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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: bucket on February 09, 2008, 02:18:44 pm

Title: Bucket's Questions
Post by: bucket on February 09, 2008, 02:18:44 pm: Don't think theres a point in making a new thread every time I can't work out a painfully easy question, so I'll just copy bec and dump them all into one thread.
Anyway

I forgot how to factorise LOL

$3x-1$ is a factor of $f(x)=3x^3 +2x^2-19x+6$ , factorise f(x).
Title: Re: Bucket's Questions
Post by: enwiabe on February 09, 2008, 02:42:11 pm: Many ways to do this. I'll be going with "by inspection"

let f(x) = 3x^3 + 2x^2 - 19x + 6

If 3x - 1 is a factor, then it can be assumed that:

(3x - 1)(x^2 + Ax - 6) = 3x^3 + 2x^2 - 19x + 6 = f(x)

If you're wondering how I got that, multiply 3x by x^2 to get the 3x^3 term. Multiply -1 by - 6 to get the +6 term. In the MIDDLE, the Ax term represents an unknown because we don't yet know what value of A will give us the two middle terms, 2x^2 and -19x.

HOWEVER, we now expand out:
let q(x) = f(x) = (3x - 1)(x^2 + Ax - 6) = 3x^3 + 3Ax^2 - 18x - x^2 - Ax + 6
= 3x^3 + (3A - 1)x^2 - (A + 18)x + 6

BUT we know that q(x) = f(x)

Therefore, 3x^3 + (3A - 1)x^2 - (A + 18)x + 6 = 3x^3 + 2x^2 - 19x + 6

equation co-efficients: 3 = 3, 3A - 1 = 2, -(A + 18) = -19, and 6 = 6

See? We obtain the correctly found co-efficients from before.
To solve the UNKNOWN 'A' co-efficient, set 3A - 1 = 2 => 3A = 3 and therefore, A = 1. You can check that A = 1 by solving equation 2: -(A + 18) = -19, therefore A + 18 = 19 => A = 1

so if A = 1 you get f(x) = q(x) = (3x-1)(x^2 + x - 6)
Using cross method: x 3
x -2

therefore f(x) = (3x - 1)(x + 3)(x - 2) = 3x^3 + 2x^2 - 19x + 6
Title: Re: Bucket's Questions
Post by: enwiabe on February 09, 2008, 02:42:38 pm: N.B. Whilst that looks very longwinded, I was explaining all steps. If you master this method it can be 3-4 lines on an exam paper.

Also, if that's the first time you've seen this method, it's a lot to digest. If you have any questions don't hesitate. This method is very powerful and gets rid of the frustration of long division.
Title: Re: Bucket's Questions
Post by: bucket on February 09, 2008, 03:34:43 pm: Wow.
No you explained it very clearly, which is surprising since most of the time people go straight to the point without explaining why they take the steps they do and leave me thinking like wtf.
Thanks for that!
Title: Re: Bucket's Questions
Post by: bucket on February 10, 2008, 12:33:50 am: Mmmm
Could someone show me the proper way to find the intercepts of a graph such as this;
$f(x) = \dfrac{2}{(x - 3)^3} + 4$
I know the general shape of the graph, just the intercepts in the answer have like numbers with cubed routes and shit, and I have no idea how they get there :P
Title: Re: Bucket's Questions
Post by: Collin Li on February 10, 2008, 12:45:53 am: y-intercept: $x=0$

$f(0) = \frac{2}{(-3)^3} + 4 = \frac{106}{27}$

x-intercept: $y=0$

$0 = \frac{2}{(x-3)^3} + 4$

$\implies \frac{1}{2} = (x-3)^3$

$\implies \sqrt[3]{\frac{1}{2}} = x-3$

$\implies x = \frac{1}{\sqrt[3]{2}} + 3$
Title: Re: Bucket's Questions
Post by: bucket on February 10, 2008, 12:47:23 am: its so simple.
aduhhhh
thanks.
Title: Re: Bucket's Questions
Post by: bucket on February 11, 2008, 09:36:57 pm: Okay this annoyed me because I was going well in this exercise until I bumped into this equation:
A transformation is defined by the matrix $\left[ \begin{array}{cc}0 & 3 \\-2 & 0\end{array} \right]\]$ . Find the equation of the image of the graph of the straight line with equation $y = 2x + 3$ under this transformation.

I think the problem is that the transformation includes a reflection in the line y=x, and I have no fucking idea what that is supposed to mean rofl.
Title: Re: Bucket's Questions
Post by: Collin Li on February 11, 2008, 09:50:56 pm: A reflection in the line $y=x$ is the inverse. That means you just swap $y$ and $x$ around in your equation. (i.e: $y = x + 3$ when reflected becomes $x = y + 3$ )
Title: Re: Bucket's Questions
Post by: Mao on February 11, 2008, 09:52:22 pm: so:

let $\left[ \begin{array}{cc}u \\v\end{array} \right]$ be the image line

$\left [ \begin{array}{cc}u \\v\end{array} \right] = \left [ \begin{array}{cc}0 & 3 \\-2 & 0\end{array} \right] \cdot \left [ \begin{array}{cc}x \\y\end{array} \right]= \left[ \begin{array}{cc}3y \\-2x \end{array} \right]$

$y=\frac{u}{3}$ and $x=-\frac{v}{2}$

subbing back into the orgininal equation:

$\frac{u}{3} = 2 \cdot -\frac{v}{2} + 3$

$v = -\frac{u}{3} + 3$

now since v is the image of y and u is the image of x, our new equation is:

$y = -\frac{x}{3} + 3$

Hope that helped :)
Title: Re: Bucket's Questions
Post by: bucket on February 11, 2008, 10:14:41 pm: wonderful =]
Title: Re: Bucket's Questions
Post by: bucket on February 19, 2008, 11:09:54 pm: Question to do with quadratics..........

The dimensions of an enclosure are shown. The perimeter of the enclosure is 160m.

(http://i7.photobucket.com/albums/y289/Taksi_/q5.jpg)

a. Find the rule for the area, $A m^2$ , of the enclosure in terms of $x$
Title: Re: Bucket's Questions
Post by: Collin Li on February 19, 2008, 11:21:16 pm: Perimeter: $2x+2y=160$

$\implies y = 80 - x$

$A = 20x + (12-x)(20-y) = 20x + (12-x)(20-80+x) = 20x + (12-x)(x-60)$

$= 20x + (-x^2 + 72x - 720)$

$\therefore\, A = -x^2 + 92x - 720$
Title: Re: Bucket's Questions
Post by: bucket on February 19, 2008, 11:23:45 pm: mm, in the answers (note these were given by the teacher so i don't know if the answers are 100% correct), it says that the solution is $-x^2+92x-720$
Title: Re: Bucket's Questions
Post by: Collin Li on February 19, 2008, 11:25:37 pm: Fixed. I got the perimeter wrong.
Title: Re: Bucket's Questions
Post by: bucket on February 19, 2008, 11:30:38 pm: ahhh thankyou so much mr coblin.
you have proven to be more help than my math teacher yet again!
Title: Re: Bucket's Questions
Post by: bucket on February 20, 2008, 12:03:18 am: HOLD ON!! lol

Why is it
(12 - x)(20 - y) and not (x - 12)(y - 20)??
Title: Re: Bucket's Questions
Post by: Collin Li on February 20, 2008, 12:11:56 am: No idea why I wrote it like that, I must have been high.

It's the same thing (mathematically) though, because you can take out a factor of -1 from both the brackets, and $(-1)\times (-1) = 1$ , haha.
Title: Re: Bucket's Questions
Post by: bucket on February 20, 2008, 12:14:11 am: haha!! oh okay, thanks =]
Title: Re: Bucket's Questions
Post by: bucket on February 21, 2008, 04:26:33 pm: Oh god this question is embarassing.
I literally slept in class during methods last year and now I'm behind on many topics, especially logarithms and exponentials

Simplify $3^4 * 9^2 * 27^3$

I'm getting $3^{14}$ as my answer but the book says the answer is $3^{17}$
Title: Re: Bucket's Questions
Post by: dcc on February 21, 2008, 04:34:35 pm: Quote from: bucket on February 21, 2008, 04:26:33 pm
Simplify $3^4 * 9^2 * 27^3$

Since $9 = 3^2$ and $27 = 3^3$

$3^4 * (3^2)^2 * (3^3)^3$

$3^4 * 3^4 * 3^9$

$3^{4 + 4 + 9}$

$3^{17}$

Remember:

$(a^b)^c = a^{b \times c}$
Title: Re: Bucket's Questions
Post by: bucket on February 21, 2008, 04:36:45 pm: ohhhhhhhhhhhhhhhhh.
how could i have not worked it out myself T_T
*bangs head with saucepan*
Title: Re: Bucket's Questions
Post by: costargh on February 21, 2008, 04:41:59 pm: Quote from: bucket on February 21, 2008, 04:36:45 pm
*bangs head with saucepan*
lol. if u ever did that id piss myself haha
Title: Re: Bucket's Questions
Post by: bucket on February 21, 2008, 07:06:24 pm: Ok, when I'm solving questions such as
"Simplify $\frac{x^3yz^{-2}*2(x^3y^{-2}z)^2}{xyz^{-1}}$ expressing your answer in positive exponent form"
I get confused whilst working the answer out because there are so many different calculations to do and I end up all over the place.
Does anyone have any tips or tricks they used to make things more methodical when solving problems such as this? Or do I just need to get used to it? lol
Title: Re: Bucket's Questions
Post by: dcc on February 21, 2008, 07:36:00 pm: Try splitting the problem into small bits, so for this problem:
Quote from: bucket on February 21, 2008, 07:06:24 pm
"Simplify $\frac{x^3yz^{-2}*2(x^3y^{-2}z)^2}{xyz^{-1}}$

First tackle the brackets, removing the square from the bracket:
$2(x^3y^{-2}z)^2$

Which when squared through becomes: $2(x^6 \times y^{-4} \times z^2)$

Then just tackle the original top line of the equation:

$x^3 \times y \times z^{-2} \times 2x^6 \times y^{-4} \times z^2$

which becomes, after adding all the powers etc. etc: $2x^9 \times y^{-3}$

Then we consider this top line and then the bottom line:
$\dfrac{2x^9 \times y^{-3}}{x \times y \times z^{-1}}$

which becomes: $2x^8 \times y^{-4} \times z$

Which when expressed in positive exponent form becomes:
$\dfrac{2x^8 \times z}{y^4}$
Title: Re: Bucket's Questions
Post by: bucket on February 21, 2008, 08:35:03 pm: sweet thanks lol.
Title: Re: Bucket's Questions
Post by: bucket on February 23, 2008, 07:38:15 pm: Simplify $2^\frac{3}{2} * 4^{-\frac{1}{4}} * 16^{-\frac{3}{4}}$

I'm getting $2^{-4} = \frac{1}{4}$ when the answer is $\frac{1}{4}$

Thanks to anyone who helps =]
Title: Re: Bucket's Questions
Post by: dcc on February 23, 2008, 08:17:31 pm: $2^\frac{3}{2} * 4^\frac{-1}{4} * 16^\frac{-3}{4}$

$2^\frac{3}{2} * (2^2)^\frac{-1}{4} * (2^4)^\frac{-3}{4}$

$2^\frac{3}{2} * 2^{-\frac{1}{2}} * 2^{-3}$

Adding the powers, we get:

$2^{-2} = \dfrac{1}{2^2} = \dfrac{1}{4}$
Title: Re: Bucket's Questions
Post by: bucket on February 23, 2008, 09:53:58 pm: thanks alot dcc
okayyy.. i have another question >.<

$\frac{3^{n-2}}{9^{1-n}} = 1$ , Solve for n.
Title: Re: Bucket's Questions
Post by: Collin Li on February 23, 2008, 09:56:29 pm: $\frac{3^{n-2}}{9^{1-n}} = 1$

$\implies \frac{3^n\cdot 3^{-2}}{9^1\cdot 9^{-n}} = 1$

$\implies \frac{3^n\cdot 9^{n}}{9^1\cdot 3^{2}} = 1$

$\implies 3^n\cdot 9^n = 9^1\cdot 3^2$

$\implies 3^n\cdot 3^{2n} = 3^2\cdot 3^2$

$\implies 3^{3n} = 3^{4}$

$\therefore 3n = 4 \implies n = \frac{4}{3}$
Title: Re: Bucket's Questions
Post by: bucket on February 23, 2008, 10:01:06 pm: ahhhhh, interesting way of doing that :p
thanks alot man
Title: Re: Bucket's Questions
Post by: Collin Li on February 23, 2008, 10:06:41 pm: Hm... it's pretty standard. A more clever approach (less 'hack at the problem' method) would be to recognise that 3 and 9 are related by a power of 2, although it's all the same:

$\frac{3^{n-2}}{9^{1-n}} = 1$

$\implies \frac{3^{n-2}}{\left(3^{1-n}\right)^2} = 1$

$\implies \frac{3^{n-2}}{3^{2-2n}} = 1$

$\implies 3^{n-2}\cdot 3^{2n-2} = 1$

$\implies 3^{3n-4} = 1$

Since $1= 3^0$ , or by taking the $\log_3$ of both sides:

$\implies 3n-4 = 0$

Or maybe you think it's strange because I didn't bring the denominator to the RHS? Meh, haha, whatever works for you - the operations you do will be essentially the same.
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 02:14:45 pm: eh
Solve $2^{x-1} * 4^{2x+1} = 32$

I'm getting $\frac{2}{5}$ and it says the answer is $\frac{4}{5}$

what am I doing wrong? lol
Title: Re: Bucket's Questions
Post by: bec on February 24, 2008, 02:35:48 pm: yay I can do this one!

So first you need to make sure everything has the same base: 2.
This gives: $2^{x-1} \times 2^{2(2x + 1)} = 2^{5}$

Now, remembering your exponential rules, you can ADD the powers on the LHS.
This gives: $2^{(x-1)+(4x +2)} = 2^{5}$

Lastly, equate the powers. (What you're doing here is taking the log base 2 of everything)
This gives: $x - 1 + 4x + 2= 5$
$5x + 1= 5$

$5x = 4$

$x = \frac {4}{5}$

hope that helps!
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 02:41:10 pm: ah... stupid little errors on my part (N)
thanks alot bec!
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 02:50:07 pm: Ok I have another one....
$8(2^{2x})-10(2^{x})+2 = 0$

The answer is 0, -2
Title: Re: Bucket's Questions
Post by: Neobeo on February 24, 2008, 03:00:48 pm: Let $A=2^x$
$8A^2-10A+2=0$
$2(A-1)(4A-1)=0$
$A=2^x=1,\frac{1}{4}$
$x=0,-2$
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 03:11:21 pm: eh.
I dont understand where 0 and -2 come from.
Title: Re: Bucket's Questions
Post by: Toothpaste on February 24, 2008, 03:19:42 pm: $A$ $=$ $2^x$

We know that $A$ $=$ $1$ and $A$ $=$ $\frac{1}{4}$ using the null factor theorem.

substituting in $A$ $=$ $2^x$ gives:
$2^x$ $=$ $1$ and $2^x$ $=$ $\frac{1}{4}$
$2^0$ = 1 therefore $x$ $=$ $1$
$2^{-2}$ = $\frac{1}{4}$ therefore $x$ $=$ $-2$ **

** $\implies$ $\log_{2} \frac{1}{4}$ $=$ $x$
change of base law
$\frac{\log_{10} \frac{1}{4}}{\log_{10} 2}$ $=$ $x$
$x$ $=$ $-2$
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 05:22:21 pm: ohh, thanks heaps man.
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 11:01:29 pm: Simplify $log_2(128) + log_3(145) - log_3(5)$ .

Ack, I don't know what to do when faced with two different bases lol.
Title: Re: Bucket's Questions
Post by: dcc on February 24, 2008, 11:22:31 pm: well, $log_2(128) = 7$ (since $2^7 = 128$ )

so another way of writing this would be:

$log_3(3^7) + log_3(145) - log_3(5)$

since $log_a(a^b) = b$

then, just continuing as you normally would, adding and dividing as necessary:

$= log_3(63423)$
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 11:28:30 pm: mm.
the question says 'simplify without using a calculator' (i didn't write it coz i didn't think it would be relevant =\)
and the answer is 9?
I'm assuming it solved for x.......?
Title: Re: Bucket's Questions
Post by: dcc on February 24, 2008, 11:37:07 pm: $log_2(128) + log_3(145) - log_3(5)$

$7 + log_3(29 * 5) - log3(5)$

$7 + log_3(29) + log_3(5) - log3(5)$

$7 + log_3(29)$

Which is as far as that one will go :P
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 11:42:40 pm: mm... i don't know how it gets 9
Title: Re: Bucket's Questions
Post by: dcc on February 24, 2008, 11:45:17 pm: the question must be wrong, because if you manually type the question into the calculator, you get an answer of roughly 10.6 ish
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 11:48:17 pm: ah ok, thanks =].

Then how do you work this one out;
$log_4(32) - log_9(27)$
Title: Re: Bucket's Questions
Post by: dcc on February 24, 2008, 11:49:56 pm: if the question was $log_2(128) + log_3(45) - log_3(5)$ then the answer would be 9 :)
Title: Re: Bucket's Questions
Post by: dcc on February 24, 2008, 11:52:01 pm: Quote from: bucket on February 24, 2008, 11:48:17 pm
$log_4(32) - log_9(27)$

$log_4(2^5) - log_9(3^3)$

Since $2 = 4^\frac{1}{2}$ and $3 = 9^\frac{1}{2}$ , we can rewrite as:

$log_4(4^\frac{5}{2}) - log_9(9^\frac{3}{2})$

$\dfrac{5}{2} - \dfrac{3}{2}$

$= 1$
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 11:53:49 pm: Quote from: dcc on February 24, 2008, 11:49:56 pm
if the question was $log_2(128) + log_3(45) - log_3(5)$ then the answer would be 9 :)
That's what I wrote! lol
How is it worked out??

Oh and thanks for solving that other problem.
Title: Re: Bucket's Questions
Post by: dcc on February 24, 2008, 11:55:41 pm: Quote from: bucket on February 24, 2008, 11:01:29 pm
Simplify $log_2(128) + log_3(145) - log_3(5)$ .

Ack, I don't know what to do when faced with two different bases lol.

thats what you wrote :)
Title: Re: Bucket's Questions
Post by: bucket on February 24, 2008, 11:57:06 pm: LOL...whoops...sorry :p
Title: Re: Bucket's Questions
Post by: dcc on February 24, 2008, 11:59:24 pm: $log_2(128) + log_3(45) - log_3(5)$

we know that $log_2(2^7) = 7$

and from our log laws:

$log_3(45) - log_3(5) = log_3(\dfrac{45}{5}) = log_3(9) = log_3(3^2) = 2$

Adding these, you get 9
Title: Re: Bucket's Questions
Post by: bucket on February 25, 2008, 01:02:40 am: Lol thanks for all of that dcc.
Title: Re: Bucket's Questions
Post by: bucket on February 25, 2008, 08:47:04 pm: uhm..
Simplify the following expressions:
$(4x^2y^2)^2 + 2(x^2y)^4$
and
$\frac{3(2x^2y^3)^4}{2x^3y^2}$

thanks in advance!
Title: Re: Bucket's Questions
Post by: Toothpaste on February 25, 2008, 08:53:11 pm: $(4x^2y^2)^2 + 2(x^2y)^4$

= $2x^8y^4 + 16x^4y^4$

= $2x^4y^4(x^4+8)$

$\frac{3(2x^2y^3)^4}{2x^3y^2}$

= $\frac{3(16x^8y^{12})}{2x^3y^2}$

= $\frac{48x^8y^{12}}{2x^3y^2}$

= $24x^5y^{10}$
Title: Re: Bucket's Questions
Post by: bucket on February 25, 2008, 10:12:03 pm: Uhm, can somebody confirm these answers?
Not that I think you're wrong toothpick, it's just that I left my text book at school and the pdf of answers on the cd starts at chapter 14 for some strange reason, and well I don't remember the answers being like that :|
You are probably right though, my memory isn't always reliable.
Title: Re: Bucket's Questions
Post by: Collin Li on February 25, 2008, 10:25:48 pm: They're correct, but the first one might not be "simplified." That's sort of subjective. The second line (expanded form) might be considered as simplified. Exam questions aren't this ambiguous - they will specify a "form" to put the answer in (e.g: "Express your answer in the form $a+\sqrt b$ ")
Title: Re: Bucket's Questions
Post by: bucket on February 25, 2008, 10:26:50 pm: Okay! Thankyou both!
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 08:13:17 pm: Express as the logarithm of a single term:
$1+log_{10}(a)-\frac{1}{3}log_{10}(b)$

edit
also;
$2log_e(x+5)=6$

and
Evaluate:
$log(\sqrt{2})+log_2(1)+2log_2(2)$
Title: Re: Bucket's Questions
Post by: Toothpaste on February 27, 2008, 09:02:17 pm: Third/last one:
$log_2(\sqrt{2})+log_2(1)+2log_2(2)$

$= log_2(\sqrt{2})+log_2(1)+log_2(4)$

$= \frac{1}{2} + 0 + 2$

$= \frac{5}{2}$

because $2^{\frac{1}{2}}=\sqrt{2}$

$2^0 = 1$

and $2^2 = 4$

alternatively:
$log_2(\sqrt{2})+log_2(1)+log_2(4)$

Use the log laws

$log_2(\sqrt{2} * 1 * 4)$
$\implies log_2(4\sqrt{2})$
$\implies log_2(4\sqrt{2}) = \frac{5}{2}$
Title: Re: Bucket's Questions
Post by: Glockmeister on February 27, 2008, 09:05:06 pm: First One

$1 + \log_{10} a - \frac{1}{3} log_{10} b$
$=\log_{10} 10 + \log_{10} a - \frac{1}{3} log_{10} b (as 1 = \log_{10} 10)$
$= \log_{10} 10a - \frac{1}{3} log_{10} b$
$= \log_{10} \frac{10a}{b^{\frac{1}{3}}}$

Second One

Note: I'm assuming you want to work out x

$2\log_{e} (x+5) = 6$
$\log_{e} (x+5) = 3$
$e^{3} = x+5$
$e^{3} - 5 = x$

PS: I hate LaTeX
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 09:38:32 pm: thanks a heap guys =]
you are a great help!
Title: Re: Bucket's Questions
Post by: AppleXY on February 27, 2008, 09:40:29 pm: Quote from: Glockmeister on February 27, 2008, 09:05:06 pm
First One

$1 + \log_{10} a - \frac{1}{3} log_{10} b$
$=\log_{10} 10 + \log_{10} a - \frac{1}{3} log{10} b (as 1 = \log_{10} 10)$
$= \log_{10} 10a - \frac{1}{3} log_{10} b$
$= \log_{10} \frac{10a}{b^{\frac{1}{3}}}$

Second One

Note: I'm assuming you want to work out x

$2\log_{e} (x+5) = 6$
$\log_{e} (x+5) = 3$
$e^{3} = x+5$
$e^{3} - 5 = x$

PS: I hate LaTeX

:O WHY?! I HATE ASCII MATH. BLOODY EWW. Srsly, you can't read anything in that format. $\LaTeX$ ftw
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 10:02:15 pm: Simplify $log_3\frac{1}{3x}$

this shit isn't clicking with me =\
Title: Re: Bucket's Questions
Post by: Collin Li on February 27, 2008, 10:17:03 pm: $\log_3\frac{1}{3x} = \log_3(3x)^{-1} = -\log_3(3x) = -\log_33 - \log_3x = -1 - \log_3x$
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 10:23:06 pm: That's what I got.
But the answer is $-x$
Title: Re: Bucket's Questions
Post by: Glockmeister on February 27, 2008, 10:28:17 pm: Cob, shouldn't that be $-1$
Title: Re: Bucket's Questions
Post by: Glockmeister on February 27, 2008, 10:30:23 pm: Quote from: AppleXY on February 27, 2008, 09:40:29 pm
Quote from: Glockmeister on February 27, 2008, 09:05:06 pm
First One

$1 + \log_{10} a - \frac{1}{3} log_{10} b$
$=\log_{10} 10 + \log_{10} a - \frac{1}{3} log{10} b (as 1 = \log_{10} 10)$
$= \log_{10} 10a - \frac{1}{3} log_{10} b$
$= \log_{10} \frac{10a}{b^{\frac{1}{3}}}$

Second One

Note: I'm assuming you want to work out x

$2\log_{e} (x+5) = 6$
$\log_{e} (x+5) = 3$
$e^{3} = x+5$
$e^{3} - 5 = x$

PS: I hate LaTeX

:O WHY?! I HATE ASCII MATH. BLOODY EWW. Srsly, you can't read anything in that format. $\LaTeX$ ftw

I'd rather just write the damn thing on paper. Took me seconds to work it out on paper, took 15 mins to typeset it onto here.
Title: Re: Bucket's Questions
Post by: Collin Li on February 27, 2008, 10:32:43 pm: Quote from: bucket on February 27, 2008, 10:23:06 pm
That's what I got.
But the answer is $-x$

That's wrong.

Quote from: Glockmeister on February 27, 2008, 10:28:17 pm
Cob, shouldn't that be $-1$

Yeah, fixed, thanks.
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 10:32:56 pm: Wouldn't it be $-log_33 + log_3x = -1 +log_3x$ ??
Because $log_amn = log_am + log_an$ right?

nevertheless, the answer is somehow -x, anybody know why?
Title: Re: Bucket's Questions
Post by: Collin Li on February 27, 2008, 10:34:48 pm: Quote from: bucket on February 27, 2008, 10:32:56 pm
Wouldn't it be $-log_33 + log_3x = -1 +log_3x$ ??
Because $log_amn = log_am + log_an$ right?

nevertheless, the answer is somehow -x, anybody know why?

No, because it is $-\left(\log_33 + \log_3x\right)$ .

The answer is wrong.
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 10:37:57 pm: ah I see.
thanks coblin.
Title: Re: Bucket's Questions
Post by: Glockmeister on February 27, 2008, 10:38:22 pm: Funnily, I did that mistake too when I was checking over that answer

Bucket: Cause remember, Bob* is not omniscient

*Back of book
Title: Re: Bucket's Questions
Post by: Neobeo on February 27, 2008, 10:40:49 pm: -x would be the answer to $log_3\frac{1}{3^x}$ though.
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 10:44:33 pm: Quote from: Glockmeister on February 27, 2008, 10:38:22 pm
Funnily, I did that mistake too when I was checking over that answer

Bucket: Cause remember, Bob* is not omniscient

*Back of book
rofl, but coblin is ;)
stupid bob...

Quote from: Neobeo on February 27, 2008, 10:40:49 pm
-x would be the answer to $log_3\frac{1}{3^x}$ though.

huh?
Title: Re: Bucket's Questions
Post by: Collin Li on February 27, 2008, 10:46:45 pm: He's saying that the book may have mistyped (or you may have misread) the question. If the problem was $\log_3\frac{1}{3^x}$ as opposed to $\log_3\frac{1}{3x}$ , then the answer would in fact be $-x$ .
Title: Re: Bucket's Questions
Post by: Glockmeister on February 27, 2008, 10:47:16 pm: Well $\log_{3} \frac{1}{3^{x}} = \log_{3} 3^{-x}$
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 10:55:01 pm: Quote from: Glockmeister on February 27, 2008, 10:47:16 pm
Well $\log_{3} \frac{1}{3^{x}} = \log_{3} 3^{-x}$

I see!

Quote from: coblin on February 27, 2008, 10:46:45 pm
He's saying that the book may have mistyped (or you may have misread) the question. If the problem was $\log_3\frac{1}{3^x}$ as opposed to $\log_3\frac{1}{3x}$ , then the answer would in fact be $-x$ .
I know that's what he was saying!
Stupid book;
(http://i7.photobucket.com/albums/y289/Taksi_/stupidbook.jpg)
maybe it did mean $log_3\frac{1}{3x}$
Title: Re: Bucket's Questions
Post by: Glockmeister on February 27, 2008, 10:59:10 pm: This wouldn't happen to be the Essential Textbook would it?
Title: Re: Bucket's Questions
Post by: bucket on February 27, 2008, 11:00:49 pm: Yeah it is, how did you know?
Title: Re: Bucket's Questions
Post by: Glockmeister on February 27, 2008, 11:16:00 pm: It's the textbook we used for Methods when I did it (last year). Kinda recognise the font.
Title: Re: Bucket's Questions
Post by: bucket on February 28, 2008, 09:10:10 pm: Solve for x:
$log_ex+log_e(3x+1)=1$
Title: Re: Bucket's Questions
Post by: enwiabe on February 28, 2008, 09:14:39 pm: 1 = log_e(e)

therefore,

log_e(x) + log_e(3x+1) = log_e(e)
log_e(x(3x+1)) = log_e(e) ... (using the log laws)
therefore, x(3x+1) = e

... cbf solving that.
Title: Re: Bucket's Questions
Post by: bucket on February 28, 2008, 09:19:42 pm: do you use the quadratic equation to solve it?
Title: Re: Bucket's Questions
Post by: Glockmeister on February 28, 2008, 09:35:30 pm: yep
Title: Re: Bucket's Questions
Post by: bucket on February 28, 2008, 09:53:14 pm: thanks man.
Title: Re: Bucket's Questions
Post by: bucket on March 01, 2008, 06:26:33 pm: Two questions =\

1) If $x=a$ and $y=c$ , express $log_{10}(\frac{100x^3y^{-\frac{1}{2}}}{y^2})$ in terms of a and c.

2) Prove that $log_{10}(\frac{ab^2}{c})+log_{10}(\frac{c^2}{ab})-log_{10}(bc)=0$

The answer for 1 is
$2+3a-\frac{5c}{2}$
Title: Re: Bucket's Questions
Post by: Neobeo on March 01, 2008, 06:32:04 pm: Quote from: bucket on March 01, 2008, 06:26:33 pm
2) Prove that $log_{10}(\frac{ab^2}{c})+log_{10}(\frac{c^2}{ab})-log_{10}(bc)=0$

$\log_{10}(\frac{ab^2}{c})+\log_{10}(\frac{c^2}{ab})-\log_{10}(bc)=\log_{10}(\frac{ab^2}{c}\times \frac{c^2}{bc}\times \frac{1}{ab}) = \log_{10}1 = 0$
Title: Re: Bucket's Questions
Post by: bucket on March 01, 2008, 06:37:47 pm: thanks neo.
Title: Re: Bucket's Questions
Post by: unknown id on March 01, 2008, 07:47:51 pm: i think you meant "If $log_{10}(x) = a$ and $log_{10}(y) = c$ "

$log_{10}(\frac{100x^3y^{-\frac{1}{2}}}{y^2})$

$= log_{10}(100x^3y^{-\frac{5}{2}})$

$= log_{10}(100) + log_{10}(x^3) + log_{10}(y^{-\frac{5}{2}})$

$= log_{10}(10^2) + log_{10}(x^3) + log_{10}(y^{-\frac{5}{2}})$

$= 2log_{10}(10) + 3log_{10}(x) + {-\frac{5}{2}}log_{10}(y)$

$= 2 + 3a - \frac{5c}{2}$
Title: Re: Bucket's Questions
Post by: bucket on March 01, 2008, 08:33:46 pm: Ah shit. I did too.
Sorry man.

Thanks for the help!
Title: Re: Bucket's Questions
Post by: bucket on March 01, 2008, 11:49:20 pm: How would you find the intercept and asymptote of the graph $2log_22x+2$
Title: Re: Bucket's Questions
Post by: Collin Li on March 01, 2008, 11:58:20 pm: $y = 2\log_22x+2$

$x\mbox{-intercept(s)}$ : $y = 0$

$0 = 2\log_22x+2$

$\implies \log_22x = -1$

$\implies 2x = 2^{-1}$

$\implies x = \frac{1}{4}$

Therefore, the x-intercept is at $\left(\frac{1}{4},\, 0\right)$

$y\mbox{-intercept}$ : $x = 0$

There are no solutions. The domain of this graph is $x > 0$ , so $x=0$ cannot be put into this equation. We know this because we know the logarithm function only accepts values greater than zero.

The asymptote is: $x=0$ . The asymptote for a typical logarithm graph is always $x=0$ unless there have been transformations applied to it (specifically, a translation parallel to the x-axis). In this case, there has been no translation, so the asymptote has not moved. The asymptote corresponds to the "edge" (not mathematical term) of the domain. For example, the domain is $x > 0$ , so the asymptote is $x=0$ .
Title: Re: Bucket's Questions
Post by: bucket on March 02, 2008, 12:19:47 am: Cheers coblin.
Title: Re: Bucket's Questions
Post by: bucket on March 08, 2008, 01:59:03 pm: $2^{x-1}=3^{x+1}$
solve for x expressing answer as the exact value.
Title: Re: Bucket's Questions
Post by: Ahmad on March 08, 2008, 02:17:41 pm: Method 1
Notice that $3 = 2^{\log_2 (3)}$

Therefore, $2^{x-1} = (2^{\log_2 (3)})^{x+1}$

$x - 1 = (x+1) \log_2 {3}$

Method 2
Take log of both sides

$\log 2^{x-1} = \log 3^{x+1}$

$(x-1) \log 2 = (x+1) \log 3$
Title: Re: Bucket's Questions
Post by: bucket on March 08, 2008, 02:28:46 pm: thanks ahmed... uhm is there any way to make x the subject?
Title: Re: Bucket's Questions
Post by: Glockmeister on March 08, 2008, 03:14:14 pm: With the second method

Divide the Left Hand Side by $(x+1)$ .
Divide the Right Hand Side by $\log 2$

You end up getting $\frac{x-1}{x+1} = \frac{\log 3}{\log 2}$

Then you can use long division to break that up.

_1____
x + 1 | x - 1
x + 1
-------
- 2

Thus $1 - \frac {2}{x+1} = \log_{2} 3 (because \frac{\log 3}{\log 2} = \log_{2} 3 )$

The rest is just simple algebraic manipulation.
Title: Re: Bucket's Questions
Post by: Neobeo on March 08, 2008, 03:55:30 pm: Traditionally, you could just expand all the terms and collect the $x$ es to one side.
Title: Re: Bucket's Questions
Post by: Glockmeister on March 08, 2008, 04:39:06 pm: True that, what didn't I think of that? lol
Title: Re: Bucket's Questions
Post by: Mao on March 08, 2008, 08:46:12 pm: if you multiplied by two on both sides:

$2^x=2 \cdot 3^{x+1}$

then take log₂ on both sides

$x=log_2 2 + log_2 3^{x+1}$

$x=1+(x+1)log_2 3$

$x=log_2 3 \cdot x + 1 + log_2 3$

:D didnt even touch rational functions :D
Title: Re: Bucket's Questions
Post by: bucket on March 08, 2008, 09:39:52 pm: ah thanks for the help guys! :D
I have another question =\

If $a log_27 = 3-log_614$ , find the value of a

note:the answer needs to be given in the approximate form, and you can only enter $log_e$ and $log_{10}$ into the calculator.
Title: Re: Bucket's Questions
Post by: Toothpaste on March 08, 2008, 09:49:08 pm: change of base law
$alog_2(7) = 3 - (\frac{log_{10}{14}}{log_{10}{6}})$

$\implies alog_2(7) = 1.52711$

$\implies a = \frac{1.52711}{log_2(7)}$

$\implies a = \frac{1.52711}{(\frac{log_{10}{7}}{log_{10}{2}})}$

$\implies \therefore a = 0.543969$
Title: Re: Bucket's Questions
Post by: bucket on March 08, 2008, 09:51:58 pm: thanks toothpick you legend.
Title: Re: Bucket's Questions
Post by: Glockmeister on March 08, 2008, 10:26:31 pm: Quote from: Mao on March 08, 2008, 08:46:12 pm
if you multiplied by two on both sides:

$2^x=2 \cdot 3^{x+1}$

then take log₂ on both sides

$x=log_2 2 + log_2 3^{x+1}$

$x=1+(x+1)log_2 3$

$x=log_2 3 \cdot x + 1 + log_2 3$

:D didnt even touch rational functions :D

That's not 100% simplifed. There's still x'es on both sides.
Title: Re: Bucket's Questions
Post by: Mao on March 08, 2008, 10:31:55 pm: pssst here u go:

$(1-log_2 3) x = 1+log_2 3$

$x=\frac{1 + log_2 3}{1- log_2 3}$

kk? :P

or $x=\frac{log_2 6}{log_2 1.5}$
Title: Re: Bucket's Questions
Post by: /0 on March 09, 2008, 03:41:47 am: Quote from: Mao on March 08, 2008, 10:31:55 pm
pssst here u go:

$(1-log_2 3) x = 1+log_2 3$

$x=\frac{1 + log_2 3}{1- log_2 3}$

kk? :P

or $x=\frac{log_2 6}{log_2 1.5}$

$=\log_{1.5} 6$

:P
Title: Re: Bucket's Questions
Post by: Neobeo on March 09, 2008, 07:26:57 am: But really, Mao has some careless arithmetical errors, which should actually give:
$x=-\log_{1.5}6$
Title: Re: Bucket's Questions
Post by: Mao on March 09, 2008, 11:38:14 am: oh gar.... i dont know how to divide two by three... :(

if you multiplied by two on both sides:

$2^x=2 \cdot 3^{x+1}$

then take log₂ on both sides

$x=log_2 2 + log_2 3^{x+1}$

$x=1+(x+1)log_2 3$

$x=log_2 3 \cdot x + 1 + log_2 3$

$(1-log_2 3) x = 1+log_2 3$

$x=\frac{1 + log_2 3}{1- log_2 3}$

$x=\frac{log_2 6}{log_2 \frac{2}{3}}$

$x=\frac{log_2 6}{-log_2 \frac{3}{2}}$

$x=\frac{log_2 6}{-log_2 1.5}$

$x=-log_{1.5} 6$

LOL at myself...
Title: Re: Bucket's Questions
Post by: Glockmeister on March 09, 2008, 02:12:32 pm: Can anyone else spot any errors in Mao's working? lol
Title: Re: Bucket's Questions
Post by: Mao on March 09, 2008, 02:41:49 pm: Quote from: Glockmeister on March 09, 2008, 02:12:32 pm
Can anyone else spot any errors in Mao's working? lol
bah it's right this time k???!

:'(
Title: Re: Bucket's Questions
Post by: bucket on March 09, 2008, 09:39:57 pm: Ok I have no idea how to work this kind of question out without using a calculator:

b) find the points of intersection of the graphs y=f(x) and y=f^-1(x)

$f(x)=2e^x-4$
and I worked out the inverse to be $f^{-1}(x)=log_e(\frac{x+4}{2})$
Title: Re: Bucket's Questions
Post by: ed_saifa on March 09, 2008, 09:41:04 pm: find the intersection with line y=x i think
Title: Re: Bucket's Questions
Post by: bucket on March 09, 2008, 09:42:50 pm: how do i do that?
do I sketch them both?
=_=
Title: Re: Bucket's Questions
Post by: ed_saifa on March 09, 2008, 09:46:51 pm: maybe. but if you're not supposed to use a calculator, then your going to have to solve it by hand.
Title: Re: Bucket's Questions
Post by: bucket on March 09, 2008, 09:49:20 pm: Yeah...

It doesn't say I can't use a calculator, but I want to know how to solve it without one, if there is a way
Title: Re: Bucket's Questions
Post by: ed_saifa on March 09, 2008, 09:49:59 pm: i have no clue. i can't seem to manipulate the algebra to get x by itself. sorry
Title: Re: Bucket's Questions
Post by: Glockmeister on March 09, 2008, 09:50:46 pm: No, you let $f(x) = f^{-1}(x)$

i.e $2e^{x} - 4 = \log_{e} \frac{x+4}{2}$
Title: Re: Bucket's Questions
Post by: /0 on March 09, 2008, 09:52:34 pm: I don't think you can solve $x=2e^x-4$ analytically. You could use Newton's method to approximate the solutions (http://en.wikipedia.org/wiki/Newton's_Method) however that is time-consuming and messy.
Title: Re: Bucket's Questions
Post by: ed_saifa on March 09, 2008, 09:53:21 pm: wouldn't that be easier than letting it equal the inverse?
Title: Re: Bucket's Questions
Post by: Collin Li on March 09, 2008, 09:53:51 pm: Intersecting any of the two with $y=x$ is the best way to go about it. The inverses of each other intersect if they intersect with their axis of symmetry, which is $y=x$ . You don't need to find the inverse this way!

However, you will need a calculator to solve it explicitly.
Title: Re: Bucket's Questions
Post by: Glockmeister on March 09, 2008, 09:54:44 pm: And not in the Methods course.
Title: Re: Bucket's Questions
Post by: Mao on March 09, 2008, 09:55:39 pm: solving in y=x is generally accepted, but sometimes there can be multiple (or infinite) solutions, such as a symmetry of y=x already exists:

$f:(0,1)\rightarrow R, where f(x)=\sqrt{1-x^2}$ (a quater circle)
which the reverse is itself... (infinite solutions)
Title: Re: Bucket's Questions
Post by: Collin Li on March 09, 2008, 09:56:14 pm: Quote from: Glockmeister on March 09, 2008, 09:54:44 pm
And not in the Methods course.

Knowing the axis of symmetry of inverses is in the Methods course. If you're referring to Newton's method, then yeah, you're right.
Title: Re: Bucket's Questions
Post by: bucket on March 09, 2008, 09:57:21 pm: ah fair go, calculator it is then O_O.
thanks all of you.
Title: Re: Bucket's Questions
Post by: Glockmeister on March 10, 2008, 02:30:36 am: Quote from: coblin on March 09, 2008, 09:56:14 pm
Quote from: Glockmeister on March 09, 2008, 09:54:44 pm
And not in the Methods course.

Knowing the axis of symmetry of inverses is in the Methods course. If you're referring to Newton's method, then yeah, you're right.

Yeah I was referring to the Newton's method, just that you guys dived in with posts whilst I was writing that.
Title: Re: Bucket's Questions
Post by: bucket on March 16, 2008, 12:42:42 pm: if $log_ey=2log_ex + 1$ find the value for y.
Title: Re: Bucket's Questions
Post by: ed_saifa on March 16, 2008, 12:45:24 pm: y= ex^2
Title: Re: Bucket's Questions
Post by: bucket on March 16, 2008, 01:04:12 pm: how did you work that out? O_O
Title: Re: Bucket's Questions
Post by: Mao on March 16, 2008, 01:28:28 pm: $log_e y=2log_e x + 1$

from the log laws:
$log_e y = log_e x^2 + log_e e$

$log_e y = log_e e \cdot x^2$

$y=e \cdot x^2$
Title: Re: Bucket's Questions
Post by: bucket on March 16, 2008, 02:20:28 pm: If $log_{10}(x-2)-3log_{10}2x=1-log_{10}Y$ , then y is equal to?
Title: Re: Bucket's Questions
Post by: Mao on March 16, 2008, 03:01:41 pm: by the same token

$log_{10}(x-2)-log_{10}(2x)^3=log_{10}10 - log_{10}y$

$log_{10}\frac{x-2}{(2x)^3} = log_{10}\frac{10}{y}$

$\frac{x-2}{(2x)^3}=\frac{10}{y}$

$y=80\frac{x^3}{x-2}$
Title: Re: Bucket's Questions
Post by: bucket on March 16, 2008, 05:18:04 pm: the inverse of function $f(x)=\frac{x+2}{x-3}$ is
Title: Re: Bucket's Questions
Post by: unknown id on March 16, 2008, 05:26:03 pm: Quote from: bucket on March 16, 2008, 05:18:04 pm
the inverse of function $f(x)=\frac{x+2}{x-3}$ is

$x = \frac{y+2}{y-3}$

$x(y-3) = y+2$

$xy-3x = y+2$

$xy-y = 3x+2$

$y(x-1) = 3x+2$

$f^{-1}(x) = \frac{3x+2}{x-1}$
Title: Re: Bucket's Questions
Post by: Mao on March 16, 2008, 05:26:39 pm: $\frac{x+2}{x-3} = \frac{x-3+5}{x-3} = 1+\frac{5}{x-3}$

now it will be a lot easier :D
Title: Re: Bucket's Questions
Post by: dcc on March 16, 2008, 05:26:56 pm: $f(x) = \dfrac{(x - 3) + 5}{x - 3} = 1 + \dfrac{5}{x-3}$

Swapping the x & y coordinates:

$x = 1 + \dfrac{5}{y - 3}$

$\dfrac{1}{y-3} = \dfrac{1}{5}(x - 1)$

$y - 3 = \dfrac{5}{x - 1}$

$y = \dfrac{5}{x-1} + 3$
Title: Re: Bucket's Questions
Post by: bucket on March 16, 2008, 06:49:28 pm: ahh are those answers the same? O_O
Title: Re: Bucket's Questions
Post by: Collin Li on March 16, 2008, 06:50:58 pm: $f^{-1}(x) = \frac{3x+2}{x-1} = \frac{3(x-1) + 5}{x-1} = 3 + \frac{5}{x-1}$

Yes (the proof is above: going from unknown's answer to dcc's answer) :)
Title: Re: Bucket's Questions
Post by: bucket on March 16, 2008, 09:39:08 pm: ohh thanks alot everyone!
Title: Re: Bucket's Questions
Post by: bucket on March 19, 2008, 09:47:52 pm: an extremely easy question.
Find $\frac{dy}{dx}$ of $y=\frac{10x^7+2x^2}{x^2}, x\not=0$
I don't understand the $x\not=0$ part.
Title: Re: Bucket's Questions
Post by: Neobeo on March 19, 2008, 10:01:37 pm: Quote from: bucket on March 19, 2008, 09:47:52 pm
an extremely easy question.
Find $\frac{dy}{dx}$ of $y=\frac{10x^7+2x^2}{x^2}, x\not=0$
I don't understand the $x\not=0$ part.

Basically they are just limiting the domain for you, in effect simplifying the equation, since you cannot divide anything by zero. Since they let $x\neq0$ , then $x^2\neq0$ , so you will not be dividing by zero. This means that you can just divide the whole thing to make $y=10x^5+2$ which should be a lot simpler.
Title: Re: Bucket's Questions
Post by: Glockmeister on March 19, 2008, 10:15:37 pm: You know, I've never thought about the $x\neq0$ in that way before. I would've had just ran straight to the quotient rule.
Title: Re: Bucket's Questions
Post by: bucket on March 19, 2008, 10:39:53 pm: That explains why I got that question wrong :P, thanks a lot neobeo.
Title: Re: Bucket's Questions
Post by: bucket on March 20, 2008, 04:53:53 pm: in a question such as the one above, how do you prove that $x\not=0$ ?
Title: Re: Bucket's Questions
Post by: Collin Li on March 20, 2008, 06:00:48 pm: Quote from: bucket on March 20, 2008, 04:53:53 pm
in a question such as the one above, how do you prove that $x\not=0$ ?

You don't prove it. It is just a specified domain. It's just telling you, do this task for the domain $x \neq 0$ .
Title: Re: Bucket's Questions
Post by: Mao on March 20, 2008, 08:27:38 pm: $y=\frac{10x^7+2x^2}{x^2}$

This is a division operation, and the absolute no-no is to divide something by 0

so $y=\frac{10x^7+2x^2}{x^2}$ is undefined when $x^2=0$

$\therefore x \not= 0$

is that what you wanted to know?
Title: Re: Bucket's Questions
Post by: AppleXY on March 20, 2008, 08:34:47 pm: Quote from: Glockmeister on March 19, 2008, 10:15:37 pm
You know, I've never thought about the $x\neq0$ in that way before. I would've had just ran straight to the quotient rule.

Yeah, using the quotient rule is not most efficient method to diff in this case.
Title: Re: Bucket's Questions
Post by: bucket on March 20, 2008, 08:35:08 pm: yeah thanks mao that makes sense.
theres questions in the exercise book asking that =\
Title: Re: Bucket's Questions
Post by: Mao on March 20, 2008, 08:46:57 pm: Quote from: AppleXY on March 20, 2008, 08:34:47 pm
Quote from: Glockmeister on March 19, 2008, 10:15:37 pm
You know, I've never thought about the $x\neq0$ in that way before. I would've had just ran straight to the quotient rule.

Yeah, using the quotient rule is not most efficient method to diff in this case.

$y=10x^5+2$ is algebraicly equivalent to $y=\frac{10x^7+2x^2}{x^2}$
but you need to include that $x \neq 0$ in your final solution

using the quotient rule avoids having to attach this restraint, as the $x^4$ denominator will imply that it is the case

however, just to be safe, always state the domain of wierd looking equations =D

another thing to be careful with is differentiability... that's a whole new can of worms :P
Title: Re: Bucket's Questions
Post by: Glockmeister on March 20, 2008, 09:42:02 pm: Yep, curves must be 'smooth' and continuous, IRRC.

I'm sure there's a more mathematical way of stating this.
Title: Re: Bucket's Questions
Post by: bucket on March 20, 2008, 10:35:26 pm: Quote from: Glockmeister on March 20, 2008, 09:42:02 pm
Yep, curves must be 'smooth' and continuous, IRRC.

I'm sure there's a more mathematical way of stating this.
rofl thats exactly what my tutor told me.
Title: Re: Bucket's Questions
Post by: bucket on March 20, 2008, 11:45:09 pm: for the function $3(x-1)^2$ , how do you find when $f'(x)>0$ ?
Title: Re: Bucket's Questions
Post by: Collin Li on March 20, 2008, 11:46:47 pm: $f'(x) = 6(x-1)$

$f'(x) > 0 \implies 6(x-1) > 0 \implies x > 1$
Title: Re: Bucket's Questions
Post by: bucket on March 21, 2008, 01:06:41 am: omg that was such a stupid question rofl.
this one i dont know what to do with either

Given that the curve $y=ax^2+\frac{b}{x}$ has a gradient of -5 at the point (2,-2), find the value of a and b
Title: Re: Bucket's Questions
Post by: Collin Li on March 21, 2008, 01:24:58 am: $\frac{dy}{dx} = 2ax - \frac{b}{x^2}$

At the point (2, -2), the gradient is -5. In other words, when $x=2$ , $\frac{dy}{dx} = -5$

$\implies -5 = 2a(2) -\frac{b}{2^2} \implies -5 = 4a - \frac{b}{4}\mbox{ ------ (1)}$

Now, to get another equation (trying to set up simultaneous equations):

The point (2, -2) on the curve tells us that when $x=2$ , $y=-2$

$\implies -2 = a(2)^2 + \frac{b}{2} \implies -2 = 4a + \frac{b}{2}\mbox{ ------ (2)}$

Subtracting (1) from (2) yields:

$-2 + 5 = \frac{b}{2} - \frac{b}{4} \implies 3 = \frac{b}{4} \implies b = 12$

Therefore, $a = -2$
Title: Re: Bucket's Questions
Post by: unknown id on March 21, 2008, 06:14:14 pm: Quote from: coblin on March 21, 2008, 01:24:58 am
Subtracting (1) from (2) yields:

$-2 + 5 = \frac{b}{2} - \frac{b}{4} \implies 3 = \frac{b}{4} \implies b = 12$

Therefore, $a = -2$

$-2 + 5 = \frac{b}{2} - \frac{-b}{4}$

$3 = \frac{3b}{4}$

$b = 4$

Thus, $a = -1$
Title: Re: Bucket's Questions
Post by: bucket on April 12, 2008, 02:14:54 am: Hey guys, I'm back again :P

I seem to be able to do these problems half way but then don't know where to go afterwards.

'Find the derivative of $4x^2(2x^2+1)^2$ with respect to x using the product rule.'

I get up to $(4x^2 \cdot8x(2x^2+1)) + ((2x^2+1)^2\cdot8x)$ but then they seem to simplify the answer quite a bit.

The answer they get is $8x(2x^2+1)(6x^2+1)$ , how do they get to this?

Is my working out completely wrong? lol
Title: Re: Bucket's Questions
Post by: ed_saifa on April 12, 2008, 08:12:17 am: 'Find the derivative of $4x^2(2x^2+1)^2$ with respect to x using the product rule.'

"Let $u= 4x^2 and v= (2x^2+1)^2$

Therefore $\frac{d}{dx} 4x^2(2x^2+1)^2 =v\frac{du}{dx} + u\frac{dv}{dx}$

$\frac{du}{dx}= 8x \frac{dv}{dx}= 8x(2x^2+1)$

Putting this all together gives

$8x(2x^2+1)^2 + 4x^2.2.4x(2x^2+1)$

$8x(2x^2+1)^2 +32x^3(2x^2+1)$

Factor out $8x$

$8x( (2x^2+1)^2 +4x^2(2x^2+1)$

Factor out $(2x^2+1)^2$

$8x(2x^2+1)(2x^2+1+4x^2)$

$8x(2x^2+1)(6x^2+1)$
Title: Re: Bucket's Questions
Post by: bucket on April 12, 2008, 03:29:26 pm: mm thanks ed
Title: Re: Bucket's Questions
Post by: bucket on May 25, 2008, 11:40:05 pm: I have a questionnn........................................

Find the derivative of $e^{4x+1}(x+1)^2$

and also of $e^{-4x}\sqrt{(x+1)}$
Title: Re: Bucket's Questions
Post by: Glockmeister on May 26, 2008, 12:32:58 am: They both use the chain rule

e.g The first one: $u=e^{4x+1}$ $v=(x+1)^{2}$

When you find out the derivative of v, you use the chain rule

$\frac{dv}{dx} = 2(x+1)$

And then you apply for the formula.

It's a similar technique for the second one.

If you can't solve it, I'll work out the solution tomorrow, when I'm a bit more awake.
Title: Re: Bucket's Questions
Post by: ell on May 26, 2008, 01:06:06 pm: For the second one:
$u=e^{-4x}$ and $v=\sqrt{x+1}=(x+1)^{1/2}$

$\frac{du}{dx}=-4e^{-4x}$ and $\frac{dv}{dx}=\frac{1}{2\sqrt{x+1}}$ (chain rule)

Apply product rule:
$\begin{align*} \frac{dy}{dx}&=v\frac{du}{dx}+u\frac{dv}{dx}\\ &=\sqrt{x+1}\cdot-4e^{-4x}+e^{-4x}\cdot\frac{1}{2\sqrt{x+1}}\\ &=\frac{-4\sqrt{x+1}}{e^{4x}}+\frac{1}{2e^{4x}\sqrt{x+1}}\\\end{align*}$

Get a common denominator:
$\begin{align*} \frac{dy}{dx}&=\frac{-4\sqrt{x+1}}{e^{4x}}\cdot\frac{2\sqrt{x+1}}{2\sqrt{x+1}}+\frac{1}{2e^{4x}\sqrt{x+1}}\\ &=\frac{-8(x+1)+1}{2e^{4x}\sqrt{x+1}}\\ &=\frac{-8x-7}{2e^{4x}\sqrt{x+1}}\\ \end{align*}$
Title: Re: Bucket's Questions
Post by: bucket on May 26, 2008, 10:51:09 pm: ahh thanks alot both of you.
x
Title: Re: Bucket's Questions
Post by: bucket on June 17, 2008, 09:33:21 pm: I have forgotten how to derive after two weeks w/o maths :S.

$f(x)=\dfrac{1}{8}(x-1)^3(8-3x)+1$ , show that $f'(x)=\dfrac{3}{8}(x-1)^2(9-4x)$ and specify when $f'(x)\geq0$
Title: Re: Bucket's Questions
Post by: Mao on June 17, 2008, 09:48:44 pm: Quote from: bucket on June 17, 2008, 09:33:21 pm
I have forgotten how to derive after two weeks w/o maths :S.

$f(x)=\dfrac{1}{8}(x-1)^3(8-3x)+1$ , show that $f'(x)=\dfrac{3}{8}(x-1)^2(9-4x)$ and specify when $f'(x)\geq0$

last thing before bed :) [hey polky :P ]

differentiation [by product rule]:

$f'(x)=\frac{1}{8}\cdot \left( 3(x-1)^2(8-3x)+(-3)(x-1)^3\right)=\frac{1}{8}[-3(x-1)+3(8-3x)](x-1)^2=\frac{3}{8}(9-4x)(x-1)^2$

when $f'(x)\geq 0$ , it implies that the function will be "positive"

looking at $f'(x)=\dfrac{3}{8}(x-1)^2(9-4x)$ , we can see three parts:

$\frac{3}{8}$ which is constant $\implies$ always positive

$(x-1)^2$ , which is a square $\implies \geq 0$

$9-4x$ , the only terms that changes sign. therefore, when this part is greater than (or equals to 0), the derivative will be positive (or 0)

$\implies x\leq \frac{9}{4}$
Title: Re: Bucket's Questions
Post by: bucket on June 17, 2008, 09:50:53 pm: hah thanks mao.
Title: Re: Bucket's Questions
Post by: bucket on June 17, 2008, 10:58:37 pm: Find the two positive numbers whose sum is 4 and such that the sum of the cube of the first and the square of the second is as small as possible.

The bolded part of the question is what's confusing me, what the hell do you equate these two numbers to if you don't know what 'as small as possible' would be?
Title: Re: Bucket's Questions
Post by: /0 on June 18, 2008, 01:17:38 am: Let the two positive numbers by x and y.

$x+y=4$

You want to minimise $x^3+y^2$ . Let $A = x^3+y^2$ (just label what u wanna minimize)

Now, from the first equation, you have $y=4-x$

$\Rightarrow A = x^3+(4-x)^2$

Now, find $\frac{\,dA}{\,dx}$ and find the nature of $A$ 's stationary points ;)
Title: Re: Bucket's Questions
Post by: bucket on June 18, 2008, 04:35:54 pm: oh :S thanks.
lol i feel stupid after that one XD
Title: Re: Bucket's Questions
Post by: bucket on June 18, 2008, 11:22:27 pm: "Find the point on the parabola $y=x^2$ that is closest to the point $(3,0)$ "
:S what the fuck do you do with this..? tangents?
Title: Re: Bucket's Questions
Post by: Glockmeister on June 18, 2008, 11:46:36 pm: I havent looked at the question in detail yet, but my hunch is that it involved the distance formula for some reason.
Title: Re: Bucket's Questions
Post by: bucket on June 18, 2008, 11:56:00 pm: but (3,0) isn't a point on the graph. it's just a point. lol
Title: Re: Bucket's Questions
Post by: ell on June 19, 2008, 01:00:52 pm: Quote from: bucket on June 18, 2008, 11:22:27 pm
"Find the point on the parabola $y=x^2$ that is closest to the point $(3,0)$ "
:S what the fuck do you do with this..? tangents?

Let $(x, y)$ be the points that are closest to $(3, 0)$ .

As Glockmeister said we need to use the distance formula. The distance between those two points are:

$D=\sqrt{(x-3)^2+(y-0)^2}$

Since we know $y=x^2$ we can sub this in:

$D=\sqrt{(x-3)^2+(x^2)^2}$

$D=\sqrt{(x-3)^2+x^4}$

We need to find when the distance is minimal, i.e. when $\frac{dD}{dx}=0$ . We can also just ignore the square root as we are going to make $\frac{dD}{dx}=0$ (since if we derive $D=\sqrt{(x-3)^2+x^4}$ then $\frac{dD}{dx}=0=\frac{1}{2}\left((x-3)^2+x^4\right)^{-\frac{1}{2}}(4x^3+2x-6)$ )

$\frac{dD}{dx}=4x^3+2x-6$

$0=4x^3+2x-6$

We need to factorise this to find solutions. Let $P(x)=4x^3+2x-6$ , then:

$P(1)=4(1)^3+2(1)-6=0$

Hence $x=1$ .

We also need to ensure that this stationary point is a minimum. Find a point before and after $x=1$ and substitute into $\frac{dD}{dx}$ to verify the nature.

$P(0)=-6$

$P(2)=30$

Drawing a gradient table: \ _ / hence minimum. (Also by $\frac{d^2D}{dx^2}=12(1)^2+2=14>0$ therefore minimum).

Substitute $x=1$ into $y=x^2$ :

$y=1$

Therefore the point closest to $(3, 0)$ on the parabola $y=x^2$ is $(1, 1)$ .
Title: Re: Bucket's Questions
Post by: bucket on June 19, 2008, 04:25:16 pm: hah thanks, thats good.
Title: Re: Bucket's Questions
Post by: bucket on June 19, 2008, 09:47:44 pm: The surface area of a cube is changing at the rate of 8cm²/s. How fast is the volume changing when the surface area is 60cm?
Title: Re: Bucket's Questions
Post by: /0 on June 19, 2008, 10:08:13 pm: Let A be surface area ( $cm^2$ ), V be volume ( $cm^3$ ), t be time (s).

You have $\frac{dA}{dt} = 8$

You need $\frac{dV}{dt}$

$\frac{dV}{dt} = \frac{dV}{dA} \times \frac{dA}{dt}$

To find $\frac{dV}{dA}$ ,

$V = \left(\sqrt{\frac{A}{6}}\right)^3 = \left(\frac{A}{6}\right)^{\frac{3}{2}}$

$\Rightarrow \frac{dV}{dA} = \frac{1}{4}\sqrt{\frac{A}{6}}$

$\therefore \frac{dV}{dt} = \frac{1}{4}\sqrt{\frac{A}{6}} \times 8 = 2\sqrt{\frac{A}{6}}$

When $A = 60cm^2$

$\frac{dV}{dt} = 2\sqrt{\frac{60}{6}} = 2\sqrt{10}$

$\therefore$ the volume is increasing at a rate of $2\sqrt{10} cm^3/s$
Title: Re: Bucket's Questions
Post by: bucket on June 19, 2008, 10:37:39 pm: perfect, thanks a lot man.
Title: Re: Bucket's Questions
Post by: bucket on June 21, 2008, 11:19:12 pm: A vessel has such a shape that when the depth of the water in it is x cm, the volume is given by the equation $V=108x + x^2$ . Water is poured in at a constant rate of $30cm/s$ . At what rate is the water rising when the depth is 8cm?
Title: Re: Bucket's Questions
Post by: lwine on June 21, 2008, 11:29:52 pm: We are looking for (I assume you mean 30cm^3/s)

$\dfrac{dx}{dt}=\dfrac{dV}{dt}\times\dfrac{dx}{dV}$ This is an application of the chain rule to relate to different rates.

$\dfrac{dV}{dx}=2x+108$ This step basically differentiates your Volume formula with respect to x.

$\dfrac{dx}{dt}=30\times\dfrac{1}{2x+108}$ Notice how the conjugate of dV/dx is used, because we require dx/dV.

When $x=8$ :

$\dfrac{dx}{dt}=30\times\dfrac{1}{16+108}=\dfrac{15}{62}cm/s$
Title: Re: Bucket's Questions
Post by: bucket on June 21, 2008, 11:32:52 pm: Ah thanks.
This question confused me, i didn't know exactly what rate of change I was looking for XD.
This helped.
Title: Re: Bucket's Questions
Post by: lwine on June 21, 2008, 11:42:40 pm: When you encounter these kind of related rates questions (there usually isn't too much variety in our textbooks), begin by writing what I call the KING EQUATION.

That is, set out exactly what rate they're looking for (whether its the change in height or the change in radius or etc; its almost always with respect to time), and write down the rate they give you (also with respect to time) and try to "fill in" the missing relation, based on the chain rule.

Then you should set out to find that missing relation, and throughout the process, keep referring to your KING EQUATION at the top to make sure you're on the right track.
Title: Re: Bucket's Questions
Post by: bucket on June 21, 2008, 11:58:23 pm: $f(x)=x^2e^x$ , Find $x:f'(x)<0$
Title: Re: Bucket's Questions
Post by: lwine on June 22, 2008, 12:06:08 am: Using the product rule:
$f'(x)=2xe^x+x^2e^x$ (As my methods teacher says: diff the first one, leave the second one alone, then diff the second one, and leave the first one alone)

$2xe^x+x^2e^x<0$

$e^x(2x+x^2)<0$ (we know e^x cannot equal 0)

$2x+x^2<0$

$x(2+x)<0$

$-2<x<0$
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 12:11:25 am: ah thanks heaps! your explanations are really easy to follow (Y).
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 12:28:27 am: gah, is it possible to find the equation of an asymptote, say for $f(x)=x+e^{-x}$ without graphing it?
Title: Re: Bucket's Questions
Post by: /0 on June 22, 2008, 12:45:20 am: Take limits as $x \to \pm \infty$ . It's an intuitive kind of thing, you need to predict how parts of the function will behave with large numbers.

As $x \rightarrow \infty$ , $e^{-x}$ becomes negligibly small, so $f(x) \rightarrow x$ .

As $x \rightarrow -\infty$ , both $x$ and $e^{-x}$ become very pronounced, and you can't really neglect either one, so there is no asymptote as $x \rightarrow -\infty$ .

So the only asymptote is $y=x$

As another example,

$y = \frac{x^3+2x+1}{(x-1)(x+3)}=\frac{8}{x+3}+\frac{1}{x-1}+x-2$

As $x \to \pm \infty$ , the x in the denominator of the fractions will be so big that they practically disappear, leaving the asymptote $y=x-2$ .
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 12:49:27 am: O_O I don't get it... lol.
Guess i'll just stick with graphing it then >.<, thanks anyway man.
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 02:51:36 pm: Given that $\frac{dy}{dx}=6x-4$ , find the approximate change in y as x increases from -2 to -1.97

and also

"A manufacturing company has a daily output y on day t of a production run given by $y=600(1-e^{-0.5t})$ . Find the instantaneous rate of change of output y with respect to t on the 10th day."

I'm so fucked for this SAC >.<.
Title: Re: Bucket's Questions
Post by: lwine on June 22, 2008, 03:45:48 pm: Mmm, I hate approximation questions.

The best formula for this scenario is $\Delta y\approx\dfrac{dy}{dx}\times\Delta x$

Considering the values given, this imples:

$\Delta y\approx(6x-4)\times\Delta x$

$\Delta y\approx\left(6(-2)-4\right)\times 0.03$

$\therefore\Delta y\approx -0.48$

The other question:

Differentiate $y=600-600e^{-0.5t}$

$\dfrac{dy}{dt}=300e^{-0.5t}$

$\text{Therefore, on the tenth day, when } t=10$

$\dfrac{dy}{dt}=300e^{-0.5\cdot 10}$

$\dfrac{dy}{dt}=2.02\text{ y per day}$
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 04:13:58 pm: ah thanks a lot man
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 07:45:18 pm: $f(x)=e^{ax}$ , find an approximation for f(h) where h is small in terms of h and a.

and I have another question, when the hell do we use the formula $f(x+h) \approx h \cdot f'(x) + f(x)$ ??
:S
Title: Re: Bucket's Questions
Post by: Mao on June 22, 2008, 08:46:10 pm: Quote from: bucket on June 22, 2008, 07:45:18 pm
$f(x)=e^{ax}$ , find an approximation for f(h) where h is small in terms of h and a.

and I have another question, when the hell do we use the formula $f(x+h) \approx h \cdot f'(x) + f(x)$ ??
:S

answered your own question :P

$f(h)=f(0+h)\approx f(0)+h\cdot f'(0)$

$\implies f(h)\approx 1+a\cdot h$
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 09:09:59 pm: wow wtf lol. thanks.
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 10:14:07 pm: $y=sin2x, x=0$ , Find the tangent at the given point.
Title: Re: Bucket's Questions
Post by: lwine on June 22, 2008, 10:26:40 pm: The tangent is always a straight line in the form $y=mx+c$

To find $m$ , the gradient, we must differentiate $y=sin2x$ and find the instantaneous rate of change at the given point.

$\dfrac{dy}{dx}=2cos2x$

$\text{let }x=0\Rightarrow\dfrac{dy}{dx}=2cos0=2=m$

$\therefore y=2x+c$

To find $c$ , we can substitute in the ordered pair at that point:
$\text{When }x=0, y=sin2(0)=0$

$\therefore \text{sub }(0,0)$

$0=2(0)+c\Rightarrow c=0$

Therefore, the tangent is:
$y=2x$
Title: Re: Bucket's Questions
Post by: bucket on June 22, 2008, 10:45:09 pm: ah thanks a tonne =]
Title: Re: Bucket's Questions
Post by: bucket on June 24, 2008, 12:45:40 am: Find $a$ for which $(1-2a)x^2+5ax+(a-1)(a-8)$ is divisible by $(x-2)$ but not $(x-1)$ .
Title: Re: Bucket's Questions
Post by: Glockmeister on June 24, 2008, 01:00:15 am: Basically (because it's late)

You let the function P(2)=0 i.e let x=2, and the equation equal zero.
Then you let P(1)=0 and then reject the solution of the P(2)=0 that matches the solution in P(1)=0
Title: Re: Bucket's Questions
Post by: bucket on June 24, 2008, 01:27:35 am: oh, never thought of that.. thanks alot man =]
Title: Re: Bucket's Questions
Post by: Glockmeister on June 24, 2008, 05:16:42 pm: Someone asked a similar question a few months before actually, I just couldn't find the post.
Title: Re: Bucket's Questions
Post by: bucket on June 24, 2008, 09:49:34 pm: Prove that $x^3+(k+1)x^2+(k-9)x-7$ is divisible by $x+1$ for all values of k.

How would you show this :s.
Title: Re: Bucket's Questions
Post by: Mao on June 24, 2008, 10:19:08 pm: Quote from: bucket on June 24, 2008, 09:49:34 pm
Prove that $x^3+(k+1)x^2+(k-9)x-7$ is divisible by $x+1$ for all values of k.

How would you show this :s.

try subbing in $x=-1$ :)
Title: Re: Bucket's Questions
Post by: bucket on June 24, 2008, 10:20:05 pm: :uglystupid2: I solved it incorrectly when I did it myself :s.
Title: Re: Bucket's Questions
Post by: Flaming_Arrow on June 24, 2008, 10:44:31 pm: $-1 + 1(k+1)-1(k-9)-7 = 0$

$-1 + k + 1 - k + 9 - 7 = 0$

$k - k + 9 - 7 = 0$

$2 = 0$

epic phail
Title: Re: Bucket's Questions
Post by: bucket on June 24, 2008, 11:11:00 pm: lol i wrote it incorrectly.
it's (k-1)

epic fail to me.
Title: Re: Bucket's Questions
Post by: Flaming_Arrow on June 24, 2008, 11:24:36 pm: Quote from: bucket on June 24, 2008, 11:11:00 pm
lol i wrote it incorrectly.
it's (k-1)

epic fail to me.

in that case

$-1 + 1(k-1)-1(k-9)-7 = 0$

$-1 + k -1 - k + 9 - 7 = 0$

$-2 + k - k + 9 - 7 = 0$

$-2+ 2 = 0$

$0 = 0$

k's will cancel out hence the value k is negligible
Title: Re: Bucket's Questions
Post by: bucket on June 24, 2008, 11:25:37 pm: yeah i worked it out after realising it was k-1.
thanks anyway lol.
Title: Re: Bucket's Questions
Post by: Flaming_Arrow on June 24, 2008, 11:27:42 pm: Quote from: bucket on June 24, 2008, 11:25:37 pm
yeah i worked it out after realising it was k-1.
thanks anyway lol.

lol np im bored so yer xD
Title: Re: Bucket's Questions
Post by: bucket on July 24, 2008, 08:01:01 pm: The curve for $\frac{dy}{dx}=e^{kx}$ is such that the tangent at $(1,e^{2})$ passes through the origin. find the gradient of the tangent and hence determine:
a. The value of K
b. The equation of the curve.
Title: Re: Bucket's Questions
Post by: Mao on July 24, 2008, 08:20:19 pm: the equation of a tangent is given by the generic equation $y_t-y_0=m\cdot (x-x_0)$

at $x=1,\; \frac{dy}{dx}=e^k$

$y_t=e^k\cdot (x-1)+e^2$

since it passes through the origin, (0,0), substituting:

$0=e^k \cdot -1 + e^2$

$\implies k=2$

antidifferentiating gives

$\implies y=\frac{1}{2}e^{2x}+\frac{e^2}{2}$
Title: Re: Bucket's Questions
Post by: Glockmeister on July 24, 2008, 08:23:23 pm: The equation you have there is the gradient equation for a particular curve, y. With this particular curve, we are told that at $(1,e^{2})$ , its tangent will intercept the origin. So:

$\mbox{Let x = 1}$
$\frac{dy}{dx} = e^x$

Now, since we know that $\frac{dy}{dx}= m_T$

$y - y_1 = m(x-x_1)$

$y - e^{2} = e^{x}(x-1)$ ...

Mao! Stop beating me to things. :(
Title: Re: Bucket's Questions
Post by: bucket on July 24, 2008, 08:36:56 pm: thankyou my darlings.
Title: Re: Bucket's Questions
Post by: bucket on July 24, 2008, 09:35:28 pm: $\int_{-2}^2 \frac{e^x+e^{-x}}{2}$

help :P
Title: Re: Bucket's Questions
Post by: Toothpaste on July 24, 2008, 09:46:00 pm: Quote from: bucket on July 24, 2008, 09:35:28 pm
$\int_{-2}^2 \frac{e^x+e^{-x}}{2}$

help :P

$\int_{-2}^2 \frac{e^x+e^{-x}}{2}dx$

= $\frac{1}{2}\int_{-2}^2 (e^x+e^{-x})dx$
= $\frac{1}{2}\int_{-2}^2 (e^x+e^{-x})dx$
= $\frac{1}{2}\int_{-2}^2 (e^x+e^{-x})dx$
= $\frac{1}{2}[-e^{-x}+e^{x}]_{-2}^2$
= $\frac{1}{2}[-e^{-2}+e^{2}-( -e^{2}+e^{-2}) ]$
= $\frac{1}{2}[2e^{2}-2e^{-2}]$
= $e^{2}-e^{-2}$
Title: Re: Bucket's Questions
Post by: bucket on July 24, 2008, 09:50:16 pm: thanksssssss
Title: Re: Bucket's Questions
Post by: Collin Li on July 24, 2008, 09:51:27 pm: You can anti-differentiate term by term (like differentiation)

How do you integrate $e^x$ ? Or $e^{ax}$ for that matter? Then, $a = -1$ .
Title: Re: Bucket's Questions
Post by: Toothpaste on July 24, 2008, 09:57:17 pm: Quote from: bucket on July 24, 2008, 09:50:16 pm
continuee....i got THAT far :P
eulers constant confuses me >.<
done :)
Title: Re: Bucket's Questions
Post by: Mao on July 25, 2008, 04:15:51 pm: Quote from: Toothpick on July 24, 2008, 09:46:00 pm
Quote from: bucket on July 24, 2008, 09:35:28 pm
$\int_{-2}^2 \frac{e^x+e^{-x}}{2}$

help :P

$\int_{-2}^2 \frac{e^x+e^{-x}}{2}dx$

= $\frac{1}{2}\int_{-2}^2 (e^x+e^{-x})dx$
= $\frac{1}{2}\int_{-2}^2 (e^x+e^{-x})dx$
= $\frac{1}{2}\int_{-2}^2 (e^x+e^{-x})dx$
= $\frac{1}{2}[-e^{-x}+e^{x}]_{-2}^2$
= $\frac{1}{2}[-e^{-2}+e^{2}-( -e^{2}+e^{-2}) ]$
= $\frac{1}{2}[2e^{2}-2e^{-2}]$
= $e^{2}-e^{-2}$

you could've done it using hyperbolic functions with MUEP/UMEP knowledge :P
Title: Re: Bucket's Questions
Post by: bucket on July 27, 2008, 03:28:30 pm: $\int _{-2}^7 (x+2)(7-x) dx$

:buck2:
Title: Re: Bucket's Questions
Post by: Toothpaste on July 27, 2008, 03:41:31 pm: Quote from: bucket on July 27, 2008, 03:28:30 pm
$\int _{-2}^7 (x+2)(7-x) dx$

:buck2:

$\int _{-2}^7 (x+2)(7-x) dx$

= $\int _{-2}^7 (-x^2+5x+14) dx$

= $[\frac{-x^3}{3}+\frac{5x}{2}+14x]_{-2}^7$

= $[\frac{-7^3}{3}+\frac{35}{2}+14(7)]-[\frac{-(-2)^3}{3}+\frac{5(-2)}{2}+14(-2)]$

= $\frac{243}{2}$
Title: Re: Bucket's Questions
Post by: bucket on July 27, 2008, 04:11:49 pm: ahh i see, so the only way to solve this is to expand the polynomial?
Title: Re: Bucket's Questions
Post by: Mao on July 28, 2008, 07:05:16 pm: Quote from: bucket on July 27, 2008, 04:11:49 pm
ahh i see, so the only way to solve this is to expand the polynomial?

you *could* use linear substitution, but that's beyond the scope of the methods course, and that use of this method for this particularly simple case would be superfluous

the easiest way is to expand the polynomial.
Title: Re: Bucket's Questions
Post by: Ahmad on July 28, 2008, 07:12:48 pm: That's one way to do it, but there are surely others, a relatively fast way I thought of for the curious:

$\int_{-2}^7 (x+2)(7-x)\, dx$

Make the substitution, $(7 + 2)u = x + 2$

$9\cdot 81 \int_{0}^{1} u(1-u)\, du = 81 \cdot 9 B(2,2) = 81\cdot 9 \frac{(\Gamma(2))^2}{\Gamma(4)} = \frac{243}{2}$

Where B(m,n) is the Beta function
Title: Re: Bucket's Questions
Post by: dcc on July 28, 2008, 08:15:31 pm: Quote from: Ahmad on July 28, 2008, 07:12:48 pm
That's one way to do it, but there are surely others, a relatively fast way I thought of for the curious:

$\int_{-2}^7 (x+2)(7-x)\, dx$

Make the substitution, $(7 + 2)u = x + 2$

$9\cdot 81 \int_{0}^{1} u(1-u)\, du = 81 \cdot 9 B(2,2) = 81\cdot 9 \frac{(\Gamma(2))^2}{\Gamma(4)} = \frac{243}{2}$

Where B(m,n) is the Beta function

Most wouldn't consider it necessary to use the Beta function to integrate a polynomial :P But then again you are Ahmad :P
Title: Re: Bucket's Questions
Post by: bucket on July 29, 2008, 12:44:54 am: lol wtffff ahmad!!!
i think ill just expand then :p
Title: Re: Bucket's Questions
Post by: bucket on August 11, 2008, 09:33:08 pm: the line with equation $y=4x+c$ is a tangent to the curve with equation $y=x^2-x-5$ . What is the value of c?
Title: Re: Bucket's Questions
Post by: /0 on August 11, 2008, 09:57:22 pm: The line $y=4x+c$ has gradient 4, so you want to find where $y=x^2-x-5$ has gradient 4.

$\frac{dy}{dx} = 2x-1$

$2x-1 = 4 \Rightarrow x = \frac{5}{2}$

$\Rightarrow y = (\frac{5}{2})^2-(\frac{5}{2})-5=-\frac{5}{4}$

So the coordinates of the tangent point are $(\frac{5}{2}, - \frac{5}{4})$ , so

$- \frac{5}{4} = 4(\frac{5}{2})+c$

$c = -\frac{45}{4}$
Title: Re: Bucket's Questions
Post by: bucket on August 11, 2008, 10:02:21 pm: love you.
Title: Re: Bucket's Questions
Post by: bucket on August 11, 2008, 10:14:46 pm: water is draining from a cone-shaped funnel at a rate of 500cm³/min. The cone has base radius of 20cm and a height of 100cm. let h be the depth of water in the funnel at the time t minutes. the rate of decrease of h in cm/min is given by..?
Title: Re: Bucket's Questions
Post by: Mao on August 11, 2008, 10:22:31 pm: $\frac{dh}{dt}=\boxed{\frac{dh}{dV}}\times \frac{dV}{dt}$

we want a relationship between h and V, i.e. volume of a cone:
$V=\frac{1}{3}\pi r^2 h$

by similar triangles, we find that $r=\frac{h}{5}$

$\implies V=\frac{\pi}{75}h^3$

$\implies \frac{dV}{dh}=\frac{\pi h^2}{25}$

$\implies \frac{dh}{dV}=\frac{25}{\pi h^2}$

$\frac{dh}{dt}=\frac{25}{\pi h^2}\times 500$

$\frac{dh}{dt}\right|_{h=100}=\frac{5}{4\pi}\approx 0.398 \mbox{cm min}^{-1}$
Title: Re: Bucket's Questions
Post by: bucket on August 16, 2008, 02:22:22 pm: The probability of a target shooter hitting the bullseye on any one shot is 0.2

What is the smallest number of shots the shooter should make to ensure a probability of more than 0.95 of hitting the bullseye at least once?

In the previous parts of the question I found that $pr(X=0)=0.3277$ and hence $pr(X\geq1)=0.6723$
Title: Re: Bucket's Questions
Post by: ell on August 16, 2008, 07:34:16 pm: Quote from: bucket on August 16, 2008, 02:22:22 pm
The probability of a target shooter hitting the bullseye on any one shot is 0.2

What is the smallest number of shots the shooter should make to ensure a probability of more than 0.95 of hitting the bullseye at least once?

$\Pr(X\geq1)>0.95$

$\implies 1 - \Pr(X<1)>0.95$

$\implies \Pr(X<1)<0.05$ (note the sign change)

Let n be the number of shots the shooter should make.

$\Pr(X<1) = \left( \begin{array}{c} n\\ 0\\ \end{array} \right)\cdot (0.2)^{0}\cdot (0.8)^{n}=(0.8)^{n}$

Now to solve $(0.8)^{n}<0.05$ , use your graphics calculator and enter $y=(0.8)^{x}$ . Go into TBLSET and make TblStart=0. Now go into TABLE and scroll down until you find the first value where Y1 is less than 0.05. It should end up being 0.04398, to which the corresponding X value is 14.

$\therefore n=14$
i.e. the smallest number of shots the shooter should make to ensure a probability of more than 0.95 at least once is 14.
Title: Re: Bucket's Questions
Post by: bucket on August 17, 2008, 03:09:31 pm: good as!!! =]
thanks man, seriously, essential maths is shit at explanations.
Title: Re: Bucket's Questions
Post by: bucket on August 18, 2008, 04:28:44 pm: Hey, with the previous question, what if it asked something like; ensure a probability of more than 0.95 of hitting the bullseye exactly twice?
Title: Re: Bucket's Questions
Post by: Mao on August 18, 2008, 04:34:43 pm: $Pr(X=2)>0.95,\; \implies \binom{n}{2}\cdot (0.2)^2\cdot (0.8)^{n-2}>0.95,\; n\ge 2$

then you'd put into the y editor $\frac{x\cdot (x-1)}{2\times 1}\cdot (0.2)^2\cdot (0.8)^{x-2}$ and look through the table to see where it gets greater than 0.95.

for this case, however, the maximum probability of hitting the bullseye twice is 0.302 at 9 or 10 trials. hence the probability fo hitting the bullseye twice is NEVER greater than 0.95 (or 1/3 to be more realistic)
Title: Re: Bucket's Questions
Post by: bucket on August 18, 2008, 04:36:17 pm: cheers.
Title: Re: Bucket's Questions
Post by: bucket on August 27, 2008, 10:50:28 pm: The length of time, X(minutes), between the arrival of customers at an autobank has a random variable with probability density function:

$ f(x) = \left\{ \begin{array}{cc} \frac{1}{5}e^{-\frac{x}{5}} & x\geq0 \\ 0 & x<0 \\ \end{array} $

Find the probability that more than 12mins elapses between successive customers given that more than 8 minutes has passed.

I translated the question into:
Find $pr(X>12|X>8)$

is that right?
if so how do I solve it now? :P I found that $pr(X>8)=0.202$

These questions that include 'given' always fuck me up :S
Title: Re: Bucket's Questions
Post by: Collin Li on August 27, 2008, 10:54:25 pm: Since the intersection of $X>8$ and $X>12$ is $X>12$ , then the formula for conditional probability becomes:

$\mbox{Pr}(X>12|X>8) = \frac{\mbox{P}(X>12)}{\mbox{P}(X>8)}$
Title: Re: Bucket's Questions
Post by: bucket on August 27, 2008, 10:57:43 pm: how did you find that as the intersection? :|
Title: Re: Bucket's Questions
Post by: Mao on August 27, 2008, 10:57:53 pm: $Pr(A|B)=\frac{Pr(A\cup B)}{Pr(B)}$

to find the conditional probability, it's the probability of BOTH events occuring, divided by the probability of the condition

i.e. in this case, X has to be >12 as well as >8, which simplifies to X>12 (because 12>8)

hence, $Pr(X>12|X>8)=\frac{Pr(X>12)}{Pr(X>8)}=\frac{\int_{12}^{\infty}f(x)\; dx}{\int_8^{\infty}f(x)\; dx}$
for the purpose of methods, if you are using an 84 or older calculator, use a very large number in place of infinity, such as 99999. This will give a fairly accurate answer, as f(x) gets quite small as it approaches to infinity. if you are using 89 or later, you are blessed with the "inifinity" key.

how to analytically evaluate those two integrals are covered in the first year course at uni =]
Title: Re: Bucket's Questions
Post by: Mao on August 27, 2008, 11:00:11 pm: think of a number line:

Code: [Select]
--------|8----|12---------> xxxxxxxxxxxxxxxxxxx (x>8) xxxxxxxxxxxxx (x>12)
the intersection of the two is fairly obvious: x>12
Title: Re: Bucket's Questions
Post by: Collin Li on August 27, 2008, 11:04:30 pm: Quote from: Mao on August 27, 2008, 10:57:53 pm
how to analytically evaluate those two integrals are covered in the first year course at uni =]

What? You can get an exact value with Methods techniques:

$\dfrac{e^{-\frac{12}{5}}}{e^{-\frac{8}{5}}} = e^{-\frac{4}{5}} = 0.449$
Title: Re: Bucket's Questions
Post by: bucket on August 27, 2008, 11:05:40 pm: thanks both of you!
and thank god i have an 89 =]

and wow :S
Title: Re: Bucket's Questions
Post by: shinny on August 27, 2008, 11:25:41 pm: Quote from: coblin on August 27, 2008, 11:04:30 pm
What? You can get an exact value with Methods techniques:
$\dfrac{e^{-\frac{12}{5}}}{e^{-\frac{8}{5}}} = e^{-\frac{4}{5}} = 0.449$

I think he's referring to 'subbing in' the infinities, which normally can't be done, but I think we're meant to know how to do that in methods with an integral like that, since it's e to a negative power, and when you sub in infinity, it just becomes zero (well..approaches).
Title: Re: Bucket's Questions
Post by: Mao on August 27, 2008, 11:46:24 pm: $Pr(X>12|X>8)=\frac{\int_{12}^{\infty}\frac{1}{5}e^{-x/5}\; dx}{\int_8^{\infty}\frac{1}{5}e^{-x/5}\; dx}$

$\implies = \frac{\lim_{t\to \infty}\left( \int_{12}^te^{-x/5}\; dx\right)}{\lim_{t\to \infty}\left( \int_{8}^te^{-x/5}\; dx\right)}$

$\implies =\frac{\lim_{t\to \infty}\left( \left[-5e^{-x/5}\right]_{12}^t\right)}{\lim_{t\to \infty}\left( \left[-5e^{-x/5}\right]_{8}^t\right)}$

$\implies =\frac{e^{-12/5}+0}{e^{-8/5}+0}=e^{-4/5}\approx 0.449$
Title: Re: Bucket's Questions
Post by: bucket on August 28, 2008, 12:16:28 am: $ f(y) = \left\{ \begin{array}{cc} Ay & 0\leq y \leq B \\ 0 & otherwise \\ \end{array} $

find A and B if $\mu=12$
Title: Re: Bucket's Questions
Post by: Collin Li on August 28, 2008, 07:54:10 am: Two unknowns: A and B

Two pieces of information:
1. The probability density function has a total area of 1
2. The mean is 12

Hence, you can solve this. Write out statements 1 and 2 in mathematical speak -- evaluating integrals and simplifying, and then use simultaneous equations (if necessary) to find A or B, then substitute it into the other equation to find the other constant.
Title: Re: Bucket's Questions
Post by: Collin Li on August 28, 2008, 08:15:43 am: 1. $\left[\frac{Ay^2}{2}\right]^B_0 = 1$

$\implies AB^2 = 2$

2. $\left[\frac{Ay^3}{3}\right]^B_0 = 12$

$\implies AB^3 = 36$

Dividing equation (2) by equation (1) yields:

$\frac{AB^3}{AB^2} = \frac{36}{2}$

$\implies B = 18$

$\implies A = \frac{1}{162}$
Title: Re: Bucket's Questions
Post by: excal on August 29, 2008, 02:51:40 am: I might just splice in how number two was derived:

Remember that the expected value of a probability density function (pdf) is $\int_{-\infty}^\infty y . f(y)\, \operatorname{d}y$ .

Because we are bound by the pdf: $0\leq y \leq B$ , we have to restrict our domain. We also know that expected value (mean) is 12.

$\implies 12 = \int_{0}^B y . f(y)\, \operatorname{d}y$

For $0\leq y \leq B$ , using $f(y) = Ay$ :

$12 = \int_{0}^B y . Ay\, \operatorname{d}y$

$12 = \int_{0}^B Ay^2 \, \operatorname{d}y$

$\implies 12 = \left[\frac{Ay^3}{3}\right]^B_0$

And then you reach coblin's step 2.
Title: Re: Bucket's Questions
Post by: bucket on August 30, 2008, 02:35:17 am: yeah, thanks both of you for your help in that last question.

Now I have NO idea how to approach this one:

$ f(x) = \left\{ \begin{array}{cc} k(a^2-x^2) & -a \leq x\leq a \\ 0 & otherwise \\ \end{array} $

b. Find the value of a which gives the standard deviation of 2.

In part a. of the question it made me find that $k=\frac{3}{4a^3}$
Title: Re: Bucket's Questions
Post by: Collin Li on August 30, 2008, 03:15:07 am: $\mbox{Var(}X\mbox{)} = \sigma^2 = \mbox{E[}X^2\mbox{]} - \left[\mbox{E(}X\mbox{)}\right]^2$

$\mbox{E[}X^2\mbox{]} = \int_{-a}^{a} kx^2(a^2-x^2)\;dx = k\left[\frac{a^2x^3}{3} - \frac{x^5}{5}\right]^a_{-a}$

$= k\left[\frac{a^5}{3} - \frac{a^5}{5} - \left(-\frac{a^5}{3} + \frac{a^5}{5}\right)\right]$

$= \frac{4a^5k}{15}$

and

$\mbox{E[}X\mbox{]} = \int_{-a}^{a} kx(a^2-x^2)\;dx = k\left[\frac{a^2x^2}{2} - \frac{x^4}{4}\right]^a_{-a}$

$= k\left[\frac{a^4}{2} - \frac{a^4}{4} - \left(\frac{a^4}{2} - \frac{a^4}{4}\right)\right]$

$= 0$

Therefore:

$\mbox{Var(}X\mbox{)} = \frac{4a^5k}{15} = \frac{a^2}{5}$ (Substituting $k$ in)

Also, $\mbox{Var(}X\mbox{)} = \sigma^2 = (2)^2 = 4$

So: $4 = \frac{a^2}{5}$

$\implies a^2 = 20$

$\implies a = \pm 2\sqrt{5}$

From the domain, since $-a < a \implies 0 < 2a \implies a > 0$

Hence: $a = 2\sqrt{5}$
Title: Re: Bucket's Questions
Post by: bucket on August 30, 2008, 01:29:16 pm: thanks heaps =]
Title: Re: Bucket's Questions
Post by: bucket on September 03, 2008, 10:43:00 pm: mmm.
how do you find the standard deviation of a normal distribution curve by just looking at it?? :S
i know the mean where the maximum occurs...

(http://i7.photobucket.com/albums/y289/Taksi_/usd-1.jpg)
Title: Re: Bucket's Questions
Post by: Collin Li on September 03, 2008, 10:45:45 pm: I guess roughly the 99.7% range is from 120 to 150 in the first one (coz that's where the curve practically touches the x-axis -- it doesn't but it looks like it), so hence 3 standard deviations from the mean is $150-135 = 15$ , and hence the standard deviation is 5.

You'd never get anything like this on the exam though, haha.
Title: Re: Bucket's Questions
Post by: bucket on September 03, 2008, 10:47:29 pm: ah cool.
thanks man =]
Title: Re: Bucket's Questions
Post by: bucket on September 15, 2008, 10:50:17 pm: in normal distrubution; are the probabilities $pr(X<2)$ and $pr(X\leq2)$ going to be the same, if the mean and the standard deviation are the same?
Title: Re: Bucket's Questions
Post by: shinny on September 15, 2008, 10:57:21 pm: Yeh they are, since
$Pr(X\leq2)=Pr(X<2)+\int_{2}^{2}f(x)dx$

$=Pr(X<2)+0$

$=Pr(X<2)$
Title: Re: Bucket's Questions
Post by: bucket on September 15, 2008, 11:06:28 pm: thanks a lot man =]
Title: Re: Bucket's Questions
Post by: bucket on September 15, 2008, 11:44:03 pm: Say that I have a normal distribution function with mean 40 and standard deviation 5.
If I want to find c for $pr(X\leq c)=0.75$ , I go into the stat/list editor, inverse>normals and type in the mean sd and the area 0.75. What do I do if I want to find $pr(x\geq c)=0.75$ ?
Title: Re: Bucket's Questions
Post by: shinny on September 15, 2008, 11:48:54 pm: $Pr(X>c)=0.75$
$1-Pr(X<c)=0.75$
$Pr(x<c)=0.25$
and then just do what u've been doing all along as usual =T

edit: oh, and to know that symmetry property, just draw yourself a graph of a normal distribution and its quite easy to work out why its the case, as well as working out some of the harder symmetry properties such as how to interpret Pr(a<Z<b) etc.
Title: Re: Bucket's Questions
Post by: bucket on September 15, 2008, 11:55:33 pm: thanks a shitload man!
good explanations :)
Title: Re: Bucket's Questions
Post by: bucket on September 16, 2008, 07:45:53 pm: "The weight if cats is normally distributed. It is known that 10% of cats weigh more than 1.8kg and 15% of cats weigh less than 1.35kg. Find the mean and the standard deviation."
Title: Re: Bucket's Questions
Post by: Collin Li on September 16, 2008, 07:56:32 pm: $\mbox{Pr}(X > 1.8) = 0.10$

$\mbox{Pr}(X < 1.35) = 0.15$

Hint (and this is how I analyse the question):
A normal distribution (like the binomial distribution) is always defined by two parameters. Here, you have two pieces of information, so you should be able to find the two unknowns. The question is, how do I get these two expressions into a form that involves the mean and the standard deviation?

Try using this:

$Z = \frac{X - \mu}{\sigma}$

So you will get:

$\mbox{Pr}\left(Z > \frac{1.8 - \mu}{\sigma}\right) = 0.10 \implies \mbox{Pr}\left(Z < \frac{1.8 - \mu}{\sigma}\right) = 0.90$

$\mbox{Pr}\left(Z < \frac{1.35 - \mu}{\sigma}\right) = 0.15$

Use invNorm(:

Quote
$a$ = invNorm(k), where $\mbox{Pr}(Z < a) = k$

Hence,

invNorm(0.90) = $\frac{1.8 - \mu}{\sigma} \approx 1.282$ (1)

invNorm(0.15) = $\frac{1.35 - \mu}{\sigma} \approx -1.036$ (2)

(1) - (2) yields:

$\frac{1.8 - 1.35}{\sigma} \approx 2.318$

$\implies \sigma \approx 0.194$

$\implies \mu \approx 1.551$
Title: Re: Bucket's Questions
Post by: bucket on September 16, 2008, 08:03:32 pm: hmm you're right but I'm confused :P. what is invNorm?
When I do inverse normals I use a program where I have to enter the area(probability), the standard deviation and the mean. :S

edit**
I worked it out =]. Thanks a lot Coblin.
Title: Re: Bucket's Questions
Post by: excal on September 16, 2008, 10:33:07 pm: Quote from: coblin on September 16, 2008, 07:56:32 pm
$\mbox{Pr}(X > 1.8) = 0.10$

$\mbox{Pr}(X < 1.35) = 0.15$

Hint (and this is how I analyse the question):
A normal distribution (like the binomial distribution) is always defined by two parameters. Here, you have two pieces of information, so you should be able to find the two unknowns. The question is, how do I get these two expressions into a form that involves the mean and the standard deviation?

Try using this:

$Z = \frac{Z - \mu}{\sigma}$

So you will get:

$\mbox{Pr}\left(Z > \frac{1.8 - \mu}{\sigma}\right) = 0.10 \implies \mbox{Pr}\left(Z < \frac{1.8 - \mu}{\sigma}\right) = 0.90$

$\mbox{Pr}\left(Z < \frac{1.35 - \mu}{\sigma}\right) = 0.15$

Use invNorm(:

Quote
$a$ = invNorm(k), where $\mbox{Pr}(Z < a) = k$

Hence,

invNorm(0.90) = $\frac{1.8 - \mu}{\sigma} \approx 1.282$ (1)

invNorm(0.15) = $\frac{1.35 - \mu}{\sigma} \approx -1.036$ (2)

(1) - (2) yields:

$\frac{1.8 - 1.35}{\sigma} \approx 2.318$

$\implies \sigma \approx 0.194$

$\implies \mu \approx 1.551$

coblin: $Z = \frac{X-\mu}{\sigma}$
Title: Re: Bucket's Questions
Post by: Collin Li on September 16, 2008, 10:35:21 pm: Quote from: Excalibur on September 16, 2008, 10:33:07 pm
coblin: $Z = \frac{X-\mu}{\sigma}$

Thanks, typo. :)

Only the formula needs to be fixed.
Title: Re: Bucket's Questions
Post by: /0 on September 17, 2008, 12:53:30 am: When do you use capitals and when don't you?

You see in the Essential Meth book it gives $z = \frac{x-\mu}{\sigma}$ instead of $Z =\frac{X-\mu}{\sigma}$ .

Also do you think any examiners will be pedantic enough penalise us for this?
Title: Re: Bucket's Questions
Post by: Collin Li on September 17, 2008, 12:55:36 am: Quote from: DivideBy0 on September 17, 2008, 12:53:30 am
When do you use capitals and when don't you?

You see in the Essential Meth book it gives $z = \frac{x-\mu}{\sigma}$ instead of $Z =\frac{X-\mu}{\sigma}$ .

Also do you think any examiners will be pedantic enough penalise us for this?

They won't be pedantic enough.

I think you should always use capitals. The convention is that capital letters denote random variables. "Z" and "X" are both random variables.
Title: Re: Bucket's Questions
Post by: /0 on September 17, 2008, 12:57:52 am: Thanks coblin... hmm, so the small letters actually stand for certain numbers and the large ones can stand for a set of numbers, that's alright... except, we don't write normal equations like $Y = 2X+3$ . Oh well I guess I'm the one being pedantic now...
Title: Re: Bucket's Questions
Post by: Collin Li on September 17, 2008, 12:59:21 am: The larger ones refer to a variable that has uncertainity, yeah. I don't think it's capital for the same reason "R" (set of real numbers) is capital.
Title: Re: Bucket's Questions
Post by: excal on September 17, 2008, 01:10:39 am: Basically, as coblin said, lower case numbers represent deterministic (non-random) variables whereas upper case represent stochastic (random) variables. It's just a notation thing, really.

You may come across this notation at some point in discrete probability: $P(X = x) = 0.5$
Title: Re: Bucket's Questions
Post by: Collin Li on September 19, 2008, 07:57:38 am: Quote from: DivideBy0 on September 17, 2008, 12:57:52 am
except, we don't write normal equations like $Y = 2X+3$ . Oh well I guess I'm the one being pedantic now...

We can. For example, if you want to find the expected profit, and you know profit is a function of $x$ , but $X$ is a random variable, then you can write: $P = f(X)$ , and it might be $P= 2X+3$ for example.

I think one reason why they have the lowercase formula is because if you have something like: $\mbox{Pr}(X = x)$ , you can only change $X$ to $Z$ by using the same manipulations on the RHS as well, hence making $z = \frac{x-\mu}{\sigma}$

For example,

$\mbox{Pr}(X = 0.5) = \mbox{Pr}\left(\frac{X - \mu}{\sigma} = \frac{0.5 - \mu}{\sigma}\right) = \mbox{Pr}\left(Z = \frac{0.5 - \mu}{\sigma}\right)$

Recognise that: $\frac{0.5 - \mu}{\sigma} = z_{0.5}$

In Methods, we are often using the lowercase formula, but fundamentally, we're using the uppercase formula, which you have to apply to both the LHS and the RHS. I hope you understand the subtle difference, but you really don't need to - most people only superficially understand this (merely changing the $X$ to a $Z$ and then using the lowercase formula on the RHS).
Title: Re: Bucket's Questions
Post by: methodsboy on September 20, 2008, 06:48:31 pm: umm i think u should re-check that question about finding the x intercept of y= 2/ (x-3)^3 + 4
u should get :
-1/2 = (x-3)^3
(-1/2) ^ (1/3) = (x-3)
(-1/2) ^ (1/3) +3 = x
to prove that this is correct, please check on a CAS calculator.
Title: Re: Bucket's Questions
Post by: shinny on September 20, 2008, 07:01:36 pm: Yeh, he did get it wrong, but thing is that question was from 7 months ago O_O coblin just forgot the negative (its the one on page one if every1 else is wondering)
Title: Re: Bucket's Questions
Post by: bucket on October 03, 2008, 11:27:21 pm: For $y=\frac{5}{3}x+kx^2-\frac{8}{9}x^3$ calculate the possible values of k that the tangents at the points with x-coords 1 and $-\frac{1}{2}$ respectively are perpendicular.
Title: Re: Bucket's Questions
Post by: ell on October 03, 2008, 11:40:49 pm: Quote from: bucket on October 03, 2008, 11:27:21 pm
For $y=\frac{5}{3}x+kx^2-\frac{8}{9}x^3$ calculate the possible values of k that the tangents at the points with x-coords 1 and $-\frac{1}{2}$ respectively are perpendicular.

$\frac{dy}{dx}=\frac{5}{3}+2xk-\frac{8}{3}x^2$

So the gradients of the two tangents can be found if we sub in 1 and $-\frac{1}{2}$ into $\frac{dy}{dx}$ . When we multiply them together they should equal -1, since they are perpendicular.

$\implies \frac{dy}{dx}=\frac{5}{3}+2k-\frac{8}{3}$

$\implies \frac{dy}{dx}=2k-1$

Now sub in $-\frac{1}{2}$ :

$\implies \frac{dy}{dx}=\frac{5}{3}-k-\frac{2}{3}$

$\implies \frac{dy}{dx}=1-k$

These two are perpendicular, so:

$(2k-1)(1-k)=-1$

$\implies -2k^2+3k-1=-1$

$\implies -2k^2+3k=0$

$\implies k(-2k+3)=0$

So $k=0$ or $k=\frac{3}{2}$
Title: Re: Bucket's Questions
Post by: Flaming_Arrow on October 03, 2008, 11:43:42 pm: u used the wrong symbol

its [/tex] not [\tex]
Title: Re: Bucket's Questions
Post by: ell on October 03, 2008, 11:45:35 pm: Quote from: chath on October 03, 2008, 11:43:42 pm
u used the wrong symbol

its [/tex] not [\tex]

ah good pick up chath, I wrote \frac so many times I was used to \, lol
Title: Re: Bucket's Questions
Post by: Flaming_Arrow on October 03, 2008, 11:46:50 pm: Quote from: ell on October 03, 2008, 11:45:35 pm
Quote from: chath on October 03, 2008, 11:43:42 pm
u used the wrong symbol

its [/tex] not [\tex]

ah good pick up chath, I wrote \frac so many times I was used to \, lol

haha xD
Title: Re: Bucket's Questions
Post by: bucket on October 03, 2008, 11:48:44 pm: cheeeeeeeeeeeeeeeeers
Title: Re: Bucket's Questions
Post by: bucket on October 06, 2008, 01:16:52 am: fdshfksdhfsdjl
Factorise:
$ a^2-b^2-(a-b)^2$

AND ALSO

$ x^2-y^2-5x-y+6$
Title: Re: Bucket's Questions
Post by: Mao on October 06, 2008, 01:23:15 am: $a^2 - b^2 - (a-b)^2 = (a+b)(a-b) - (a-b)(a-b) = (a+b - (a-b))(a-b) = 2b(a-b)$

$x^2 - y^2 -5x - y + 6 = \left(x^2 - 2\frac{5}{2}x + \frac{25}{4}\right) - \frac{25}{4} - \left(y^2 + 2\frac{1}{2}y+\frac{1}{4}\right) + \frac{1}{4}+6 = \left(x-\frac{5}{2}\right) ^2 - \left(y+\frac{1}{2}\right)^2 = (x+y-2)(x-y-3)$
Title: Re: Bucket's Questions
Post by: bucket on October 09, 2008, 12:57:43 am: from the 2007** exam 2:
the simultaneous equations $mx+12y=24$ and $3x+my=m$ have a unique solution for:

answer is m: R\{-6,6}

i didnt understand what they were saying in the assessment report :S

edit**
Title: Re: Bucket's Questions
Post by: dekoyl on October 09, 2008, 01:26:20 am: I know this isn't really helping but which question in 2007 Exam 2? I can't seem to find it. I'd like to have a look at what I did as I've done this exam. Interesting question raised, bucket. :)
Title: Re: Bucket's Questions
Post by: humph on October 09, 2008, 01:31:30 am: Quote from: bucket on October 09, 2008, 12:57:43 am
from the 2008 exam 2:
the simultaneous equations $mx+12y=24$ and $3x+my=m$ have a unique solution for:

answer is m: R\{-6,6}

i didnt understand what they were saying in the assessment report :S
If $m=0$ then clearly $x=0,y=2$ is the unique solution. So assume that $m \neq 0$ .
Then we multiply the first expression by $3$ and the second by $m$ such that the simultaneous equations read $3mx+36y=72$ and $3mx+m^2 y=m^2$ . Subtracting the former from the latter, we find that $(m^2 - 36)y=m^2 - 72$ .
Now if $m=\pm 6$ then this is saying that $0 = -36$ , which is impossible. Thus there cannot be a solution if $m=\pm 6$ .
If $m \neq \pm 6$ , then we can divide through by $m^2 - 36$ to find that $y = \frac{m^2 - 72}{m^2 - 36}$ , which we can then substitute back into the original equation to find the corresponding $x$ -coordinate $x = \frac{12m}{m^2 - 36}$ . This is the unique solution for $m \in \mathbb{R} \setminus \{-6,6\}$ .

Graphically, the reason why the simultaneous equations don't have a solution when $m=\pm 6$ is because the lines are parallel at that point: substituting these values of $m$ back into the original equations and making $y$ the subject yields the pairs of equations $y=\mp \frac{x}{2} + 2$ and $y = \mp \frac{x}{2} + 1$ , which, as they have different $y$ -intercepts, are clearly parallel.

On this thought, you could really just rearrange them to be the equations of lines. If $m=0$ then clearly $x=0,y=2$ is the unique solution. So assume that $m \neq 0$ . Then the original equations can be rearranged to read $y = -\frac{m}{12}x + 2$ and $y=-\frac{3}{m}x + 1$ . These have the same gradient when $-m/12 = -3/m$ , which is the same as when $m=\pm 6$ . In this case, however, they have different $y$ -intercepts, and so are clearly parallel, and in Euclidean geometry, parallel lines never intersect, so there is no solution. If, on the other hand, $m \neq \pm 6$ , then the lines aren't parallel. As straight non-parallel lines intersect exactly once in Euclidean geometry, it follows that there exists a unique solution given by the coordinates of the point of intersection.
Title: Re: Bucket's Questions
Post by: bucket on October 09, 2008, 01:58:55 am: Wow. Very well explained humphdogg, thanks a lot :).

Quote from: dekoyl on October 09, 2008, 01:26:20 am
I know this isn't really helping but which question in 2007 Exam 2? I can't seem to find it. I'd like to have a look at what I did as I've done this exam. Interesting question raised, bucket. :)
This was question 5 in the multiple choice... seems a bit much for a multi choice question. :S but I guess people who understand the theory well would pick it up quickly by elimination.
Title: Re: Bucket's Questions
Post by: dekoyl on October 09, 2008, 02:11:39 am: Quote from: bucket on October 09, 2008, 01:58:55 am
This was question 5 in the multiple choice... seems a bit much for a multi choice question. :S but I guess people who understand the theory well would pick it up quickly by elimination.
VCAA 2007 Exam 2? Because Q5 of multiple choice is an antidifferentiation question.
Title: Re: Bucket's Questions
Post by: bucket on October 09, 2008, 02:20:12 am: http://www.vcaa.vic.edu.au/vce/studies/mathematics/cas/pastexams/2007/2007mmCAS2.pdf
Title: Re: Bucket's Questions
Post by: dekoyl on October 09, 2008, 02:35:44 am: Oh! I do non-CAS hehe. Thanks for clearing things up.
Title: Re: Bucket's Questions
Post by: bucket on October 13, 2008, 12:34:08 am: This is from 2007 exam 1. =\

$P$ is a point on the line $2x+y-10=0$ such that the length of $OP$ , the line segment from the origin $O$ to $P$ , is a minimum. Find the coordinates of $P$ and this minimum length.
Title: Re: Bucket's Questions
Post by: dekoyl on October 13, 2008, 12:50:22 am: This is what I did. Sorry if this doesn't exactly make sense.
You want the line that is perpendicular to the point P. (Shortest distance).

$y = -2x + 10$

$dy/dx = -2$ This is the gradient of the tangent.

$\implies 1/2$ is the gradient of the normal. (This is what we want as this crosses P).

Using $y = mx + c$ and substituting $(0,0)$ (origin) we find that the equation of the normal is $y= x/2$

Now just make $x/2 = -2x + 10$

You can find $x$ from solving the above then sub into either equation to find $y$ coordinate.

Then using $\sqrt{(x_2 - x_1)^{2} + (y_2 - y_1)^{2}}$ (distance formula), you can find the distance between O and P.

Hope I was able to help. :)
Title: Re: Bucket's Questions
Post by: bucket on October 13, 2008, 12:54:30 am: ah, thanks a lot man =]
Title: Re: Bucket's Questions
Post by: Collin Li on October 13, 2008, 12:57:47 am: Another approach:

$y= 10 - 2x$

Therefore the distance of the line from the origin at any $x$ is the distance between the points:

$(0,\,0)$ and $(x,\, 10-2x)$

Let $D$ be this distance:

$D = \sqrt{ (x)^2 + (10-2x)^2 }$

$\implies D = \sqrt{ x^2 + 4x^2 - 40x + 100 }$

$\implies D = \sqrt{5 (x^2 - 8x + 20)}$

$\therefore \frac{dD}{dx} = \frac{5(2x - 8)}{2\sqrt{5 (x^2 - 8x + 20)}}$

I know the stationary point will be a minimum, because the original function is the square root of a parabola with a local minimum. Since the square root function is monotonic increasing (1 to 1, and always has a positive gradient), I can conclude that the original function, $D$ , will have a local minimum too.

Minimum occurs at: $\frac{dD}{dx} = 0 \implies 2x - 8 = 0 \implies x = 4$

Therefore, minimum distance: $D(4) = \sqrt{5 (16 - 32 + 20) } = 2\sqrt{5}$ and,

Coordinates of $P$ : $(4, 2)$
Title: Re: Bucket's Questions
Post by: dekoyl on October 13, 2008, 01:00:21 am: Quote from: coblin on October 13, 2008, 12:57:47 am
Another approach: ...
Ah nice approach Coblin. My mind isn't as flexible.
I'll keep that method hot in my mind. Might come in useful some time.
Title: Re: Bucket's Questions
Post by: Collin Li on October 13, 2008, 01:01:19 am: Quote from: dekoyl on October 13, 2008, 01:00:21 am
Quote from: coblin on October 13, 2008, 12:57:47 am
Another approach: ...
Ah nice approach Coblin. My mind isn't as flexible.
I'll keep that method hot in my mind. Might come in useful some time.

Your method is more clever, IMO.

I presented my method because it's the "hack method" approach. You don't need to have a good grasp of geometry to do it this way. ;)
Title: Re: Bucket's Questions
Post by: phagist_ on October 13, 2008, 08:32:33 pm: Quote from: bucket on October 13, 2008, 12:34:08 am
This is from 2007 exam 1. =\

$P$ is a point on the line $2x+y-10=0$ such that the length of $OP$ , the line segment from the origin $O$ to $P$ , is a minimum. Find the coordinates of $P$ and this minimum length.
goddamit I remember this question.. But I added the distance up wrong and got $\sqrt{18}$ .. so frustrating haha, I knew as soon as I came out of the exam as well..

/random
Title: Re: Bucket's Questions
Post by: dcc on October 13, 2008, 10:44:01 pm: Quote from: bucket on October 13, 2008, 12:34:08 am
This is from 2007 exam 1. =\

$P$ is a point on the line $2x+y-10=0$ such that the length of $OP$ , the line segment from the origin $O$ to $P$ , is a minimum. Find the coordinates of $P$ and this minimum length.

Another take on dekoyl's idea of perpendicular lines:

(this is mainly for specialist students, and is quite useful when solving vector questions in specialist)

define $\vec{P}(x) = \vec{OP} = x \vec{i} + y \vec{j} = x \vec{i} + \left(10 - 2x\right)\vec{j}$ . This vector defines the line from the origin to a point P on the line [yex]y = 10 - 2x[/tex].

also, we find: $\vec{\dot{P}}(x) = \vec{i} - 2 \vec{j}$ is the vector of the tangent of the curve $y = 10 - 2x$ at point P.

We wish to find the situation where the line $\vec{OP}$ and the curve are normals to each other (i.e, they are perpendicular)

Sowe wish to find when $\vec{P}(x) \cdot \vec{\dot{P}}(x) = 0$ (remembering our dot product identities)

$\left(x\right)\left(1\right) + \left(10 - 2x\right)\left(-2\right)= 0 \ \therefore\ x = 4$

so $\vec{P}(4) = 4 \vec{i} + 2 \vec{j} \implies | \vec{P}(4) | = \sqrt{4^2 + 2^2} = \boxed{2\sqrt{5}}$ is the minimum distance of the line OP and P has the coordinates $\left(4,\ 2\right)$ .
Title: Re: Bucket's Questions
Post by: bucket on October 17, 2008, 12:34:03 am: Ok, so my methods teacher kind of didn't bother to teach our class anything about 'general solutions of circular function equations' and since one of those questions popped up in the MAV 2008 exam I though I'd try to teach them to myself...having a tiiiinnyyy bit of difficulty with some questions though.

~~1. The general solution for an equation is $n\pi + (-1)^n \cdot sin^{-1}(\frac{1}{2})$ Find the solutions for the interval $[-2\pi, 2\pi]$~~
nevermind...

~~2. Find the general solution for $2cos(2x+\frac{\pi}{4})=\sqrt{2}$ and use this to find the solutions for the interval $[-2\pi ,2\pi]$~~
dw about this one anymore :P

thanks glockmeister
Title: Re: Bucket's Questions
Post by: Glockmeister on October 17, 2008, 12:50:46 am: Solved on IRC
Title: Re: Bucket's Questions
Post by: dekoyl on October 17, 2008, 12:55:51 am: Sorry could you post the solution here, Glockmeister (doesn't have to be LaTeX)? Thanks.
I'm not on IRC so missed it:(
Title: Re: Bucket's Questions
Post by: Glockmeister on October 17, 2008, 12:58:40 am: Sure

1. The general solution for an equation is $n\pi + (-1)^n \cdot sin^{-1}(\frac{1}{2})$ Find the solutions for the interval $[-2\pi, 2\pi]$

The one thing about general solutions is that it is an equation that gives us the answers for a trig function where the domain is $\mathsbb{R}$ . But here, we want a range of answers within a restricted domain. First step however, is to change that sin inverse

$n\pi + (-1)^n \cdot sin^{-1}(\frac{1}{2}) n \in \mathsbb{Z}$

$= n\pi + (-1)^n \cdot \frac{\pi}{6}) n \in \mathsbb{Z}$

N is an interger so what we do is to let n = a series whole numbers. The answers that you get can not go beyond the domain $[-2\pi, 2\pi]$ . If you were to do the sums, you would find that n = -1, 0 , 1. Put those numbers in and you should get $x = \frac{-7\pi}{6}, \frac{\pi}{6}, \frac{5\pi}{6}$

2 Find the general solution for $2cos(2x+\frac{\pi}{4})=\sqrt{2}$ and use this to find the solutions for the interval $[-2\pi ,2\pi]$

$2cos(2x+\frac{\pi}{4})=\sqrt{2}$

$cos(2x+\frac{\pi}{4})=\frac{\sqrt{2}}{2}}$

$2(x+\frac{\pi}{8})= 2k\pi \pm \frac{\pi}{4} k \in \mathsbb{Z}$

$x + \frac{\pi}{8} = k\pi \pm \frac{\pi}{8}$

$\therefore x = k \pi or x = k\pi - \frac{\pi}{4}$

And then you would od the same thing as 1, subsitute integers which give an answer that fit the domain
Title: Re: Bucket's Questions
Post by: dekoyl on October 17, 2008, 01:21:55 am: ^ Thank you, good sir. I don't recall seeing a question like this so it's a great help. :)
Title: Re: Bucket's Questions
Post by: bucket on October 17, 2008, 01:29:23 am: actually, with the second one, the general form is:
$2n\pi \pm cos^{-1}(a)$

So when you get $cos(2x+\frac{\pi}{4}) = \frac{\sqrt{2}}{2}$
the next step would be $[2(x+\frac{\pi}{8})]= 2n\pi \pm \frac{\pi}{4}$
then you get x on its own:
$x=\pi n + 0 or \pi n -\frac{\pi}{4} n \in \mathsbb{z}$

and then you proceed to sub in $n=-1, 0, 1, 2$ and you get the answers $-\frac{5\pi}{4}, -\pi , -\frac{\pi}{4}, 0, \frac{3\pi}{4}, \pi , \frac{7\pi}{4}$
apparently this is only in the CAS course (which Glock didn't do :S)
Title: Re: Bucket's Questions
Post by: Glockmeister on October 17, 2008, 04:38:55 pm: Yeah, that would be right... I was thinking I did something wrong when I was asleep last night, that x value felt so wrong.
Title: Re: Bucket's Questions
Post by: bucket on October 20, 2008, 11:29:58 pm: How do you manually integrate equations which have like, two types of functions?
ie. $e^{2x}sin(3x)$ or $log_e(sin(x))$ ?

Mm, I can't seem to find any instructions on how to this in my textbook, it just says to use the calculator :S
Are we even required to know how to do this manually?
Title: Re: Bucket's Questions
Post by: dekoyl on October 20, 2008, 11:34:56 pm: Quote from: bucket on October 20, 2008, 11:29:58 pm
How do you manually integrate equations which have like, two types of functions?
ie. $e^{2x}sin(3x)$ or $log_e(sin(x))$ ?

Mm, I can't seem to find any instructions on how to this in my textbook, it just says to use the calculator :S
Are we even required to know how to do this manually?
Simply, no. However, you may be required to integrate it if it's part of a "hence" question.
Title: Re: Bucket's Questions
Post by: Glockmeister on October 20, 2008, 11:37:18 pm: For integrating I would not thing that you would need to know how to do that (If I remember right, this is integration by parts, which isn't even in the spech course)

Of course if this is part of a 'differenitate this, hence find the integral of that' question, then you know they steps required for this technique
Title: Re: Bucket's Questions
Post by: Collin Li on October 20, 2008, 11:44:49 pm: Find the derivative of $e^{2x}\cos(3x)$ and $e^{2x}\sin(3x)$ , and hence find $\int e^{2x}\sin(3x) \, dx$

The question should tell you what derivatives you've gotta find. Have a go at that... it should work (I think).

No clue how to do the second one, haha.
Title: Re: Bucket's Questions
Post by: bucket on October 20, 2008, 11:52:03 pm: I dont even know how to do that lol
Title: Re: Bucket's Questions
Post by: Collin Li on October 21, 2008, 12:00:28 am: $\frac{d}{dx}\left[e^{2x}\cos(3x)\right] = 2e^{2x}\cos(3x) - 3e^{2x}\sin(3x)$

$\frac{d}{dx}\left[e^{2x}\sin(3x)\right] = 2e^{2x}\sin(3x) + 3e^{2x}\cos(3x)$

Integrating both sides of these two results:

$e^{2x}\cos(3x) + C_1= 2 \int e^{2x}\cos(3x) \, dx - 3 \int e^{2x}\sin(3x) \, dx$

$e^{2x}\sin(3x) + C_2 = 2 \int e^{2x}\sin(3x) \, dx + 3 \int e^{2x}\cos(3x) \, dx$

Subtracting 3 times the top equation from 2 times the bottom equation to eliminate $\int e^{2x}\cos(3x) \, dx$ yields:

$2e^{2x}\sin(3x) - 3e^{2x}\cos(3x) + 2C_2 - 3C_1 = 4 \int e^{2x}\sin(3x) \, dx + 9 \int e^{2x}\sin(3x) \, dx$

$\implies 2e^{2x}\sin(3x) - 3e^{2x}\cos(3x) + 2C_2 - 3C_1 = 13 \int e^{2x}\sin(3x) \, dx$

$\implies \frac{1}{13}\left[2e^{2x}\sin(3x) - 3e^{2x}\cos(3x)\right] + C = \int e^{2x}\sin(3x) \, dx$ , where $C = \frac{1}{13}\left(2C_2 - 3C_1\right)$

It's unlikely you'll get that though.

It's much more likely you'll get something like this:

Find the derivative of $xe^x$ and hence find $\int xe^x \, dx$

Solution

$\frac{d}{dx}\left(xe^x\right) = e^x + xe^x$

Integrating both sides:

$\implies xe^x + C = \int e^x \, dx + \int xe^x \, dx$

$\implies xe^x + C = e^x + \int xe^x \, dx$

$\implies xe^x - e^x + C = \int xe^x \, dx$
Title: Re: Bucket's Questions
Post by: dekoyl on October 21, 2008, 12:05:01 am: Quote from: coblin on October 20, 2008, 11:44:49 pm
Find the derivative of $e^{2x}\cos(3x)$ and $e^{2x}\sin(3x)$ , and hence find $\int e^{2x}\sin(3x) \, dx$

The question should tell you what derivatives you've gotta find. Have a go at that... it should work (I think).

Hmmm I'm going around in circles with this one.
I come up with either:
$\int 2e^{2x}sin(3x).dx = e^{2x}sin(3x) - \int 3e^{2x}cos(3x).dx + c$

or

$-3\int e^{2x}sin(3x).dx = e^{2x}cos(3x) - \int 2e^{2x}cos(3x).dx +c$

~~Just thought I'd attempt it. Haven't done Maths for a while :(~~

Ah solving simultaneously. I didn't see that :(
Title: Re: Bucket's Questions
Post by: Glockmeister on October 21, 2008, 12:05:40 am: Quote from: coblin on October 20, 2008, 11:44:49 pm

No clue how to do the second one, haha.

I know how to do that

$log_e(sin(x))$

$Put \int (log_e(sin(x)),x) into CAS calculator$

$Calcuator says \int (log_e(sin(x)),dx$

Therefore answer is $\int (log_e(sin(x)),dx$

Hmm... even the CAS calculator can't do it. Must bring out the big guns. Ahmad, Neobeo, where are ya?
Title: Re: Bucket's Questions
Post by: bucket on October 21, 2008, 12:13:03 am: ahhhh thanks a heap!
edit* lol at that glockmeister!!
I made that question up btw, I think i might have seen it the other way around, ie. $sin(log_e(x))$
Title: Re: Bucket's Questions
Post by: Mao on October 21, 2008, 12:39:37 am: that $\int \log_e(\sin (x))\; dx$ screams of "NON ELEMENTARY"
even taylor expansion will have terribly short radius of convergence [it has quite a few asymptotes, too many for my liking in fact]
that means you don't worry about it :)

$\int \sin(\log_e(x))\; dx$ is doable, and quite a few other functions (usually products of two functions), via the method of integration by parts, it is the reverse chain rule which can be quite freaky, but in short it is not required by MM [not even at SM level].
Title: Re: Bucket's Questions
Post by: bucket on October 21, 2008, 01:11:45 am: Lol, thank god.

Anyway new question, bit fuzzy with inequalities :S.

"Find the set of values for x for which $|x^2-5x|\geq 6$ "
Title: Re: Bucket's Questions
Post by: dekoyl on October 21, 2008, 02:35:13 am: First, consider the $x^{2} - 5x$ part. We want to find the intersection of this with $y = 6$

$x^{2} - 5x = 6$

$x^{2} - 5x - 6 = 0$

$(x - 6)(x + 1) = 0$

$\therefore x\geq6$ or $x\leq-1 \implies$ First set of answers (look at the graph if you're not sure).

$y=6$ also intercepts near the stationary point of the graph at...

$x^{2} - 5x = -6$ (I do this because we can flip it on the x-axis to get what we want later)

$(x-3)(x-2)=0$

$\therefore x\geq2$ or $x\leq3$

$\therefore |x^2 - 5x| \geq 6$ when $2\leq x \leq3$ and $x\geq6$ or $x\leq-1$

Sorry if I made any mistakes but hope you got the gist of it.

Fruck 2:34am, need to memorise 80 Latin words and yet I'm still browsing VN. :(
Title: Re: Bucket's Questions
Post by: Mao on October 21, 2008, 09:26:49 am: $|x^2-5x|\ge 6\implies x^2-5x\ge 6\; \mbox{or}\; x^2-5x\le -6$

in the first case, $x^2-5x-6\ge 0\implies (x-6)(x+1)\ge 0 \implies x\le -1\; \mbox{or} \; x\ge 6$

in the second case, $x^2-5x+6\le 0 \implies (x-3)(x-2)\le 0 \implies 2\le x\le 3$

hence, the solution is $(-\infty,-1]\cup [2,3]\cup [6,\infty)$

dekoyl: your "2>x<3" is a little stuffed :P
Title: Re: Bucket's Questions
Post by: dekoyl on October 21, 2008, 05:45:04 pm: Quote from: Mao on November 18, 1973, 06:02:40 pm
dekoyl: your "2>x<3" is a little stuffed :P
Ah thanks. :-[

You presented your answer in a much neater way *takes note*.
Title: Re: Bucket's Questions
Post by: bucket on October 28, 2008, 12:24:51 am: Mmm, I recently came across a binomial distribution problem in an exam 1 trial paper, and I realised I have no fucking idea how to evaluate them.
So, how would you evaluate something such as...say...
$^{5}C_2 \cdot (\frac{3}{10})^2 \cdot(\frac{7}{10})^3$
This isn't from the actual question but I came up with an expression similar to this.
Title: Re: Bucket's Questions
Post by: shinny on October 28, 2008, 12:50:43 am: Well I assume your main problem is the $^{5}C_2$
Basically;
$\because ^{n}C_r=\frac{n!}{r!(n-r)!}$

$\therefore ^{5}C_2=\frac{5!}{2! \times 3!}$

$=\frac{5 \times 4}{2}$

$=10$

As for the two powers, just do it by hand I guess =\
Title: Re: Bucket's Questions
Post by: bucket on October 28, 2008, 01:36:26 am: yeah the ncr thing was what was fucking me up, thanks a lot man.
Title: Re: Bucket's Questions
Post by: Mao on October 28, 2008, 10:34:35 pm: people often think of combinatorics in the difficult way.

this is my way of deciphering it:

10C5 = 10*9*8*7*6 / 1*2*3*4*5

i.e. counting down from 10, get 5 numbers. then counting up from 1, get 5 numbers.
the "n" controls where you start up the top, and the "r" controls how many numbers you count

so 20C3 = 20*19*18/1*2*3, 7C2 = 7*6/1*2
Title: Re: Bucket's Questions
Post by: dekoyl on October 28, 2008, 10:36:05 pm: Quote from: Mao on October 28, 2008, 10:34:35 pm
people often think of combinatorics in the difficult way.

this is my way of deciphering it:

10C5 = 10*9*8*7*6 / 1*2*3*4*5

i.e. counting down from 10, get 5 numbers. then counting up from 1, get 5 numbers.
the "n" controls where you start up the top, and the "r" controls how many numbers you count

so 20C3 = 20*19*18/1*2*3, 7C2 = 7*6/1*2
I was taught the same way last year but I've forgotten it. Thanks for bringing it up Mao.
Title: Re: Bucket's Questions
Post by: bucket on November 03, 2008, 07:15:06 pm: Hey I was doing a trial exam and I got the answer $\sqrt{20}$ and when i looked up the answer it was corrext, but expressed as $2\sqrt{5}$ . Do you think I would be marked incorrect if it were in the real exam? The question did not say "in simplest form".
Title: Re: Bucket's Questions
Post by: Collin Li on November 03, 2008, 07:17:53 pm: Yeah, that seems acceptable.
Title: Re: Bucket's Questions
Post by: Mao on November 03, 2008, 08:46:24 pm: Quote from: bucket on November 03, 2008, 07:15:06 pm
Hey I was doing a trial exam and I got the answer $\sqrt{20}$ and when i looked up the answer it was corrext, but expressed as $2\sqrt{5}$ . Do you think I would be marked incorrect if it were in the real exam? The question did not say "in simplest form".

I've heard from MAV that they do need the simplest surd, but not the simplest fraction

i.e. $\sqrt{20}$ is not acceptable, but $\frac{2}{4}$ is...
Title: Re: Bucket's Questions
Post by: bucket on November 03, 2008, 08:53:22 pm: meh, i never seem to remember to simplify surds :(
Title: Re: Bucket's Questions
Post by: bucket on November 05, 2008, 10:43:51 pm: Mmm, whenever I do a trial exam I always fuck up when i need to manually plot a trig graph. Does anybody have any tricks for doing these? I have the most difficulty determining endpoints and the y intercept whenever the graph is translated in any direction.
Title: Re: Bucket's Questions
Post by: fredrick on November 05, 2008, 10:47:07 pm: Quote from: bucket on November 05, 2008, 10:43:51 pm
Mmm, whenever I do a trial exam I always fuck up when i need to manually plot a trig graph. Does anybody have any tricks for doing these? I have the most difficulty determining endpoints and the y intercept whenever the graph is translated in any direction.
I find the intercepts and max/min values and then endpoints and go from there, but i agree i hate skecting those
Title: Re: Bucket's Questions
Post by: bucket on November 05, 2008, 10:51:33 pm: they take up like 10 fucking minutes and then i end up getting them wrong anyway =_= lol
Title: Re: Bucket's Questions
Post by: dekoyl on November 05, 2008, 10:59:32 pm: Quote from: bucket on November 05, 2008, 10:43:51 pm
Mmm, whenever I do a trial exam I always fuck up when i need to manually plot a trig graph. Does anybody have any tricks for doing these? I have the most difficulty determining endpoints and the y intercept whenever the graph is translated in any direction.
I have troubles with the y-ints as well. I sometimes have trouble seeing if the maximum occurs on the right or left side of the y-axis.
Title: Re: Bucket's Questions
Post by: /0 on November 05, 2008, 11:00:38 pm: Quote from: bucket on November 05, 2008, 10:43:51 pm
Mmm, whenever I do a trial exam I always fuck up when i need to manually plot a trig graph. Does anybody have any tricks for doing these? I have the most difficulty determining endpoints and the y intercept whenever the graph is translated in any direction.

I think it gets easier when you split it up into different steps. For instance, I would normally do:

If $f(x) = a\sin{(b(x+c))}+d$ ,

- I will lightly draw the lines y = d, y = d+a, y = d-a on the graph paper. This gives me boundaries which I cannot step over, as well as the middle.

- Then I solve f(x) = 0 for x, marking in the intercept points. (I generally find this the hardest part of graphing; not so much the solving, but more the scaling of the graph) A neat trick is if d = 0, then the solutions will be $\frac{\pi}{b}$ away from each other. You can also work these out by applying transformations to your standard graph of $\sin{x}$ .

- Plug x in for endpoints and y-intercept, marking these points on the graph. (I would have thought this would be a simpler step o.O)

- Now it's a simple game of connect the dots. If you have done it right, everything should fall in place. If not, recheck your arithmetic. Check transformations regardless, just to be sure.

- If after everything it still doesn't look right, move on and come back to it later.
Title: Re: Bucket's Questions
Post by: bucket on November 05, 2008, 11:11:33 pm: Thats a good approach to it. The problem I have with endpoints is when I get something such as $5cos(\frac{2\pi}{3})$ . I know it's probably very dumb but I don't know how to work these out when they are not the values I'm used to!!
Title: Re: Bucket's Questions
Post by: fredrick on November 05, 2008, 11:32:21 pm: Quote from: bucket on November 05, 2008, 11:11:33 pm
Thats a good approach to it. The problem I have with endpoints is when I get something such as $5cos(\frac{2\pi}{3})$ . I know it's probably very dumb but I don't know how to work these out when they are not the values I'm used to!!
$\frac{2\pi}{3}$ is in the second quadrant and cos is negative in that quadrant. So just convert it to first quadrant and you get $-5cos(\frac{\pi}{3})$
Title: Re: Bucket's Questions
Post by: danieltennis on November 05, 2008, 11:35:35 pm: Quote from: fredrick on November 05, 2008, 11:32:21 pm
Quote from: bucket on November 05, 2008, 11:11:33 pm
Thats a good approach to it. The problem I have with endpoints is when I get something such as $5cos(\frac{2\pi}{3})$ . I know it's probably very dumb but I don't know how to work these out when they are not the values I'm used to!!
$\frac{2\pi}{3}$ is in the second quadrant and cos is negative in that quadrant. So just convert it to first quadrant and you get $-5cos(\frac{\pi}{3})$

Another way is just to convert $\frac{2\pi}{3}$ into degrees
Title: Re: Bucket's Questions
Post by: Mao on November 05, 2008, 11:49:50 pm: Quote from: bucket on November 05, 2008, 10:43:51 pm
Mmm, whenever I do a trial exam I always fuck up when i need to manually plot a trig graph. Does anybody have any tricks for doing these? I have the most difficulty determining endpoints and the y intercept whenever the graph is translated in any direction.

draw the most basic function, then follow the steps of transformations.

normally, you only need to draw one quarter of the graph, as the functions are all periodic and perfectly symmetrical.

as for the table of exact values, remember this:

for $\theta = \left\{ 0,\frac{\pi}{6},\frac{\pi}{4},\frac{\pi}{3},\frac{\pi}{2}\right\}$

$\sin \theta = \left\{ \frac{\sqrt{0}}{2},\frac{\sqrt{1}}{2},\frac{\sqrt{2}}{2},\frac{\sqrt{3}}{2},\frac{\sqrt{4}}{2}\right\}$
and cos is the reverse of this order.
now just gotta remember how to use the unit circle and you have all the angles you need.
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 12:27:41 am: independent events vs mutually exclusive, whats the difference again and how do you find the probabilities of the intersection of these events.
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 12:33:26 am: If $A$ and $B$ are mutually exclusive, the probability of their intersection is zero.

Quote
$\mbox{Pr}(A \cap B) = 0$

If they are independent, then given one of them (conditional probability), the probability of the other is the same as the probability of the other alone. A consequence of this is that the probability of their intersection is the product of both the individual probabilities.

Quote
$\mbox{Pr}(A | B) = \mbox{Pr}(A)$

Since $\mbox{Pr}(A | B) = \frac{\mbox{Pr}(A \cap B)}{\mbox{Pr}(B)}$

$\implies \mbox{Pr}(A \cap B) = \mbox{Pr}(A | B) \cdot \mbox{Pr}(B) = \mbox{Pr}(A) \cdot \mbox{Pr}(B)$
Title: Re: Bucket's Questions
Post by: Glockmeister on November 06, 2008, 12:35:18 am: Independent events (I'm assuming that's what you want because independent variables is a further/other sciences topic) are events where the probability of one event occuring does not affect the probability of another even occuring. The forumula is below

$\Pr(A \cap B)=\Pr(A)\Pr(B) \!\,$

Mutually exclusive events are events where one event can not occur alongside another event i.e $\Pr(A \cap B) = 0$ Thus the forumula is

$Pr (A \cup B) = Pr(A) + Pr(B)$
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 12:38:16 am: thanks heaps guys
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 12:43:12 am: edit* I want to rephrase the question to be more general :P

How do I find $pr(A\cup B)$ when A and B are independent? :S
Title: Re: Bucket's Questions
Post by: fredrick on November 06, 2008, 12:43:57 am: add them?
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 12:44:38 am: So it's the same as the mutually exclusive events?
Title: Re: Bucket's Questions
Post by: Glockmeister on November 06, 2008, 12:50:24 am: yeah, you would need to have $Pr(A)$ and $Pr(B)$ to be able to work out $Pr(A \cup B)$
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 12:56:53 am: In general: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) + \mbox{Pr}(A \cap B)$

Since, for independent events: $\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \cdot \mbox{Pr}(B)$

Then: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) + \mbox{Pr}(A) \cdot \mbox{Pr}(B)$
Title: Re: Bucket's Questions
Post by: Mao on November 06, 2008, 01:08:43 am: Quote from: coblin on November 06, 2008, 12:56:53 am
In general: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) - \mbox{Pr}(A \cap B)$

Since, for independent events: $\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \cdot \mbox{Pr}(B)$

Then: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) - \mbox{Pr}(A) \cdot \mbox{Pr}(B)$

corrected.
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 01:14:28 am: oh I see now, thanks... this type of probability always confuses me =_=
Title: Re: Bucket's Questions
Post by: cara.mel on November 06, 2008, 09:18:07 am: Quote from: Mao on November 06, 2008, 01:08:43 am
Quote from: coblin on November 06, 2008, 12:56:53 am
In general: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) - \mbox{Pr}(A \cap B)$

Since, for independent events: $\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \cdot \mbox{Pr}(B)$

Then: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) - \mbox{Pr}(A) \cdot \mbox{Pr}(B)$

corrected.

at least you're not deleting incorrect posts any more, and taking all the credit by editing one line of someone elses work, kinda a good thing your power is gone eh?
Title: Re: Bucket's Questions
Post by: Mao on November 06, 2008, 11:23:25 am: Quote from: caramel on November 06, 2008, 09:18:07 am
Quote from: Mao on November 06, 2008, 01:08:43 am
Quote from: coblin on November 06, 2008, 12:56:53 am
In general: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) - \mbox{Pr}(A \cap B)$

Since, for independent events: $\mbox{Pr}(A \cap B) = \mbox{Pr}(A) \cdot \mbox{Pr}(B)$

Then: $\mbox{Pr}(A \cup B) = \mbox{Pr}(A) + \mbox{Pr}(B) - \mbox{Pr}(A) \cdot \mbox{Pr}(B)$

corrected.

at least you're not deleting incorrect posts any more, and taking all the credit by editing one line of someone elses work, kinda a good thing your power is gone eh?

lol, i delete my own incorrect posts because they don't contribute anything.
and i plan to delete this post when coblin fix his post, so...
GG
Title: Re: Bucket's Questions
Post by: cara.mel on November 06, 2008, 11:28:41 am: You've deleted coblin's posts before on a maths board. It is disgusting behaviour.

Why the hell would I care what you do to your own posts? When you infringe upon the rights of others is when I get angry.
Title: Re: Bucket's Questions
Post by: Mao on November 06, 2008, 11:42:06 am: Quote from: caramel on November 06, 2008, 11:28:41 am
You've deleted coblin's posts before on a maths board. It is disgusting behaviour.

Why the hell would I care what you do to your own posts? When you infringe upon the rights of others is when I get angry.

I had chocolate ice cream last friday. You should condemn me because there are still kids in Africa in poverty.
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 11:52:10 am: fuck off caramel, stop picking fights
Title: Re: Bucket's Questions
Post by: cara.mel on November 06, 2008, 11:59:34 am: He picked fights about a month before I started fighting back.
If you stay quiet (I originally ignored them hoping he would stop), he does it more.
Now I speak up against him, he stops following my posts and bagging them. Because he doesn't want to look worse than he already is.

I am not staying quiet against this bully. Someone has to stand up to him.
Title: Re: Bucket's Questions
Post by: cara.mel on November 06, 2008, 12:01:16 pm: Quote from: Mao on November 06, 2008, 11:42:06 am
Quote from: caramel on November 06, 2008, 11:28:41 am
You've deleted coblin's posts before on a maths board. It is disgusting behaviour.

Why the hell would I care what you do to your own posts? When you infringe upon the rights of others is when I get angry.

I had chocolate ice cream last friday. You should condemn me because there are still kids in Africa in poverty.

Why would I condemn you for that?
You didn't steal the ice cream. You didn't claim you made the ice cream. You didn't do anything bad.
Title: Re: Bucket's Questions
Post by: cara.mel on November 06, 2008, 12:10:28 pm: http://vcenotes.com/forum/index.php/topic,6047.msg74147.html#msg74147
http://vcenotes.com/forum/index.php/topic,6325.msg76895.html#msg76895
http://vcenotes.com/forum/index.php/topic,6198.msg75352.html#msg75352
http://vcenotes.com/forum/index.php/topic,6198.msg75369.html#msg75369
http://vcenotes.com/forum/index.php/topic,6198.msg75498.html#msg75498

The fact of the matter is that he was attacking me and picking fights for ~2 weeks (if you read through, you will find no provoking on my behalf) before I said anything back. Most of the community stays silent. I will not tolerate his bullshit.
THis is how you get treated when you do Mao a big favour. I would suggest no one copies my mistake. Don't show him kindness because it bites you back on the arse. Hard.
Title: Re: Bucket's Questions
Post by: droodles on November 06, 2008, 12:21:07 pm: CARAMEL WITH INVASION THREAD FIGHITNIG ~!~!~!

CHOCORATE ICED-CREAM LINKING TO AFRICAN CHILD ?~!~ NEW LEVEL ATTACK
Title: Re: Bucket's Questions
Post by: Mao on November 06, 2008, 12:21:37 pm: (http://nissaninfiniticlub.net/photopost/data/500/4455normal_Internet-SeriousBusiness.jpg)

this thread is begging to be split.
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 12:22:28 pm: Quote from: caramel on November 06, 2008, 11:59:34 am
He picked fights about a month before I started fighting back.
If you stay quiet (I originally ignored them hoping he would stop), he does it more.
Now I speak up against him, he stops following my posts and bagging them. Because he doesn't want to look worse than he already is.

I am not staying quiet against this bully. Someone has to stand up to him.
stfu, fight with him elsewhere. you're not fighting back, you're looking for reasons to fight with him.
hey enwiabe, she's clearly trolling, ban??
Title: Re: Bucket's Questions
Post by: Mao on November 06, 2008, 12:29:11 pm: Quote from: caramel on November 06, 2008, 12:10:28 pm
http://vcenotes.com/forum/index.php/topic,6047.msg74147.html#msg74147
http://vcenotes.com/forum/index.php/topic,6325.msg76895.html#msg76895
http://vcenotes.com/forum/index.php/topic,6198.msg75352.html#msg75352
http://vcenotes.com/forum/index.php/topic,6198.msg75369.html#msg75369
http://vcenotes.com/forum/index.php/topic,6198.msg75498.html#msg75498

The fact of the matter is that he was attacking me and picking fights for ~2 weeks (if you read through, you will find no provoking on my behalf) before I said anything back. Most of the community stays silent. I will not tolerate his bullshit.
THis is how you get treated when you do Mao a big favour. I would suggest no one copies my mistake. Don't show him kindness because it bites you back on the arse. Hard.

oh, btw, well done. you've quoted me from two threads
one of them is the exam uploading thread..... [we all know how the collective student community feel about that: the more the merrier]. if you be the side that stays silent, then in that thread I can say with fair confidence that my contention is shared by many.
the second is the abortion thread..... the attack was aimed at brendan/coblin's arguments [and indeed all rationalists]

so that brings me to ask: is this what you think hostility is?
Title: Re: Bucket's Questions
Post by: enwiabe on November 06, 2008, 12:30:54 pm: children, children, no more of this public bickering. Take it to a private channel. Don't make me reprimand you guys :(
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 12:34:12 pm: moving back onto the topic; how do I deal with this question :p
$2sin^2x=1$
Title: Re: Bucket's Questions
Post by: Mao on November 06, 2008, 12:35:57 pm: $\implies \sin^2 (x)=\frac{1}{2}$

$\implies \sin(x)=\pm \frac{1}{\sqrt{2}}$
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 12:37:31 pm: thanks
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 01:03:21 pm: Quote from: Mao on November 06, 2008, 11:42:06 am
Quote from: caramel on November 06, 2008, 11:28:41 am
You've deleted coblin's posts before on a maths board. It is disgusting behaviour.

Why the hell would I care what you do to your own posts? When you infringe upon the rights of others is when I get angry.

I had chocolate ice cream last friday. You should condemn me because there are still kids in Africa in poverty.

OT:

That's the fundamental difference between the idea of human "rights" and what is a sustainable set of rights. When people believe that kids in Africa have some inherent "right" to food, although it is well-intentioned, it implies they have the right to coerce others to attain that food. On the other hand, libertarians believe that the set of rights that humans have are based on what they cannot do - that is, they cannot infringe on the rights of others. When you produce rights based on what they can do, it means they have some command over resources that they do not own, and it is not viable in a world with scarce and finite resources.

It's the contest between positive rights and negative rights.

That said, however, when you post on a private forum, you don't have the right to have your post permenantly etched into the server. It is to the discretion of the owner (and whoever the owner delegates to moderate).

Don't be silly about the rationalist thing. I've made a few posts explaining the flexibility of rationalism, and I've explained to you on IRC how rational behaviour encompasses any kind of value-input (including yours!). The only reason why you love quipping "rationalist, rationalist!" is because you used to be able to fool people into thinking that Brendan and I have a heart of stone. If you actually read our posts, you would see we acknowledge that other opinions are well-intentioned but we show how they do not actually get them to that point.
Title: Re: Bucket's Questions
Post by: bec on November 06, 2008, 01:21:25 pm: $\therefore x= \frac {\sqrt {7sin(t)} }{e^{19}}$

Oh sorry, am I in the wrong place?
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 03:28:22 pm: Hmm, does anybody have any questions which are similar to "consider equation f(x). Find the tangent which touches this curve and passes through the origin"?
I need some practice with these questions...I've only done like 3 in my whole life >.<.
Title: Re: Bucket's Questions
Post by: bec on November 06, 2008, 03:43:30 pm: Yup, I was just doing one then!

Let $f(x)=4x^3 + 1$ . Find the equation of the tangent to f(x) that passes through the origin.
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 03:45:43 pm: neap 2006? :p
lol thanks anyway bec :)
Title: Re: Bucket's Questions
Post by: bec on November 06, 2008, 04:07:04 pm: haha i tried
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 05:07:29 pm: "State the maximal domain for $\sqrt{\frac{x+2}{x-3}}$ "
In a previous part of this question I sketched $\frac{x+2}{x-3}$ ...do I use that graph to help me or do I work it out algebraically?
Title: Re: Bucket's Questions
Post by: nerd on November 06, 2008, 05:14:55 pm: Well, all you need to do is look at the graph and see all the points where y is negative and NOT count those in the domain.

Remember that you can't take the square root of a negative number, but all other values are ok!
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 05:15:57 pm: oh thanks =]
Title: Re: Bucket's Questions
Post by: shinny on November 06, 2008, 05:16:42 pm: And don't forget to not include the asymptote, but to include the x-intercepts.
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 05:21:49 pm: You've got two main conditions that need to be satisfied (the intersection of these two):

1) $x\neq 3$

2) $\frac{x+2}{x-3} \geq 0$

The second one is a bit more complicated:

Either (i) or (ii) are acceptable domains (i.e: the union of these domains constitute the 2nd condition)

i) if $x>3$ , then $x+3 > 0$ , hence: $x+2 \geq 0 \implies x \geq -2 \therefore \; x > 3$

ii) if $x<3$ , then $x+3 < 0$ , hence: $x+2 \leq 0 \implies x \leq -2 \therefore \; x \leq -2$

So, the maximal domain is the intersection of:

1) $x\neq 3$ , and

2) the union of $x > 3$ and $x \leq -2$

So you should get all real values excluding the region $(-2, 3]$
Title: Re: Bucket's Questions
Post by: fredrick on November 06, 2008, 05:23:08 pm: Quote from: coblin on November 06, 2008, 05:21:49 pm
You've got two main conditions that need to be satisfied (the intersection of these two):

1) $x\neq 3$

2) $\frac{x+2}{x-3} \geq 0$

The second one is a bit more complicated:

Either (i) or (ii) are acceptable domains (i.e: the union of these domains constitute the 2nd condition)

i) if $x>3$ , then $x+3 > 0$ , hence: $x+2 \geq 0 \implies x \geq -2 \therefore \; x > 3$

ii) if $x<3$ , then $x+3 < 0$ , hence: $x+2 \leq 0 \implies x \leq -2 \therefore \; x \leq -2$

So, the maximal domain is the intersection of:

1) $x\neq 3$ , and

2) the union of $x > 3$ and $x \leq -2$

So you should get all real values excluding the region $(-2, 3]$
or you can sketch the graph and pick the values above the x-axis
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 05:23:41 pm: Same thing. You're solving for the same numbers essentially. This is an analytical approach as opposed to a graphical approach.
Title: Re: Bucket's Questions
Post by: rowshan on November 06, 2008, 08:58:57 pm: Coblin, howd you go from here

Quote from: coblin on November 06, 2008, 05:21:49 pm
You've got two main conditions that need to be satisfied (the intersection of these two):

Either (i) or (ii) are acceptable domains (i.e: the union of these domains constitute the 2nd condition)

i) if $x>3$ , then $x+3 > 0$ , hence: $x+2 \geq 0 \implies x \geq -2 \therefore \; x > 3$

ii) if $x<3$ , then $x+3 < 0$ , hence: $x+2 \leq 0 \implies x \leq -2 \therefore \; x \leq -2$

So, the maximal domain is the intersection of:

2) the union of $x > 3$ and $x \leq -2$

So you should get all real values excluding the region $(-2, 3]$

to here?
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 09:01:15 pm: Where to where?

That should be written as R\(-2,3]. I said it in words.

$\frac{x+2}{x+3} \geq 0$ can be simplified by multiplying both sides by $x+3$ . However, it's not that simple for an inequality. If we multiply by a negative number, it changes the sign, which is why I have explored both cases.
Title: Re: Bucket's Questions
Post by: cara.mel on November 06, 2008, 09:48:44 pm: Three threads.
You weren't attacking brendan and coblin, you were attacking my post alone. You didn't answer my question and used it just to mock me (possibly you don't have an answer to it, you seem incapable of explaining your contention)

Bucket: I can't fight with him elsewhere as he has blocked all methods of private correspondance. Quite useful when you borrow something and NEED TO GIVE IT BACK.
Title: Re: Bucket's Questions
Post by: rowshan on November 06, 2008, 10:14:46 pm: yeah i read your answer properly coblin thanks a shit load =D
just a thought...how come we wernt taught this inequality stuff in earlier years?
Title: Re: Bucket's Questions
Post by: lacoste on November 06, 2008, 10:19:15 pm: Quote from: droodles on November 06, 2008, 12:21:07 pm
CHOCORATE ICED-CREAM LINKING TO AFRICAN CHILD ?~!~ NEW LEVEL ATTACK

you pisser!! made me laugh LAUGH OUT LOUD,,, reminds me of pokemon
Title: Re: Bucket's Questions
Post by: bucket on November 06, 2008, 10:45:40 pm: Quote from: caramel on November 06, 2008, 09:48:44 pm
Three threads.
You weren't attacking brendan and coblin, you were attacking my post alone. You didn't answer my question and used it just to mock me (possibly you don't have an answer to it, you seem incapable of explaining your contention)

Bucket: I can't fight with him elsewhere as he has blocked all methods of private correspondance. Quite useful when you borrow something and NEED TO GIVE IT BACK.
...get out of my question thread =_=
Title: Re: Bucket's Questions
Post by: vce01 on November 06, 2008, 10:57:22 pm: Quote from: bec on November 06, 2008, 03:43:30 pm
Yup, I was just doing one then!

Let $f(x)=4x^3 + 1$ . Find the equation of the tangent to f(x) that passes through the origin.

can someone explain how to do this? thanks
Title: Re: Bucket's Questions
Post by: onlyfknhuman on November 06, 2008, 11:00:40 pm: what does it mean by passes through the origin? i thougth that meant passing through (0,0).
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 11:02:35 pm: Quote from: vce01 on November 06, 2008, 10:57:22 pm
Quote from: bec on November 06, 2008, 03:43:30 pm
Yup, I was just doing one then!

Let $f(x)=4x^3 + 1$ . Find the equation of the tangent to f(x) that passes through the origin.

can someone explain how to do this? thanks

$f'(x) = 12x^2$

We are looking for a tangent: $y = mx + c$ , which passes through the origin, $(0,0)$ .

Therefore: $0 = m(0) + c \implies c = 0$

$\implies y = mx$

What else do we know? At the point when the tangent touches the curve, $(x_1, y_1)$ :

1) $y_1 = 4x_1^3 + 1$

2) $m = 12x_1^2$

Substituting this into: $y = mx$ at $(x_1, y_1)$

$\implies y_1 = mx_1$

$\implies 4x_1^3 + 1 = (12x_1^2)x_1$

$\implies 4x_1^3 + 1 = 12x_1^3$

$\implies x_1^3 = \frac{1}{8}$

$\implies x_1 = \frac{1}{2}$

Substituting this into equation 2:

$m = 12x_1^2 = \frac{12}{4} = 3$

Therefore, the tangent is $y=3x$
Title: Re: Bucket's Questions
Post by: onlyfknhuman on November 06, 2008, 11:04:41 pm: ah fuck im such a retard lol

so pro coblin
Title: Re: Bucket's Questions
Post by: Collin Li on November 06, 2008, 11:06:47 pm: Quote from: onlyfknhuman on November 06, 2008, 11:04:41 pm
ah fuck im such a retard lol

so pro coblin

It's a sort of difficult question. Always think of these kind of questions as a quest to assemble the pieces of information you have to find what you want.
Title: Re: Bucket's Questions
Post by: vce01 on November 06, 2008, 11:07:59 pm: Quote from: coblin on November 06, 2008, 11:02:35 pm
Quote from: vce01 on November 06, 2008, 10:57:22 pm
Quote from: bec on November 06, 2008, 03:43:30 pm
Yup, I was just doing one then!

Let $f(x)=4x^3 + 1$ . Find the equation of the tangent to f(x) that passes through the origin.

can someone explain how to do this? thanks

$f'(x) = 12x^2$

We are looking for a tangent: $y = mx + c$ , which passes through the origin, $(0,0)$ .

Therefore: $0 = m(0) + c \implies c = 0$

$\implies y = mx$

What else do we know? At the point when the tangent touches the curve, $(x_1, y_1)$ :

1) $y_1 = 4x_1^3 + 1$

2) $m = 12x_1^2$

Substituting this into: $y = mx$ at $(x_1, y_1)$

$\implies y_1 = mx_1$

$\implies 4x_1^3 + 1 = (12x_1^2)x_1$

$\implies 4x_1^3 + 1 = 12x_1^3$

$\implies x_1^3 = \frac{1}{8}$

$\implies x_1 = \frac{1}{2}$

Substituting this into equation 2:

$m = 12x_1^2 = \frac{12}{4} = 3$

Therefore, the tangent is $y=3x$

awesome, thanks :)
Title: Re: Bucket's Questions
Post by: bucket on November 07, 2008, 12:00:02 am: This is from 2007 exam 1, i don't know why I have trouble with this =_=
Solve $f(x)=sin\frac{2\pi x}{3} = -\frac{\sqrt{3}}{2}$ for $x\epsilon [0,3]$
and

Let $R \rightarrow R, g(x)=3f(x-1)+2$ Find the smallest possible value of x of which g(x) is a maximum.
Title: Re: Bucket's Questions
Post by: ed_saifa on November 07, 2008, 12:17:48 am: $f(x)=sin\frac{2\pi x}{3} = -\frac{\sqrt{3}}{2}$ for $x\epsilon [0,3]$
Adjust the domain so $0\le x \le 3$ goes to $0\le \frac{2\pi x}{3} \le {2\pi$
$\frac{2\pi x}{3}= \pi + \frac{\pi}{3} , 2\pi - \frac{\pi}{3}$
Just rearrange to solve for x.

$g(x)=3f(x-1) + 2$
From above $sin\frac{2\pi x}{3}$ has a period 3. So from this, and by drawing a quick graph, it has a maximum at $x= \frac{3}{4}$
For $g(x)$ the vertical translations don't affect x value in this case. Since it's been translated 1 unit to the right, the maximum must be $\frac{3}{4} + 1=\frac{7}{4}$
Title: Re: Bucket's Questions
Post by: bucket on November 07, 2008, 07:30:34 am: ahhhh i get it, except for one part; how did you change the domain?
I was able to solve it without doing so but I'm just curious :p
...I know the exam is in like, 1.5 hours lol but I'm just curious :) If anyone is awake to show me I'd be grateful!
Title: Re: Bucket's Questions
Post by: onlyfknhuman on November 07, 2008, 08:03:29 am: You dont really need to adjust the domain.

But you see the original domain is [0,3]

the 0<x<3

u times 2pi/3 into x
whatever you to do x
you to '3' and '0'

therefore

2pi/3 x 3
= 6pi /3
= 2pi
therefore 0<2pix/3 <2pi

Bucket is that really you in the picture? i really wanna punch you in the face LOL :D
Title: Re: Bucket's Questions
Post by: onlyfknhuman on November 07, 2008, 08:03:47 am: OMG OMG OMG OMG THE EXAM IS IN 1 HOUR!
Title: Re: Bucket's Questions
Post by: bucket on November 07, 2008, 01:30:21 pm: It's not me lol.
www.yearbookyourself.com
Title: Re: Bucket's Questions
Post by: onlyfknhuman on November 07, 2008, 02:23:23 pm: LOL COOL.
Title: Re: Bucket's Questions
Post by: bucket on November 08, 2008, 06:02:57 pm: How do you do this?? Note this is from a CAS exam.

[IMG]http://img375.imageshack.us/img375/6826/questionxt9.th.jpg[/img][IMG]http://img375.imageshack.us/images/thpix.gif[/img]
Title: Re: Bucket's Questions
Post by: danieltennis on November 08, 2008, 06:07:08 pm: Quote from: bucket on November 08, 2008, 06:02:57 pm
How do you do this?? Note this is from a CAS exam.

[IMG]http://img375.imageshack.us/img375/6826/questionxt9.th.jpg[/img][IMG]http://img375.imageshack.us/images/thpix.gif[/img]

HAHA. I just did this question.
The best way is to solve for a,b,c,d,e on ur cas using matrices.
We know that a=.1 and c=3.9
Therefore, c=39 x .1
c = 3.9
Hence, c=39a
So the answer is D.
Title: Re: Bucket's Questions
Post by: methodsboy on November 08, 2008, 06:13:54 pm: Quote from: bucket on November 08, 2008, 06:02:57 pm
How do you do this?? Note this is from a CAS exam.

[IMG]http://img375.imageshack.us/img375/6826/questionxt9.th.jpg[/img][IMG]http://img375.imageshack.us/images/thpix.gif[/img]

wow! which exam is that from?
Title: Re: Bucket's Questions
Post by: bucket on November 08, 2008, 06:14:28 pm: geez that takes a lot of time for a multi choice q :S

its from tssm cas 2008
Title: Re: Bucket's Questions
Post by: danieltennis on November 08, 2008, 06:14:56 pm: I reckon. It took me like 3 minutes.
Title: Re: Bucket's Questions
Post by: methodsboy on November 08, 2008, 06:17:58 pm: if that was on the exam i would simply guess and move on..
Title: Re: Bucket's Questions
Post by: bucket on November 08, 2008, 09:56:34 pm: The graphs of $y=kx-3$ intersects the graph $y=x^2+8x$ at two distinct points for:

a. $k=11$
b. $k>8+2\sqrt{3}$ or $k<8-2\sqrt{3}$
c. $5\leq k \leq 6$
d. $8-2\sqrt{3} \leq k \leq 8+2\sqrt{3}$
e. $k=5$

How do I work this out quickly?
Title: Re: Bucket's Questions
Post by: fredrick on November 08, 2008, 10:15:16 pm: make the 2 equations equal. Take all to one side. Apply Discriminant and it mut be >0
Title: Re: Bucket's Questions
Post by: bucket on November 08, 2008, 10:23:44 pm: oh. simple enough. thanks.
Title: Re: Bucket's Questions
Post by: bucket on November 08, 2008, 11:20:46 pm: Hmm, I really am not good with inequalities.
Here is my working for this question.

$kx-3=x^2+8x$

$kx-3-x^2-8x=0$

$-x^2+(k-8)x-3=0$

$\Delta >0$ to have two solutions

$\therefore (k-8)^2+12 > 0$

This is where I stuff up.

$(k-8)^2 > 12$

$k-8 >\pm 2\sqrt{3}$

$k > 8+2\sqrt{3}$ or $8-2\sqrt{3}$ I know I've done something wrong here with the inequality sign lol.
The correct answer is $k>8+2\sqrt{3}$ or $k<8-2\sqrt{3}$ . Where does the < sign come in? :P
Title: Re: Bucket's Questions
Post by: nerd on November 08, 2008, 11:29:21 pm: Its best to draw the graoh of $y=(k-8)^2$ and $y=12$ , see where they intersect and see where $y=(k-8)^2$ is greater than $y=12$ .

Remember, when you have non-linear inequalities, you must draw the graph, or at the very least visualise it, in order to solve the inequality!
Title: Re: Bucket's Questions
Post by: bucket on November 08, 2008, 11:32:47 pm: Oh yeah we can use our calculators!
Lol thanks a lot man.
Title: Re: Bucket's Questions
Post by: bucket on November 08, 2008, 11:55:57 pm: Is { $x:cos^2(x)+2cos(x)=0$ }= asking for solutions? lol.
Title: Re: Bucket's Questions
Post by: ell on November 08, 2008, 11:59:04 pm: Quote from: bucket on November 08, 2008, 11:55:57 pm
Is { $x:cos^2(x)+2cos(x)=0$ }= asking for solutions? lol.

yup, its asking for the values of x. also : (colon) means 'such that'
Title: Re: Bucket's Questions
Post by: onlyfknhuman on November 09, 2008, 12:01:13 am: Bucket im curious what you got for this, cause i did this earlier today and i got

$Cos(x) = 0 and Cos(x) = -2$

And the answer stated that it was $Cos(x) = 0$ alone. i think
Title: Re: Bucket's Questions
Post by: ell on November 09, 2008, 12:03:14 am: $\cos x$ can't equal -2.
Title: Re: Bucket's Questions
Post by: nerd on November 09, 2008, 12:18:13 am: Think about the graph of of $y=\cos x$ . For any value of $x$ , $y$ it can't be -2 (the range is [-1,1]) so the only solution is $\cos x=0$
Title: Re: Bucket's Questions
Post by: bucket on November 09, 2008, 09:09:33 pm: "If c=20 calculate the probability that a type A butterfly is misclassified and a type B butterfly is misclassified."

WTF IS C???
Title: Re: Bucket's Questions
Post by: Mao on November 09, 2008, 09:23:48 pm: sounds like cont-random/normal dist.
what's the full question?
Title: Re: Bucket's Questions
Post by: bucket on November 09, 2008, 09:40:14 pm: Thats part G of a question 4 in here.