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anonymous

  • 5 years ago

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

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  1. bahrom7893
    • 5 years ago
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    WOOT MY FAVE TOPIC! OPTIMIZATION!

  2. bahrom7893
    • 5 years ago
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    let me refresh my memory real quick.

  3. anonymous
    • 5 years ago
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    so glad you can help i've been stuck for like 2 hours.

  4. bahrom7893
    • 5 years ago
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    what was the distance formula again? sorry haven't done this in almost 2 years

  5. bahrom7893
    • 5 years ago
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    nevermind ill get one of my review books

  6. bahrom7893
    • 5 years ago
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    k workin on it

  7. anonymous
    • 5 years ago
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    sorry the distance formula is \[D= \sqrt{(x1-x2) ^{2}+(y1-y2)^{2}}\]

  8. bahrom7893
    • 5 years ago
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    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?

  9. bahrom7893
    • 5 years ago
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    I mean the problems i used to solve were find the minimum from curve to a point

  10. anonymous
    • 5 years ago
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    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

  11. bahrom7893
    • 5 years ago
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    k

  12. anonymous
    • 5 years ago
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    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.)

  13. anonymous
    • 5 years ago
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    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

  14. bahrom7893
    • 5 years ago
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    im getting x = 3 too

  15. anonymous
    • 5 years ago
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    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

  16. anonymous
    • 5 years ago
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    Look at the graphs of those 2 equations and you'll see what why it doesn't make sense

  17. bahrom7893
    • 5 years ago
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    k brb lemme find my calculator

  18. bahrom7893
    • 5 years ago
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    the minimum is 1.5

  19. bahrom7893
    • 5 years ago
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    wait no

  20. anonymous
    • 5 years ago
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    how did you get that answer?

  21. bahrom7893
    • 5 years ago
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    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

  22. bahrom7893
    • 5 years ago
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    what class is this?

  23. anonymous
    • 5 years ago
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    multivariable calc

  24. anonymous
    • 5 years ago
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    my teacher is ridiculous. we've never been over anything like this... and i have no idea what i am doing.

  25. bahrom7893
    • 5 years ago
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    im checkin the stuff, i think i gotta get my other textbook, i took calc 3 last semester.. worst class ever.

  26. anonymous
    • 5 years ago
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    it's been better then linear algebra if you ask me... just wish i had a better teacher

  27. anonymous
    • 5 years ago
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    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

  28. anonymous
    • 5 years ago
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    sorry to peace like this

  29. bahrom7893
    • 5 years ago
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    okay where can i send the stuff to?

  30. anonymous
    • 5 years ago
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    you could put it on here, or email me at vballninja@gmail.com

  31. bahrom7893
    • 5 years ago
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    k

  32. anonymous
    • 5 years ago
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    thanks so much!!!!!

  33. bahrom7893
    • 5 years ago
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    np, ill do my best

  34. bahrom7893
    • 5 years ago
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    i found some stuff, looking thru it

  35. bahrom7893
    • 5 years ago
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    we were differentiatin it wrong

  36. anonymous
    • 5 years ago
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    how are we supposed to differentiate it?

  37. bahrom7893
    • 5 years ago
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    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

  38. bahrom7893
    • 5 years ago
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    hey can u go on twiddla? it'll be easier to write there i think

  39. bahrom7893
    • 5 years ago
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    http://www.twiddla.com/494934

  40. bahrom7893
    • 5 years ago
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    hey jessica click on the link above, im waitin for u on twiddla

  41. bahrom7893
    • 5 years ago
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    ill wait here too just in case

  42. anonymous
    • 5 years ago
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    sorry i'm in the meeting now so i'm like trying to be discrete

  43. bahrom7893
    • 5 years ago
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    oh okay, fine then just wait till its over, i got plenty of time right now

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