A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • This Question is Open
  1. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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 \[\Large z = 0\]

  2. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  3. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

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

  4. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  5. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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)

  6. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  7. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  8. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    If that link didn't work.

  9. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  10. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  11. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  12. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  13. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  14. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    you factor out the dx

  15. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  16. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  17. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  18. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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..

  19. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  20. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    I just googled part of the question.

  21. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  22. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    it can help to look at functions on wolfram alpha

  23. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  24. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  25. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  26. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

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

  27. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  28. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    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

  29. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  30. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  31. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    z and y*

  32. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  33. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  34. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  35. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    The constraint will still hold.

  36. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  37. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  38. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    no the constraint restricts it to the x, y plane

  39. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  40. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  41. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  42. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    It can be inferred from the constraint that z = 0

  43. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  44. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  45. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  46. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    It is confined to the x,y plane

  47. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  48. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  49. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  50. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    the constraint restricts what you are looking at

  51. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  52. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    to ignore that is to ignore the constraint

  53. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  54. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  55. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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.

  56. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  57. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  58. myininaya
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  59. Michele_Laino
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    z=0, is a work hypothesis @myininaya

  60. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  61. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  62. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  63. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    No we cannot conclude that

  64. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  65. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  66. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  67. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  68. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  69. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  70. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    ∂z = 0

  71. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  72. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  73. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    e^0 = 1

  74. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  75. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    because the answerer chose z = 0

  76. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  77. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  78. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  79. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  80. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    the constraint is a two variable equation

  81. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    thus z is not a variable it is a constant

  82. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  83. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    no because dw/dz = 0 according to the constraint

  84. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    that is the derivative of a constant

  85. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    thus z is a constant

  86. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    so you have to pick one

  87. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  88. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  89. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  90. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  91. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  92. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  93. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  94. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    The derivative will just change by a constant value

  95. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  96. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  97. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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)

  98. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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.

  99. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  100. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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.

  101. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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?

  102. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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.

  103. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  104. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  105. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  106. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  107. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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

  108. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    Sorry for the confusion, anyways sleep now.

  109. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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.

  110. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

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

  111. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    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.

  112. Australopithecus
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 4

    with the question. Anyways sleep now

  113. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Will do.

  114. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  115. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Ahh yes, makes sense. Was quite ambiguous.

  116. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  117. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

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

  118. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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 !

  119. unknownunknown
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    Yep definitely, everyone here was a great help.

  120. theEric
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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: \(dz=0\).

  121. phi
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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.

  122. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    dz=0

  123. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.