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anonymous

  • one year ago

hey @parthkohli

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  1. anonymous
    • one year ago
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    when i say 30 i mean 40

  2. ParthKohli
    • one year ago
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    what does "reducing the length of the curve" mean? I mean you can draw the shortest possible path from (10,10) to (30,0) that also happens to be a straight line.

  3. anonymous
    • one year ago
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    yea but it needs to be smooth

  4. anonymous
    • one year ago
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    so polynomial

  5. ParthKohli
    • one year ago
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    what does that mean? a linear function is a polynomial too. what exactly do you mean by smooth?

  6. anonymous
    • one year ago
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    mena no rough edges and its curve

  7. anonymous
    • one year ago
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    like continuous function

  8. ParthKohli
    • one year ago
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    ahhh, better.

  9. anonymous
    • one year ago
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    is there function that looks like the black line i drew

  10. ParthKohli
    • one year ago
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    so you need a polynomial function that makes a smooth transition and is also the shortest

  11. anonymous
    • one year ago
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    so it has curve to smothly attach to another track but at the smae shorter length

  12. anonymous
    • one year ago
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    yes

  13. anonymous
    • one year ago
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    but at the same time *

  14. ParthKohli
    • one year ago
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    what's the context of this question? it's really specific

  15. anonymous
    • one year ago
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    i have done the question let me send the link

  16. anonymous
    • one year ago
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    but what i am trying to figure out is the shorter length than this

  17. ParthKohli
    • one year ago
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    sorry, my OS stopped working.

  18. anonymous
    • one year ago
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    so on xy plane i want a function that give me length <27.37

  19. anonymous
    • one year ago
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    |dw:1443516744812:dw|

  20. anonymous
    • one year ago
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    ok

  21. ParthKohli
    • one year ago
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    You're permitted to use calculators right?\[g(x) = ax^4 + bx^3 + cx^2 + dx + e\]\[10^4 a + 10^3 b + 10^2 c + 10 d + e = 10\]\[30^4 a + 30^3 b + 30^2 c + 30 d + e=0\]\[\int_{10}^{30} g(x)dx=200\]\[4a(10)^3 + 3b(10)^2 + 2c(10) + d(10) =0\]Now the last one is tricky.

  22. anonymous
    • one year ago
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    yess

  23. ParthKohli
    • one year ago
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    You know how to calculate arc length using integrals right?

  24. anonymous
    • one year ago
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    yea but i use wolfram alpha normaly u mind using it sending the link plz

  25. ParthKohli
    • one year ago
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    Because you really are looking to minimise arc-length given the above constraints.

  26. anonymous
    • one year ago
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    ok wait do i do the same steps as i did for my previous function find a,b,c,d

  27. anonymous
    • one year ago
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    and then put them in an expression and use arc length to get length

  28. ParthKohli
    • one year ago
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    I think Wolfram is gonna have some good fun with this one.

  29. anonymous
    • one year ago
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    heheh

  30. ParthKohli
    • one year ago
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    Here, you have a, b, c, d and e.

  31. anonymous
    • one year ago
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    yea kinda forgot about e

  32. anonymous
    • one year ago
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    thank you so much for ur help

  33. ParthKohli
    • one year ago
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    I'm curious to see how you'll input so much into Wolfram. It's scary.

  34. anonymous
    • one year ago
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    what if i find an expression then too

  35. ParthKohli
    • one year ago
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    sorry, the last equation is\[4a(10^3) + 3b(10^2 ) + 2c(10) + d = \color{red}1\]

  36. anonymous
    • one year ago
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    ok

  37. ParthKohli
    • one year ago
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    wow the expression is way too long... you'll have to be subscribe to pro

  38. anonymous
    • one year ago
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    i have subscribe pro

  39. ParthKohli
    • one year ago
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    wow, cool.

  40. baru
    • one year ago
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    @ParthKohli for arguments sake, (specifics not necessary) i would write the function for arc length squared lets call it L=f(a b c d e) now what?to minimize, we need to substitute variables in terms of each other so that they represent their inter-dependencies right? how will you do that? you have four other equations ? or am i way off, and talking crap?

  41. ParthKohli
    • one year ago
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    yes - we have four equations. in the arc length function, we'll eliminate four variables and will be left with a function that is in terms of only one variable.

  42. baru
    • one year ago
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    is that a general thing, as in if there are n unknowns and (n-1) equations we can bring it down to 1 variable?

  43. ParthKohli
    • one year ago
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    I suppose. Let's take a smaller example.\[f(x,y,z) = x+y-z\]\[x=y+1\]\[z = -x\]\[f(x,y,z) =x+(x-1) - (-x) = 3x - 1\]

  44. ParthKohli
    • one year ago
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    Three variables, two equations. Now we can minimise the function with one-variable very easily.

  45. baru
    • one year ago
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    ok. thanks :)

  46. baru
    • one year ago
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    @ParthKohli one other thing, i get why you have picked an equation with 5 unknowns since we have 5 pieces of information, but the equation could just as easily been \[ax^7 + bx^6 +cx^5 +dx +e\] or something else equally arbitrary so why specifically have you chosen the one that you have?

  47. ParthKohli
    • one year ago
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    that is a very good observation, one that I did give some thought to but skipped. unfortunately, it may be impossible to find *the* polynomial that satisfies the above constraints because the degree may go as high as... anything we want it to reach. the degree-seven polynomial you wrote may not be the best degree-seven polynomial to satisfy the conditions because it's missing some terms and unless we're sure while solving for the seventh-degree that those terms turn out to be zero, we'll never be sure. so choose a degree-four polynomial adds certainty to what we get - we can be sure that we'll get the best fourth-degree polynomial. of course there are polynomials much better than that one, but we can only go as far as we can look.

  48. ParthKohli
    • one year ago
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    ultimately, we're looking for a polynomial that resembles a straight-line the most in the region x = 10 to x = 30, so we can add as many terms as we like, but we can't go that far - can we?

  49. baru
    • one year ago
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    ah..makes sense. thanks a lot!!

  50. ParthKohli
    • one year ago
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    by the way, we still can approach this problem through multivariable calculus for polynomials that have more terms than five. that way, we can get an even closer polynomial.

  51. anonymous
    • one year ago
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    hey its not making sense

  52. ParthKohli
    • one year ago
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    thanks for the question - I really had to think about it to come up with an answer.

  53. anonymous
    • one year ago
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    5 variables how do i get values for 5 unknown variable

  54. ParthKohli
    • one year ago
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    plug it into WolframAlpha

  55. anonymous
    • one year ago
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    i only have 4 equation

  56. anonymous
    • one year ago
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    i found a ,b, c ,d but not e

  57. ParthKohli
    • one year ago
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    were you able to find a, b, c, d in terms of e?

  58. anonymous
    • one year ago
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    \[0.0001852x ^{4}-0.01246x ^{3}+0.2076x ^{2}-0.01555x+e\]

  59. anonymous
    • one year ago
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    http://math.bd.psu.edu/~jpp4/finitemath/4x4solver.html

  60. ParthKohli
    • one year ago
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    wait what? you were able to uniquely determine a, b, c, d?!

  61. anonymous
    • one year ago
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    from here

  62. ParthKohli
    • one year ago
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    it's not a 4x4 equation so

  63. anonymous
    • one year ago
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    yea i thought so help plz

  64. ParthKohli
    • one year ago
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    plug the stuff into Wolfram and it should give you all other variables in terms of one variable

  65. baru
    • one year ago
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    @ayeshaafzal221 where did you find this question?

  66. anonymous
    • one year ago
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    it didnt

  67. anonymous
    • one year ago
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    its my practise question for exams

  68. ParthKohli
    • one year ago
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    what did it return?

  69. ParthKohli
    • one year ago
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    you don't have a Wolfram pro subscription do you?

  70. anonymous
    • one year ago
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    on my ipad yes

  71. anonymous
    • one year ago
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    even on my ipad not working

  72. ParthKohli
    • one year ago
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    there's this data entry or something feature which you can use

  73. anonymous
    • one year ago
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    ok

  74. anonymous
    • one year ago
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    or u know the values i have found cant i just sub it into any equation and get the e value

  75. ParthKohli
    • one year ago
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    no, you can't do that. it's not a 4x4 equation so you most likely input the wrong equations.

  76. anonymous
    • one year ago
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    yea true i am gonna go old fashion , and do em manually

  77. baru
    • one year ago
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    @ParthKohli , when you say multi-variable calculus approach: are you talking about Lagrange multipliers?(i've just reached this topic)

  78. ParthKohli
    • one year ago
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    Obviously. Yes.

  79. baru
    • one year ago
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    alright..thanks

  80. anonymous
    • one year ago
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    @ParthKohli srry i am disturbing u again when u say we have five variables and one of them was minimum arc length what do u mean?

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