4U Maths Question Thread

[vikasarkalgud]

OMG, thankyou so much for helping with the +C common denominator error I had. I wasn't going to sleep if I couldn't get that solution.

Still a little fuzzy on the +C idea though, a little embarrassing as a 4u student, but was the problem here that I was trying to multiply +C by a variable/unknown term? coz its okay to do if its just a constant like "2", or "-e^0.1" since C represents any constant anyway?

[RuiAce]

OMG, thankyou so much for helping with the +C common denominator error I had. I wasn't going to sleep if I couldn't get that solution.

Still a little fuzzy on the +C idea though, a little embarrassing as a 4u student, but was the problem here that I was trying to multiply +C by a variable/unknown term? coz its okay to do if its just a constant like "2", or "-e^0.1" since C represents any constant anyway?

You introduce a new constant, that depends on the old constant.

So say that your original constant was \(C\). Your new constant would be a new arbitrary constant, say \(C_0\) instead. The idea is that whilst \(C\) is arbitrary, you can't use the same \(C\) to represent two different things
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Basically what happened was that when you multiplied everything else by a term, namely the \(k(1-\alpha)\), you didn't multiply \(C\) to it at all. That's no longer mathematically correct.

[kaustubh.patel]

Hey rui need some help. I've been given this volumes question and also the answer but am really confused how it works. I can do the second part but i'm not sure what to do for the first.

[RuiAce]

Hey rui need some help. I've been given this volumes question and also the answer but am really confused how it works. I can do the second part but i'm not sure what to do for the first.

The thickness in your case is obviously just \( \delta x\).




The one thing to watch out for here is really just the height. Everything else is very routine in the method of cylindrical shells so if you have any problem with it you'll need to elaborate.

[clovvy]

I swear I see this equation all the time with implicit questions, but can't do it T_T

[RuiAce]

I swear I see this equation all the time with implicit questions, but can't do it T_T

This is a lot that you're throwing in all at once. What progress have you made on every part in question. Or alternatively would you like us to focus on certain bits.

There's no point in spoon-feeding every single bit.

[clovvy]

This is a lot that you're throwing in all at once. What progress have you made on every part in question. Or alternatively would you like us to focus on certain bits.
Only the first part of c I could do..  Without knowing the intercepts and all that I cannot progress with it (I need to know how to get it), once I can recognize critical points or intercepts than perhaps I can work out there rest (I wasn't being specific there sorry)
There's no point in spoon-feeding every single bit.

The only thing I could do is differentiate but I cannot recognize the intercepts or anything like that..  It is a weird equation to work with

[RuiAce]

For now, these are sketch solutions (which you can ask step by step for more details if necessary). Whilst you cannot tell the intercepts 'by inspection', they are very easy to find if we return to old techniques we were taught in Year 11. The question's difficulty really kicks in once you reach d) iii), with the very unusual deduction of a self intersecting curve.

Which certainly is not MX2 material. But this question wasn't meant to be usual MX2 difficulty anyway given that it was the last one in the Sydney Grammar book.

[mxrylyn]

Heyy

Just reading about the rule of LIATE for integration on my AN book.

It seems really useful, but I couldn't understand how to use it or weather it helps me pick the U or the V'

Edit:
After reading it a few more times I realised that it clearly says that the rule is to pick which one to differentiate, which is the U.

Sorry, & Thank you

[arii]

Harder 3U topic. Appeared in HSC 4-unit 1989 past paper.

I had a bit of trouble doing the questions from (c) onwards.

[vikasarkalgud]

Hey Rui, mechanics question - vertical motion

essentially, x = 1/2k (ln(g/(g-kv^2))
Terminal velocity, vT = sqrt(g/k)
I have to show distance particle must fall to reach velocity 1/2vT is x = 1/2k(ln(2/3))............I'm getting x = 1/2k(ln(4/3)) am I wrong?

[RuiAce]

Harder 3U topic. Appeared in HSC 4-unit 1989 past paper.

I had a bit of trouble doing the questions from (c) onwards.

The first few parts only require 2U concepts.


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(and is the main reason why it was in such an old paper, but would not be considered suitable exam content anymore.)



That it is tangential to \(PQ\) can be proven in a similar way, because the horizontal distance from the vertical line \(x=1\) to \(S(6,7)\) is also 5. The proper write-up is left as your exercise.

[RuiAce]

Hey Rui, mechanics question - vertical motion

essentially, x = 1/2k (ln(g/(g-kv^2))
Terminal velocity, vT = sqrt(g/k)
I have to show distance particle must fall to reach velocity 1/2vT is x = 1/2k(ln(2/3))............I'm getting x = 1/2k(ln(4/3)) am I wrong?

1/2k is interpreted to mean \( \frac{1}{2} k\). Did you mean 1/(2k)?

Although regardless, I am also getting 4/3 assuming you do need \(v = \frac12 \sqrt{\frac{g}{k}} \). In fact \( \ln \frac23 \) would be weird, because that'd imply displacement was allowed to be negative for a 4U resisted motion question.

[vikasarkalgud]

1/2k is interpreted to mean \( \frac{1}{2} k\). Did you mean 1/(2k)?

Although regardless, I am also getting 4/3 assuming you do need \(v = \frac12 \sqrt{\frac{g}{k}} \). In fact \( \ln \frac23 \) would be weird, because that'd imply displacement was allowed to be negative for a 4U resisted motion question.

thanks, that's all i needed, source I use is often incorrect in their answers, similar thing happened couple of weeks ago and you verified for me. Thankyou so much, saves me a lot of time that might've been wasted trying to fix a correct answer

[scienceislife]

Why is the domain of cis(theta) restricted to between -pi and pi?