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March 29, 2024, 08:09:09 am

Author Topic: Mathematics Question Thread  (Read 1296904 times)  Share 

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RuiAce

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Re: Mathematics Question Thread
« Reply #645 on: October 19, 2016, 11:41:33 pm »
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Hey guys, i was wondering if anyone could help give me iron out some stuff: e.g

locus and parabola for y^2=4ax, is the directrix y=-a and the focus (x+ or - a, y). For some reason i cant grasp the y^2 side despite knowing how the x^2 side works

This may be hard to explain but how would i graph velocity graphs ? I can work out the intercepts and roots, but i get stumped when wondering how to draw them and i dont necessarily have a question on hand

The sum and product of roots and the variations outside of alpha + beta, alpha beta, there is one of them which i cant exactly remember ( which i know isnt helpful in trying to get help for) but its a more 'complex' one that requires a bit more algebraic manipulation

Thanks
Recall that the parabola x^2=4ay has directrix y=-a and focus (0,a)

Hence, the parabola y^2=4ax has directrix x=-a and focus (a,0)
_________

Sorry but yes you will need to find a question. What you're saying isn't coming through. If you just meant derivative curve sketching that's somewhere in this thread already.
________


Deng

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Re: Mathematics Question Thread
« Reply #646 on: October 20, 2016, 12:02:55 am »
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This example isnt the best example because i know how to graph this one, essentially the motion questions where i have to graph the velocity as part of it, also for part V for questions like that when would i know i would have to integrate or i can use t = 0;x=2 , t=1;x=4 etc

Thanks again !
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RuiAce

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Re: Mathematics Question Thread
« Reply #647 on: October 20, 2016, 12:12:29 am »
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This example isnt the best example because i know how to graph this one, essentially the motion questions where i have to graph the velocity as part of it, also for part V for questions like that when would i know i would have to integrate or i can use t = 0;x=2 , t=1;x=4 etc

Thanks again !
Still not sure what the problem is. If you're given the equation then like you said, you basically know how to draw it.

You need to further clarify exactly what you mean by "you don't know how to sketch it". If you're given the actual equation of the velocity, you will virtually always know when and how to sketch it.

That one is basically sketching y=3x2-9x+6 except with y switched for v and x switched for t
_________________________

Part (v) can be made slightly easier with the answer to part (i). The idea is that the displacement takes into account direction, but the distance does not. So you need to watch out for whenever the particle turns around.

Of course, you CAN use integration for part (v). You just don't need to.
_________________________

Also, if that is a past HSC question, consider checking worked solutions.
« Last Edit: October 20, 2016, 12:15:35 am by RuiAce »

Deng

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Re: Mathematics Question Thread
« Reply #648 on: October 20, 2016, 12:42:18 am »
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For the equation (y-k)^2= 4a(x-h)^2
Would the focus be (k + a, h) and the directix y= k -a

So for the y^2 side would it essentially be the x value shifting up and down for the focus and the directix would be the x value minus the focal lenght?

Thanks
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RuiAce

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Re: Mathematics Question Thread
« Reply #649 on: October 20, 2016, 08:10:47 am »
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For the equation (y-k)^2= 4a(x-h)^2
Would the focus be (k + a, h) and the directix y= k -a

So for the y^2 side would it essentially be the x value shifting up and down for the focus and the directix would be the x value minus the focal lenght?

Thanks
Recall that for the parabola (x-h)^2 = 4a(y-k)
The vertex is at (h,k).
Since the parabola concaves UP, the focus is a units ABOVE the vertex, and the directrix is a units BELOW the vertex.
i.e. (h, k+a) and y=k-a

I never memorised these formulae because there was no point. I used common sense

So for the parabola (y-k)^2 = 4a(x-h)
The vertex is still at (h,k).
The parabola concaves to the RIGHT now. So the focus is a units TO THE RIGHT of the vertex, and the directrix is a units to the LEFT of the vertex
i.e. (h+a, k) and x=h-a

epherbertson

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Re: Mathematics Question Thread
« Reply #650 on: October 20, 2016, 09:19:21 am »
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This is a question from the 2005 paper that no one in my class has figured out yet so just looking for a bit of help to conquer this mess.

Thanks,
Emily  :-\

RuiAce

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Re: Mathematics Question Thread
« Reply #651 on: October 20, 2016, 10:47:53 am »
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This is a question from the 2005 paper that no one in my class has figured out yet so just looking for a bit of help to conquer this mess.

Thanks,
Emily  :-\
Already done in post #466

SimplyNikhil

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Re: Mathematics Question Thread
« Reply #652 on: October 20, 2016, 11:32:34 am »
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 ???

BPunjabi

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Re: Mathematics Question Thread
« Reply #653 on: October 20, 2016, 11:48:15 am »
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Could you please include each working step

http://pasteboard.co/h2n5xrD5Q.png

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Deng

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Re: Mathematics Question Thread
« Reply #654 on: October 20, 2016, 11:49:18 am »
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@SimplyNikhil ( not sure how to tag your post )

But to find the area it would be top curve - bottom curve
ln2x-lnx
By index laws that would become ln2
Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve

Hope i answered your question
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BPunjabi

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Re: Mathematics Question Thread
« Reply #655 on: October 20, 2016, 11:50:34 am »
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@SimplyNikhil ( not sure how to tag your post )

But to find the area it would be top curve - bottom curve
ln2x-lnx
By index laws that would become ln2
Since ln2 is a constant you can integrate it to xln2 and you just simply sub in b and a to solve

Hope i answered your question

Lol I was thinking that too but then I choked and was like what the fuck? do we integrate ln2 or divide now??
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Deng

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Re: Mathematics Question Thread
« Reply #656 on: October 20, 2016, 11:52:11 am »
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@BPunjabi

How i solve these ( which may not be correct, Rui may correct me after if im wrong)

Equation is Sin(2x+pi/3)

To find the where it crosses the x axis
let 2x+pi/3=0
2x = -pi/3
x = -pi/6

Therefore when we inspect A,B,C,D
A and B are eliminated
And between C and D , D is -pi/6

Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me
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Deng

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Re: Mathematics Question Thread
« Reply #657 on: October 20, 2016, 11:56:03 am »
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Hey guys, i was wondering on how i would identify the second derivative for this q

The first derivative would be f'(a) < 0 since the gradient is negative
I just cant picture in my head what the second derivate would look like ( some reason i picture a horizontal line since f(x) looks like a parabola to me
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RuiAce

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Re: Mathematics Question Thread
« Reply #658 on: October 20, 2016, 12:00:15 pm »
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???
This is partly wrong. Please wait for a correction. This question is actually much harder than what it looks like.


\
Everything that's wrong starts from about here. But I'm not going to delete this post. I'm going to KEEP this mistake here, but also post up a new solution.



« Last Edit: October 20, 2016, 12:08:35 pm by jamonwindeyer »

BPunjabi

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Re: Mathematics Question Thread
« Reply #659 on: October 20, 2016, 12:05:41 pm »
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@BPunjabi

How i solve these ( which may not be correct, Rui may correct me after if im wrong)

Equation is Sin(2x+pi/3)

To find the where it crosses the x axis
let 2x+pi/3=0
2x = -pi/3
x = -pi/6

Therefore when we inspect A,B,C,D
A and B are eliminated
And between C and D , D is -pi/6

Again, not sure if this is the best way to do it but that is how ive always done it and it has worked for me
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