Author Topic: 3U Maths Question Thread (Read 1236519 times) Tweet Share

fun_jirachi · « **Reply #3585 on:** August 10, 2018, 07:24:50 am »

hahahahaha ok, thanks for the advice!!!

Abdul_k · « **Reply #3586 on:** August 11, 2018, 03:46:16 pm »

Hi, I'm hoping that someone could help me with this question?

RuiAce · « **Reply #3587 on:** August 11, 2018, 04:48:59 pm »

Quote from: Abdul_k on August 11, 2018, 03:46:16 pm

Hi, I'm hoping that someone could help me with this question?

$\text{Choose 5 out of 12 for the first team: }\binom{12}{5}\\ \text{and then choose 5 out of the remaining 7 for the second: }\binom75.\\ \text{Also, since the teams are of the same size, we need a correction factor of }\frac{1}{2!}.\\ \text{So there are }\binom{12}{5} \binom75 \frac{1}{2!}\text{ possible games.}$
$\text{In the event that the twins are against each other}\\ \text{we just send one twin into each team.}\\ \text{Note however, that by doing so, the teams have been made }\textbf{distinguishable}.$
$\text{So we choose 4 out of 10 for the first team}\\ \text{and 4 out of the remaining 6 for the second team}\\ \text{and no additional correction factor is required.}$
$\text{Hence the probability is }\frac{\binom{10}4 \binom64}{\binom{12}5 \binom75 \frac1{2!}}$

key to success · « **Reply #3588 on:** August 15, 2018, 12:18:09 pm »

Hello, could someone help me with this question please?

Find the eqaution of the chord of contact to the parabola x^2=16y from the point (6,-4).
Please!!
THANK YOU!!

RuiAce · « **Reply #3589 on:** August 15, 2018, 12:29:46 pm »

Quote from: key to success on August 15, 2018, 12:18:09 pm

Hello, could someone help me with this question please?

Find the eqaution of the chord of contact to the parabola x^2=16y from the point (6,-4).
Please!!
THANK YOU!!

The equation of the chord of contact is on your formula sheet.

key to success · « **Reply #3590 on:** August 15, 2018, 02:57:15 pm »

Quote from: RuiAce on August 15, 2018, 12:29:46 pm

The equation of the chord of contact is on your formula sheet.

ohh its as simple as subbing into the eqn?? thankss matee!!

key to success · « **Reply #3591 on:** August 15, 2018, 05:09:26 pm »

help with the second part pleasee- thank you!

jamonwindeyer · « **Reply #3592 on:** August 15, 2018, 06:58:40 pm »

Quote from: key to success on August 15, 2018, 05:09:26 pm

help with the second part pleasee- thank you!

Hi! First, midpoint of PQ:

$x_X=\frac{2ap+2aq}{2}=a(p+q)\\y_X=\frac{ap^2+aq^2}{2}=\frac{a}{2}(p^2+q^2)$

Now $XT$ is perpendicular to the directrix, meaning it is a vertical line. So, $T$ has the same x-coordinate as the midpoint from above, and a y-coordinate corresponding with the directrix.

$x_T=a(p+q)\\y_T=-a$

Now W is the midpoint of XT:

$x=a(p+q)\\y=\frac{\frac{a}{2}(p^2+q^2)-a}{2}$

Now it is just about eliminating the two parameters. The trick here is the usual trick for this sort of question:

$(p^2+q^2)=(p+q)^2-2pq=(p+q)^2+2\ \text{(from the first part)}$

And we know that:

$x=a(p+q)\\p+q=\frac{x}{a}\\\therefore(p+q)^2=\frac{x^2}{a^2}$

Take all of that and substitute into the y-coordinate for W:

$y=\frac{\frac{a}{2}(p^2+q^2)-a}{2}=\frac{\frac{a}{2}(p+q)^2+2\frac{a}{2}-a}{2}=\frac{\frac{a}{2}\frac{x^2}{a^2}+a-a}{2}=\frac{\frac{x}{2a^2}}{2}=\frac{x}{4a^2}\\\therefore x=4a^2y$

Subject to little errors, but that's the rough idea!!

key to success · « **Reply #3593 on:** August 15, 2018, 09:52:59 pm »

Quote from: jamonwindeyer on August 15, 2018, 06:58:40 pm

Hi! First, midpoint of PQ:

$x_X=\frac{2ap+2aq}{2}=a(p+q)\\y_X=\frac{ap^2+aq^2}{2}=\frac{a}{2}(p^2+q^2)$

Now $XT$ is perpendicular to the directrix, meaning it is a vertical line. So, $T$ has the same x-coordinate as the midpoint from above, and a y-coordinate corresponding with the directrix.

$x_T=a(p+q)\\y_T=-a$

Now W is the midpoint of XT:

$x=a(p+q)\\y=\frac{\frac{a}{2}(p^2+q^2)-a}{2}$
Now it is just about eliminating the two parameters. The trick here is the usual trick for this sort of question:

$(p^2+q^2)=(p+q)^2-2pq=(p+q)^2+2\ \text{(from the first part)}$

And we know that:

$x=a(p+q)\\p+q=\frac{x}{a}\\\therefore(p+q)^2=\frac{x^2}{a^2}$
Take all of that and substitute into the y-coordinate for W:

$y=\frac{\frac{a}{2}(p^2+q^2)-a}{2}=\frac{\frac{a}{2}(p+q)^2+2\frac{a}{2}-a}{2}=\frac{\frac{a}{2}\frac{x^2}{a^2}+a-a}{2}=\frac{\frac{x}{2a^2}}{2}=\frac{x}{4a^2}\\\therefore x=4a^2y$

Subject to little errors, but that's the rough idea!!

Ah okay- thank you so much!

key to success · « **Reply #3594 on:** August 16, 2018, 08:10:20 pm »

Hello, does anyone know whether everytime we use the equations,
v^2 = n^2 (a^2 - (x-xo)^2)
v = an(cosnt)
a = -an^2 (sinnt)

do we need to derive them from the basic equations given on the formula sheet?
Note that I'm doing the catholic trial and my school strictly follows the criteria given by the CSSA.

Thanks

jamonwindeyer · « **Reply #3595 on:** August 17, 2018, 10:15:10 pm »

Quote from: key to success on August 16, 2018, 08:10:20 pm

Hello, does anyone know whether everytime we use the equations,
v^2 = n^2 (a^2 - (x-xo)^2)
v = an(cosnt)
a = -an^2 (sinnt)

do we need to derive them from the basic equations given on the formula sheet?
Note that I'm doing the catholic trial and my school strictly follows the criteria given by the CSSA.

Thanks

The second and third ones aren't always true for SHM, so you have to be careful! I wouldn't be using those directly. The first one however is definitely (to my understanding) okay to quote, it certainly was when I did my HSC

dermite · « **Reply #3596 on:** August 18, 2018, 11:02:36 am »

hi there, i need some help with this question. thanks!

RuiAce · « **Reply #3597 on:** August 18, 2018, 11:18:02 am »

Quote from: dermite on August 18, 2018, 11:02:36 am

hi there, i need some help with this question. thanks!

$\text{This particular one can be done with only algebra.}\\ \begin{align*}P(x)&= 2(x-3)^2(x+3)^2\\ &= 2[(x-3)(x+3)]^2\\ &= 2(x^2-9)^2\\ &= 2(x^4 - 18x^2 + 81)\end{align*}$
Therefore D.

dermite · « **Reply #3598 on:** August 18, 2018, 12:19:35 pm »

Quote from: RuiAce on August 18, 2018, 11:18:02 am

$\text{This particular one can be done with only algebra.}\\ \begin{align*}P(x)&= 2(x-3)^2(x+3)^2\\ &= 2[(x-3)(x+3)]^2\\ &= 2(x^2-9)^2\\ &= 2(x^4 - 18x^2 + 81)\end{align*}$
Therefore D.

is it the only way of doing this question?
since it is a mcq, could the process of elimination be used instead?

S200 · « **Reply #3599 on:** August 18, 2018, 12:31:09 pm »

Quote from: dermite on August 18, 2018, 12:19:35 pm

is it the only way of doing this question?
since it is a mcq, could the process of elimination be used instead?

If you were going to do elimination, wouldn't you need a function with an x^2 term in it anyway?
And to get that function you would have to have pretty much done Rui's way already?

Search the forums now!

We have moved!

Author Topic: 3U Maths Question Thread (Read 1236519 times) Tweet Share

fun_jirachi

Re: 3U Maths Question Thread

Abdul_k

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

key to success

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

key to success

Re: 3U Maths Question Thread

key to success

Re: 3U Maths Question Thread

jamonwindeyer

Re: 3U Maths Question Thread

key to success

Re: 3U Maths Question Thread

key to success

Re: 3U Maths Question Thread

jamonwindeyer

Re: 3U Maths Question Thread

dermite

Re: 3U Maths Question Thread

RuiAce

Re: 3U Maths Question Thread

dermite

Re: 3U Maths Question Thread

S200

Re: 3U Maths Question Thread

Recent Posts

User:
Password: