## 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.

1. Agent_A

2. ganeshie8

#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

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

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

The sign is positive, but decreasing....?

6. ganeshie8

nope, remember how first derivative can be used to tell whether a function is increasing/decreasing from single variable calculus ?

7. Agent_A

Somewhat, yes, but muddy.

8. ganeshie8

First derivative at a particular point on the curve gives the slope of tangent line at that point |dw:1435034181715:dw|

9. ganeshie8

Notice that the slope of tangent line will be "positive" in the interval in which the function is "increasing" |dw:1435034301743:dw|

10. ganeshie8

Also the slope of tangent line will be "negative" in the interval in which the function is "decreasing" |dw:1435034404265:dw|

11. ganeshie8

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

In our specific example, the function is "decreasing" along rows, so $$f_x$$ must be negative.

13. Agent_A

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

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

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

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

17. Agent_A

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

Just pick any one row and forget about everything else.

19. Agent_A

I'm doing two problems at once.

20. Agent_A

Sorry for the mess.

21. ganeshie8

thats okay :)

22. Agent_A

Thanks for understanding.... I'm just so stressed right now.

23. ganeshie8

Just answer me this, in the below highlighted row, what do you notice about the values ? |dw:1435035389370:dw|

24. Agent_A

Decreasing, as we go from left to right.

25. ganeshie8

Yes. so $$f_x(-2,~0)~~\lt ~0$$ |dw:1435035518143:dw|

26. ganeshie8

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

Increasing....

28. ganeshie8

good, so $$f_y(-2,~1)\gt 0$$ |dw:1435035643341:dw|

29. Agent_A

Ohhhhhhh I get the pattern.

30. Agent_A

is f_xx true, then?

31. Agent_A

and f_yy

32. ganeshie8

sry i don't seem to remember, il review quick and get back meantime try other problems maybe..

33. Agent_A

Okay!

34. Agent_A

Thanks!

35. ganeshie8

@Michele_Laino

36. ganeshie8

I'm thinking of working the first differences but not entirely sure

37. Michele_Laino

we can try to write the taylor series up to the second derivative of f, with for example y= const @ganeshie8

38. Michele_Laino

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

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

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

|dw:1435037234174:dw|

42. Agent_A

If it helps, the answer is letter B.

43. Agent_A

For number 15. I just don't have the solutions.

44. Agent_A

Answers to the other numbers: 15) B 16) A 17) E 18) A 19) B

45. ganeshie8

Yeah we cannot tell anything about mixed partials i guess so B makes sense to me

46. ganeshie8

|dw:1435037903299:dw|

47. Michele_Laino

we have checked four conditions, so I think option D. Am I right?

48. Michele_Laino

namely: the first and fourth conditions are false; the second and the third conditions are true

49. Michele_Laino

and we can not infer anything on the fifth condition

50. ganeshie8

hmm yeah since the first and fourth conditions are false we may say that only two inferences are true

51. ganeshie8

#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

|dw:1435038498718:dw|

53. ganeshie8
54. Agent_A

Thanks! How were those numbers selected? Was it through the symmetric difference quotient?

55. ganeshie8

Ohh no, i guess we should use symmetric difference quotient then

56. Agent_A

Ohhh yeah, I just am confused on how to plug the values in :P.

57. Agent_A

*am just

58. ganeshie8

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

59. Agent_A

Nice! Thanks!

60. ganeshie8

#17 just follows from #16

61. ganeshie8

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

62. ganeshie8

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 = $$\langle -2,~3 \rangle$$

63. Agent_A

I see!

64. Agent_A

Ohhhhhh it's becoming clear to me.... I hope. Haha. I feel like it is.

65. Agent_A

It really was just a matter of where to look, on the chart....

66. Agent_A

How do we do number 18? I'm used to solving for it with the equation given.

67. Agent_A

Do we divide <-2, 3> by $\sqrt{(-2)^2+(3)^2}$?

68. ganeshie8

remember, the function value doesn't change when you stay on a specific level curve ?

69. Agent_A

I think so.

70. Agent_A

When you're along one level curve, yes....

71. ganeshie8

Also recall the fact that gradient vector is perpendicular to the level curve

72. Agent_A

Wait, never mind what I said about the root.

73. Agent_A

Oh yes, it is.

74. ganeshie8

|dw:1435040370688:dw|

75. ganeshie8

|dw:1435040574730:dw|

76. ganeshie8

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

77. Agent_A

Ohhhh, yes

78. ganeshie8

An easy way to work it if you don't happen to remember the rotation matrix is to simply use the dot product

79. ganeshie8

we already know that gradient = $$\langle -2, 3\rangle$$ dot that with each of the given options and see which one evaluates to $$0$$

80. Agent_A

Of course. Yes. I did not think about using the given answers.

81. Agent_A

Wow.

82. Agent_A

Wait, hmmm what is z?

83. Agent_A

I can't use the dot product without a z-value....

84. Agent_A

Ah well. I still get zero, so I guess you can use it without a z-value....

85. Agent_A

I used 0 for the z-value.

86. Agent_A

Oh yeah.... -__-

87. Agent_A

oops

88. ganeshie8

no wait, i got confused

89. ganeshie8

nope, that is correct. #18 is about directional derivative, $$u$$ is a vector in $$xy$$ plane, so $$z=0$$

90. Agent_A

:)

91. ganeshie8

#19 should be easy, just work the tangent plane approximation

92. Agent_A

And linearization is just: $L(x, y) = f(x_0, y_0)+f_x(x_0, y_0)(x-x_0)+f_y(x_0, y_0)(y-y_0)$?

93. ganeshie8

that should do

94. ganeshie8

or you may use the gradient

95. ganeshie8

normal vector of the required tangent plane is $$\langle f_x,~f_y,~-1 \rangle$$

96. ganeshie8

so the tangent plane is $f_x x+f_yy-z = k$ you cna work the $$k$$ value by plugging in the point $$(-2,~1,~33)$$

97. ganeshie8

whichever way you feel easy... upto you

98. Agent_A

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 $f_x = 33$

99. ganeshie8

hey still here ?

100. ganeshie8

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 $$k$$

101. Agent_A

Hi! I'm back.

102. Agent_A

Ohhhhh okay. That's another method I never knew about. Thanks! That should do it! Let me try that right now.

103. Agent_A

This is a simpler method! I like it! Thank You very much, @ganeshie8 ! Hope I do well on my exam!

104. ganeshie8

I'm sure you will do great! good luck!

105. Agent_A

Thank You!