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Hello I am having trouble understand the following in my notes;

It was an example;

Q: Is $$v=K(-yi+xj)$$ a conservative vector field?

A:

If it was, then $$K(-yi+xj)=\nabla \psi$$

Here is where I am confused;

Next line says

$$\partial \psi/\partial x= -Ky$$ which implies $\psi(x,y)=-Kxy+f(y)$

$$\partial \psi / \partial y= Kx$$ which implies $\psi(x,y)=Kxy+g(x)$

and so we are able to conclude that it is not conservative from inspection.

But how did we do the partial derivatives. I thought the partial of \psi w.r.t x would mean we treat everything except x constant, etc. Can anyone please help to explain this to me?

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  • $\begingroup$ $\nabla\psi=\frac{\partial\psi}{\partial x}\vec i+\frac{\partial\psi}{\partial y}\vec j$. $\endgroup$ – amd Feb 18 '16 at 20:56
  • $\begingroup$ @amd can you elaborate please. does this mean I take derivative wrt to I and j compontent like whatever is with i is the x? $\endgroup$ – PersonaA Feb 18 '16 at 21:12
  • $\begingroup$ It means that the coefficients of $\vec i$ and $\vec j$ in $\nabla\psi$ are equal to the partial derivatives of $\psi$, so you just match them up to the corresponding components of $v$. $\endgroup$ – amd Feb 18 '16 at 21:16
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You're right that when we take the partial derivative of $\psi$ w.r.t. $x$, we treat everything besides $x$ as constant. That's why

$$\frac{\partial}{\partial x}\left(-Kxy + f(y)\right)= \frac{\partial}{\partial x}(-Kxy) + \frac{\partial}{\partial x}(f(y)) = -Ky$$

because if some function $f$ doesn't have $x$ as an argument, it doesn't care at all about $f$, so the change in $f$ w.r.t $x$ is zero.

Treating $y$ as constant when you take a derivative doesn't mean that you won't have $y$'s left in the result. It's just like how

$$\frac{d}{dx} 2x = 2$$

$2$ is constant with respect to $x$ (it's constant with respect to everything) but that doesn't mean the derivative won't have any $2$'s in it.

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  • $\begingroup$ I understand that part but I don't understand is the first direction, how d/dx partial of k(-yi+xj) gives -Ky $\endgroup$ – PersonaA Feb 18 '16 at 20:58
  • $\begingroup$ Ah, I see. It's not that, but rather that $\nabla f = (\frac{df}{dx},\frac{df}{dy})$ . So because we have $K(-yi+xj) = \nabla \psi$, we get $\frac{d\psi}{dx} = \text{the first component of $K(-yi+xj)$} = -Ky$. $\endgroup$ – Eli Rose -- REINSTATE MONICA Feb 18 '16 at 21:39
  • $\begingroup$ So your saying if we have it like I do above ie already written in terms of gradient then we don't even take derivative because we already have them? $\endgroup$ – PersonaA Feb 18 '16 at 21:43
  • $\begingroup$ @PersonaA: Yep. You took all those derivatives in the gradient. $\endgroup$ – Eli Rose -- REINSTATE MONICA Feb 18 '16 at 22:02

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