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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: brightsky on August 25, 2010, 08:23:35 pm

Title: brightsky's Noob-tastic Question Thread
Post by: brightsky on August 25, 2010, 08:23:35 pm: The area of the region enclosed by the graphs of $y = x^2 (x - 1)$ and $y = -x^3 + x$ . Somehow I keep getting 0 but it's obviously not...:(
Title: Re: brightsky's Noob-tastic Question Thread
Post by: Russ on August 25, 2010, 08:29:28 pm: I can't graph it because I can't find my TI84, but if you're using a calculator it may be summing it and getting 0 that way. Try doing the sections individually?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: andy456 on August 25, 2010, 08:48:49 pm: $(\int_{-0.5}^{0}x^2(x-1)dx-\int_{-0.5}^{0}-x^3+x dx)+(\int_{0}^{1}-x^3+x dx-\int_{0}^{1}x^2(x-1)dx)$

$= \frac{37}{96}$
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on August 25, 2010, 09:14:14 pm: Thanks andy. :)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on September 25, 2010, 12:10:12 pm: Consider $f:\mathbb{R} \backslash \{0\} \to \mathbb{R}$ where $f(x) = \log_e|x|$ and $g:\left(-\infty, -\frac{5}{2}\right) \to \mathbb{R}$ where $g(x) = 2x + 5$ . Find the domain for $f(g(x))$ .

My working is that $-\infty < g(x) < 0$ , and $g(x) \neq 0$ , so $x \neq -\frac{5}{2}$ . However, $f(g(x)) = \log_e|2x + 5|$ , so $x$ remains in $\left(-\infty, -\frac{5}{2}\right)$ .

But the answer given is $x \in (-\infty, 0) \backslash \{-\frac{5}{2}\}$ .

Any help appreciated.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: m@tty on September 25, 2010, 12:30:09 pm: Maximal domain of f is $\mathbb{R}\setminus\{0\}$

Range of g is $(-\infty, 0)$

Need to restrict domain of g such that it's range is a subset of the domain of f.

But it already is. So, as you said, domain stays as $\left(-\infty,-\frac{5}{2}\right)$ .

I don't know why they have given that answer, they have extended the domain past the initial constraints put on g... ??
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on September 25, 2010, 12:39:32 pm: Thanks m@tty! Just for reference, the question is MAV 2006 Exam 1 Question 4 a.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: stonecold on September 25, 2010, 06:58:33 pm: Yeah, I don't know what they've done there!
Title: Re: brightsky's Noob-tastic Question Thread
Post by: 8039 on September 25, 2010, 08:52:30 pm: Quote from: brightsky on August 25, 2010, 08:23:35 pm
The area of the region enclosed by the graphs of $y = x^2 (x - 1)$ and $y = -x^3 + x$ . Somehow I keep getting 0 but it's obviously not...:(

First find the points of intersection by equaling the graphs to eachother, -0.5, 0 and 1. It would help to do a rough sketch of both graphs... then you could see that for the -0.5 to 0 $y = x^2 (x - 1)$ is on top, then from 0 to 1 is on the bottom of the other graph. I'm sure you can figure out the rest haha
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on October 09, 2010, 11:21:51 pm: How do you find a write $\sin(5x)$ in terms of $\sin(x)$ and/or $\cos(x)$ ? My teacher said there is a way using De Moivre's theorem, but I don't know how...
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on October 09, 2010, 11:24:42 pm: Haha, here's some inspiration: http://vcenotes.com/forum/index.php/topic,12907.msg144763.html#msg144763

What you are finding should follow the same train of thought (almost the same!)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on October 09, 2010, 11:41:35 pm: Oh wait...me so stupid..

$\sin(5x) = \sin(x + 4x) = \sin(x)\cos(4x) + \sin(4x)\cos(x) = \sin(x) (\cos^2(2x) - \sin^2(2x)) + (2\sin(2x)\cos(2x)) \cos(x) = \sin(x) [(\cos^2(x) - \sin^2(x))]^2 - (2\sin(x)\cos(x))^2 + (4\sin(x)\cos(x))(\cos^2(x) - \sin^2(x)) \cos(x)$ then expand...

I'm still skeptical about this as I'm sure there is a more elegant solution that can be used for all simplifications of $\sin(nx)$ . My teacher said to work with $z^5$ but not sure how.

Quote from: TrueTears on October 09, 2010, 11:24:42 pm
Haha, here's some inspiration: http://vcenotes.com/forum/index.php/topic,12907.msg144763.html#msg144763

What you are finding should follow the same train of thought (almost the same!)

Thanks for that TT! :) I'm not surehow effective Euler's formula would be though, or else there is some crazy algebra work?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on October 09, 2010, 11:56:54 pm: You can keep repeatedly using it until you are happy to stop.

If you are interpreting it to be simplified to strictly sin(x) and cos(x) you should be able to still do it by repeatedly simplifying with Eulers, I haven't been bothered to try this but see where it leads you.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on October 10, 2010, 12:12:52 am: Quote from: brightsky on October 09, 2010, 11:41:35 pm
I'm sure there is a more elegant solution that can be used for all simplifications of $\sin(nx)$ .
Yup, I'll let you do the working but i came up with this idea, obtain the cos(nx) and sin(nx) from (cos(x)+isin(x))^n using Binomial expansion, see what you get :D
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on October 10, 2010, 12:24:39 am: Ah, smart.... ;D

Hmm..any hints on what I do with the cos(nx)?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: Linkage1992 on October 10, 2010, 01:28:10 am: yeah i think the answers wrong, because that's what I got too...
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on October 10, 2010, 01:30:16 am: It's quite a misleading question, but defying all initial intuition, Jess is taking without replacement (as kamil suggested in the other post). Taking that into account, you have the required answer.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on January 18, 2011, 08:47:10 pm: I was just like to know, is there an efficient way to approach questions that ask you to find the intersection between two circles? It seems really draining and error-prone to use the substitution method. :/
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on February 11, 2011, 10:34:07 pm: Find the range of $y = \frac{2x - 5}{x^2 + 6}$ . My approach was:

$y(x^2 + 6) = 2x - 5$

$x^2y + 6y = 2x - 5$

$yx^2 - 2x + (6y + 5) = 0$

We want $\Delta = 4 - 4y(6y+5) \geq 0$ , which yields $-1 \leq y \leq \frac{1}{6}$ .

This method seems a bit dodgy. Just like to know, is this usually how one would approach these questions, or is there a better more generic way of doing it?

Thanks.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: dooodyo on February 11, 2011, 10:41:12 pm: Hey Brightsky,

How did you get -1 < k < (1/6) as the range?

I thought that you could just resolve the equation into partial
fractions and then simply just find the range from the new
equation?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on February 11, 2011, 10:44:55 pm: Quote from: dooodyo on February 11, 2011, 10:41:12 pm
Hey Brightsky,

How did you get -1 < k < (1/6) as the range?

I thought that you could just resolve the equation into partial
fractions and then simply just find the range ?

Nah, I don't think you can simplify it into partial fractions. I got that by multiplying both sides by (x^2 + 6) and then solving the resulting quadratic to find the restrictions for y. But seriously, it's not an obvious method, and it just seems very obscure. I'm sure there is a better way to do these types of questions (a general idea that can be applied to all 'range' finding).
Title: Re: brightsky's Noob-tastic Question Thread
Post by: kamil9876 on February 11, 2011, 11:31:29 pm: I think that's a good solution. There is no real way for ALL functions but I will show you an approach that works for differentiable functions where the domain is a CLOSED interval $[a,b]$ . Just find the minimum and maximum using differentiation (don't forget to compare them with the endpoints) and then the range is simply [m,M] since your function is continous (google intermediate value theorem and extreme value theorem if you want). Now this approach doesn't generalise nicely to an open interval or to the whole or $\mathbb{R}$ (like in your case) or even worse it is crap for something like $1/x$ . But what you can do if the domain is whole of $\mathbb{R}$ is the following if your lucky: if the limits as x goes to infinity or -infinity are in $[m,M]$ (m is the smallest local minimum, M is the largest local maximum that you get from differentiating(if they exist)) then you can conclude that the range is indeed $[m,M]$ .

This should work for your example as the limits are 0 and should be inside the [m,M] that you get from differentiating (note that $[m,M]$ is always a subset of the range if the function is differentiable over an interval (no gaps like 1/x which has gap at x=0))
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 13, 2011, 02:40:59 pm: How to find the domain and range of:

$\sqrt{x +3} + \frac{x}{\sqrt{2x - 1}}$

More specifically in regards to the domain,

I got $x + 3 \ge 0 \implies x \ge -3$ and $2x-1>0 \implies x > \frac{1}{2}$ and solve simultaneously, but if you take the $\sqrt{x+3}$ up into the denominator, i.e.

$\frac{\sqrt{x+3}\sqrt{2x-1} + x}{\sqrt{2x-1}}$ , letting $(x+3)(2x-1) \ge 0$ gives you different solutions...am I missing something?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on March 13, 2011, 02:55:04 pm: I suck at these questions, but:
$Let \ f(x) = \sqrt{x +3} \ and \ g(x)=\frac{x}{\sqrt{2x - 1}}$
$Domf \ \cap \ Domg$
$[-3, \infty ) \ \cap \ ( \frac{1}{2}, \infty)$
$Dom=( \frac{1}{2}, \infty)$

I think for domains/ranges, you always should use the original function and not rearrange it

EDIT: stupid error fixed
Title: Re: brightsky's Noob-tastic Question Thread
Post by: xZero on March 13, 2011, 03:01:53 pm: Quote from: Rohitpi on March 13, 2011, 02:55:04 pm
I suck at these questions, but:
$Let \ f(x) = \sqrt{x +3} \ and \ g(x)=\frac{x}{\sqrt{2x - 1}}$
$Domf \ U \ Domg$
$[3, \infty ) \ U \ ( \frac{1}{2}, \infty)$
$Dom = [3, \infty )$

I think for domains/ranges, you always should use the original function and not rearrange it

shouldnt it be $[-3,\infty) \ U \ (\frac{1}{2},\infty)$

if we combine it,
$\frac{\sqrt{x+3}\sqrt{2x-1} + x}{\sqrt{2x-1}}$
you'll get $(x+3)(2x-1)\ge 0$
$x\ge -3$ and $x\ge \frac{1}{2}$

now if both brackets are negative, you can still get $\ge 0$ but if we consider the denominator, $(2x-1)> 0$ , hence both brackets must be positve. So we change it to $x>\frac{1}{2}$ , which is the same as the answer we get before we combine it into a fraction
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 13, 2011, 03:05:01 pm: Thanks for that, yeah that's what I got, but I was a bit skeptical about $(x+3)(2x-1) \ge 0$ giving two solutions. I think I'm getting confused over something trivial/obvious, but my brain isn't working...:/

Quote from: Rohitpi on March 13, 2011, 02:55:04 pm

I think for domains/ranges, you always should use the original function and not rearrange it

What if the original question was to find the domain/range of $\frac{ \sqrt{(x+3)(2x-1)} + x}{\sqrt{2x-1}}$ ?

Also any help on the range?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 13, 2011, 03:10:19 pm: Quote from: xZero on March 13, 2011, 03:01:53 pm
Quote from: Rohitpi on March 13, 2011, 02:55:04 pm
I suck at these questions, but:
$Let \ f(x) = \sqrt{x +3} \ and \ g(x)=\frac{x}{\sqrt{2x - 1}}$
$Domf \ U \ Domg$
$[3, \infty ) \ U \ ( \frac{1}{2}, \infty)$
$Dom = [3, \infty )$

I think for domains/ranges, you always should use the original function and not rearrange it

shouldnt it be $[-3,\infty) \ U \ (\frac{1}{2},\infty)$

if we combine it,
$\frac{\sqrt{x+3}\sqrt{2x-1} + x}{\sqrt{2x-1}}$
you'll get $(x+3)(2x-1)\ge 0$
$x\ge -3$ and $x\ge \frac{1}{2}$

now if both brackets are negative, you can still get $\ge 0$ but if we consider the denominator, $(2x-1)> 0$ , hence both brackets must be positve. So we change it to $x>\frac{1}{2}$ , which is the same as the answer we get before we combine it into a fraction

The quadratic inequation gives $x \le -3$ and $x \ge \frac{1}{2}$ .
Title: Re: brightsky's Noob-tastic Question Thread
Post by: xZero on March 13, 2011, 03:14:30 pm: really?
$(x+3)(2x-1)\ge 0$

$(x+3)\ge 0$ and $(2x-1)\ge 0$

$x\ge -3$ and $x\ge \frac{1}{2}$

now we can consider when they are both negative since 2 negatives will become positive, hence greater than 0. however (2x-1) must be greater than 0 so in this case we can only consider when both brackets are positve, hence the answer
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on March 13, 2011, 03:19:58 pm: Quote from: xZero on March 13, 2011, 03:14:30 pm
really?
$(x+3)(2x-1)\ge 0$

$(x+3)\ge 0$ and $(2x-1)\ge 0$

$x\ge -3$ and $x\ge \frac{1}{2}$

I thought you were not able to solve quadratic inequations like that. I though a mental sketch graph was necessary... Which would make $x \le -3$ and $x \ge \frac{1}{2}$ correct.

EDIT: Wolfram Alpha agrees too
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 13, 2011, 03:20:16 pm: Quote from: xZero on March 13, 2011, 03:14:30 pm
really?
$(x+3)(2x-1)\ge 0$

$(x+3)\ge 0$ and $(2x-1)\ge 0$

$x\ge -3$ and $x\ge \frac{1}{2}$

now we can consider when they are both negative since 2 negatives will become positive, hence greater than 0. however (2x-1) must be greater than 0 so in this case we can only consider when both brackets are positve, hence the answer

Yeah, but we must consider the double negative case as well. The case you just gave is basically just one solution, i.e.g $x \ge \frac{1}{2}$ . The other one is redundant.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on March 13, 2011, 03:23:09 pm: Quote from: brightsky on March 13, 2011, 02:40:59 pm
But if you take the $\sqrt{x+3}$ up into the denominator, i.e.

$\frac{\sqrt{x+3}\sqrt{2x-1} + x}{\sqrt{2x-1}}$ , letting $(x+3)(2x-1) \ge 0$ gives you different solutions...am I missing something?

Then the domain of the numerator still has to intersect with the domain of the denominator, cancelling one of the solutions of the numerator
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 13, 2011, 03:26:38 pm: Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on March 13, 2011, 03:27:40 pm: Quote from: brightsky on March 13, 2011, 03:26:38 pm
Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$

Yes, I think thats how to do it as a combined fraction

Remember for all sum/difference/product functions:
$Domf \ \cap \ Domg$
Title: Re: brightsky's Noob-tastic Question Thread
Post by: xZero on March 13, 2011, 03:30:32 pm: Quote from: brightsky on March 13, 2011, 03:26:38 pm
Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$

thats basically what i did before oO
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 13, 2011, 03:31:56 pm: Quote from: Rohitpi on March 13, 2011, 03:27:40 pm
Quote from: brightsky on March 13, 2011, 03:26:38 pm
Hmmm...so:

$2x-1 > 0$ ....[1]

$(x+3)(2x-1) \ge 0$ ....[2]

From [1], $x > \frac{1}{2}$

From [2], $x \le -3$ , $x \ge \frac{1}{2}$

Solving simultaneously, we only get $x > \frac{1}{2}$

Yes, I think thats how to do it as a combined fraction

Remember for all sum/difference/product functions:
$Domf \ \cap \ Domg$

Omg..haha I just realised x >= -3 was redundant from my first post. aiya >< Thanks Rohitpi and ZeroX!
Title: Re: brightsky's Noob-tastic Question Thread
Post by: xZero on March 13, 2011, 03:34:08 pm: Quote from: Rohitpi on March 13, 2011, 03:19:58 pm
Quote from: xZero on March 13, 2011, 03:14:30 pm
really?
$(x+3)(2x-1)\ge 0$

$(x+3)\ge 0$ and $(2x-1)\ge 0$

$x\ge -3$ and $x\ge \frac{1}{2}$

I thought you were not able to solve quadratic inequations like that. I though a mental sketch graph was necessary... Which would make $x \le -3$ and $x \ge \frac{1}{2}$ correct.

EDIT: Wolfram Alpha agrees too

well you're correct if the denominator doesn't exist, however in this question $x\le -3$ was not valid as both brackets must be positive, thats why i skipped that step
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on March 13, 2011, 03:36:56 pm: Oh, sorry for doubting it then! :)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: VCE247 on March 14, 2011, 07:35:21 pm: Brightsky is this methods 3+4 O_O
Title: Re: brightsky's Noob-tastic Question Thread
Post by: nacho on March 14, 2011, 07:43:15 pm: Quote from: VCE247 on March 14, 2011, 07:35:21 pm
Brightsky is this methods 3+4 O_O
yea,
brightsky is Derreck Ha's successor.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: m@tty on March 14, 2011, 08:06:13 pm: Quote from: nacho on March 14, 2011, 07:43:15 pm
Quote from: VCE247 on March 14, 2011, 07:35:21 pm
Brightsky is this methods 3+4 O_O
yea,
brightsky is Derreck Ha's successor.

I believe it goes like this:

Derrick Ha < James Lu < brightsky ;) haha
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on March 14, 2011, 08:09:30 pm: Quote from: m@tty on March 14, 2011, 08:06:13 pm
I believe it goes like this:

Derrick Ha < James Lu < brightsky ;) haha

+1, I think thats about right
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 14, 2011, 08:38:22 pm: lolwut?! I think putting thushan in the place of that guy called "brightsky" would be more applicable. This is the height of overestimation.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: vea on March 14, 2011, 08:43:38 pm: Quote from: brightsky on March 14, 2011, 08:38:22 pm
lolwut?! I think putting thushan in the place of that guy called "brightsky" would be more applicable. This is the height of overestimation.

True, brightsky completes his VCE in 2013...
So: Derrick Ha<James Lu<thushan<brightsky
Title: Re: brightsky's Noob-tastic Question Thread
Post by: iNerd on March 14, 2011, 08:44:30 pm: Quote from: vea on March 14, 2011, 08:43:38 pm
Quote from: brightsky on March 14, 2011, 08:38:22 pm
lolwut?! I think putting thushan in the place of that guy called "brightsky" would be more applicable. This is the height of overestimation.

True, brightsky completes his VCE in 2013...
So: Derrick Ha<James Lu<thushan<brightsky
;D.

Sorry brightsky but that was extremely smart.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on March 15, 2011, 08:31:12 pm: Can someone clarify, does R^+ / R^- include 0 or not? I remember reading something about this on VN but I can't find it. :/
Title: Re: brightsky's Noob-tastic Question Thread
Post by: iNerd on March 15, 2011, 08:31:56 pm: Quote from: brightsky on March 15, 2011, 08:31:12 pm
Can someone clarify, does R^+ / R^- include 0 or not? I remember reading something about this on VN but I can't find it. :/
From my knowledge it's neither. Don't go on me though -.-
Title: Re: brightsky's Noob-tastic Question Thread
Post by: luken93 on March 15, 2011, 08:36:32 pm: R+ = (0, infinity)
R- = (-infinity, 0)
R+ u R- = R\{0}
R = (-infinity,0) u {0} u (0, infinity)
Source: wolfram
Title: Re: brightsky's Noob-tastic Question Thread
Post by: iNerd on March 15, 2011, 08:39:28 pm: Quote from: luken93 on March 15, 2011, 08:36:32 pm
R+ = (0, infinity)
R- = (-infinity, 0)
R+ u R- = R\{0}
R = (-infinity,0) u {0} u (0, infinity)
Source: wolfram
I'm confused? Doesn't circle bracket mean not included? So was I right? R+ or R-, neither includes 0?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: luken93 on March 15, 2011, 08:42:16 pm: Quote from: ATAR on March 15, 2011, 08:39:28 pm
Quote from: luken93 on March 15, 2011, 08:36:32 pm
R+ = (0, infinity)
R- = (-infinity, 0)
R+ u R- = R\{0}
R = (-infinity,0) u {0} u (0, infinity)
Source: wolfram
I'm confused? Doesn't circle bracket mean not included? So was I right? R+ or R-, neither includes 0?
correct
Title: Re: brightsky's Noob-tastic Question Thread
Post by: burbs on March 15, 2011, 08:44:46 pm: yes.

R does not specifically have 0 which is why luken has done U{0}

if a graph has a domain of like [0,infinity) you could say R+ U{0}
Title: Re: brightsky's Noob-tastic Question Thread
Post by: m@tty on March 15, 2011, 08:53:14 pm: It depends, sometimes $\mathbb{R}^+$ is taken as non-negative real numbers, while at other times it means positives only.

I would say when answering questions take it as the positive reals and if needed go $\mathbb{R}^+ \cup \{0\}$ just to be unambiguous.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on April 07, 2011, 05:14:39 pm: Quote from: burbs on March 15, 2011, 08:44:46 pm
yes.

R does not specifically have 0 which is why luken has done U{0}

if a graph has a domain of like [0,infinity) you could say R+ U{0}
$\mathbb{R}$ is defined to contain 0.

http://en.wikipedia.org/wiki/Real_number
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on April 18, 2011, 09:30:17 pm: Just another question, how does one reflect any graph about a certain line ax + by + c = 0?
So for instance, if I wanted to reflect the curve y = sqrt(x) about the line y = 1/2 x + 5, what would the equation of the resulting graph be?
Thanks. :)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: xZero on April 18, 2011, 09:50:50 pm: We did something similar in my maths unit, basically you rotate the line y = 1/2 x + 5 so it's parallel to the x-axis, do the same rotation with the line ax + by + c = 0, reflect it by the rotated line which is parallel to the x-axis (easy) then rotate it back. Not sure if theres a easier method
Title: Re: brightsky's Noob-tastic Question Thread
Post by: moekamo on April 18, 2011, 10:16:20 pm: when rotating(CCW) about another line that makes an angle of $\phi$ with the positive x axis, you apply the matrix transformation:
$\begin{bmatrix} x' \\ y' \end{bmatrix} =\begin{bmatrix} \cos 2\phi & -\sin 2\phi \\ -\sin 2 \phi & -\cos 2\phi \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

you would then solve for x, y or eliminate the somehow and put them into the equation y=sqrt(x) and simplify, which im not sure if it can be done or not... maybe someone else with awesome skills can tell you if its possible...
Title: Re: brightsky's Noob-tastic Question Thread
Post by: enpassant on April 18, 2011, 10:32:16 pm: Quote from: brightsky on April 18, 2011, 09:30:17 pm
Just another question, how does one reflect any graph about a certain line ax + by + c = 0?
So for instance, if I wanted to reflect the curve y = sqrt(x) about the line y = 1/2 x + 5, what would the equation of the resulting graph be?
Thanks. :)

Using high school maths
find the intersection of the two lines and the angle between them
the reflection is a line through the intersection and make the same angle on the other side
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on April 19, 2011, 01:44:26 pm: Quote from: enpassant on April 18, 2011, 10:32:16 pm
Quote from: brightsky on April 18, 2011, 09:30:17 pm
Just another question, how does one reflect any graph about a certain line ax + by + c = 0?
So for instance, if I wanted to reflect the curve y = sqrt(x) about the line y = 1/2 x + 5, what would the equation of the resulting graph be?
Thanks. :)

Using high school maths
find the intersection of the two lines and the angle between them
the reflection is a line through the intersection and make the same angle on the other side

Which two lines are you referring to?
Quote from: moekamo on April 18, 2011, 10:16:20 pm
when rotating(CCW) about another line that makes an angle of $\phi$ with the positive x axis, you apply the matrix transformation:
$\begin{bmatrix} x' \\ y' \end{bmatrix} =\begin{bmatrix} \cos 2\phi & -\sin 2\phi \\ -\sin 2 \phi & -\cos 2\phi \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$

you would then solve for x, y or eliminate the somehow and put them into the equation y=sqrt(x) and simplify, which im not sure if it can be done or not... maybe someone else with awesome skills can tell you if its possible...

I saw this matrix on Wikipedia, but I'm not sure how it is derived. And where does the given linear equation about which you reflect it come into play?
Quote from: xZero on April 18, 2011, 09:50:50 pm
We did something similar in my maths unit, basically you rotate the line y = 1/2 x + 5 so it's parallel to the x-axis, do the same rotation with the line ax + by + c = 0, reflect it by the rotated line which is parallel to the x-axis (easy) then rotate it back. Not sure if theres a easier method

Hmm, never thought of that...seems like a good way.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on April 19, 2011, 10:39:55 pm: In the notation $f: R \to R$ , what does the second R mean?
Also with the transformation $T: R^2 \to R^2$ , what do the first and second R^2s mean respectively?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: luken93 on April 19, 2011, 11:00:39 pm: The second R is known as the co-domain, and it specifies the field of y-values that f(x) allows. It's different to the range, because the range is the corresponding values of f(x), whereas R is more of a restriction of values... Oh, and for methods it'll always be R

Can't remember what the R^2 means though XD

EDIT: Read this for R²
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on April 19, 2011, 11:02:03 pm: Quote from: luken93 on April 19, 2011, 11:00:39 pm
The second R is known as the co-domain, and it specifies the field of y-values that f(x) allows. It's different to the range, because the range is the corresponding values of f(x), whereas R is more of a restriction of values... Oh, and for methods it'll always be R

Can't remember what the R^2 means though XD

ahh makes sense, thanks luken!
Title: Re: brightsky's Noob-tastic Question Thread
Post by: kamil9876 on April 19, 2011, 11:22:54 pm: Quote from: luken93 on April 19, 2011, 11:00:39 pm

EDIT: Read this for R²

I highly doubt it, he probably means $\mathbb{R}^2$ just like he means $\mathbb{R}$ instead of $R$ .

It is the set of pairs of real numbers $(a,b)$ so basically your $T$ is a function which sends points on the cartesian plane to some points on a cartesian plane. Example: if $T$ was rotation by pi/2 anti-clockwise about origin then $T( (0,1) )=(1,0)$ (remember rotation matrices, rings a bell?)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: luken93 on April 19, 2011, 11:43:50 pm: Quote from: kamil9876 on April 19, 2011, 11:22:54 pm
Quote from: luken93 on April 19, 2011, 11:00:39 pm

EDIT: Read this for R²

I highly doubt it, he probably means $\mathbb{R}^2$ just like he means $\mathbb{R}$ instead of $R$ .

It is the set of pairs of real numbers $(a,b)$ so basically your $T$ is a function which sends points on the cartesian plane to some points on a cartesian plane. (example, if $T$ was rotation be pi/2 clockwise then $T( (0,1) )=(1,0)$ (remember rotation matrices, rings a bell?)
haha, it was the only thing I found and it seemed kinda relevant :P
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on April 20, 2011, 03:03:34 pm: Quote from: brightsky on April 19, 2011, 10:39:55 pm
In the notation $f: R \to R$ , what does the second R mean?
Also with the transformation $T: R^2 \to R^2$ , what do the first and second R^2s mean respectively?

I always thought $\mathbb{R}^2$ referred to the Cartesian plane (real by real). So, $T: \mathbb{R}^2 \to \mathbb{R}^2$ would mean transformations in the Cartesian plane?

EDIT: didn't see kamil's post, my bad :(
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on April 24, 2011, 04:51:04 pm: In general $\mathbb{R}^n$ is just the set of all ordered n-tuples $(a_1, a_2, \cdots, a_n)$
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on April 24, 2011, 05:31:55 pm: Quote from: TrueTears on April 24, 2011, 04:51:04 pm
In general $\mathbb{R}^n$ is just the set of all ordered n-tuples $(a_1, a_2, \cdots, a_n)$

n-tuples?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on April 24, 2011, 05:33:39 pm: Quote from: brightsky on April 24, 2011, 05:31:55 pm
Quote from: TrueTears on April 24, 2011, 04:51:04 pm
In general $\mathbb{R}^n$ is just the set of all ordered n-tuples $(a_1, a_2, \cdots, a_n)$

n-tuples?

I too was confused...

From wikipedia:
Quote from: Wikipedia
In mathematics and computer science, a tuple is an ordered list of elements. In set theory, an (ordered) n-tuple is a sequence (or ordered list) of n elements, where n is a positive integer.

...

Tuples are usually written by listing the elements within parentheses "( )" and separated by commas; for example, (2,7,4,1,7) denotes a 5-tuple.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on April 24, 2011, 05:34:36 pm: eg, your cartesian coordinates are all 2-tuples (x,y)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on April 24, 2011, 05:40:25 pm: Quote from: TrueTears on April 24, 2011, 05:34:36 pm
eg, your cartesian coordinates are all 2-tuples (x,y)

Are there other 'tuples' that aren't associated with axes? It's hard to imagine, say, a 4-tuple, for instance, if that makes sense..
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on April 24, 2011, 05:45:45 pm: The axes you guys relate to 2,3-tuples are just geometric interpretations/functions. The idea of $\mathbb{R}^n$ is a concept in its own right, it doesn't need to have any (visible) geometric significance.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: luffy on April 24, 2011, 06:04:24 pm: I may have completely misunderstood you, but are you referring to dimensions?

For example, 2-tuples will produce a 2-dimensional graph. (i.e. 2 axes)

3-tuples will produce 3-dimensional graphs. (3-axes)

So, in theory, n-tuples would produce an n-dimensional graph? (n-axes)

Or am I completely confused?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: Andiio on April 24, 2011, 06:15:43 pm: Quote from: luffy on April 24, 2011, 06:04:24 pm
I may have completely misunderstood you, but are you referring to dimensions?

For example, 2-tuples will produce a 2-dimensional graph. (i.e. 2 axes)

3-tuples will produce 3-dimensional graphs. (3-axes)

So, in theory, n-tuples would produce an n-dimensional graph? (n-axes)

Or am I completely confused?

I think you're right, because if we have a 2-tuple, i.e. (x,y) ordered pair, we will naturally have 2 axes (an 'x' axis and 'y' axis) formed along with it.
Not sure if I'm right though :P
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on April 24, 2011, 06:23:51 pm: Quote from: luffy on April 24, 2011, 06:04:24 pm
I may have completely misunderstood you, but are you referring to dimensions?

For example, 2-tuples will produce a 2-dimensional graph. (i.e. 2 axes)

3-tuples will produce 3-dimensional graphs. (3-axes)

So, in theory, n-tuples would produce an n-dimensional graph? (n-axes)

Or am I completely confused?
Depends on your definition of dimensions, the general definition is that the dimension of a vector space (in this case $\mathbb{R}^n$ ) is the number of vectors that make up a basis in that vector space.

Thus we have $\dim(\mathbb{R}^n) = n$

There is no such thing that: "n-tuples produce a n-dimensional graph", the italic terms you used are too informal and not precise enough.

The correct way to interpret is that the dimension of the vector space $\mathbb{R}^n$ is n.

I suggest you guys just treat it as notation for now and don't worry about the details, you need linear algebra knowledge to understand what I mean. Thinking too deeply at this stage without fundamentals will only confuse you :)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: luffy on April 24, 2011, 06:31:59 pm: Quote from: TrueTears on April 24, 2011, 06:23:51 pm
Quote from: luffy on April 24, 2011, 06:04:24 pm
I may have completely misunderstood you, but are you referring to dimensions?

For example, 2-tuples will produce a 2-dimensional graph. (i.e. 2 axes)

3-tuples will produce 3-dimensional graphs. (3-axes)

So, in theory, n-tuples would produce an n-dimensional graph? (n-axes)

Or am I completely confused?
Depends on your definition of dimensions, the general definition is that the dimension of a vector space (in this case $\mathbb{R}^n$ ) is the number of basis in that vector space.

Thus we have $\dim(\mathbb{R}^n) = n$

There is no such thing that: "n-tuples produce a n-dimensional graph", the italic terms you used are too informal and not precise enough.

The correct way to interpret is that the dimension of the vector space $\mathbb{R}^n$ is n.

I suggest you guys just treat it as notation for now and don't worry about the details, you need linear algebra knowledge to understand what I mean. Thinking too deeply at this stage without fundamentals will only confuse you :)

Haha - Fair enough. I was just using colloquial language to understand it.

I think you just spiked everyone's interest by the mention of the word "tuples".

Maths gets amazingly repetitive, and so we seek completely new knowledge to satisfy our 'hunger' :P
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on April 24, 2011, 06:42:18 pm: That's great! I encourage you guys to read further if you have the time. If you are interested to really delve deeper, please take time to read the book, download link is below.

http://www.megaupload.com/?d=E8WXXE2Q
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on April 30, 2011, 10:56:21 pm: What's the general idea behind how to prove concurrency using vectors?
More specifically, how do you prove that the medians of a triangle are concurrent?
And from there, how do you prove that the centroid divides each median in the ratio 2:1? (Vectors or an easier way using simple geometry)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 01, 2011, 04:25:03 pm: bump.

and also, just to confirm, when we say something is twice the size of another thing, we are referring to the area of the "other thing" being twice the area of the original right? not the length, or width or radius?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: TrueTears on May 01, 2011, 04:29:08 pm: Quote from: brightsky on May 01, 2011, 04:25:03 pm
bump.

and also, just to confirm, when we say something is twice the size of another thing, we are referring to the area of the "other thing" being twice the area of the original right? not the length, or width or radius?
they should really define "size", or it should be clear from the context... from my knowledge, i don't think there's a set definition for size, in general, in mathematics? depends on the situation.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 01, 2011, 04:31:00 pm: Quote from: TrueTears on May 01, 2011, 04:29:08 pm
Quote from: brightsky on May 01, 2011, 04:25:03 pm
bump.

and also, just to confirm, when we say something is twice the size of another thing, we are referring to the area of the "other thing" being twice the area of the original right? not the length, or width or radius?
they should really define "size", or it should be clear from the context... from my knowledge, i don't think there's a set definition for size, in general, in mathematics? depends on the situation.

I got a question that reads as follows:

"Place a square inside a circle and a circle inside the square and so on. How many squares do you have to place before the next circle is smaller than one third the size of the original circle?"

Not sure if it's talking about area or radius in this case..
Title: Re: brightsky's Noob-tastic Question Thread
Post by: nacho on May 01, 2011, 04:35:22 pm: the size would probably refer to area, i imagine if they wanted radius, they would explicitly state so

luken raises a fair point though !
Title: Re: brightsky's Noob-tastic Question Thread
Post by: luken93 on May 01, 2011, 04:37:54 pm: I would say radius, but as TT says I think it's more of a problem specific definition. I only say radius because the radius is half the length of the square - which is an important characteristic in solving?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 02, 2011, 09:28:05 pm: I read somewhere that a cube, when divided by 7, would always give remainders -1, 0 or 1. Is there a good proof for this, instead of simply trying numbers from 1 - 7?

Also any help on the previous question:
Quote
What's the general idea behind how to prove concurrency using vectors?
More specifically, how do you prove that the medians of a triangle are concurrent?
And from there, how do you prove that the centroid divides each median in the ratio 2:1? (Vectors or an easier way using simple geometry)

would be appreciated. Thanks!!
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 03, 2011, 07:37:03 pm: How would you work out how many digits are in 2^(2009)?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: syn14 on May 04, 2011, 08:49:07 pm: Quote from: brightsky on May 03, 2011, 07:37:03 pm
How would you work out how many digits are in 2^(2009)?

Not sure if you're looking for the precise number of digits, but a good way of approximating it is as follows.

2^10 = 1024

log10 (1024) is close to 3.

log10 (2^2009) = 200log10 (2^10) + log10 (2^9) which is roughly 603. Alternatively you could use the floor function and stuff...

Quote from: brightsky on May 02, 2011, 09:28:05 pm
I read somewhere that a cube, when divided by 7, would always give remainders -1, 0 or 1. Is there a good proof for this, instead of simply trying numbers from 1 - 7?

Yeh this is true and can be fairly easily proven by modulo arithmetic. I obviously don't know how to use Latex but the basic gist of it is to take the numbers 1 to 7 in mod 7 and cube the number and its remainder. Any number in mod 7 will be congruent to one of 1~7 and do the same thing to see it'll have one of those 3 remainders.
Title: Re: brightsky's Noob-tastic Question Thread
Post by: yawho on May 04, 2011, 10:24:51 pm: Quote
What's the general idea behind how to prove concurrency using vectors?
More specifically, how do you prove that the medians of a triangle are concurrent?
And from there, how do you prove that the centroid divides each median in the ratio 2:1? (Vectors or an easier way using simple geometry)

I have seen something on vector proofs of concurrency of medians, altitudes and perpendicular bisectors on itute.
[/quote]
Title: Re: brightsky's Noob-tastic Question Thread
Post by: Ahmad on May 08, 2011, 02:19:58 pm: Quote from: brightsky on May 02, 2011, 09:28:05 pm
What's the general idea behind how to prove concurrency using vectors?
More specifically, how do you prove that the medians of a triangle are concurrent?
And from there, how do you prove that the centroid divides each median in the ratio 2:1? (Vectors or an easier way using simple geometry)

Recall in a triangle a median is a line connecting a vertex to the midpoint of the edge opposite the vertex, so that a triangle has 3 median lines. We should make sure we know what it means for the medians of a triangle to be concurrent. It means that the 3 medians intersect at a single point, for example we might imagine that each pair of medians intersects at a different point, so that there are 3 possible intersection points (3 ways to form a pair of medians), we have to prove that this cannot happen.

Let's set things up. First let ABC be a triangle and let A', B', C' be the midpoint of the edges opposite A, B and C respectively, so that the medians are AA', BB', CC'. Furthermore let's call the point at which the medians AA' and BB' intersect P. I've sketched a diagram of our setup below.

(http://i202.photobucket.com/albums/aa132/ahmad0/centroid.png)

What would we need to show? Well, we need to show that the lines AA' and CC' intersect at P, and the lines BB' and CC' intersect at P. Both of these are true if and only if the median CC' passes through P, and that's what we'll aim to show.

Next we should convert the statement "the median CC' passes through P" into a statement about vectors if we hope to prove it using vectors. Another way to phrase the statement is that the points C, P and C' are collinear, and this is something we should be familiar with and be able to recognise as being the same as the (vector) statement that CC' = k CP for some constant k (i.e. CC' and CP are parallel).

We've now set up the framework and we know what we need to prove. The next bit, which is actually proving the statement relies only on technique (if you have experience there isn't much to think about).

Our aim is to write CC' and CP in terms of only AB' and AC' (i.e. the minimum number of variables which completely define the problem, we could've chosen other quantities like AB and CB, as long as there are two).

We have CC' = -2 AB' + AC'
However it seems more difficult to get CP, at this stage we should bear in mind the information we have and must use if we are to solve the problem, which is that AA' = a AP and BB' = b AP for some constants a, b (which is to say the points A, P, A' and B, P, B' are each collinear).

Using this we find that: CP = CA + AP = -2 AB' + a AA' = -2 AB' + a (AC' + AB') = (a - 2) AB' + a AC'

Even having done this we note that we haven't got enough information to deduce that CC' = k CP for some constant k since we have this problematic constant of a. The problem is that we haven't used BB' = b AP in any of our equations and if we think about it a little we find that we could've found CP in another way by going through the path C-A-B-P.

If we do that we get CP = -2 AB' + 2 AC' + BP = -2 AB' + 2 AC' + b (-2 AC' + AB') = (b - 2) AB' + (2 - 2b) AC'

Now we have two different expressions for CP which are CP = (b - 2) AB' + (2 - 2b) AC' = (a - 2) AB' + a AC'. Since AB' and AC' are not parallel we can equate the coefficients of AB' and AC' (this is to say that AB' and AC' are linearly independent, intuitively they form a coordinate system for the plane). Hence we have the simultaneous equations b - 2 = a - 2 and 2 - 2b = a, which we solve to get a = b = 2/3.

Going back to our expression for CP but putting in a = 2/3 or b = 2/3 we have CP = -4/3 AB' + 2/3 AC', comparing this to our expression for CC' we have CC' = 3/2 CP, so that our constant k from before is 3/2. And we've proved what we wanted, and along the way we've essentially obtained the fact that the centroid, which we can now confidently say is the well-defined point P, divides each median in the ratio 2 : 1 (follows from our constant k = 3/2, do you see how?).
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 12, 2011, 11:23:07 pm: Thanks Ahmad!!! Amazing explanation!

I got another question, if tanx = a, find sinx cosx in terms of a. I'm looking for an algebraic solution, without the need to draw triangles.

Thanks!
Title: Re: brightsky's Noob-tastic Question Thread
Post by: xZero on May 12, 2011, 11:32:44 pm: $tan(x)=a$

$\frac{sin(x)}{cos(x)}=a$

$\frac{sin^2(x)}{cos^2(x)}=a^2$

$\frac{sin^2(x)}{1-sin^2(x)}=a^2$

$sin^2(x)=a^2-a^2sin^2(x)$

$sin^2(x)+a^2sin^2(x)=a^2$

$sin^2(x)(1+a^2)=a^2$

$sin^2(x)=\frac{a^2}{1+a^2}$

$sin(x)=\pm\frac{a}{\sqrt{1+a^2}}$ (depending on the quadrant)

$cos^2(x)=1-sin^2(x)$

$cos^2(x)=1-\frac{a^2}{1+a^2}$

$cos^2(x)=\frac{1}{1+a^2}$

$cos(x)=\pm\frac{1}{\sqrt{1+a^2}}$ (depending on the quadrant)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 12, 2011, 11:40:15 pm: woah never thought of that, thanks xzero!
Title: Re: brightsky's Noob-tastic Question Thread
Post by: kenny_ on May 17, 2011, 12:01:30 am: hey brightsky, if you don't mind me asking are you in year 10?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: pi on May 17, 2011, 08:03:28 pm: Quote from: kenny_ on May 17, 2011, 12:01:30 am
hey brightsky, if you don't mind me asking are you in year 10?

Yeh, he is. He's pretty damn brilliant :)
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 19, 2011, 08:54:09 pm: I've got a few questions:

(1) How does one derive a general formula for the surface area of an oblique cylinder?
(2) How about an oblique cone?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 19, 2011, 08:57:56 pm: Also, I was wondering if anyone can explain this application of the Cavalieri principle:

http://en.wikipedia.org/wiki/Cavalieri%27s_principle#Spheres

Many thanks! [Solved - thanks gossamer!]
Title: Re: brightsky's Noob-tastic Question Thread
Post by: brightsky on May 21, 2011, 09:05:07 am: Trivial thing, but when asked to express an answer in form, say, a + b ln(c), are we able to write something like 3 - 2 ln(4), or do we have to write 3 + (-2ln(4))?

Similarly, when asked to express answer in form y = (x-x1)(x-x2)(x-x3), are we able to write y = (x - 2)(x-3)^2 or do we have to write y = (x-2)(x-3)(x-3)?
Title: Re: brightsky's Noob-tastic Question Thread
Post by: jane1234 on May 21, 2011, 10:55:25 am: 1. I think 3 - 2ln(4) is fine.

2. I'd write it as y= (x-2)(x-3)(x-3) because thats the form they want, not y = (x-x1)(x-x2)^2