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Author Topic: 3U Maths Question Thread  (Read 1230382 times)  Share 

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jamonwindeyer

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Re: 3U Maths Question Thread
« Reply #165 on: May 25, 2016, 09:34:31 pm »
+1
You know I sat your MX1 paper right :P

I do, you mentioned it a little why ago I'm pretty sure (during that conversation when I mentioned that it was an easier paper than usual)  :o

amandali

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Re: 3U Maths Question Thread
« Reply #166 on: May 27, 2016, 05:22:34 pm »
0
h

how to do ques 2 thanks

katherine123

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Re: 3U Maths Question Thread
« Reply #167 on: May 27, 2016, 06:04:44 pm »
0
question: a die is rolled 100 times. What is the most likely number of sixes that will be thrown?
Ans: 16

am i supposed to use the greatest coefficient for these kind of ques cuz the answer i got was wrong using this method

nCr*(1/6)^r*(5/6)^(n-r) divided by  nC(r-1)*(1/6)^(r-1)*(5/6)^(n-r+1)

or am i supposed to do  (1/6)*100


jakesilove

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Re: 3U Maths Question Thread
« Reply #168 on: May 27, 2016, 11:17:47 pm »
+2
(Image removed from quote.)h

how to do ques 2 thanks

Hey!

This was a bit of a tricky question, and I actually think I formulated my logic wrong (I assumed that you can only form a parcel from the shirts made in a specific day, but then the maximum number of shirts that could be used from the sister is 20, so that doesn't quite work out). However, I did get to the right answer! I hate probability, but I hope this helps you out a bit :)



Jake
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jakesilove

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Re: 3U Maths Question Thread
« Reply #169 on: May 27, 2016, 11:29:33 pm »
+2
question: a die is rolled 100 times. What is the most likely number of sixes that will be thrown?
Ans: 16

am i supposed to use the greatest coefficient for these kind of ques cuz the answer i got was wrong using this method

nCr*(1/6)^r*(5/6)^(n-r) divided by  nC(r-1)*(1/6)^(r-1)*(5/6)^(n-r+1)

or am i supposed to do  (1/6)*100

Hey!

This is a really tough one, and you could really do it a number of ways. I would start by saying that, as a matter of probabilities, each value should come up 16 times (16*6=96). This is if the die was perfectly weighted. Then, we can take a quick look at the remaining four rolls. Basically, if it is more likely than not that a six comes up in 4 rolls, then it is more likely than not that a six will be rolled 17 times in 100 rolls. This calculation is just a simple binomial, and I'm not fabulous with LaTex so won't attempt it.

Overall, it will be 1 minus the probability of rolling NO sixes in four rolls. The problem is that, when I compute the probability of rolling a 6 in 4 rolls, I get 0.482. Therefore, 1-0.482=0.518, which is greater than 50%. By that calculation, you should be slightly more likely to roll 17 6s than 16 6s. Someone may have to point out a mistake that I made!

Jake
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FallonXay

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Re: 3U Maths Question Thread
« Reply #170 on: June 03, 2016, 03:55:59 pm »
0
Hii!
I'm having trouble with the following question, any help would be much appreciated  :)

Thanks!

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katherine123

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Re: 3U Maths Question Thread
« Reply #171 on: June 03, 2016, 04:22:14 pm »
0
1. particle has a=2-2x, initially v=4m/s ,  x=0
a)find v in terms of x

what i got was v=(16+4x-2x^2)^1/2   how do i know whether v is positive or negative
b) find the greatest v
do i normally just let a=0 for min/max velocity
what do i have to do if the ques asks for max/min acceleration

particle starts from the origin and has 
v=cos^2(x)
a= -2sin(x)*cos^3(x)
x= tan^-1(t)   inverse
where 0<=x<π
Describe the motion of particle from its initial position to its limiting position  (2 marks)

the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2
Im not sure what's required for questions that ask me to describe the motion
the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate   

jakesilove

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Re: 3U Maths Question Thread
« Reply #172 on: June 03, 2016, 04:59:12 pm »
+2
Hii!
I'm having trouble with the following question, any help would be much appreciated  :)

Thanks!

Hey!

I've attached my solution before. This is just a straightforward application of the derivative formula for inverse trigonometric functions: you need to be extremely comfortable with these kinds of questions!

Hope this helps :)



Jake
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jakesilove

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Re: 3U Maths Question Thread
« Reply #173 on: June 03, 2016, 05:06:14 pm »
+3
1. particle has a=2-2x, initially v=4m/s ,  x=0
a)find v in terms of x

what i got was v=(16+4x-2x^2)^1/2   how do i know whether v is positive or negative
b) find the greatest v
do i normally just let a=0 for min/max velocity
what do i have to do if the ques asks for max/min acceleration

particle starts from the origin and has 
v=cos^2(x)
a= -2sin(x)*cos^3(x)
x= tan^-1(t)   inverse
where 0<=x<π
Describe the motion of particle from its initial position to its limiting position  (2 marks)

the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2
Im not sure what's required for questions that ask me to describe the motion
the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate

Hey! Can't say I fully understand what you're actually asking, but I'll have a go.

I assume you're asking how to tell whether, when you square root v, you should make the equation positive or negative. Essentially, you just sub a point in and see which one works; in this case, you were right to choose the positive route as initially the particle is moving to the right (velocity is positive, so it is assumed to be moving right).

In terms of finding the maximum and minimum velocity/acceleration etc. for questions like this, you usually let the particle be at its endpoints or center. Unlike classical motion (ie, 2U stuff), it becomes increasingly difficult to just differentiate for t and find a standard max/min. Instead, we just have to remember that maximum velocity will occur when the particle is at the center of motion, and maximum acceleration will occur when the particle is at its endpoints (ie. either side of its motion).

Your second question is absolutely right; all that's being asked is a general description of displacement, velocity and acceleration! Plus, you did it in a succinct and mathematically accurate way :)

Jake
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Studying a combined Advanced Science/Law degree at UNSW

FallonXay

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Re: 3U Maths Question Thread
« Reply #174 on: June 03, 2016, 05:16:08 pm »
0
Hey!

I've attached my solution before. This is just a straightforward application of the derivative formula for inverse trigonometric functions: you need to be extremely comfortable with these kinds of questions!

Hope this helps :)

(Image removed from quote.)

Jake

ok, thanks. I understand this much, but i'm having trouble further simplifying it. The final answer provided in the textbook is -1/√ (9-x2). How did the 3 in the denominator cancel out?
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RuiAce

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Re: 3U Maths Question Thread
« Reply #175 on: June 03, 2016, 05:19:35 pm »
+2
ok, thanks. I understand this much, but i'm having trouble further simplifying it. The final answer provided in the textbook is -1/√ (9-x2). How did the 3 in the denominator cancel out?

By simply expanding it in.



« Last Edit: June 03, 2016, 05:22:05 pm by RuiAce »

FallonXay

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Re: 3U Maths Question Thread
« Reply #176 on: June 03, 2016, 05:23:12 pm »
0
By simply expanding it in.





ohhh, i see!!! Thank you so much!  ;D
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RuiAce

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Re: 3U Maths Question Thread
« Reply #177 on: June 03, 2016, 05:26:31 pm »
+3
1. particle has a=2-2x, initially v=4m/s ,  x=0
a)find v in terms of x

what i got was v=(16+4x-2x^2)^1/2   how do i know whether v is positive or negative
b) find the greatest v
do i normally just let a=0 for min/max velocity
what do i have to do if the ques asks for max/min acceleration

particle starts from the origin and has 
v=cos^2(x)
a= -2sin(x)*cos^3(x)
x= tan^-1(t)   inverse
where 0<=x<π
Describe the motion of particle from its initial position to its limiting position  (2 marks)

the graph i get for displacement vs time is an inverse tangent curve with range: 0<=y<π/2
Im not sure what's required for questions that ask me to describe the motion
the answer i wrote was: the particle is moving to the right of origin with a speed that is increasing at a decreasing rate

Your answer to the second question is, as Jake pointed out, correct. To make it clearer for you:




For the first question, this is a classic example of a question that I HATE because it gives you insufficient information to properly derive an expression for the sign of the velocity.

In your case, it is safest to just say that
= + sqrt(16+4x-2x2) if v≥0
= - sqrt(16+4x-2x2) if v≤0



« Last Edit: June 03, 2016, 05:35:07 pm by RuiAce »

amandali

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Re: 3U Maths Question Thread
« Reply #178 on: June 05, 2016, 08:58:38 am »
0
why is the answer 2/5

RuiAce

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Re: 3U Maths Question Thread
« Reply #179 on: June 05, 2016, 09:24:37 am »
+1
« Last Edit: June 05, 2016, 09:28:22 am by RuiAce »