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Agent_A

  • one year ago

Please answer all the questions (and in a neat format). Thanks! (see photo) Need all the answers at a glance.

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  1. Agent_A
    • one year ago
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  2. ganeshie8
    • one year ago
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    #15 For \(f_x\) we keep \(y\) fixed and see how the function changes as we change \(x\) so simply look at rows

  3. Michele_Laino
    • one year ago
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    yes! we have to apply this formula: \[\Large {f_x}\left( {{x_0},{y_0}} \right) \cong \frac{{f\left( {{x_0} + \Delta x,{y_0}} \right) - f\left( {{x_0},{y_0}} \right)}}{{\Delta x}}\]

  4. ganeshie8
    • one year ago
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    In each row, as \(x\) is increasing, notice that the function is decreasing. |dw:1435033219984:dw| In light of that, what can you say about the sign of \(f_x\) ?

  5. Agent_A
    • one year ago
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    The sign is positive, but decreasing....?

  6. ganeshie8
    • one year ago
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    nope, remember how first derivative can be used to tell whether a function is increasing/decreasing from single variable calculus ?

  7. Agent_A
    • one year ago
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    Somewhat, yes, but muddy.

  8. ganeshie8
    • one year ago
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    First derivative at a particular point on the curve gives the slope of tangent line at that point |dw:1435034181715:dw|

  9. ganeshie8
    • one year ago
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    Notice that the slope of tangent line will be "positive" in the interval in which the function is "increasing" |dw:1435034301743:dw|

  10. ganeshie8
    • one year ago
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    Also the slope of tangent line will be "negative" in the interval in which the function is "decreasing" |dw:1435034404265:dw|

  11. ganeshie8
    • one year ago
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    That means the sign of first derivative can be used to know if a function is increasing/decreasing : \(\large f'(c)~\gt~0~~\iff~~\text{f(x) is increasing at x=c}\) \(\large f'(c)~\lt~0~~\iff~~\text{f(x) is decreasing at x=c}\)

  12. ganeshie8
    • one year ago
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    In our specific example, the function is "decreasing" along rows, so \(f_x \) must be negative.

  13. Agent_A
    • one year ago
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    Sorry, I don't mean to rush you, but if I may, I just want to get straight to the point, just because of the interest of time. I have a final tomorrow and a lot of these concepts are not so clear to me, and I know that, but I'm trying to save time, so I can study all of them, even though I don't have a profound understanding (just because of time, like I said). I just wanted to ask, I am familiar with the zero-gradient, but not when I have to use a contour plot, without any equation to derive it. How would this work?

  14. Agent_A
    • one year ago
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    Your help's been very generous, @ganeshie8 . Sorry if I sound a little blunt. I'm just rushing to study as much as I can, because I have a final in less than 15 hours.

  15. ganeshie8
    • one year ago
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    I can understand, but there is no way to proceed further when you don't remember the relation between first derivative and increasing/decreasing functions. The question is asking exactly that. You cannot skip

  16. Agent_A
    • one year ago
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    OOOPS. Sorry, I was referring to another problem, when I said "zero-gradient". So that wouldn't make sense here.

  17. Agent_A
    • one year ago
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    Okay, I'm just totally confused now. Ughhhh. I've done this before, but the table just confuses me a lot. I don't know where to look.

  18. ganeshie8
    • one year ago
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    Just pick any one row and forget about everything else.

  19. Agent_A
    • one year ago
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    I'm doing two problems at once.

  20. Agent_A
    • one year ago
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    Sorry for the mess.

  21. ganeshie8
    • one year ago
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    thats okay :)

  22. Agent_A
    • one year ago
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    Thanks for understanding.... I'm just so stressed right now.

  23. ganeshie8
    • one year ago
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    Just answer me this, in the below highlighted row, what do you notice about the values ? |dw:1435035389370:dw|

  24. Agent_A
    • one year ago
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    Decreasing, as we go from left to right.

  25. ganeshie8
    • one year ago
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    Yes. so \(f_x(-2,~0)~~\lt ~0 \) |dw:1435035518143:dw|

  26. ganeshie8
    • one year ago
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    Next, pick any one column and tell me what do you notice about the values as you go from bottom to top |dw:1435035560839:dw|

  27. Agent_A
    • one year ago
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    Increasing....

  28. ganeshie8
    • one year ago
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    good, so \(f_y(-2,~1)\gt 0\) |dw:1435035643341:dw|

  29. Agent_A
    • one year ago
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    Ohhhhhhh I get the pattern.

  30. Agent_A
    • one year ago
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    is f_xx true, then?

  31. Agent_A
    • one year ago
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    and f_yy

  32. ganeshie8
    • one year ago
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    sry i don't seem to remember, il review quick and get back meantime try other problems maybe..

  33. Agent_A
    • one year ago
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    Okay!

  34. Agent_A
    • one year ago
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    Thanks!

  35. ganeshie8
    • one year ago
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    @Michele_Laino

  36. ganeshie8
    • one year ago
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    I'm thinking of working the first differences but not entirely sure

  37. Michele_Laino
    • one year ago
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    we can try to write the taylor series up to the second derivative of f, with for example y= const @ganeshie8

  38. Michele_Laino
    • one year ago
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    like this: \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)x + \frac{{{x^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]

  39. Michele_Laino
    • one year ago
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    oops.. \[\large f\left( {x,{y_0}} \right) = f\left( {{x_0},{y_0}} \right) + {f_x}\left( {{x_0},{y_0}} \right)\Delta x + \frac{{{{\left( {\Delta x} \right)}^2}}}{2}{f_{xx}}\left( {{x_0},{y_0}} \right)\]

  40. ganeshie8
    • one year ago
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    Ahh so the finite differences actually works \(f_{yy}(-2,1)~= 0\) because the first differences are constant along a column : |dw:1435036964859:dw|

  41. ganeshie8
    • one year ago
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    |dw:1435037234174:dw|

  42. Agent_A
    • one year ago
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    If it helps, the answer is letter B.

  43. Agent_A
    • one year ago
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    For number 15. I just don't have the solutions.

  44. Agent_A
    • one year ago
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    Answers to the other numbers: 15) B 16) A 17) E 18) A 19) B

  45. ganeshie8
    • one year ago
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    Yeah we cannot tell anything about mixed partials i guess so B makes sense to me

  46. ganeshie8
    • one year ago
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    |dw:1435037903299:dw|

  47. Michele_Laino
    • one year ago
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    we have checked four conditions, so I think option D. Am I right?

  48. Michele_Laino
    • one year ago
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    namely: the first and fourth conditions are false; the second and the third conditions are true

  49. Michele_Laino
    • one year ago
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    and we can not infer anything on the fifth condition

  50. ganeshie8
    • one year ago
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    hmm yeah since the first and fourth conditions are false we may say that only two inferences are true

  51. ganeshie8
    • one year ago
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    #16 should be easy, just approximate the partials at (-2, 1) and take the magnitude \[\large \|\nabla f\| = \sqrt{{f_x}^2+{f_y}^2}\]

  52. ganeshie8
    • one year ago
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    |dw:1435038498718:dw|

  53. Agent_A
    • one year ago
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    Thanks! How were those numbers selected? Was it through the symmetric difference quotient?

  54. ganeshie8
    • one year ago
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    Ohh no, i guess we should use symmetric difference quotient then

  55. Agent_A
    • one year ago
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    Ohhh yeah, I just am confused on how to plug the values in :P.

  56. Agent_A
    • one year ago
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    *am just

  57. ganeshie8
    • one year ago
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    Yes symmetric difference quotient also gives the same answer http://www.wolframalpha.com/input/?i=sqrt%28%28%2832.41-33.61%29%2F%282*0.1%29%29%5E2%2B%28%2833.9-32.1%29%2F%282*0.1%29%29%5E2%29

  58. Agent_A
    • one year ago
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    Nice! Thanks!

  59. ganeshie8
    • one year ago
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    #17 just follows from #16

  60. ganeshie8
    • one year ago
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    direction is \(\langle -6, ~9\rangle \) http://www.wolframalpha.com/input/?i=%3C%2832.41-33.61%29%2F%282*0.1%29%2C+%2833.9-32.1%29%2F%282*0.1%29%3E

  61. ganeshie8
    • one year ago
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    Notice that the direction \(\langle -6, ~9\rangle \) is same as \(k\langle -6, ~9\rangle \) let \(k=\frac{1}{3}\) and we get direction = \(\langle -2,~3 \rangle \)

  62. Agent_A
    • one year ago
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    I see!

  63. Agent_A
    • one year ago
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    Ohhhhhh it's becoming clear to me.... I hope. Haha. I feel like it is.

  64. Agent_A
    • one year ago
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    It really was just a matter of where to look, on the chart....

  65. Agent_A
    • one year ago
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    How do we do number 18? I'm used to solving for it with the equation given.

  66. Agent_A
    • one year ago
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    Do we divide <-2, 3> by \[\sqrt{(-2)^2+(3)^2}\]?

  67. ganeshie8
    • one year ago
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    remember, the function value doesn't change when you stay on a specific level curve ?

  68. Agent_A
    • one year ago
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    I think so.

  69. Agent_A
    • one year ago
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    When you're along one level curve, yes....

  70. ganeshie8
    • one year ago
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    Also recall the fact that gradient vector is perpendicular to the level curve

  71. Agent_A
    • one year ago
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    Wait, never mind what I said about the root.

  72. Agent_A
    • one year ago
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    Oh yes, it is.

  73. ganeshie8
    • one year ago
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    |dw:1435040370688:dw|

  74. ganeshie8
    • one year ago
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    |dw:1435040574730:dw|

  75. ganeshie8
    • one year ago
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    just observe that rotating the gradient vector by 90 degrees gives you the tangent vector to level curve and in that direction the function remains constant

  76. Agent_A
    • one year ago
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    Ohhhh, yes

  77. ganeshie8
    • one year ago
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    An easy way to work it if you don't happen to remember the rotation matrix is to simply use the dot product

  78. ganeshie8
    • one year ago
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    we already know that gradient = \(\langle -2, 3\rangle\) dot that with each of the given options and see which one evaluates to \(0\)

  79. Agent_A
    • one year ago
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    Of course. Yes. I did not think about using the given answers.

  80. Agent_A
    • one year ago
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    Wow.

  81. Agent_A
    • one year ago
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    Wait, hmmm what is z?

  82. Agent_A
    • one year ago
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    I can't use the dot product without a z-value....

  83. Agent_A
    • one year ago
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    Ah well. I still get zero, so I guess you can use it without a z-value....

  84. Agent_A
    • one year ago
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    I used 0 for the z-value.

  85. Agent_A
    • one year ago
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    Oh yeah.... -__-

  86. Agent_A
    • one year ago
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    oops

  87. ganeshie8
    • one year ago
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    no wait, i got confused

  88. ganeshie8
    • one year ago
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    nope, that is correct. #18 is about directional derivative, \(u\) is a vector in \(xy\) plane, so \(z=0\)

  89. Agent_A
    • one year ago
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    :)

  90. ganeshie8
    • one year ago
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    #19 should be easy, just work the tangent plane approximation

  91. Agent_A
    • one year ago
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    And linearization is just: \[L(x, y) = f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)\]?

  92. ganeshie8
    • one year ago
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    that should do

  93. ganeshie8
    • one year ago
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    or you may use the gradient

  94. ganeshie8
    • one year ago
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    normal vector of the required tangent plane is \(\langle f_x,~f_y,~-1 \rangle \)

  95. ganeshie8
    • one year ago
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    so the tangent plane is \[f_x x+f_yy-z = k\] you cna work the \(k\) value by plugging in the point \((-2,~1,~33)\)

  96. ganeshie8
    • one year ago
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    whichever way you feel easy... upto you

  97. Agent_A
    • one year ago
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    I tried it. I thought it was a plug-and-chug method, but again, I can't seem to find the values of \[f_x , f_y\]. I don't think \[f_x = 33\]

  98. ganeshie8
    • one year ago
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    hey still here ?

  99. ganeshie8
    • one year ago
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    From #16 we have \[\nabla f = \langle f_x,~f_y \rangle = \langle -6,9 \rangle \] so the tangent plane is simply \[-6x+9y-z=k\] plugin the point \((-2,1,33)\) and solve \(k\)

  100. Agent_A
    • one year ago
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    Hi! I'm back.

  101. Agent_A
    • one year ago
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    Ohhhhh okay. That's another method I never knew about. Thanks! That should do it! Let me try that right now.

  102. Agent_A
    • one year ago
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    This is a simpler method! I like it! Thank You very much, @ganeshie8 ! Hope I do well on my exam!

  103. ganeshie8
    • one year ago
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    I'm sure you will do great! good luck!

  104. Agent_A
    • one year ago
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    Thank You!

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