Recent content by Dvsdvs

isn't \nabla.\nabla = \nabla^2= d^2/dx^2+d^2/dy^2+etc. ugh the text never comes out right why is line 1 not equal to negative line 3?

but i can't calculate the flux at the origin. as the vector is undefined...so how do u take dot product of this \nabladot F when F doesn't even exist at the origin

oooo alright i see this now thanks!

and this Use the Divergence theorem to show that the flux will be the same even if the curve C is not a circle, but a reasonably nice curve enclosing the origin (closed, orientable, simple (i.e. doesn't cross itself), simply connected, enclosing the origin). But this only makes things worse...

The 2D vector F = kq((x)/(x^2+y^2), (y)/(x^2+y^2)) Show that this vector eld is both conservative, and divergence free away from the origin (0; 0). Done dot product=0 and cross product=0 Find the Find the flux of this vector field over the circle C given by x2+y2 = a2 also says that...

alrite so i said flux=\int\int\nabladotF dA. since i just proved that dot product = 0. I am left with \int\int (0) dx dy over the domain: x^2 + Y^2=a^2 sooo, what now

well surface is x^2+y^2=a^2. and for curl ill add a 0 vector for z and the del operator will be (d/dx, d/dy, d/dz) making it R3. and i actually have to prove its conservative away from the region. so i will take \nabla X F in that order yeah i realize i have it swapped before hand. I don't...

yeah thank you. would you know if it would help to convert to polar coordinates when calculating the flux?

Hello, this is pretty straightforward. I need to take cross product: kq(x/(x^2+y^2),y/(x^2+y^2)) x (d/dx,d/dy) since kq is a scalar can i just leave it outside the calculation until the very end and for now just calculate the cross product of the two and once i get a definite answer...

correct me if I'm wrong but i think it's the second one with the xyz being distributed

this is a very quick question. my teacher wants me to prove that ((kq)/(sqroot(x^2+y^2+z^2))) (x,y,z) is conservative. by this does it mean that F=((kq)/(sqroot(x^2+y^2+z^2)))i+((kq)/(sqroot(x^2+y^2+z^2)))j+((kq)/(sqroot(x^2+y^2+z^2)))k or does it she mean that...

lol yeah forgot to check formulas. its sinX + sinY = 2sin[ (X + Y) / 2 ] cos[ (X - Y) / 2 ] and it will prob work with both parts ill check it tomorrow. Also, f(x,y) has a singularity on x+y. questions wants proof that it is/ it is not removable. Is it true that it is removable by plugging...

f(x,y)=(sinx+siny)/(x+Y) as (x,y) approaches (0,0) and then for part II (pi/3,-pi/3) I know that sin(x+Y)/(x+y) would=1 by some simple tweaks. But in my problem, the 2 sins on the numerator are confusing me a little. Since x and y are approaching the same point on the first limit can i say...

oh wooow i feel so stupid really. yeah so dot product = 0 b/c they are perpendicular...Also the overall question for the exercise was to show that if speed is constant. then show that E does now work along the path of the particle. For this it means that //v(t)// is constant which is to say...

nice. thank you very much. How do i show that v.(qE+(v x B)=qE.v in other words why is it that v.(V x b)=0. i know this is probably an elementary question but I am not too good at working with vectors.