# Math Help - Curl...

1. ## Curl...

Show that the component of ∇φ(x, y) in the i direction of the surface φ(x, y) = exp−(x^2 + y^2) is:

−(√2)/e

at the point r = 1/(√2) (i + j).

Hi. Im having some trouble with the question above. Not really sure where to go with this, but as a starter I know that the curl = dφ/dx i + dφ/dy j. I got these to be:

dφ/dx = -2x exp-(x^2 + y^2)
dφ/dy = -2y exp-(x^2 + y^2)

Thanks

2. Originally Posted by Croc
Show that the component of ∇φ(x, y) in the i direction of the surface φ(x, y) = exp−(x^2 + y^2) is:

−(√2)/e

at the point r = 1/(√2) (i + j).

Hi. Im having some trouble with the question above. Not really sure where to go with this, but as a starter I know that the curl = dφ/dx i + dφ/dy j. I got these to be:

dφ/dx = -2x exp-(x^2 + y^2)
dφ/dy = -2y exp-(x^2 + y^2)

Thanks

Why in "abstract algebra", for hollie mollie's sake?!! . This belongs in the calculus section.
Besides this some symbols didn't show correctly...the first one was nabla = the gradient of $\phi$ , or what?

Tonio

3. Originally Posted by Croc
Show that the component of ∇φ(x, y) in the i direction of the surface φ(x, y) = exp−(x^2 + y^2) is:

−(√2)/e

at the point r = 1/(√2) (i + j).

Hi. Im having some trouble with the question above. Not really sure where to go with this, but as a starter I know that the curl = dφ/dx i + dφ/dy j. I got these to be:

dφ/dx = -2x exp-(x^2 + y^2)
dφ/dy = -2y exp-(x^2 + y^2)

Thanks
As Tonio said, this does not belong in the "Linear Algebra and Abstract Algebra" section but in the Calculus section- but I'm not going to bang my head against the wall!

I hope you do not know "that the curl= dφ/dx i + dφ/dy j"! That's the gradient, $\nabla \phi$ or grad $\phi$. "Curl" is an operator on vector valued functions: $\nabla\times\vec{f}$.

You should also know that $D_{\vec{v}} f(x,y)= \nabla f\cdot \vec{v}$ as long as $\vec{v}$ is a unit vector- and $\vec{i}$ is, of course, a unit vector. That is, the answer to your problem is just $\nabla \phi\cdot\vec{i}= \frac{\partial \phi}{\partial x}$, evaluated, of course, at $x= \frac{1}{\sqrt{2}}$, $y= \frac{1}{\sqrt{2}}$.