3U Maths Question Thread

[diggity]

Can i say t = x/(dx/dt)? (If its important, when t=0 X=0)

[RuiAce]

Can i say t = x/(dx/dt)? (If its important, when t=0 X=0)

I don't see why this is true? A simple counterexample here would be \(x = t^2\), which would give \(RHS = \frac{t^2}{2t} = \frac{t}{2} \neq t\).

I sort of understand your explanation, but I don't get what you did in the fraction.

His explanation is definitely valid. The question wants you to focus on a specific person, which is basically equivalent to assuming that the person to be chosen has been pre-defined.

What he did in the fraction could be simplified though. The idea is that because we're just focusing on the one person, \( \frac{1}{\binom{30}{12}} \) is enough. You can ignore the factorial-like argument if you do not understand it.

[diggity]

That's weird, I swear I've seen this work before... maybe it was just a coincidence? My thinking is to liken it to a d,s,t triangle, i.e d = st like in physics, where dx/dt is s, d is x, and t is.. t.

you know what, maybe it didn't work because dx/dt isn't constant, which just occured to me now. I think that's why it worked last time. My bad, I feel dumb now

[akjen]

Help with v, please
ANSWER: 15⁰<θ<45^o OR 71^o34^'<θ<75 (NOT sure if this answer is correct btw)
Thanks in advance!

[DrDusk]

Help with v, please
ANSWER: 15⁰<θ<45^o OR 71^o34^'<θ<75 (NOT sure if this answer is correct btw)
Thanks in advance!

When proving the part before, you end up with



However if we want it to hit the wall we must change the equality to less than or equal to as we need it to have a 'y' displacement of less than 20 when the x displacement is 40 for it to hit the wall. We therefore get







If we were to substitute something like u = tanx then we could solve it like a normal 3u inequality which I'm sure your familiar with. Doing so we get.


Now all we need to do is solve the other quadratic:



Putting together the solutions from the first and second quadratic get you the answer

[akjen]

When proving the part before, you end up with



However if we want it to hit the wall we must change the equality to less than or equal to as we need it to have a 'y' displacement of less than 20 when the x displacement is 40 for it to hit the wall. We therefore get







If we were to substitute something like u = tanx then we could solve it like a normal 3u inequality which I'm sure your familiar with. Doing so we get.


Now all we need to do is solve the other quadratic:



Putting together the solutions from the first and second quadratic get you the answer

Thanks!

[Youssefh_]

Can someone please assist me on how to address this question, I never quite understood the exact procedure in doing questions like these
thank you

[fun_jirachi]

Hi there!

Basically, we have the identity that


Try applying it to some other questions! The gist of this is just to make the inside of the sine the same as the denominator somehow, then subbing in 1.

Hope this helps

EDIT: Alternatively, since this is a multiple choice question, you could just punch in a really small value of x into that expression in your calculator, and you'd get the answer regardless.

[Mcnelson]

Hi, I have a question about the 2015 HSC Maths ext1 Q14 c.
The question states that two players (A and B) are playing a series of games in which they are equally likely to win.

For part (i) it made sense that for A to win 5 games we would use binomial probability for A winning four of the first six games and then multiplying it by the 1/2 once  more to account for the 5th win on the 7th game.

I then used the similar method to make an expression for a player winning in at most 7 games (in part (ii)): P= 6C4*(1/2)^7 + 5C4*(1/2)^6 + 4C4*(1/2)^5

So then for part (iii) can we just use the expression in part (ii) and just substitute (n+1) wins in place of the 5 wins to prove :
nCn*2^n + (n+1)Cn*2^(n-1) + (n+2)Cn*2^(n-2) + .... + 2nCn = 2^2n  and why would this be equal to the probability of A winning the prize??

[fun_jirachi]

Welcome to the forums!

Not sure what you mean by 'just substitute (n+1) wins in place of the five wins' - n is just some arbitrary integer for the purpose of generalising the result in iii). If you could explain what you mean, might help your understanding

Also, the expression is also not necessarily the probability of A winning; note that a probability must be between 0 and 1, and the RHS is clearly bigger than 1, since 2²ⁿ=1 when n=0, which isn't a positive integer. Rather, the question asks us to generalise the result in ii) and apply it to prove another, different result.

Also, there are a few things to note before answering this question:
- To win n+1 times, someone must have won by the (2n+1)^th game; this should be reasonably simple to surmise
- To win n+1 times, we must consider the first n games that person wins, plus however many the other wins during that process, and take it as given that the person winning n games up to that point wins the next game.

Starting from the beginning, let's have A win n+1 games, all in a row. From i), we see that he wins the first n games, then wins the (n+1)^th game in the expression below:

Similarly, if A takes n+2 games to win n+1 of them, we have:

Continuing until the (2n+1)^th game, we have the chance of A winning n+1 games:


But clearly, A is just as likely to win n+1 games as B is, since the chance of each of them winning is equal. Thus, the above expression is equal to half, and we can manipulate the expression from there.



And we are done

Hope this helps

[spnmox]

Six men and six women are to be seated at a round table. In how many different ways can they be seated if men and women alternate?

Could someone please explain this? I've always had trouble with combinations questions.

Also:
A particle is moving in simple harmonic motion. The displacement of the particle is x and its velocity, v, is given by the equation v^2 = n^2 ( 2kx − x^2), where n and k are constants. The particle is initially at x = k. Which function, in terms of time t, could represent the motion of the particle?
The answer is x = k sin (nt) + k. I'm just not sure how to work out the amplitude.

[fun_jirachi]

Six men and six women are to be seated at a round table. In how many different ways can they be seated if men and women alternate?

Could someone please explain this? I've always had trouble with combinations questions.

Also:
A particle is moving in simple harmonic motion. The displacement of the particle is x and its velocity, v, is given by the equation v^2 = n^2 ( 2kx − x^2), where n and k are constants. The particle is initially at x = k. Which function, in terms of time t, could represent the motion of the particle?
The answer is x = k sin (nt) + k. I'm just not sure how to work out the amplitude.

Hey there!

For the first question, consider seating the women first: if this happens, we seat 6 women around a table, and there are 5! ways of doing so.
Now, we have 6 seats in between each woman to seat a man. Note that each seat is now different, and rotating the arrangements of the men actually matters, because each seat indicates a different position between a different pair of women. As such, to seat the first man, there are 6 possibilities, then 5 for the second, and so on ie. 6! to seat the men.

As such, the answer is 5! x 6!

Basically here, we complete the square.
v^2 = n^2 (2kx - x^2)
v^2 = n^2 (k^2 - (k-x)^2)

Now, we know that this is in the form v^2=n^2(a^2-(x-b)^2), we know that the centre of motion is k, the amplitude is also k, and since the particle is at k when t equals zero, we use the sine function and append the k on the end, such that x=ksin(nt)+k.

Hope this helps

[spnmox]

Hey there!

For the first question, consider seating the women first: if this happens, we seat 6 women around a table, and there are 5! ways of doing so.
Now, we have 6 seats in between each woman to seat a man. Note that each seat is now different, and rotating the arrangements of the men actually matters, because each seat indicates a different position between a different pair of women. As such, to seat the first man, there are 6 possibilities, then 5 for the second, and so on ie. 6! to seat the men.

As such, the answer is 5! x 6!

Basically here, we complete the square.
v^2 = n^2 (2kx - x^2)
v^2 = n^2 (k^2 - (k-x)^2)

Now, we know that this is in the form v^2=n^2(a^2-(x-b)^2), we know that the centre of motion is k, the amplitude is also k, and since the particle is at k when t equals zero, we use the sine function and append the k on the end, such that x=ksin(nt)+k.

Hope this helps

Oh okay, and why don't we need to multiply by two in the combinations question - for the option of seating men first? Does it not matter?

[fun_jirachi]

It doesn't matter whether men are seated first or women are seated first; consider the smaller case of two men A and B, and two women C and D. If we seat the women first, then seat the men, we have _C_D. We then seat A and B in 2! ways. Now, consider if we sat A and B first; we'd have _A_B, with 2! ways to seat C and D. Note that here, in the first case, we have ACBD or BCAD or some rotation of the two. Similarly, in the second case, we have CADB and DACB, which you'll notice are just rotations. The same principle can be applied to any number of people provided the number in each group is the same, and they alternate, and there are two groups. Fundamentally you could've sat the men first, but it doesn't matter because the combination doesn't 'tell' you who you sat down first

Hope this makes sense

[spnmox]

thanks! it makes sense, yeah

hey guys, for the attached graph, why is the equation of motion for x for the particle from B = wtcosbeta, and not wtcosbeta+d? I thought Vtcostheta could only be used if the particle started at the origin.