Partial Differentiation -- If w=x+y and s=(x^3)+xy+(y^3), find 𝝏w/𝝏s

[Robin64]

Homework Statement

If w=x+y and s=(x^3)+xy+(y^3), find 𝝏w/𝝏s

Relevant Equations

𝝏w/𝝏s=(𝝏w/𝝏x)*(𝝏x/𝝏s)+(𝝏w/𝝏y)*(𝝏y/𝝏s)

𝝏w/𝝏x=1

and then I wasn't sure about 𝝏x/𝝏s, so I tried implicitly differentiating s:

1=(3x^2)(𝝏x/𝝏s)+y(𝝏x/𝝏s)+x(𝝏y/𝝏s)+(3y^2)(𝝏y/𝝏s)

And then I shaved my head in frustration.

 

[Charles Link]

If you are taking a partial derivative, what parameter is being held constant?

 

[haruspex]

Is this the whole question, word for word? It doesn't make sense to me.
If s varies it is because x and or y varies, but a given change in s can be achieved by changes in x and y in different ways, and these lead to different changes in w.

 

[Robin64]

I forgot this part of the question. Apologies: (x^2)*y+x(y^2)=t

 

[Charles Link]

Is $dt=0$? i.e. is the partial derivative w.r.t. $s$ at fixed $t$? You need to be more clear please.

 

[Robin64]

This is all I know.

 

[haruspex]

This is all I know.

It could have been made clearer in the question, but from the full text one can guess that the partial wrt s means t constant, and mutatis mutandis.

 

[Robin64]

I apologize for that. Nothing else has flummoxed in calc, but whatever reason, this has done just that.

 

[haruspex]

I apologize for that. Nothing else has flummoxed in calc, but whatever reason, this has done just that.

Are you ok to go ahead now?

 

[Robin64]

Not really.

 

[haruspex]

Not really.

What equation do you get if you $\partial/\partial s$ the equation for t?

 

[Delta2]

Guys i don't understand why you interfere with $t$. You can evaluate $\frac{\partial x}{\partial s}$ directly as $$\frac{\partial x}{\partial s}=\frac{1}{\frac{\partial s}{\partial x}}$$ you ll just find it as function of x and y instead of s and t.
Same logic applies to $\frac{\partial y}{\partial s}$.

 

[Robin64]

t is just t(x,y). I don't see how 𝝏t/𝝏s help me.

 

[Robin64]

And frankly I don't see how to calculate 𝝏t/𝝏s

 

[haruspex]

Guys i don't understand why you interfere with $t$. You can evaluate $\frac{\partial x}{\partial s}$ directly as $$\frac{\partial x}{\partial s}=\frac{1}{\frac{\partial s}{\partial x}}$$ you ll just find it as function of x and y instead of s and t.
Same logic applies to $\frac{\partial y}{\partial s}$.

You can't do that with partials.
x=r cos(θ)
$\partial x/\partial r=\cos(\theta)$
$r^2=x^2+y^2$
$\partial r/\partial x=2x$
The flaw is that something different is being held constant in the two cases.

 

[Charles Link]

What equation do you get if you $\partial/\partial s$ the equation for t?

alternatively (for the OP) you can set $dt=0$.

 

[Delta2]

You can't do that with partials.
x=r cos(θ)
$\partial x/\partial r=\cos(\theta)$
$r^2=x^2+y^2$
$\partial r/\partial x=2x$
The flaw is that something different is being held constant in the two cases.

Yes you can. You have a mistake in the last line, you get that $$2r\frac{\partial r}{\partial x}=2x\Rightarrow\frac{\partial r}{\partial x}=\frac{x}{r}=\cos\theta$$

 

[Delta2]

Oh sorry , should have prove that $\frac{\partial r}{\partial x}=\frac{1}{\cos\theta}$, maybe you are right @haruspex .

 

[Robin64]

alternatively (for the OP) you can set $dt=0$.

When you say let dt=0, let the partial of t with respect to x plus the partial of t with respect to y=0?

 

[haruspex]

Yes you can. You have a mistake in the last line, you get that $$2r\frac{\partial r}{\partial x}=2x\Rightarrow\frac{\partial r}{\partial x}=\frac{x}{r}=\cos\theta$$

Yes, that was careless, but as you saw the corrected version makes my point.

 

[Robin64]

alternatively (for the OP) you can set $dt=0$.

So if dt=0 then:

(2xy+y^2)dx+(2xy+x^2)dy=0...?

 

[haruspex]

When you say let dt=0, let the partial of t with respect to x plus the partial of t with respect to y=0?

No, @Charles Link and I are saying the same thing in different ways. Differentiate the expression for t partially wrt s.
The key is that if there are two independent variables u and v then $\frac{\partial f}{\partial u}$ is shorthand for $\frac{\partial f}{\partial u}|_{v}$, i.e. the change in f as u varies but v stays constant. In particular, $\frac{\partial v}{\partial u}=0$ by definition.

In my Cartesian/polar example, the partial derivative wrt x implicitly means holding y constant, whereas the partial wrt r means holding theta constant.

 

[Robin64]

t isn't a function of s.

 

[haruspex]

t isn't a function of s.

No matter.. $\partial t/\partial s$ means the change in t if s changes but t doesn't, i.e. 0.
It produces what you got in post #21 except that, usefully, you have $\partial x/\partial s$ in place of dx and $\partial y/\partial s$ in place of dy.

 

[Robin64]

Ok, I see that.

 

[Robin64]

then I solve for 𝝏x/𝝏s and 𝝏y/𝝏s?

 

[Delta2]

So @haruspex, under suitable assumptions on what is being held constant (in your example if we keep $\theta$ constant) can we use the equation $$\frac{\partial x}{\partial s}=\frac{1}{\frac{\partial s}{\partial x}}$$

 

[Robin64]

Adding to my confusion is this. The solution to the problem is below, and I have no idea how to get there. Can anyone provide more illumination? I looked for resources that describe the application of the chain rule to these types of partial derivatives, but I can find nothing. Boas' "Mathematical Methods in the Physical Sciences" is less than helpful. Is there a YouTube video or a book that better describes how to approach a problem such as this one?

 

[haruspex]

then I solve for 𝝏x/𝝏s and 𝝏y/𝝏s?

In principle, you can differentiate both the expression for t and the expression for s partially wrt s to get two equations involving 𝝏x/𝝏s and 𝝏y/𝝏s, and hence solve for those. But it will get very messy.

A much easier way here is to rewrite the expression for s in terms of w and t, eliminating x and y. It takes a little ingenuity, but not much work.

 

[haruspex]

So @haruspex, under suitable assumptions on what is being held constant (in your example if we keep $\theta$ constant) can we use the equation $$\frac{\partial x}{\partial s}=\frac{1}{\frac{\partial s}{\partial x}}$$

If you elect to keep the same variable constant in both derivatives then you don't need partials at all. Just differentiate normally, but treating that variable as a constant.
E.g. if we set θ to be constant and write y = x tan(θ), we have 𝝏y/𝝏x = dy/dx = tan(θ) and 𝝏x/𝝏y = dx/dy = cot(θ).

Btw, the Cartesian/polar example can be viewed graphically. Draw a vector r and a corresponding vector x on the x axis. The partial derivative of one wrt the other is represented by the component of the one in the direction of the other. You can see that in each case you multiply by cos theta.

 

[Charles Link]

In principle, you can differentiate both the expression for t and the expression for s partially wrt s to get two equations involving 𝝏x/𝝏s and 𝝏y/𝝏s, and hence solve for those. But it will get very messy.

I did it this way to verify the answer of post 28, and it's not too difficult if you use Kramer's rule type solution to the two linear equations.

 

[Delta2]

So here in this problem when we take the partial derivatives $\frac{\partial x}{\partial s},\frac{\partial y}{\partial s}$ we assume that we keep t constant?
And when calculating $\frac{\partial x}{\partial t},\frac{\partial y}{\partial t}$ we assume that we keep s constant?

 

[haruspex]

So here in this problem when we take the partial derivatives $\frac{\partial x}{\partial s},\frac{\partial y}{\partial s}$ we assume that we keep t constant?
And when calculating $\frac{\partial x}{\partial t},\frac{\partial y}{\partial t}$ we assume that we keep s constant?

Yes. As I wrote, this could have been made clearer, but it is fairly standard that the reader is expected to figure out how the different sets of variables relate, and hence what is being kept constant in each partial derivative.

 

[Delta2]

Yes. As I wrote, this could have been made clearer, but it is fairly standard that the reader is expected to figure out how the different sets of variables relate, and hence what is being kept constant in each partial derivative.

Ok fine but something doesn't look quite right to me. The chain rule is
$$1=\frac{\partial s}{\partial s}=\frac{\partial s}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial s}{\partial y}\frac{\partial y}{\partial s}$$.
In the above when we calculate for example $\frac{\partial s}{\partial x}$ we keep constant y, but when we calculate $\frac{\partial x}{\partial s}$ we keep constant t. Isn't that contradicting or it's that how the chain rule is supposed to work?

 

[haruspex]

Ok fine but something doesn't look quite right to me. The chain rule is
$$1=\frac{\partial s}{\partial s}=\frac{\partial s}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial s}{\partial y}\frac{\partial y}{\partial s}$$.
In the above when we calculate for example $\frac{\partial s}{\partial x}$ we keep constant y, but when we calculate $\frac{\partial x}{\partial s}$ we keep constant t. Isn't that contradicting or it's that how the chain rule is supposed to work?

If you take x and y there as Cartesian coordinates and s as the r of polar coordinates your equation gives $1=\cos^2(\theta)+\sin^2(\theta)$.
How about that?

 

[Delta2]

Something just doesn't look quite right to me, I guess because we change on the fly what is supposed to be kept constant, but anyway I guess that's how the chain rule is supposed to work.

 

[Robin64]

Thanks, all, for the help. Using Kramer's rule to solve for the answer made it nice and easy.

 

[Office_Shredder]

I think the polar coordinates example demonstrating you can flip the partial derivatives is misleading because small moves in r and theta are orthogonal to each other, or maybe was just a lucky choice of coordinates.

Let t=x, s=x+y. Then $\partial s / \partial x = 1$, but when you flip the coordinates you get x=t,y=s-t, and $\partial x / \partial s= 0$.

The proper thing to do is to write down the Jacobian matrix, and then when you flip the coordinates you can just invert the Jacobian.

 

[Charles Link]

Ok fine but something doesn't look quite right to me. The chain rule is
$$1=\frac{\partial s}{\partial s}=\frac{\partial s}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial s}{\partial y}\frac{\partial y}{\partial s}$$.
In the above when we calculate for example $\frac{\partial s}{\partial x}$ we keep constant y, but when we calculate $\frac{\partial x}{\partial s}$ we keep constant t. Isn't that contradicting or it's that how the chain rule is supposed to work?

In this problem, it is something of the form $s=u_a v_a+u_b v_b$. In this case $\frac{\partial{s}}{\partial{s}}=u_a (\frac{\partial{v_a}}{\partial{s}})+v_a (\frac{ \partial{u_a}}{\partial{s}}) + (similarly \, for \, b)$. The partials on $u$ and $v$ are done with the chain rule. (We also have $s=s(u_a, v_a, u_b,v_b)$, and we apply the chain rule in taking the partial w.r.t. $s$, e.g. first taking the partial w.r.t. $u_a$, and treating the others as constants, etc., and then taking partial w.r.t. $v_a$, etc.).

 

[Delta2]

In this problem, it is something of the form $s=u_a v_a+u_b v_b$. In this case $\frac{\partial{s}}{\partial{s}}=u_a (\frac{\partial{v_a}}{\partial{s}})+v_a (\frac{ \partial{u_a}}{\partial{s}}) + (similarly \, for \, b)$. The partials on $u$ and $v$ are done with the chain rule. (We also have $s=s(u_a, v_a, u_b,v_b)$, and we apply the chain rule in taking the partial w.r.t. $s$, e.g. first taking the partial w.r.t. $u_a$, and treating the others as constants, etc., and then taking partial w.r.t. $v_a$, etc.).

Hm, sorry I just can't understand what are the $u_a,v_a,u_b,v_b$. Don't we just take $s=x^3+xy+y^3$ and apply the chain rule as in post #34 to get 1 equation and then also do similar $$0=\frac{\partial t}{\partial s}=\frac{\partial t}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial t}{\partial y}\frac{\partial y}{\partial s}$$ to get the other equation with $t=x^2y+xy^2$

 

[Charles Link]

I think it is something we have all done so often that we do it almost automatically. The partials are difficult to write out in Latex, so let me do implicit differentials as the first step: $dt=3x^2 \, dx+x\, dy+y \, dx+3y^2 \, dy$. You then make each differential a partial w.r.t. $s$.

 

[haruspex]

I think the polar coordinates example demonstrating you can flip the partial derivatives is misleading

Which post are you referring to?

 

[Office_Shredder]

Which post are you referring to?

#17, though I now realize #18 is observing that it didn't work, so I guess my example was not helpful.

 

[haruspex]

I did it this way to verify the answer of post 28, and it's not too difficult if you use Kramer's rule type solution to the two linear equations.

Do you mean Cramer's rule?

 

[Charles Link]

Do you mean Cramer's rule?

Did I misspell it? I remember it from my second course in high school algebra, which was about 50 years ago. I googled it now, and yes, I misspelled it.

 

[haruspex]

Did I misspell it? I remember it from my second course in high school algebra, which was about 50 years ago. I googled it now, and yes, I misspelled it.

Ok. A case of Kramer v. Cramer.

I solved it this way:
s looks a bit like ( x+y)³, so try
$s=x^3+xy+y^3=w^3-3x^2y-3xy^2+xy$
$=w^3-3t+xy$
$t=xy(x+y)=xyw$
$xy=\frac tw$
$s=w^3-3t+\frac tw$
Now diff wrt s.

 

[Charles Link]

Ok. A case of Kramer v. Cramer.

In the Dustin Hoffman movie, "Kramer vs. Kramer", it is spelled with a "K". LOL :)

 

[haruspex]

In the Dustin Hoffman movie, "Kramer vs. Kramer", it is spelled with a "K". LOL :)

But what I hadn't noticed is that the movie title misspells versus as "vs.". That's the plural form; it should just be "v."

 

[Delta2]

Ok. A case of Kramer v. Cramer.

I solved it this way:
s looks a bit like ( x+y)³, so try
$s=x^3+xy+y^3=w^3-3x^2y-3xy^2+xy$
$=w^3-3t+xy$
$t=xy(x+y)=xyw$
$xy=\frac tw$
$s=w^3-3t+\frac tw$
Now diff wrt s.

Nice work indeed. With a little bit of algebra you get one equation for s that can give us both partials of w when differentiating with respect to s or t. Very clever shortcut, well done!.

 

[anuttarasammyak]

Thnks to @harupsex #46
s=w^3-3t+\frac{t}{w}
Differentiating it
ds-(3w^2-\frac{t}{w^2})dw+(3-\frac{1}{w})dt=0
So we get
\frac{\partial w}{\partial s}|_t=\frac{1}{3w^2-\frac{t}{w^2}}
\frac{\partial w}{\partial t}|_s=\frac{3-\frac{1}{w}}{3w^2-\frac{t}{w^2}}