unknownunknown
  • unknownunknown
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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jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
unknownunknown
  • unknownunknown
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
Michele_Laino
  • 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 \[\Large z = 0\]
Australopithecus
  • 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

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Michele_Laino
  • Michele_Laino
the equation of that variety is: \[\Large {x^2}y + x{y^2} = 1\]
Australopithecus
  • 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
unknownunknown
  • 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)
myininaya
  • myininaya
http://www.mit.edu/~hlb/tempdir/done/MIT18_02SC_pb_42_comb.pdf take a look at this
myininaya
  • myininaya
this seems like basically the same pdf but they have z=0
myininaya
  • myininaya
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
myininaya
  • myininaya
If that link didn't work.
unknownunknown
  • 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
Australopithecus
  • 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
Australopithecus
  • Australopithecus
(xe^y + xe^z + ye^z)(0) = 0 because dz = 0
Australopithecus
  • Australopithecus
but yeah you can see that on that pdf that was posted
unknownunknown
  • 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?
Australopithecus
  • Australopithecus
you factor out the dx
Australopithecus
  • Australopithecus
then you just divide both sides by dx and you get dw/dx
myininaya
  • 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
Australopithecus
  • 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
unknownunknown
  • 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..
Australopithecus
  • Australopithecus
That isn't really an error, you can infer it from the function
myininaya
  • myininaya
I just googled part of the question.
unknownunknown
  • 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.
Australopithecus
  • Australopithecus
it can help to look at functions on wolfram alpha
Australopithecus
  • Australopithecus
most calc text books have a list of multivariable formulas and their specific shapes
myininaya
  • 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
Australopithecus
  • Australopithecus
yeah most questions asked by profs are online if you dig deep enough or you can find ones similar
myininaya
  • myininaya
There are some people who are way more expert than me at being an expert googler
unknownunknown
  • 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.
Michele_Laino
  • 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
unknownunknown
  • 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?
unknownunknown
  • unknownunknown
Eg: constraint x=3, this would still allow for all values of x and y.
unknownunknown
  • unknownunknown
z and y*
Australopithecus
  • Australopithecus
z = 0, because it is not present in the equation. This constraint is restricted to the x, y plane
unknownunknown
  • unknownunknown
A function in the xy plane right? Which would extend to + and - infinity in the z plane with that shape.
unknownunknown
  • 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.
unknownunknown
  • unknownunknown
The constraint will still hold.
Australopithecus
  • Australopithecus
If there was a z variable present in it but there isn't, its an equation of 2 variables
unknownunknown
  • 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?
Australopithecus
  • Australopithecus
no the constraint restricts it to the x, y plane
Australopithecus
  • Australopithecus
by the constraint z=0 and dw/dz = 0
unknownunknown
  • 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?
Australopithecus
  • 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
Australopithecus
  • Australopithecus
It can be inferred from the constraint that z = 0
Australopithecus
  • Australopithecus
Although, I do agree it might have been nice if they showed it more so in the answer.
unknownunknown
  • unknownunknown
So where in the original pdf is the constraint z=0?
Australopithecus
  • Australopithecus
Look at the graph of the constraint http://www.wolframalpha.com/input/?i=xy^2+%2B+xy+%3D+1+graph
Australopithecus
  • Australopithecus
It is confined to the x,y plane
Australopithecus
  • Australopithecus
since the constraint is confined to the x,y plane z =0
unknownunknown
  • unknownunknown
But we can set z to be anything in that graph, and it would extend into space, and still be correct.
Australopithecus
  • 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
Australopithecus
  • Australopithecus
the constraint restricts what you are looking at
Australopithecus
  • Australopithecus
The constraint says, dw/dz = 0 it also says z = 0
Australopithecus
  • Australopithecus
to ignore that is to ignore the constraint
Australopithecus
  • Australopithecus
The question is asking for a slope in terms of the constraint so to ignore the constraint would give you an incorrect answer
unknownunknown
  • unknownunknown
Hmm, so we're not just constraining the x and y variables in w? We must also then assume z=0?
Australopithecus
  • 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.
Australopithecus
  • Australopithecus
Sorry wrong photo, here this can be helpful for identifying multivariate functions
unknownunknown
  • 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?
myininaya
  • 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
Michele_Laino
  • Michele_Laino
z=0, is a work hypothesis @myininaya
Australopithecus
  • Australopithecus
Yes you are right sorry, I think they just picked z = 0 arbitrarily, sorry you are right it goes +/- z axis infinitely
Australopithecus
  • Australopithecus
I apologize I made a mistake, good this is cleared up now though
unknownunknown
  • unknownunknown
Ahh interesting, thanks though @Australopithecus for the insights. So can we conclude the original pdf contains an error by omitting that z=0?
Australopithecus
  • Australopithecus
No we cannot conclude that
Australopithecus
  • Australopithecus
They just assumed you would pick z = 0, but you could pick any z value
unknownunknown
  • 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?
Australopithecus
  • Australopithecus
they didn't need to specify because in terms of the constraint, dz = 0
Australopithecus
  • Australopithecus
so you could sub z into it but it would all become zero anyways so why would you
unknownunknown
  • 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.
Australopithecus
  • Australopithecus
no no, I am just using d instead of daron because I am lazy dont get confused
Australopithecus
  • Australopithecus
∂z = 0
Australopithecus
  • 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
unknownunknown
  • 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?
Australopithecus
  • Australopithecus
e^0 = 1
unknownunknown
  • unknownunknown
Yes, but why would you even put e^0?
Australopithecus
  • Australopithecus
because the answerer chose z = 0
unknownunknown
  • unknownunknown
But we're then evaluating a partial, when its not asking that.
Australopithecus
  • Australopithecus
like you were saying you could look at this slope at different values of z
unknownunknown
  • 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.
Australopithecus
  • Australopithecus
z is a constant, in your example x is not a constant
Australopithecus
  • Australopithecus
the constraint is a two variable equation
Australopithecus
  • Australopithecus
thus z is not a variable it is a constant
unknownunknown
  • unknownunknown
z isn't a constant, still a variable in dw / dx. We can input any value of z right.
Australopithecus
  • Australopithecus
no because dw/dz = 0 according to the constraint
Australopithecus
  • Australopithecus
that is the derivative of a constant
Australopithecus
  • Australopithecus
thus z is a constant
Australopithecus
  • Australopithecus
so you have to pick one
unknownunknown
  • unknownunknown
Okay so going back to that then, assuming dw/dz = 0, and z is a constant. Why do we make that assumption again?
Australopithecus
  • 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?
Australopithecus
  • 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
unknownunknown
  • 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.
Australopithecus
  • 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
unknownunknown
  • unknownunknown
If you wanted to allow all values of z in the constraint, how would you then express that?
Australopithecus
  • Australopithecus
you are looking at a larger three variable function constrained by a two variable function
Australopithecus
  • Australopithecus
The derivative will just change by a constant value
Australopithecus
  • Australopithecus
You are applying a constraint because you don't want to look at the whole functions derivative
unknownunknown
  • 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?
Australopithecus
  • 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)
Australopithecus
  • 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.
unknownunknown
  • 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?
Australopithecus
  • 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.
unknownunknown
  • 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?
Australopithecus
  • 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.
Australopithecus
  • Australopithecus
I am going to sleep now, more people should be on tomorrow. Maybe they can help you with this better than I can.
unknownunknown
  • unknownunknown
Alright, you've given me a lot to ponder. Thanks for your help and time, it's much appreciated.
unknownunknown
  • 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.
unknownunknown
  • unknownunknown
slight correction * (partial derivatives we arrive at 0, implying they are constant)
Australopithecus
  • 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
Australopithecus
  • Australopithecus
Sorry for the confusion, anyways sleep now.
unknownunknown
  • 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.
Australopithecus
  • Australopithecus
if @ganeshie8 has anything to say on this I would love to hear it, he or she is better at mathematics than I.
Australopithecus
  • 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.
Australopithecus
  • Australopithecus
with the question. Anyways sleep now
unknownunknown
  • unknownunknown
Will do.
ganeshie8
  • 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
unknownunknown
  • unknownunknown
Ahh yes, makes sense. Was quite ambiguous.
unknownunknown
  • unknownunknown
Still confused why they have a set of pdfs with all the correct values, yet they're not using them.
unknownunknown
  • unknownunknown
@myininaya managed to find all these pdfs that seem to have a lot of the errors fixed
ganeshie8
  • 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 !
unknownunknown
  • unknownunknown
Yep definitely, everyone here was a great help.
theEric
  • 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: \(dz=0\).
phi
  • 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.
anonymous
  • anonymous
dz=0

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