Find the minimum distance between the curves y=-x^2 and y=(6-x)^2

- jessica

Find the minimum distance between the curves y=-x^2 and y=(6-x)^2

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- katieb

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- bahrom7893

WOOT MY FAVE TOPIC! OPTIMIZATION!

- bahrom7893

let me refresh my memory real quick.

- jessica

so glad you can help i've been stuck for like 2 hours.

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## More answers

- bahrom7893

what was the distance formula again? sorry haven't done this in almost 2 years

- bahrom7893

nevermind ill get one of my review books

- bahrom7893

k workin on it

- jessica

sorry the distance formula is \[D= \sqrt{(x1-x2) ^{2}+(y1-y2)^{2}}\]

- bahrom7893

okay well how far did u get in this problem? I've never done problems with curves. Only distance to the point. Do u have anoy worked out examples?

- bahrom7893

I mean the problems i used to solve were find the minimum from curve to a point

- jessica

Right that's where I'm having the problem
so I've created a formula by subtracting the two curves from eachother... hold on i'll put all my work on here, i thought i had gotten to an answer, but my answer didn't make sense so now i'm back at square 1

- bahrom7893

k

- jessica

Let D(x) = | [(6 – x)^2] - [-x^2] | be the
function representing the Distance between those two given functions.
Both of the given functions are continuous.
So D(x) is continuous.
If D(x) is 0 at some point, then that’s our minimal distance.
Otherwise, D(x) is never zero.
And so it would be always strictly positive or strictly negative for any x.
Can it ever be true that those two curves cross, making D(x) = 0?
We’d like to solve –x^2 = (6-x)^2.
This gives –x^2 = 36 – 12x + x^2.
Or x^2 – 6x + 18 = 0.
This yields only complex roots, and so the curves never cross for any x.
Therefore, either D(x) is always negative, or it is always positive.
(It can’t go between being positive and negative or else it would have to be zero at some point.)
Let’s pick a test point, some easy point to evaluate D(x).
Like D(0). D(0) = 36.
So D(x) will always be positive.
Moreover, we want to minimize D(x), and we can drop the absolute value bars now.
[If D(x) was always negative, we would then want to maximize D(x) to find the minimal distance (I don’t know if that makes sense to you).]
So minimize it using calculus. Find the critical points of D’(x).
Then test to see whether those critical points are minima
(x = c is a minimum if f’(c) = 0 (or x = c is a cusp) and
f’(x) < 0 for x slightly less than c and f’(x) > 0 for x slightly bigger than c).
Then choose the smallest minimum. (I think there’s only one minimum in this case.)

- jessica

I found my minimum to be x=3
but then i didn't know if that three needed to be plugged into the original equations, and then the distance found between those two points. That didn't make sense becuase the points were (3,9) and (3,-9) and the distance from that is way far

- bahrom7893

im getting x = 3 too

- jessica

Does that thought process make sense though?
What does the 3 mean... Do I just find D(3)? 18 doesn't seem like the right answer

- jessica

Look at the graphs of those 2 equations and you'll see what why it doesn't make sense

- bahrom7893

k brb lemme find my calculator

- bahrom7893

the minimum is 1.5

- bahrom7893

wait no

- jessica

how did you get that answer?

- bahrom7893

nevermind i just looked at the graph and my first guess was 1.5, but then if u move up along one and down the other u'll see that there are maybe closer point

- bahrom7893

what class is this?

- jessica

multivariable calc

- jessica

my teacher is ridiculous. we've never been over anything like this... and i have no idea what i am doing.

- bahrom7893

im checkin the stuff, i think i gotta get my other textbook, i took calc 3 last semester.. worst class ever.

- jessica

it's been better then linear algebra if you ask me... just wish i had a better teacher

- jessica

i actually have to go to a meeting, and i'll be back in like an hour... but i really really appreciate any help that you could throw out there for me

- jessica

sorry to peace like this

- bahrom7893

okay where can i send the stuff to?

- jessica

you could put it on here, or email me at vballninja@gmail.com

- bahrom7893

k

- jessica

thanks so much!!!!!

- bahrom7893

np, ill do my best

- bahrom7893

i found some stuff, looking thru it

- bahrom7893

we were differentiatin it wrong

- jessica

how are we supposed to differentiate it?

- bahrom7893

http://in.answers.yahoo.com/question/index?qid=20100218090950AAGvf4z
see there it's like we need first to find derivative with respect to x1 then x2

- bahrom7893

hey can u go on twiddla? it'll be easier to write there i think

- bahrom7893

http://www.twiddla.com/494934

- bahrom7893

hey jessica click on the link above, im waitin for u on twiddla

- bahrom7893

ill wait here too just in case

- jessica

sorry i'm in the meeting now so i'm like trying to be discrete

- bahrom7893

oh okay, fine then just wait till its over, i got plenty of time right now

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