May 22, 2019, 11:20:28 pm

1 Member and 3 Guests are viewing this topic.

#### RuiAce

• HSC Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8322
• "All models are wrong, but some are useful."
• Respect: +2228
##### Re: 4U Maths Question Thread
« Reply #2280 on: May 14, 2019, 11:04:31 am »
0
Hello,
I am confused with this question in the attachment below. I know how to find the area of the triangle (stated in my working out), but I don't know why Area (OLM) is independent of the point P. Can anyone please explain this thoroughly? Thanks
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.)

#### david.wang28

• MOTM: MAR 19
• Trendsetter
• Posts: 178
• Always do everything equanimously
• Respect: +5
##### Re: 4U Maths Question Thread
« Reply #2281 on: May 14, 2019, 11:10:28 am »
0
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)

Dream course(s):
Nursing (UTS)
Education (maths or physics) (USYD)
Engineering/or any form of science (UNSW or UTS)

#### david.wang28

• MOTM: MAR 19
• Trendsetter
• Posts: 178
• Always do everything equanimously
• Respect: +5
##### Re: 4U Maths Question Thread
« Reply #2282 on: May 21, 2019, 05:28:32 pm »
0
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):
Nursing (UTS)
Education (maths or physics) (USYD)
Engineering/or any form of science (UNSW or UTS)

#### esteban

• Fresh Poster
• Posts: 1
• Respect: 0
##### Re: 4U Maths Question Thread
« Reply #2283 on: 16 hours ago »
0
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

• Trailblazer
• Posts: 30
• “Sugar Peas!”
• Respect: +1
##### Re: 4U Maths Question Thread
« Reply #2284 on: 4 hours ago »
0
Currently stuck on these recurrence relation questions.
Any ideas?

Cheers!

#### RuiAce

• HSC Lecturer
• Honorary Moderator
• Great Wonder of ATAR Notes
• Posts: 8322
• "All models are wrong, but some are useful."
• Respect: +2228
##### Re: 4U Maths Question Thread
« Reply #2285 on: 4 hours ago »
+1
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

• Trailblazer
• Posts: 30
• “Sugar Peas!”
• Respect: +1
##### Re: 4U Maths Question Thread
« Reply #2286 on: 3 hours ago »
0
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

• MOTM: AUG 18
• Forum Obsessive
• Posts: 403
• All doom and Gloom.
• Respect: +241
##### Re: 4U Maths Question Thread
« Reply #2287 on: 3 hours ago »
0
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]
HSC 2019: Physics | Chemistry | English Advanced | Maths Extension 1 | Maths Extension 2

#### not a mystery mark

• Trailblazer
• Posts: 30
• “Sugar Peas!”
• Respect: +1
##### Re: 4U Maths Question Thread
« Reply #2288 on: 3 hours ago »
0
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.