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

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

Mathematics
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WOOT MY FAVE TOPIC! OPTIMIZATION!
let me refresh my memory real quick.
so glad you can help i've been stuck for like 2 hours.

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what was the distance formula again? sorry haven't done this in almost 2 years
nevermind ill get one of my review books
k workin on it
sorry the distance formula is \[D= \sqrt{(x1-x2) ^{2}+(y1-y2)^{2}}\]
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?
I mean the problems i used to solve were find the minimum from curve to a point
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
k
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.)
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
im getting x = 3 too
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
Look at the graphs of those 2 equations and you'll see what why it doesn't make sense
k brb lemme find my calculator
the minimum is 1.5
wait no
how did you get that answer?
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
what class is this?
multivariable calc
my teacher is ridiculous. we've never been over anything like this... and i have no idea what i am doing.
im checkin the stuff, i think i gotta get my other textbook, i took calc 3 last semester.. worst class ever.
it's been better then linear algebra if you ask me... just wish i had a better teacher
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
sorry to peace like this
okay where can i send the stuff to?
you could put it on here, or email me at vballninja@gmail.com
k
thanks so much!!!!!
np, ill do my best
i found some stuff, looking thru it
we were differentiatin it wrong
how are we supposed to differentiate it?
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
hey can u go on twiddla? it'll be easier to write there i think
http://www.twiddla.com/494934
hey jessica click on the link above, im waitin for u on twiddla
ill wait here too just in case
sorry i'm in the meeting now so i'm like trying to be discrete
oh okay, fine then just wait till its over, i got plenty of time right now

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