Undergrad Question about Constrained Differentials

[cwill53]

https://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

In part 3 of this example, after dy was expressed in terms of dx, can someone explain to me why z in the total differential equation dw was set to zero? It doesn't make sense to me at the moment.

 

[cwill53]

@BvU Did they make a mistake here?

Also, shouldn’t dy be negative?

 

[fresh_42]

https://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

In part 3 of this example, after dy was expressed in terms of dx, can someone explain to me why z in the total differential equation dw was set to zero? It doesn't make sense to me at the moment.

I don't see anything set to zero. I see $ze^y\,dx = c\,dx = 0$. $ze^y$ is a constant $c$ from the point of view of the differential operator $dx$ since it does not depend on $x$.

 

[PeroK]

https://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

In part 3 of this example, after dy was expressed in terms of dx, can someone explain to me why z in the total differential equation dw was set to zero? It doesn't make sense to me at the moment.

I can't make much sense of that either. Perhaps they forgot to say that they wanted $\frac{\partial w}{\partial x} \big{|}_{z = 0}$. That's what they seem to have calculated anyway.

 

[cwill53]

I don't see anything set to zero. I see $ze^y\,dx = c\,dx = 0$. $ze^y$ is a constant $c$ from the point of view of the differential operator $dx$ since it does not depend on $x$.

The way z is set to zero here doesn’t make sense to me. Also the fact that dy isn’t negative.

 

[PeroK]

Also the fact that dy isn’t negative.

That looks plain wrong.

 

[cwill53]

I can't make much sense of that either. Perhaps they forgot to say that they wanted ∂w∂x|z=0. That's what they seem to have calculated anyway.

Why would they even want that though lol? Is it true that if I didn’t set $\frac{\partial w}{\partial x}$ and $\frac{\partial w}{\partial y}$equal to zero, that I would get the correct answer? I also don't even see the point in this as

$$dw=\frac{\partial w}{\partial x}dx+\frac{\partial w}{\partial y}dy+\frac{\partial w}{\partial z}dz$$

$\frac{\partial w}{\partial x}$ is already a term here.

 

[fresh_42]

View attachment 268306
The way z is set to zero here doesn’t make sense to me. Also the fact that dy isn’t negative.

Didn't see that one. Yes, t´his looks wrong as written. I agree with @PeroK that it probably should have been $\left. \dfrac{d}{dx}\right|_{z=0}$ instead.

 

[cwill53]

Didn't see that one. Yes, t´his looks wrong as written. I agree with @PeroK that it probably should have been $\left. \dfrac{d}{dx}\right|_{z=0}$ instead.

I’m still a bit confused by that. Why would they need to do this at all?

 

[fresh_42]

I’m still a bit confused by that. Why would they need to do this at all?

How is $z$ defined in (1)? It is not defined in (3), and in (2) not independent. So what is it?
And I have trouble to read these tiny formulas. Type it out please.

 

[cwill53]

How is $z$ defined in (1)? It is not defined in (3), and in (2) not independent. So what is it?
And I have trouble to read these tiny formulas. Type it out please.

So, (1) says

1. Find the total differential for $w=zxe^{y}+xe^{z}+ye^{z}$.
$$dw=(ze^{y} + e^{z})dx + (zxe^{y} + e^{z})dy + (xe^{y} + xe^{z} + ye^{z})dz$$

(2) asks us to write dw in terms of dt. That part is unrelated to the confusion.

3. Now suppose $w$ is as above and $x^{2}y+y^{2}x=1.$ Assuming x is the independent variable, find $\frac{\partial w}{\partial x}$.

They proceed to say the following (after mistakenly writing $dy=\frac{2xy+y^{2}}{x^{2}+2xy}dx$ instead of $dy=-\frac{2xy+y^{2}}{x^{2}+2xy}dx$):

 

[cwill53]

@fresh_42 They end up arriving at

$$\frac{\partial w}{\partial x}=\frac{x^{2}+4xy+y^{2}}{x^{2}+2xy}$$

 

[BvU]

I think the given solution is incorrect. Nowhere is given that $z=0$.

 

[PeroK]

What I wrote is all the problem says, z is not specifically defined to be anything.

I would make the assumption that that page is unreliable. It specifically sets $z = 0$ that is clear.

 

[fresh_42]

@fresh_42 They end up arriving at

$$\frac{\partial w}{\partial x}=\frac{x^{2}+4xy+y^{2}}{x^{2}+2xy}$$

I have had another look on the original pdf, and meanwhile I think that the error comes in with the constraint. Maybe they implicitly meant by $x^2y+y^2x=1$ that $z=0$. That's the only way this all makes sense to me.

 

[PeroK]

I have had another look on the original pdf, and meanwhile I think that the error comes in with the constraint. Maybe they implicitly meant by $x^2y+y^2x=1$ that $z=0$. That's the only way this all makes sense to me.

I suspect whoever wrote that page confused $dz = 0$ to get the partial derivative wrt $x$ and $z = 0$. In any case, I can't see much value in trying to make sense of poor material.