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

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  • #1
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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.
 
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Answers and Replies

  • #2
Charles Link
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If you are taking a partial derivative, what parameter is being held constant?
 
  • #3
haruspex
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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.
 
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  • #4
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I forgot this part of the question. Apologies: (x^2)*y+x(y^2)=t
 
  • #5
Charles Link
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Is ##dt=0 ##? i.e. is the partial derivative w.r.t. ## s ## at fixed ## t ##? You need to be more clear please.
 
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  • #7
haruspex
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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.
 
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  • #8
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I apologize for that. Nothing else has flummoxed in calc, but whatever reason, this has done just that.
 
  • #9
haruspex
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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?
 
  • #10
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Not really.
 
  • #11
haruspex
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Not really.
What equation do you get if you ##\partial/\partial s## the equation for t?
 
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  • #12
Delta2
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Guys i dont 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}##.
 
  • #13
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t is just t(x,y). I don't see how 𝝏t/𝝏s help me.
 
  • #14
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And frankly I don't see how to calculate 𝝏t/𝝏s
 
  • #15
haruspex
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Guys i dont 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(ΞΈ)
##\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.
 
  • #16
Charles Link
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What equation do you get if you ##\partial/\partial s## the equation for t?
alternatively (for the OP) you can set ## dt=0 ##.
 
  • #17
Delta2
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You can't do that with partials.
x=r cos(ΞΈ)
##\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 $$2r\frac{\partial r}{\partial x}=2x\Rightarrow\frac{\partial r}{\partial x}=\frac{x}{r}=\cos\theta$$
 
  • #18
Delta2
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Oh sorry , should have prove that ##\frac{\partial r}{\partial x}=\frac{1}{\cos\theta}##, maybe you are right @haruspex .
 
  • #19
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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?
 
  • #20
haruspex
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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.
 
  • #21
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alternatively (for the OP) you can set ## dt=0 ##.
So if dt=0 then:

(2xy+y^2)dx+(2xy+x^2)dy=0.....?
 
  • #22
haruspex
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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.
 
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  • #23
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t isn't a function of s.
 
  • #24
haruspex
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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.
 
  • #25
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Ok, I see that.
 

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