ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: enpassant on April 08, 2008, 04:32:39 pm

Title: minimum distance *challenge*
Post by: enpassant on April 08, 2008, 04:32:39 pm: find the minimum distance between y=e^(x-1/2) and y=ln(x)+1/2.
Title: Re: minimum distance
Post by: Mao on April 08, 2008, 05:18:14 pm: the distance between $y=e^{x-\frac{1}{2}}$ and $y=ln(x)+\frac{1}{2}$ can be modelled by

$f(x)=e^{x-\frac{1}{2}}-ln(x)-\frac{1}{2}$

to find the minimum, you would find the derivative:

$f'(x)=e^{x-\frac{1}{2}}-\frac{1}{x}$

$0=\frac{e^x}{\sqrt{e}}-\frac{1}{x}$

$xe^x-\sqrt{e}=0$

at this point, it becomes impossible to solve (using knowledge from the methods course), hence using calculator:

$x\approx 0.766$

hence the minimum distance will be:

$|f(0.766)|=|e^{0.766-\frac{1}{2}}-ln(0.766)-\frac{1}{2}|\approx 1.07$
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 05:22:41 pm: I think you misunderstood the question.
Title: Re: minimum distance
Post by: Mao on April 08, 2008, 05:23:54 pm: Quote from: evaporade on April 08, 2008, 05:22:41 pm
I think you misunderstood the question.
please specify

EDIT:
OH! i see

that is DEFINITELY not part of the methods course.

if you need help with these questions (presumably not from VCE), here's the place:

http://vcenotes.com/forum/index.php/board,55.0.html
VCE Notes > Tertiary Education > Faculties > Science > Mathematics
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 06:21:03 pm: you don't do this in tertiary
Title: Re: minimum distance
Post by: Mao on April 08, 2008, 06:29:26 pm: Quote from: evaporade on April 08, 2008, 06:21:03 pm
you don't do this in tertiary
and you dont do this in methods

go look in the study design and tell me whereabouts is a methods student required to find minimum non-verticle distance between two functions.

and:
http://vcenotes.com/forum/index.php/topic,2754.msg35083.html#msg35083
take a deep breath buddy
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 06:33:38 pm: go look in the study design and tell me whereabouts is a methods student required to find 1+1=2

Moderator action: Removed edit-war and supposed double post (see http://vcenotes.com/forum/index.php/topic,1644.0.html)
Title: Re: minimum distance
Post by: Mao on April 08, 2008, 06:39:31 pm: Quote from: evaporade on April 08, 2008, 06:33:38 pm
go look in the study design and tell me whereabouts is a methods student required to find 1+1=2
it doesnt, because:

Quote from: http://en.wikipedia.org/wiki/Invalid_proof#Proof_that_1_.3D_0
Proof that 1 = 0

Take the statement

$x = 1$

Taking the derivative of each side,

$\frac{d}{dx}x = \frac{ d}{dx}1$

The derivative of x is 1, and the derivative of 1 is 0. Therefore,

$1 = 0$

Q.E.D.

so?
Title: Re: minimum distance
Post by: Toothpaste on April 08, 2008, 06:41:36 pm: Quote from: evaporade on April 08, 2008, 06:33:38 pm
go look in the study design and tell me whereabouts is a methods student required to find 1+1=2

http://www.vcaa.vic.edu.au/prep10/csf/index.html
http://www.vcaa.vic.edu.au/prep10/csf/klas/index.html#mathematics

CSF.
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 06:42:11 pm: if x=1, x is a constant, d/dx(x) = d/dx (1) =0.
Title: Re: minimum distance
Post by: Mao on April 08, 2008, 06:45:45 pm: dude you are so totally missing the point.

stop before u make a total fool of yourself
Title: Re: minimum distance
Post by: Neobeo on April 08, 2008, 06:54:02 pm: Quote from: enpassant on April 08, 2008, 04:32:39 pm
find the minimum distance between y=e^(x-1/2) and y=ln(x)+1/2.

$\frac{\sqrt{2}}{2}$

Mao is right. This knowledge is not required for the methods course.
Title: Re: minimum distance
Post by: AppleXY on April 08, 2008, 06:54:48 pm: Isn't the minimum distance the integral? Umm yeah, hmm, but its the minimum.

EDIT: HOWD YOU DO THAT NEOMAN :D
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 06:57:24 pm: Well done Neobeo.
A year 12 student who has a good understanding of gradient functions, distance between two points, and finding minimum can do the question.
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 07:14:21 pm: Too easy for Ahmad
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 07:18:02 pm: Sorry if I offended anyone while 'mathematising' in a 'non-mathematian' way
Title: Re: minimum distance
Post by: evaporade on April 08, 2008, 07:50:14 pm: For this problem it can also be solved by recognising the two functions are inverse of each other, and find the shortest distance between a point and its image.
Title: Re: minimum distance
Post by: Mao on April 08, 2008, 10:00:21 pm: further posts in this topic:
please read http://vcenotes.com/forum/index.php/topic,2754.msg35177.html#msg35177 before proceding.

thank you