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anonymous
 5 years ago
Find the minimum distance between the curves y=x^2 and y=(6x)^2
anonymous
 5 years ago
Find the minimum distance between the curves y=x^2 and y=(6x)^2

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bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0WOOT MY FAVE TOPIC! OPTIMIZATION!

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0let me refresh my memory real quick.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so glad you can help i've been stuck for like 2 hours.

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0what was the distance formula again? sorry haven't done this in almost 2 years

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0nevermind ill get one of my review books

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry the distance formula is \[D= \sqrt{(x1x2) ^{2}+(y1y2)^{2}}\]

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0okay 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
 5 years ago
Best ResponseYou've already chosen the best response.0I mean the problems i used to solve were find the minimum from curve to a point

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Right 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Let 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 = (6x)^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.)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I 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
 5 years ago
Best ResponseYou've already chosen the best response.0im getting x = 3 too

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Does 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Look at the graphs of those 2 equations and you'll see what why it doesn't make sense

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0k brb lemme find my calculator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how did you get that answer?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0nevermind 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my teacher is ridiculous. we've never been over anything like this... and i have no idea what i am doing.

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0im checkin the stuff, i think i gotta get my other textbook, i took calc 3 last semester.. worst class ever.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it's been better then linear algebra if you ask me... just wish i had a better teacher

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry to peace like this

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0okay where can i send the stuff to?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you could put it on here, or email me at vballninja@gmail.com

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0i found some stuff, looking thru it

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0we were differentiatin it wrong

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0how are we supposed to differentiate it?

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0http://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
 5 years ago
Best ResponseYou've already chosen the best response.0hey can u go on twiddla? it'll be easier to write there i think

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0hey jessica click on the link above, im waitin for u on twiddla

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0ill wait here too just in case

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry i'm in the meeting now so i'm like trying to be discrete

bahrom7893
 5 years ago
Best ResponseYou've already chosen the best response.0oh okay, fine then just wait till its over, i got plenty of time right now
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