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September 16, 2019, 01:06:45 pm

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#### david.wang28

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##### Re: 4U Maths Question Thread
« Reply #2280 on: May 14, 2019, 11:10:28 am »
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The point $P$ depends entirely on its parameter $t$. Since your expression for the area does not have this parameter appearing, i.e. it is independent of whatever $t$ is, equivalently speaking it is independent of the point $P$.

(Note that $c$ is assumed to be a constant, not some parameter that is allowed to vary.)
Ahhh, I get it now. I didn't quite get this concept since the book does not clearly mention the parameter stuff with that question. Thanks Rui!
HSC 2019: English Advanced(77) (Forgot everything), Chemistry, Physics, Maths Extension 1(35) (repeating), Maths Extension 2, Business Studies(80) (screw this)

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#### david.wang28

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##### Re: 4U Maths Question Thread
« Reply #2281 on: May 21, 2019, 05:28:32 pm »
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Hello,
I am stuck on Q22 in the link below. I have done some of the question, but I don't get why PRQS is a rhombus unless t^2 = 1 (see attachment). Can anyone please help me out with the remaining part of the question? Thanks
HSC 2019: English Advanced(77) (Forgot everything), Chemistry, Physics, Maths Extension 1(35) (repeating), Maths Extension 2, Business Studies(80) (screw this)

Dream course(s):
Science/Nursing (USYD)
Education (maths or physics) (USYD)
Engineering (flexible) (UNSW or UTS)

#### esteban

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##### Re: 4U Maths Question Thread
« Reply #2282 on: May 22, 2019, 06:53:48 am »
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The midpoint of RS and  the midpoint of PQ are both (ct,c/t). If the diagonals of a quadrilateral are perpendicular bisectors of each other, then that quadrilateral is a rhombus.

#### not a mystery mark

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##### Re: 4U Maths Question Thread
« Reply #2283 on: May 22, 2019, 06:47:59 pm »
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Currently stuck on these recurrence relation questions.
Any ideas?

Cheers!

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2284 on: May 22, 2019, 07:00:11 pm »
+3
Currently stuck on these recurrence relation questions.
Any ideas?

Cheers!
Remember to also use the add/subtract or multiply/divide same thing trick for recurrence relations, not just integration by parts. The first question is relatively similar to what I did in the April lecture.
\begin{align*}I_n &= \int_0^1 (1-x^r)^n \,dx\\ &= \left[x (1-x^r)^n \right]_0^1 - \int_0^1 x\cdot -rx^{r-1} n (1-x^r)^{n-1}\,dx\\ &= - nr \int_0^1 -x^r(1-x^r)^{n-1} \\ &= -nr \int_0^1 \left[ (1-x^r) - 1 \right] (1-x^r)^{n-1}\\ &= -nr \int_0^1 (1-x^r)^n - (1-x^r)^{n-1}\,dx\\ &= -nr \left( I_n - I_{n-1} \right)\\ \therefore (nr+1) I_n &= nr I_{n-1}\end{align*}

#### not a mystery mark

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##### Re: 4U Maths Question Thread
« Reply #2285 on: May 22, 2019, 07:41:03 pm »
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Remember to also use the add/subtract or multiply/divide same thing trick for recurrence relations, not just integration by parts. The first question is relatively similar to what I did in the April lecture.
\begin{align*}\therefore (nr+1) I_n &= nr I_{n-1}\end{align*}

Oh my god, you're an absolute weapon. This logic makes sense. I knew I shouldn't have missed the April lectures.

Just a few questions, where did the [x(1-x^r)^n] bit go? and, How do you know when to use the add/subtract trick?

Thank you so much!!

#### fun_jirachi

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##### Re: 4U Maths Question Thread
« Reply #2286 on: May 22, 2019, 07:52:45 pm »
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The [x(1-x^r)^n] bit equates to zero when subbing in the boundaries, so it basically just disappears

Add/subtract trick comes down mostly to intuition. In this case, seeing that you have an nr out the front of the integral and xr just hanging around, with part of In-1 should tell you that you should manipulate this in some shape or form to the original, especially since you have an nr+1 coefficient for In. Basically, you're looking for an nr x (In-1 - In) to manipulate the integral to find the result. A good way of thinking about it is that if you have a result, think about what you're working towards and think about how you might get to that result. It really just comes down to practice and intuition
Failing everything, but I'm still Flareon up.

HSC 2018: Modern History [88] | 2U Maths [98]
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#### not a mystery mark

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##### Re: 4U Maths Question Thread
« Reply #2287 on: May 22, 2019, 07:57:12 pm »
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The [x(1-x^r)^n] bit equates to zero when subbing in the boundaries, so it basically just disappears

Ahh cheers, man! Can't wait for you to state rank this course haha. Also an absolute weapon.

#### louisaaa01

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##### Re: 4U Maths Question Thread
« Reply #2288 on: May 24, 2019, 06:40:36 pm »
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#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2289 on: May 24, 2019, 07:02:19 pm »
+3
What attempts have you made so far at this problem? And what is the answer we are aiming for?

#### louisaaa01

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##### Re: 4U Maths Question Thread
« Reply #2290 on: May 24, 2019, 07:24:19 pm »
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The answer aims for: α = tan-12

So far, I've considered similar triangles in Solid A, discovering that the area of the hollowed square = a2-x2

However, I'm slightly confused as to how to find the area of the slice in Solid B- if you could please help with this, that would be much appreciated!

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2291 on: May 24, 2019, 07:31:26 pm »
+3

I considered the side view briefly but realised it was getting me nowhere. The top view became the nicer one apparently, because you can draw the radii of the circle wherever you feel is convenient.

#### not a mystery mark

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##### Re: 4U Maths Question Thread
« Reply #2292 on: May 29, 2019, 11:21:13 am »
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Heyy, back with a volume question. Still trying to get my head around cylindrical shell since we haven't done it in class yet.

The area bounded by the curve y = e^(x^(2)), the lines x=1 and y=1 is rotated about the y-axis. Use the method of cylindrical shells to calculate the volume of the solid of revolution formed.

I got the answer pi(0.5e^2 -e-1) but I am pretty sure that isn't right since it returns a negative value...

#### RuiAce

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##### Re: 4U Maths Question Thread
« Reply #2293 on: May 29, 2019, 12:02:16 pm »
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Heyy, back with a volume question. Still trying to get my head around cylindrical shell since we haven't done it in class yet.

The area bounded by the curve y = e^(x^(2)), the lines x=1 and y=1 is rotated about the y-axis. Use the method of cylindrical shells to calculate the volume of the solid of revolution formed.

I got the answer pi(0.5e^2 -e-1) but I am pretty sure that isn't right since it returns a negative value...
You should've arrived at $\delta V =2\pi x\left(e^{x^2} - 1 \right)\delta x$ and $V = 2\pi \int_0^1 xe^{x^2} - x\,dx$. (If you got that then you probably made an accident with the actual integration)