10

Vx = f(TOI)
Thus Vx = f(TOI) = F(h)

------------------------
This means that if time of impact is 3 seconds. There will be a UNIQUE H, for time of impact of 10 seconds, there will be a UNIQUE H.


That means if TOI = 3 seconds
Vx = W/3 + 10

If TOI = 10 seconds
Vx = W/10 + 10


H can be anything, TOI is a function of H, and Vx is a function of TOI (which is a function of H) and W.

HENCE.

Vx is a function of H and W.

 

[k02033]

There are infinite solutions.
Take case 1 for example.
Imagine L to be 10m.
Take W to be 0.

What does Vx has to be to hit bird ? 10m/s

Ok there are 3 conditions that the stone must satisfy

(1) it must be at (W,L) at t=T1
(2) it must have a maximum height of 2L
(3) it must hit the bird at t=2T2-T1

So in your case 1, the stone is thrown at (0,0) at t=0 but it must be at (0,10) at T1 to satisfy (1), which means the stone must be launched with U=0, and since U=0 the stone cannot possibility satisfy (3), this is a demonstration of what i said in my previous post. So your case 1 is invalid.

No. For any specified W and L, there will be one Vx and Vy which causes the stone to hit the bird. Given Vx > 10.

Ok since you have chosen W and L to be arbitrary parameters, i first ask you to use W, and L that are not the W and L my 6 equations yield and then try and find solutions for U and V such that the stone still satisfies all 3 conditions. I guarantee this is not possible.

Then i ask W and L to be the unique values that my 6 equations suggests. Now i ask you to find U and V such that the stone still satisfies all 3 conditions and they also has to be different to the unique values suggested by my 6 equations. Again my bet is that this is also not possible.

These 2 results means there is only a set of unique solutions.

If you still need more proof, i'll try to make a more rigour proof at the end of my exams.

Yes i need more proof, those you gave are unsatisfactory .

 

[random-1005]

Ok there are 3 conditions that the stone must satisfy

(1) it must be at (W,L) at T1
(2) it must a maximum height of 2L
(3) it must hit the bird at T2

So in your case 1, the stone is thrown at (0,0) at t=0 but it must be at (0,10) at T1 to satisfy (1), which means the stone must be launched with U=0, and since U=0 the stone cannot possibility satisfy (3), this is a demonstration of what i said in my previous post. So your case 1 is invalid.



Ok since you have chosen W and L to be arbitrary parameters, i specify W and L to be the unique values that my 6 equations suggests. Now i ask you to find U and V such that the stone still satisfies all 3 conditions and they also has to be different to the unique values suggested by my 6 equations. My bet is that this is not possible, i.e. there is only a set of unique solutions.



Yes i need more proof, those you gave are unsatisfactory .

ok take it easy, i think its enough of that question .

 

[k02033]

1. A given height of pole, H specifies the time of impact. (As 2H is max height).
That is TOI = f(h)


2. X distance between bird and stone is.

W + 10t - Vx*t = distance between bird and stone.

Take impact( x distance between bird and stone = 0)

Vx*t = W + 10t
Vx = W/t + 10

Vx = f(TOI)
Thus Vx = f(TOI) = F(h)

------------------------
This means that if time of impact is 3 seconds. There will be a UNIQUE H, for time of impact of 10 seconds, there will be a UNIQUE H.


That means if TOI = 3 seconds
Vx = W/3 + 10

If TOI = 10 seconds
Vx = W/10 + 10


H can be anything, TOI is a function of H, and Vx is a function of TOI (which is a function of H) and W.

HENCE.

Vx is a function of H and W.

You are not checking if those values corresponds to a stone that satisfies all 3 conditions. You are just setting parameters and blindly solving other variables in terms of those parameters. You are not checking if those solutions answers the question.

 

[k02033]

http://community.boredofstudies.org/13/mathematics-extension-1/246100/projectile-motion.html read my post in this thread.

That is what you are trying to implement, but in this clearing the wall question, the restrictions are lighter, hence infinite solutions with T as the arbitrary parameter.

 

[bouncing]

Question: Simplify nC(r+1) / (nCr)

the latex equation editor is just above the quick post button its like near the very bottom of the page


nC(r+1)/nCr
= {n!/(r+1)![n-(r+1)]!}/{n!/r!(n-r)!}
={r!(n-r)!}/{(r+1)!(r-n-1)!}
=n-r/r+1

is that right :\

 

[undalay]

You are not checking if those values corresponds to a stone that satisfies all 3 conditions. You are just setting parameters and blindly solving other variables in terms of those parameters. You are not checking if those solutions answers the question.

Which part do you disagree with?

1. For some pole height, H, (Given the restriction max height = 2H), you can find TOI (time of impact) in the y plane.

2. This TOI in y plane must = TOI in x plane. That is TOI = f(H)

3. TOI occurs when displacement from origin of stone = displacement from origin of bird.

4. That is Vx*TOI = w + 10*TOI

Vx = w/TOI + 10

Vx = f(w,H)

Which part does not make sense? If there is a specific set you disagree with I can flesh it out.

I already proved to you that there is not only 1 unique solution by giving you 2 Vx solutions

 

[random-1005]

the latex equation editor is just above the quick post button its like near the very bottom of the page


nC(r+1)/nCr
= {n!/(r+1)![n-(r+1)]!}/{n!/r!(n-r)!}
={r!(n-r)!}/{(r+1)!(r-n-1)!}
=n-r/r+1

is that right :\

yes (n-r)/(r+1), good job

just testing n!= (n-1)! x n

thats a good formula to know, especially for the greatest coeffiecent question ( if they ever ask that question in the hsc)

 

[k02033]

I already proved to you that there is not only 1 unique solution by giving you 2 Vx solutions

But you have not shown that the stone with those V will satisfy the conditions prescribed by the question.

Which part do you disagree with?

1. For some pole height, H, (Given the restriction max height = 2H), you can find TOI (time of impact) in the y plane.

2. This TOI in y plane must = TOI in x plane. That is TOI = f(H)

3. TOI occurs when displacement from origin of stone = displacement from origin of bird.

4. That is Vx*TOI = w + 10*TOI

Vx = w/TOI + 10

Vx = f(w,H)

Which part does not make sense? If there is a specific set you disagree with I can flesh it out.

Sorry but could you use U, V, W, L, T1, T2 as your variables?

Could you elaborate those first two points?

And most importantly, take any set of your solutions that does not satisfy my 6 equations and demonstrate mathematically that your set of solutions do indeed reflect a scenario in which the stone satisfies these 3 conditions:

(1) it must be at (W,L) at t=T1
(2) it must have a maximum height of 2L
(3) it must hit the bird at t=2T2-T1

If you can do this, you have proven me wrong.

 

[undalay]

Sorry but could you use U, V, W, L, T1, T2 as your variables?

Could you elaborate those first two points?

And most importantly, take any set of your solutions that does not satisfy my 6 equations and demonstrate mathematically that your set of solutions do indeed reflect a stone that satisfies these 3 conditions:

(1) it must be at (W,L) at t=T1
(2) it must have a maximum height of 2L
(3) it must hit the bird at t=2T2-T1

If you can do this, you have proven me wrong.

You are far complicating your problem with your variables.
I have posted up a proof. It is quite simple. If you need any more explanation post.

 

[nikkifc]

just testing n!= (n-1)! x n

thats a good formula to know, especially for the greatest coeffiecent question ( if they ever ask that question in the hsc)

And induction

 

[k02033]

You are far complicating your problem with your variables.

That's not true, considering those variables is necessary for the stone to meet the constraints. Which is the point i am trying to get at, i.e. the solutions you have in mind do not meet those constraints, therefore not answering the question at hand.

I don't open attachments here, post your work as a regular post please.

 

[undalay]

That's not true, considering those variables is necessary for the stone to meet the constraints.

I don't open attachments here, post your work as a regular post please.

No. You open it. I took a long time typing it up neatly. I can't copy and paste it.

Here a really really raelly relaly basic logic scenario.
Object 1 displacement is (x1,y1)
Object 2 displacement is (x2,y2)
A hit occurs only when x1=x2, and y1=y2
Where x = Vxt

Take object 1 = stone, object 2 = bird.
Let's say im standing below the pole.

What must Vx be in order to hit the bird? Well it MUST be 10. If this isn't instantly obvious to you ... give up on math. (Otherwise x1 will not equal x2 - ever)

Now lets say the pole is 1 m away. Then obviously the stone must go faster than the bird in order "catch up.

So in these two cases the Vx is different.
If you want the proper proof. Look at word docu. I took the effort to type it up. Your own fault if you dont want to look at it.

 

[bouncing]

lol stupid bird question causing so many issues

gotta revise my bionomial! seems like ive forgotten alot
for finding independent term do you do let Tk+1= independent?


sometimes i do it the lazy/easy way where you just find first like 3 terms and its usually in a patter and you just count from there.. but that wont work if its like ^3532423 or something

 

[bouncing]

No. You open it. I took a long time typing it up neatly. I can't copy and paste it.

Here a really really raelly relaly basic logic scenario.
Object 1 displacement is (x1,y1)
Object 2 displacement is (x2,y2)
A hit occurs only when x1=x2, and y1=y2
Where x = Vxt

Take object 1 = stone, object 2 = bird.
Let's say im standing below the pole.

What must Vx be in order to hit the bird? Well it MUST be 10. If this isn't instantly obvious to you ... give up on math. (Otherwise x1 will not equal x2 - ever)

Now lets say the pole is 1 m away. Then obviously the stone must go faster than the bird in order "catch up.

So in these two cases the Vx is different.
If you want the proper proof. Look at word docu. I took the effort to type it up. Your own fault if you dont want to look at it.

open it, i just did and i dont even know what you guys are talking about - dw its not a virus of any sort LOL

 

[random-1005]

lol stupid bird question causing so many issues

gotta revise my bionomial! seems like ive forgotten alot
for finding independent term do you do let Tk+1= independent?


sometimes i do it the lazy/easy way where you just find first like 3 terms and its usually in a patter and you just count from there.. but that wont work if its like ^3532423 or something

tk+1 = nCk (a)^k (b)^(n-k)

using exponent laws you will get it to some constant times a power of x, and you let the power be zero, give me a few minutes and ill put up the soln

 

[random-1005]

ok so we have (3x^2 +2x^-1) ^12

t k+1 = 12Ck * (3x^2)^k * (2x^-1)^12-k
= 12Ck * (3^k) * ( x^2k) * (2)^(12-k) *(x^(k-12))
= 12Ck * (x^(3k-12)) * (3^k) *(2)^(12-k)

set power of x to zero

therefore k=4

t5 = 12C4 * 3^4 * (2^8)

fairly straight foward method, just indice laws, raise a power to another power, multiply powers, when multiplying same base add the powers

also if they had something like "Show there is no independant term in the expansion of......." you would follow same procedure but you will find that k is not an integer when you set the power to zero.

ANOTHER NOTE: if for example it was ( 3x^2 - 2/x)^12 you would have to be VERY careful with the negative sign and make sure you apply the power to the negative sign.

 

[undalay]

That's not true, considering those variables is necessary for the stone to meet the constraints. Which is the point i am trying to get at, i.e. the solutions you have in mind do not meet those constraints, therefore not answering the question at hand.

I don't open attachments here, post your work as a regular post please.

My mistake, I misinterpreted the question.

I interpreted the part "a stone is thrown so that it will hit a bird at the top of a pole." to be hitting the bird as it flies away. Rather than literally on top of the pole.