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November 01, 2024, 10:53:53 am

Author Topic: 4U Maths Question Thread  (Read 708072 times)  Share 

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aadharmg

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Re: 4U Maths Question Thread
« Reply #1830 on: April 21, 2018, 06:26:45 pm »
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I'm trying to teach myself a bit of Volumes and in the first exercise I've kind of stumbled upon this question (highlighted in image), I took multiple routes in order to get to the answer but I think I went wrong somewhere each time in terms of my interpretation of the solid. There are no worked solutions/examples, hence I require your help. Thank You.

RuiAce

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Re: 4U Maths Question Thread
« Reply #1831 on: April 21, 2018, 07:28:50 pm »
+2
I'm trying to teach myself a bit of Volumes and in the first exercise I've kind of stumbled upon this question (highlighted in image), I took multiple routes in order to get to the answer but I think I went wrong somewhere each time in terms of my interpretation of the solid. There are no worked solutions/examples, hence I require your help. Thank You.

aadharmg

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Re: 4U Maths Question Thread
« Reply #1832 on: April 21, 2018, 07:47:39 pm »
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Thank you so much!

Dragomistress

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Re: 4U Maths Question Thread
« Reply #1833 on: April 28, 2018, 05:14:08 pm »
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For conics, is it worth trying to remember the many equations? Or should they all be derived on the spot?

Also, do people derive the chord of contact?

RuiAce

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Re: 4U Maths Question Thread
« Reply #1834 on: April 28, 2018, 05:28:39 pm »
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For conics, is it worth trying to remember the many equations? Or should they all be derived on the spot?

Also, do people derive the chord of contact?
The exam will either tell you to derive it on the spot (including the chord of contact, yes), or they'll just give you the formula.

vikasarkalgud

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Re: 4U Maths Question Thread
« Reply #1835 on: May 06, 2018, 01:03:50 pm »
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hey, need help for this mechanics questions. very close to the solution they provided but not exactly it. just need help with part d)

A submarine traavels underwater with a constant driving force of mF (where m is the mass of the submarine). Water pressure exerts a resistance to its motion of v^2 per unit mass ( v is velocity and v > 0)

a) Show that the equation of motion is d^2x/dt^2 = F - v^2 --------------------------Done
b) Show that the terminal velocity is sqrt(F)-------------------------------Done
c) The cruising speed is half the terminal velocity. How far does the submarine need to travel to reach its cruising speed? (Ans: 1/2ln(4/3))------Done
d) How long does it take to reach its cruising speed? (So close!!!!!!)

answer is t = 1/(2sqrt(F)ln3) -------------------------I'm getting ln3 on numerator!!!!!!!!!!!! so triggered

Would appreciate if someone able to show their method if they get the solution, so I can compare and see where I might have gone wrong. Ty!!!

RuiAce

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Re: 4U Maths Question Thread
« Reply #1836 on: May 06, 2018, 01:30:52 pm »
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hey, need help for this mechanics questions. very close to the solution they provided but not exactly it. just need help with part d)

A submarine traavels underwater with a constant driving force of mF (where m is the mass of the submarine). Water pressure exerts a resistance to its motion of v^2 per unit mass ( v is velocity and v > 0)

a) Show that the equation of motion is d^2x/dt^2 = F - v^2 --------------------------Done
b) Show that the terminal velocity is sqrt(F)-------------------------------Done
c) The cruising speed is half the terminal velocity. How far does the submarine need to travel to reach its cruising speed? (Ans: 1/2ln(4/3))------Done
d) How long does it take to reach its cruising speed? (So close!!!!!!)

answer is t = 1/(2sqrt(F)ln3) -------------------------I'm getting ln3 on numerator!!!!!!!!!!!! so triggered

Would appreciate if someone able to show their method if they get the solution, so I can compare and see where I might have gone wrong. Ty!!!
Your answer is definitely correct.

vikasarkalgud

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Re: 4U Maths Question Thread
« Reply #1837 on: May 06, 2018, 01:39:53 pm »
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TY SO MUCH. relieved.

aadharmg

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Re: 4U Maths Question Thread
« Reply #1838 on: May 08, 2018, 12:07:35 am »
+1
Hey, had an issue with a mechanics question. I'm probably missing a very obvious part, but I've tried many times and I keep leading myself to these weeeird integrals. Has to be much easier. Someone please help.

RuiAce

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Re: 4U Maths Question Thread
« Reply #1839 on: May 08, 2018, 07:51:12 am »
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Hey, had an issue with a mechanics question. I'm probably missing a very obvious part, but I've tried many times and I keep leading myself to these weeeird integrals. Has to be much easier. Someone please help.



kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #1840 on: May 10, 2018, 03:41:36 am »
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A bit of help with an integration que please.

jazzycab

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Re: 4U Maths Question Thread
« Reply #1841 on: May 10, 2018, 12:03:52 pm »
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A bit of help with an integration que please.

You can evaluate that integral using integration by parts:


I'm not sure what you're expected to know in the HSC 4U syllabus as far as limits are concerned.
We have an additional condition that I haven't used as yet. That is, \(s\) is a positive constant. Thus, as \(N\) approaches \(+\infty, sN>0\Rightarrow e^{sN}>0\); \(e^{sN}\) is increasing.
So when \(N\) is large, as \(N\) increases \(e^{sN}\) increases a lot faster than \(sN+1\) increases. Essentially this tells us that \(\lim_{N\rightarrow\infty}\left(\frac{sN+1}{e^{sN}}\right)=0\) (we could show this readily using L'Hopital's rule if it is in the course).
So we end up with:

kaustubh.patel

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Re: 4U Maths Question Thread
« Reply #1842 on: May 10, 2018, 03:57:45 pm »
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Dude Jazzycab you're actually a massive help, and i'm a bit weak with limits but i understood the denomitor is exponentially greater then numerator so the whole fraction approached 0. Thanks a lot man this is very nicely done.

vikasarkalgud

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Re: 4U Maths Question Thread
« Reply #1843 on: May 10, 2018, 07:50:27 pm »
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hey rui, another mechanics question,

A particle of unit mass moves in a straight line against a resistance kv^alpha (0<alpha<1), where v is the speed in m/s. It starts at a speed of 3000 m/s and this drops to 1500 m/s in one second. After a further 2 seconds the particle comes to rest. Find alpha, to 2 d.p (Answer: 0.42)

I attached where i got up to, giving about 0.86. I got 3 = (3000)^(1-alpha).Idk where i went wrong :(

RuiAce

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Re: 4U Maths Question Thread
« Reply #1844 on: May 10, 2018, 08:01:17 pm »
+3
hey rui, another mechanics question,

A particle of unit mass moves in a straight line against a resistance kv^alpha (0<alpha<1), where v is the speed in m/s. It starts at a speed of 3000 m/s and this drops to 1500 m/s in one second. After a further 2 seconds the particle comes to rest. Find alpha, to 2 d.p (Answer: 0.42)

I attached where i got up to, giving about 0.86. I got 3 = (3000)^(1-alpha).Idk where i went wrong :(

This transition wasn't justified. You multiplied out \(k (1-\alpha)\) but you didn't multiply that out to \(C\) as well.