Question

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.

Answer 1

http://ocw.mit.edu/courses/mathematics/18-02sc-multivariable-calculus-fall-2010/2.-partial-derivatives/part-c-lagrange-multipliers-and-constrained-differentials/session-42-constrained-differentials/MIT18_02SC_pb_42_comb.pdf

Answer 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
\[\Large z = 0\]

Answer 3

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

Answer 4

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

Answer 5

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

Answer 6

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)

Answer 7

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

Answer 8

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

Answer 9

https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CCsQFjACahUKEwjo8d7H3IbGAhUJQZIKHd2SACE&url=http%3A%2F%2Fwww.mit.edu%2F~hlb%2Ftempdir%2Fdone%2FMIT18_02SC_pb_42_comb.pdf&ei=twN5Vai6IomCyQTdpYKIAg&usg=AFQjCNHK6DNTMbqqMnTJbstkHoLh2l0neQ&sig2=ejRcNzP0UPL4XWLz-xT4EA

Answer 10

If that link didn't work.

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

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?

Answer 16

you factor out the dx

Answer 17

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

Answer 18

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

Answer 19

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

Answer 20

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..

Answer 21

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

Answer 22

I just googled part of the question.

Answer 23

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

Answer 24

it can help to look at functions on wolfram alpha

Answer 25

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

Answer 26

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

Answer 27

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

Answer 28

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

Answer 29

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.

Answer 30

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

Answer 31

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?

Answer 32

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

Answer 33

z and y*

Answer 34

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

Answer 35

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

Answer 36

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.

Answer 37

The constraint will still hold.

Answer 38

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

Answer 39

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?

Answer 40

no the constraint restricts it to the x, y plane

Answer 41

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

Answer 42

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?

Answer 43

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

Answer 44

It can be inferred from the constraint that z = 0

Answer 45

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

Answer 46

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

Answer 47

Look at the graph of the constraint

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

Answer 48

It is confined to the x,y plane

Answer 49

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

Answer 50

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

Answer 51

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

Answer 52

the constraint restricts what you are looking at

Answer 53

The constraint says,

dw/dz = 0
it also says z = 0

Answer 54

to ignore that is to ignore the constraint

Answer 55

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

Answer 56

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

Answer 57

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.

Answer 58

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

Answer 59

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?

Answer 60

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

Answer 61

z=0, is a work hypothesis @myininaya

Answer 62

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

Answer 63

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

Answer 64

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

Answer 65

No we cannot conclude that

Answer 66

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

Answer 67

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?

Answer 68

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

dz = 0

Answer 69

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

Answer 70

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.

Answer 71

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

Answer 72

∂z = 0

Answer 73

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

Answer 74

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?

Answer 75

e^0 = 1

Answer 76

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

Answer 77

because the answerer chose z = 0

Answer 78

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

Answer 79

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

Answer 80

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.

Answer 81

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

Answer 82

the constraint is a two variable equation

Answer 83

thus z is not a variable it is a constant

Answer 84

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

Answer 85

no because

dw/dz = 0 according to the constraint

Answer 86

that is the derivative of a constant

Answer 87

thus z is a constant

Answer 88

so you have to pick one

Answer 89

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

Answer 90

It is not an assumption,

take the partial derivative of:

\[x^2y + y^2x = 1\]

in terms of ∂w/∂z

what do you get?

Answer 91

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

Answer 92

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.

Answer 93

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

Answer 94

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

Answer 95

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

Answer 96

The derivative will just change by a constant value