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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: bec on November 08, 2008, 04:40:29 pm

Title: Equating gradients
Post by: bec on November 08, 2008, 04:40:29 pm: Does anyone have/can anyone think up any questions that require us to find the value of a point on a graph by equating gradients?
I can't find any examples of it (otherwise I would put one up here so that I actually make sense...)

Thanks
Title: Re: Equating gradients
Post by: Collin Li on November 08, 2008, 05:06:42 pm: That's a pretty ambiguous request? Haha. What do you mean?
Title: Re: Equating gradients
Post by: shinny on November 08, 2008, 05:11:32 pm: Your request actually fits perfectly into any question that involves finding stationary points =P In those questions, you're just equating gradient to zero and finding those points ahah. For other values, it's the same process so I don't see why you need any specific practise.
Title: Re: Equating gradients
Post by: danieltennis on November 08, 2008, 05:20:17 pm: I think she could be talking about hybrid functions and making the various functions continous with the same gradients?
Title: Re: Equating gradients
Post by: bec on November 08, 2008, 05:37:59 pm: haha yeah I know, that was a pretty ambiguous request - I wasn't even totally sure of what I was looking for. I found an example of what i mean though:

(http://i291.photobucket.com/albums/ll304/foskeyr/untitled.jpg)

Ever seen a question like that?
Title: Re: Equating gradients
Post by: dekoyl on November 08, 2008, 05:40:35 pm: Mind posting rest of the question bec? :)

And no I haven't seen one like that so I'd like to attempt it. Thanks.
Title: Re: Equating gradients
Post by: danieltennis on November 08, 2008, 05:46:28 pm: Quote from: bec on November 08, 2008, 05:37:59 pm
haha yeah I know, that was a pretty ambiguous request - I wasn't even totally sure of what I was looking for. I found an example of what i mean though:

(http://i291.photobucket.com/albums/ll304/foskeyr/untitled.jpg)

Ever seen a question like that?

This is from one of the VCAA exams. Not to sure what year. Maybe 2003?
Title: Re: Equating gradients
Post by: onlyfknhuman on November 08, 2008, 05:48:14 pm: What the gook is that, geesus fug no idea :-[
Title: Re: Equating gradients
Post by: bec on November 08, 2008, 05:48:30 pm: Oh sorry, it got cut off! I'm on a rolllllll tonight...

The question is VCAA 2005 CAS exam 1, Q4

"Show that $\frac {f(-p)+f(p)}{2}=f(0)$ "
Title: Re: Equating gradients
Post by: dekoyl on November 08, 2008, 05:49:30 pm: Quote from: onlyfknhuman on November 08, 2008, 05:48:14 pm
What the gook is that
Gook. GOOK?

Quote from: danieltennis on November 08, 2008, 05:46:28 pm

This is from one of the VCAA exams. Not to sure what year. Maybe 2003?

Thank you daniel.

Edit: Ah thanks bec!
Title: Re: Equating gradients
Post by: Mao on November 08, 2008, 07:53:31 pm: Quote from: bec on November 08, 2008, 05:48:30 pm
Oh sorry, it got cut off! I'm on a rolllllll tonight...

The question is VCAA 2005 CAS exam 1, Q4

"Show that $\frac {f(-p)+f(p)}{2}=f(0)$ "

mmm i see. equating the gradient of AB and BC will yield this relationship.

from my knowledge, they may ask you [though it is very rare] for a composite function that "join smoothly"

if you want to prepare for these, grab a random function [say $y=\log_e (x+2)$ ] , pick a domain [say $[2,\infty)$ ], pick another function type [say $p(x)=ax^2+bx+c$ ], now try to find values of those variables that makes the graphs join smoothly.

after you've done that a couple times, you'll have pretty good grasp of it.
hope that's what you wanted.
Title: Re: Equating gradients
Post by: bec on November 08, 2008, 08:07:58 pm: Yeah thanks Mao but that's not really what I meant. I don't even know what I meant really, it was (again) just a note I had written to myself a while ago that "equating gradients" was something I had to work on.

So um...disregard this whole thread...? haha
Title: Re: Equating gradients
Post by: droodles on November 08, 2008, 08:13:41 pm: dekoyl just derive the word with respect to 'O' and it won't be bad
Title: Re: Equating gradients
Post by: shinny on November 08, 2008, 08:17:01 pm: $\frac{d}{do}(gook)=\frac{d}{do}(gko^{2})$
$\implies \frac{d}{do}(gook)=2gko$

Like that droodles?
Title: Re: Equating gradients
Post by: Collin Li on November 08, 2008, 08:17:50 pm: 2GOK
Title: Re: Equating gradients
Post by: Mao on November 08, 2008, 08:21:51 pm: Quote from: shinjitsuzx on November 08, 2008, 08:17:01 pm
$\frac{d}{do}(gook)=\frac{d}{do}(gko^{2})$
$\implies \frac{d}{do}(gook)=2gko$

Like that droodles?

provided neither "g" and "k" are functions of "o"

if they both are:

$\frac{d}{do}(gook) = \frac{d}{do}(o^2)\cdot gk + \frac{d}{do}(gk)\cdot o^2 = 2ogk + \left(\frac{dg}{do}k+\frac{dk}{do}g\right)\cdot o^2$

$\implies \frac{d}{do}(gook) = o\cdot \left( 2gk + ok\frac{dg}{do} + og\frac{dk}{do}\right)$

wait... too far?
Title: Re: Equating gradients
Post by: shinny on November 08, 2008, 08:23:11 pm: Yes Mao, you clearly took it too far.
Title: Re: Equating gradients
Post by: Mao on November 08, 2008, 08:24:31 pm: :P
Title: Re: Equating gradients
Post by: droodles on November 08, 2008, 08:25:05 pm: ARE YOU A VICTIM OF RACIAL SLUR?

DO YOU ENJOY MATHS?

WELL HERE'S A SOLUTION...DERIVE THEM AWAY!

Quote from: Mao on November 08, 2008, 08:21:51 pm
provided neither "g" and "k" are functions of "o"

if they both are:

$\frac{d}{do}(gook) = \frac{d}{do}(o^2)\cdot gk + \frac{d}{do}(gk)\cdot o^2 = 2ogk + \left(\frac{dg}{do}k+\frac{dk}{do}g\right)\cdot o^2$

$\implies \frac{d}{do}(gook) = o\cdot \left( 2gk + ok\frac{dg}{do} + og\frac{dk}{do}\right)$

NOW YOUR PROBLEMS ARE 'SOLVED'
Title: Re: Equating gradients
Post by: bec on November 08, 2008, 08:26:08 pm: oh no he went there
Title: Re: Equating gradients
Post by: Mao on November 08, 2008, 08:29:45 pm: completely random, but here's an insight I had when I was deriving 'gook':

the multiple products rule:

$\mbox{Let }f(x)=f_1 (x) \cdot f_2 (x) \cdot f_3(x) \cdots$

$\frac{df}{dx} = \frac{df_1}{dx}\cdot \frac{f(x)}{f_1(x)} + \frac{df_2}{dx}\cdot \frac{f(x)}{f_2(x)} + \frac{df_3}{dx}\cdot \frac{f(x)}{f_3(x)} + \cdots$

thank you for your time.

which means, the derivative of gook can be re-expressed as

$\frac{d}{do}(gook) = (gook) \cdot \left( 2o^{-1} + g^{-1} \cdot \frac{dg}{do} + k^{-1}\cdot \frac{dk}{do}\right)$

it appears the racial slur is invincible..... :'(
Title: Re: Equating gradients
Post by: shinny on November 08, 2008, 08:32:22 pm: I PREFER INTEGRATING RACIAL TENSION INTO THIS FORUM THANK YOU VERY MUCH

$\int (gook) do=\int (gko^{2}) do$

$=[\frac{1}{3}gko^{3}]$

$\int \frac{1}{3}gko^{3} do=[\frac{1}{12}gko^{4}]$

$...$

$=[\frac{1}{9000}gko^{9000}]$

EQUALS
GOOOOOOOOOOOK
Well maybe not quite.

Quote from: bec on November 08, 2008, 08:07:58 pm
So um...disregard this whole thread...? haha
And yes, this is what bec gets for derailing her own thread.
Title: Re: Equating gradients
Post by: droodles on November 08, 2008, 08:36:43 pm: apparently at mhs the students get erections after they see 50/50 for any of their sacs

shinsjuxusijxxxznarutofannerd confirm?
Title: Re: Equating gradients
Post by: shinny on November 08, 2008, 08:43:12 pm: Quote from: droodles on November 08, 2008, 08:36:43 pm
apparently at mhs the students get erections after they see 50/50 for any of their sacs
You should some students' reaction when they LOSE a mark.

Quote from: droodles on November 08, 2008, 08:36:43 pm
shinsjuxusijxxxznarutofannerd confirm?

Where'd the Naruto come from? o_O Clearly that description fits ed_saifa better.
Title: Re: Equating gradients
Post by: Toothpaste on November 08, 2008, 08:45:27 pm: 真実
translate: jp -> en
then find romaji.
Title: Re: Equating gradients
Post by: vce08 on November 08, 2008, 08:52:20 pm: 50/50 is clearly standard at MHS