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Author Topic: 3U Maths Question Thread  (Read 1230270 times)  Share 

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jakesilove

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Re: 3U Maths Question Thread
« Reply #330 on: July 17, 2016, 01:55:15 pm »
+1
dx/dt= 4, dy/dt=4t
dy/dx= dy/dt * dx/dt  = 4t * 1/4 = t --> t=3, dy/dx=3

Thanks High Tide! That's the other method I alluded to at the end of my response; clearly quicker if you are more comfortable with this methodology.
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RuiAce

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Re: 3U Maths Question Thread
« Reply #331 on: July 17, 2016, 02:16:54 pm »
+1
dx/dt= 4, dy/dt=4t
dy/dx= dy/dt * dx/dt  = 4t * 1/4 = t --> t=3, dy/dx=3


Thanks High Tide! That's the other method I alluded to at the end of my response; clearly quicker if you are more comfortable with this methodology.
Could've just asked. It does have a name you know :P
« Last Edit: July 17, 2016, 02:21:51 pm by RuiAce »

conic curve

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Re: 3U Maths Question Thread
« Reply #332 on: July 17, 2016, 09:16:06 pm »
0
Could someone here please help me with this question:

I've seen the working out for this question but I'm struggling to understand the theory behind it. It would be lovely if someone could solve it for me and explain to me the theory behind it (because that's what I'm struggling with)

Greatly appreciated

Thanks  ;D

RuiAce

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Re: 3U Maths Question Thread
« Reply #333 on: July 17, 2016, 09:40:48 pm »
+1
Could someone here please help me with this question:

I've seen the working out for this question but I'm struggling to understand the theory behind it. It would be lovely if someone could solve it for me and explain to me the theory behind it (because that's what I'm struggling with)

Greatly appreciated

Thanks  ;D


conic curve

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Re: 3U Maths Question Thread
« Reply #334 on: July 17, 2016, 09:50:46 pm »
0



Um, did you take the restriction into account because that's what confused me the most about this question?

RuiAce

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Re: 3U Maths Question Thread
« Reply #335 on: July 17, 2016, 10:04:32 pm »
+1
Um, did you take the restriction into account because that's what confused me the most about this question?
Mathematicians call these types of proofs "Proofs without loss of generality"

This means that the proof most certainly lacks rigour in that some assumptions had to be made to arrive at the conclusion. However, the idea is that because the proof is so similar to the other cases, making such an assumption does not damage the final result.

The Pythagorean identity assumes that both sine and cosine are positive simultaneously.
The triangle proof produces an even greater flaw in that it assumes we will always be in the first quadrant.
« Last Edit: July 17, 2016, 10:06:08 pm by RuiAce »

amandali

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Re: 3U Maths Question Thread
« Reply #336 on: July 19, 2016, 02:08:03 am »
0

how to do ques 4a) 
ans is 5:28pm to 9:12 pm

jakesilove

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Re: 3U Maths Question Thread
« Reply #337 on: July 19, 2016, 10:10:04 am »
+2
(Image removed from quote.)
how to do ques 4a) 
ans is 5:28pm to 9:12 pm

Hey! So we need to start out by recognising the general form of an equation like this.



Now, we spend some time filling in all the blanks. Firstly, we know that A is the amplitude, which will just be half the distance from high tide to low tide. So,



Now, we know that period is calculated by


so


Taking period in hours (and therefore time in hours). We solve this, and plug in what we have



Now, we can see that the center of motion is halfway between high and low tide. This is the value for c, so we can plug that in easily



To find b, we have to sub stuff in. It's up to you how you do this part; I think it's easiest to let t=0 at low tide (8.2m) and solve from there. You just need to remember that, at high tide, t doesn't equal 7:20pm, it will equal 6 hours of 15 minutes.





So, the overall equation is


From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way!

Jake
« Last Edit: July 19, 2016, 10:11:48 am by jakesilove »
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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #338 on: July 19, 2016, 10:13:27 am »
+2
(Image removed from quote.)
how to do ques 4a) 
ans is 5:28pm to 9:12 pm

Edit: Jake snuck in, but I did it differently, so I'll post this anyway. Both are correct!

Hey amandali! This one is tough because we have to form the function for simple harmonic motion ourselves. Let's do that first, and there are multiple ways. The amplitude is given by halving the difference between low tide and high tide, that is, 3.2 metres. The equilibrium position is given by the halfway point between those, meaning, 11.4 metres. The period can be calculated by considering the fact that it there is 6 hours and 15 minutes (375 minutes) between low and high tide, therefore, the period of the harmonic motion is double that, 750 minutes. Therefore;



We put all of that data together to get a function for height in terms of time, which as implied, is simple harmonic:



For this version, t=0 at 1:05pm. Note that I use the cosine function so that it can start at the extrema instead of the equilibrium as required. There are multiple ways to form this function! This is just my way  ;D

So we need times when the height is greater than 13.3 metres, so, let's solve the equality first.



We find that those (remember t=0 is 1:05pm) correspond to the answers you gave  ;D hope this helps!! The funny thing about this question is that there is no calculus, just testing how well you can form the simple harmonic motion expressions  ;D

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Re: 3U Maths Question Thread
« Reply #339 on: July 19, 2016, 10:13:29 am »
+2






Don't forget arccos just means cos-1


So, the overall equation is


From there, I'll leave it to you. All you need to do is set the height equal to 13.3 and find times t for which the relationship is true. Make sure to find two times. Those will be two 'hour' values, which you must add to the initial 1:05pm. Once you've done a bunch of questions like this, it becomes pretty easy, because you always answer it in the same way!

Jake

I thought you disappeared so I started doing the question anyway :P it was too late for me to realise that wait this one is actually a toughie. Anyway I worked with minutes so it's slightly different. I also didn't get to check my working out fully yet

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #340 on: July 19, 2016, 10:15:29 am »
+1






Don't forget arccos just means cos-1

I thought you disappeared so I started doing the question anyway :P it was too late for me to realise that wait this one is actually a toughie. Anyway I worked with minutes so it's slightly different. I also didn't get to check my working out fully yet

Your working is fine. And we seriously need a system...

RuiAce

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Re: 3U Maths Question Thread
« Reply #341 on: July 19, 2016, 10:16:58 am »
+1
Your working is fine. And we seriously need a system...
He messaged me about the qn haha I was intending to just step out of the way, but then it was as if he vanished into thin air

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #342 on: July 19, 2016, 10:21:04 am »
+1
He messaged me about the qn haha I was intending to just step out of the way, but then it was as if he vanished into thin air

I need to be included in these messages!  8)

amandali

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Re: 3U Maths Question Thread
« Reply #343 on: July 19, 2016, 01:09:07 pm »
0

i dont get ques 4ii) 
the ans is written below

jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #344 on: July 19, 2016, 02:08:21 pm »
+1
(Image removed from quote.)
i dont get ques 4ii) 
the ans is written below

Hey again!

This is a Bernoulli Trial, so we could consider the different ways it can happen using the Binomial distribution. However, we have to consider the fact that the number of trials changes based on the events to that point. That said, for me, it is easier to break it down into the 3 possibilities: 3 sets played, 4 sets played, or 5 sets played.

3 sets played is an easy one, Novak must win in straight sets:



For the rest, we need to adjust our thinking. If they are playing until three sets are won, and we are considering Novak winning, then that means Novak must win the last game.

So, for when we want to consider the probability of winning in 4 sets, we only need to consider a Bernoulli Trial with 3 sets, because we know the fourth will be won by Novak.



So we take the appropriate term from here, and multiply by 2/3 to consider the probability of Novak winning the 4th set. So it's binomial distribution for the first 3 sets, then just basic probability in the fourth, because we know Novak needs to win it:



This corresponds to the second term of your answer (I think there is a dictation error in the solution, because it out by a multiple of 2/3.

We do the same thing for winning in 5 sets; consider the binomial probability of Novak winning 2 sets out of 4 (since we don't know the order), and then multiplying by 2/3 to consider the probability of him winning in the 5th. This corresponds to the last term of your answer, and if you add everything together, you get the answer  ;D I hope that helps!!  ;D