Author Topic: minimum distance challenge (Read 3950 times) Tweet Share

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.

Mao · « **Reply #1 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:

$|f(0.766)|=|e^{0.766-\frac{1}{2}}-ln(0.766)-\frac{1}{2}|\approx 1.07$

evaporade · « **Reply #2 on:** April 08, 2008, 05:22:41 pm »

I think you misunderstood the question.

Mao · « **Reply #3 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

evaporade · « **Reply #4 on:** April 08, 2008, 06:21:03 pm »

you don't do this in tertiary

Mao · « **Reply #5 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

evaporade · « **Reply #6 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)

Mao · « **Reply #7 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?

Toothpaste · « **Reply #8 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.

evaporade · « **Reply #9 on:** April 08, 2008, 06:42:11 pm »

if x=1, x is a constant, d/dx(x) = d/dx (1) =0.

Mao · « **Reply #10 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

Neobeo · « **Reply #11 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.

AppleXY · « **Reply #12 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

evaporade · « **Reply #13 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.

evaporade · « **Reply #14 on:** April 08, 2008, 07:14:21 pm »

Too easy for Ahmad

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