## unknownunknown one year ago Is this another error, or is there something I'm missing? For question 3, why not just use partial dw / partial x = (ze^y + e^z) ? Why is it setting values to 0 such as in (0 + e0) ? What has that answer got to do with anything? Thanks.

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1. unknownunknown
2. Michele_Laino

since your professor, want to evaluate $\Large \frac{{\partial w}}{{\partial x}}$ along a variety, which lies on the plane xy, and the cartesian equation of the xy-plane is $\Large z = 0$

3. Australopithecus

I am pretty sure it is because there is no z term in the constraint, thus z = 0 also by taking the partial derivative in terms of z of the constraint you get dz=0

4. Michele_Laino

the equation of that variety is: $\Large {x^2}y + x{y^2} = 1$

5. Australopithecus

From question one you know what dw equals so you can apply the constraint and write it in terms of the partial derivative in respect to x

6. unknownunknown

And then how does the third part equal the fourth part? From dx + 2xy+y^2/(x^2+2xy) = (x^2+4xy+y^2)/(x^2+2xy)

7. myininaya

http://www.mit.edu/~hlb/tempdir/done/MIT18_02SC_pb_42_comb.pdf take a look at this

8. myininaya

this seems like basically the same pdf but they have z=0

9. myininaya
10. myininaya

11. unknownunknown

Wow @myininaya, that's great, how did you find that? I'd say 50% of the pdfs in this course have obvious things like that omitted

12. Australopithecus

There is no z term in the constraint, therefore z = 0, the partial derivative of the constraint in respect to z is equal to 0 so you sub that into the equation you wrote for dw in the first question

13. Australopithecus

(xe^y + xe^z + ye^z)(0) = 0 because dz = 0

14. Australopithecus

but yeah you can see that on that pdf that was posted

15. unknownunknown

And then how does the third part equal the fourth part, I can see the numerator being x^2+4xy+y^2, but the denominator there wouldn't make sense solving for dx?

16. Australopithecus

you factor out the dx

17. Australopithecus

then you just divide both sides by dx and you get dw/dx

18. myininaya

$dw=1 dx+\frac{2xy+y^2}{x^2+2xy} dx \\ dw=\frac{x^2+2xy}{x^2+2xy} dx+\frac{2xy+y^2}{x^2+2xy} dx \\ dw=(\frac{x^2+2xy}{x^2+2xy}+\frac{2xy+y^2}{x^2+2xy} )dx$ what he said you can combine the fractions now

19. Australopithecus

When taking the total derivative of the constraint you can find what dy equals then you can sub that into the equation from the first question so you have it only in terms of dx and dw

20. unknownunknown

Ahh yep I see it. Makes sense now. But that z=0 not being included, and the third term not being cancelled in the original pdf, those errors are rampant in this course. Yet the pdf you posted myinanaya seems without errors. How did you locate that? I wonder if they exist for all the other pdfs..

21. Australopithecus

That isn't really an error, you can infer it from the function

22. myininaya

I just googled part of the question.

23. unknownunknown

Seems so. But at this level this topic is entirely new to me, so the explanation there that z=0 would have been crucial.

24. Australopithecus

it can help to look at functions on wolfram alpha

25. Australopithecus

most calc text books have a list of multivariable formulas and their specific shapes

26. myininaya

this is the exact thing I typed into google "3. Now suppose w is as above and x2y + y2x = 1. Assuming x is the independent variable," and it was the 3rd pdf/link thingy

27. Australopithecus

yeah most questions asked by profs are online if you dig deep enough or you can find ones similar

28. myininaya

There are some people who are way more expert than me at being an expert googler

29. unknownunknown

I can see now the z=0 term isn't necessary, but the thing is is that most of the pdf's actually contain proper errors. I've posted quite a few of them in the Multivar MIT section here. It seems strange there is a correct pdf here, why they wouldn't include it in the actual course site.

30. Michele_Laino

more precisely, the variety which is considered by your teacher is defined by the subsequent cartesian equations: $\Large \left\{ \begin{gathered} {x^2}y + x{y^2} = 1 \hfill \\ z = 0 \hfill \\ \end{gathered} \right.$ @unknownunknown

31. unknownunknown

Actually not sure I understand entirely the inference if z=0 isn't specified. If it isn't specified in the constrain, this could mean z would be a plane and could contain all values of z no?

32. unknownunknown

Eg: constraint x=3, this would still allow for all values of x and y.

33. unknownunknown

z and y*

34. Australopithecus

z = 0, because it is not present in the equation. This constraint is restricted to the x, y plane

35. unknownunknown

A function in the xy plane right? Which would extend to + and - infinity in the z plane with that shape.

36. unknownunknown

Taking that constraint, if I make z = 5, or z = 119, or anything, it is still a valid constraint. My z value can be anything.

37. unknownunknown

The constraint will still hold.

38. Australopithecus

If there was a z variable present in it but there isn't, its an equation of 2 variables

39. unknownunknown

But our main function does have a z variable, so as long as it intersects in the x & y point, the z can be anything no?

40. Australopithecus

no the constraint restricts it to the x, y plane

41. Australopithecus

by the constraint z=0 and dw/dz = 0

42. unknownunknown

If it's specified as z=0 yes, but in the original pdf it isn't right? Are you sure we can infer that if it isn't specified? For example imagine any function in 3 space, and we restrict it to x=3. At x=3 we should still have a plane of all values in the zy plane located at x=3 no?

43. Australopithecus

The original pdf is right, because the constraint is confined to z=0, or specifically the xy plane just like the equation x^2 + y^2 = 1 is constrained to the x, y plane

44. Australopithecus

It can be inferred from the constraint that z = 0

45. Australopithecus

Although, I do agree it might have been nice if they showed it more so in the answer.

46. unknownunknown

So where in the original pdf is the constraint z=0?

47. Australopithecus

Look at the graph of the constraint http://www.wolframalpha.com/input/?i=xy^2+%2B+xy+%3D+1+graph

48. Australopithecus

It is confined to the x,y plane

49. Australopithecus

since the constraint is confined to the x,y plane z =0

50. unknownunknown

But we can set z to be anything in that graph, and it would extend into space, and still be correct.

51. Australopithecus

No the constraint forbids that, because you are looking at the slope in terms of the constraint and the constraint doesn't ever leave the x,y plane thus z always equals 0

52. Australopithecus

the constraint restricts what you are looking at

53. Australopithecus

The constraint says, dw/dz = 0 it also says z = 0

54. Australopithecus

to ignore that is to ignore the constraint

55. Australopithecus

The question is asking for a slope in terms of the constraint so to ignore the constraint would give you an incorrect answer

56. unknownunknown

Hmm, so we're not just constraining the x and y variables in w? We must also then assume z=0?

57. Australopithecus

We are constraining all variables, we are not assuming z = 0, in terms of this constraint z always equals 0 because it is a two variable equation. There are no assumptions.

58. Australopithecus

Sorry wrong photo, here this can be helpful for identifying multivariate functions

59. unknownunknown

So how is the constraint different say, from setting z=3 as a level and viewing the function as a contout plot? This would then give the xy plane at z=3. A constraint of x=3 wouldn't be a level of the zy plane? Are they two different concepts?

60. myininaya

hey @Australopithecus I haven't been saying anything because I have been thinking about this constraint problem for awhile now. Let me see if you agree. The constraint x^2y+y^2x=1 shows that z doesn't depend on x and only y does. So z can be anything that isn't related to x. And if z is not related to x then it can't be related to y since y and x share a relationship. so dz/dx=0 but I don't have a good reason yet to conclude z=0 dz/dx=0 implies z=constant but I don't know how to explain why the constant would be zero sorry not trying to disagree or anything I'm still trying to process why they didn't specify that z=0 in the pdf

61. Michele_Laino

z=0, is a work hypothesis @myininaya

62. Australopithecus

Yes you are right sorry, I think they just picked z = 0 arbitrarily, sorry you are right it goes +/- z axis infinitely

63. Australopithecus

I apologize I made a mistake, good this is cleared up now though

64. unknownunknown

Ahh interesting, thanks though @Australopithecus for the insights. So can we conclude the original pdf contains an error by omitting that z=0?

65. Australopithecus

No we cannot conclude that

66. Australopithecus

They just assumed you would pick z = 0, but you could pick any z value

67. unknownunknown

But, the correct form would be as it is in the first line.. dw = (ze^y + e^z)dx, since this term allows us to pick anything. But then in the next line, it picks 0 without specifying it as such. The first line would allow us to pick anything, the second assumes 0, and it gives an equality there. That is not an error?

68. Australopithecus

they didn't need to specify because in terms of the constraint, dz = 0

69. Australopithecus

so you could sub z into it but it would all become zero anyways so why would you

70. unknownunknown

So it's asking to find $\partial dw / \partial dx$ , by definition the coefficient of dx must be that partial. Not (0 + 1) as it says.

71. Australopithecus

no no, I am just using d instead of daron because I am lazy dont get confused

72. Australopithecus

∂z = 0

73. Australopithecus

the partial derivative of the constraint in terms of z is ∂w/∂z = 0 therefore ∂z =0 thus, (xe^y + xe^z + ye^z)(0) = 0 because ∂z = 0 so you could sub a z value into that but it would still become equal to 0

74. unknownunknown

Yeah that's fine for the third term. But looking at the first term where we have dx. In the first line that is correct, as it is partial w/ partial x, yet in the second line it spontaneously changes to 1?

75. Australopithecus

e^0 = 1

76. unknownunknown

Yes, but why would you even put e^0?

77. Australopithecus

because the answerer chose z = 0

78. unknownunknown

But we're then evaluating a partial, when its not asking that.

79. Australopithecus

like you were saying you could look at this slope at different values of z

80. unknownunknown

Say we take the derivative of some f(x) = x^2, we get 2x. We can't then say f'(x) = 0 because we decided to arbitrarily evaluate the derivative at 0.

81. Australopithecus

z is a constant, in your example x is not a constant

82. Australopithecus

the constraint is a two variable equation

83. Australopithecus

thus z is not a variable it is a constant

84. unknownunknown

z isn't a constant, still a variable in dw / dx. We can input any value of z right.

85. Australopithecus

no because dw/dz = 0 according to the constraint

86. Australopithecus

that is the derivative of a constant

87. Australopithecus

thus z is a constant

88. Australopithecus

so you have to pick one

89. unknownunknown

Okay so going back to that then, assuming dw/dz = 0, and z is a constant. Why do we make that assumption again?

90. Australopithecus

It is not an assumption, take the partial derivative of: $x^2y + y^2x = 1$ in terms of ∂w/∂z what do you get?

91. Australopithecus

you get ∂w/∂z = 0 hence the constraint says that z is a constant, its slope does not change, therefore it is not a variabl

92. unknownunknown

So then your first graph would be correct instead of the second graph? Since in the second z isn't a constant but taking many different values.

93. Australopithecus

They are both actually the same graph, just one is in 2D. You could look at that graph at any level of z and it would look the same though. If z was a variable in the constraint this would not be the case. It is the same deal with this derivative

94. unknownunknown

If you wanted to allow all values of z in the constraint, how would you then express that?

95. Australopithecus

you are looking at a larger three variable function constrained by a two variable function

96. Australopithecus

The derivative will just change by a constant value

97. Australopithecus

You are applying a constraint because you don't want to look at the whole functions derivative

98. unknownunknown

yes but, here z isn't the evaluation, it's just another independent variable right? w is the evaluation. So say the constraint is x=3, would this mean that we set y=0 and z=0, and only take that one point, x=3?

99. Australopithecus

It isn't really a variable it is a constant in terms of this constraint. This is what the derivative would look like if z could be all values ∂w/∂x = (z + e^z)(rest of derivative)

100. Australopithecus

they restricted it though to 0 arbitrarily, it wasn't incorrect to do so because z is a constant. You can set it to anything the slope would only change by a coefficient.

101. unknownunknown

So say we have f(x,y,z) = x*y*z, and constraint g(x,y,z)=x=3 then f(x,y,z) = 3*c1*c2, (where c1,c2 are constants), instead of 3*y*z?

102. Australopithecus

x would be the constant by that constraint, the other variables would not, since you are pretty much saying x = 3. If you are arguing that by the constraint that ∂w/∂z = 0 does not mean that z is a constant by the constraint I have to disagree.

103. unknownunknown

Okay so, taking that function then w(x,y,z), and assuming g(x,y,z) = x^2y+y^2x=1, how is the logic different there by setting z as a constant instead of leaving it unconstrained?

104. Australopithecus

I do not follow how the constraint you provided is different from the constraint in the original question. Look by the constraint: ∂w/∂z = 0 This means that according to the constraint the value of z remains constant. This means the slope of w does not change is respect to z. Thus you can pick an arbitrary value of z.

105. Australopithecus

I am going to sleep now, more people should be on tomorrow. Maybe they can help you with this better than I can.

106. unknownunknown

Alright, you've given me a lot to ponder. Thanks for your help and time, it's much appreciated.

107. unknownunknown

For anyone reading this, I'd greatly appreciate an explanation. If this logic is true (that z is a constant), then for any constraint that doesn't include the other variables, if we take their partial derivatives, we also arrive at a constant. Then f(x,y,z) for any constraint x=3, must also be 3*c1*c2 (c1,c2 are constants). Any insights as to why this isn't the case if this holds in the original pdf, are welcomed.

108. unknownunknown

slight correction * (partial derivatives we arrive at 0, implying they are constant)

109. Australopithecus

Sorry I was half asleep still am but I am incorrect ∂w/∂z = 0 implies w does not change in respect z by the constraint but that does not imply that z is a constant for the slope ∂w/∂x so yes z is a variable, to answer your question yes I believe that is mistake that they did not include the constraint. Their answer would be correct if they stated that they had constrained z to equal 0

110. Australopithecus

Sorry for the confusion, anyways sleep now.

111. unknownunknown

Ahh @Australopithecus, you saved me an entire day of thought. Thanks very much for letting me know before bed! That clears it up then, that it's a mistake. Ok take care.

112. Australopithecus

if @ganeshie8 has anything to say on this I would love to hear it, he or she is better at mathematics than I.

113. Australopithecus

One more note, when you feel satisfied with the answer (I would give it a day maybe to see what others say) but make sure to use the rate a qualified helper button when you are done.

114. Australopithecus

with the question. Anyways sleep now

115. unknownunknown

Will do.

116. ganeshie8

Yeah the question has slight ambiguity. If $$x^2y+y^2x=1$$ is a $$3$$ dimensional surface, then the answer must contain $$z$$ terms also. Looks they are assuming the constraint is a two dimensional curve in $$xy$$ plane, then setting $$z=0$$ makes sense

117. unknownunknown

Ahh yes, makes sense. Was quite ambiguous.

118. unknownunknown

Still confused why they have a set of pdfs with all the correct values, yet they're not using them.

119. unknownunknown

@myininaya managed to find all these pdfs that seem to have a lot of the errors fixed

120. ganeshie8

Wow! that sucks... getting to know that there is a mistake in the question itself after pondering over it for several hours... But happy that we took time to understand the partials thoroughly and nailed down the actual issue !

121. unknownunknown

Yep definitely, everyone here was a great help.

122. theEric

Hey, all! I'm really excited because I think that I can set this one to rest. Though I know that I'm not aware of all the technicalities involved, this one relies on the wording. I agree whole-heartedly that we cannot assume that $$z=0$$. That is not part of the constraint. The "constraint" just gives extra information, and we are then looking at a more specific case of the possibilities of our $$x$$ and $$y$$. The $$z$$ is not involved, and so it is limited as it was before, or maybe a little more so since we're looking at a specific case. Here is why that term on the right disappears: The wording for #3 has, "Assuming x is the independent variable..." So, $$x$$ is "the" one that's changing. The $$z$$ is not the one that we are considering changing, so $$dz=0$$. The long story short: $$dz=0$$.

123. phi

It does not make sense to set z=0 in the answer to the question as asked. If the question states z=0, then it does make sense.

124. anonymous

dz=0