At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
How familiar are you with the divergence and curl formulas? If you are, use the formulas to find the divergence and curl respectively of the function, then substitute 1 in for x, y and z. If not, I'll post the formulas.
please help me out with the formulas im very new to this topic
Sure. The divergence of a function is the derivative with respect to x of the i component, and similarly d/dy of the j component and d/dz of the k component. The curl is a little bit more tricky. Let me try and use latex.
derivative of d/dx= 1-2y? d/dy=2z? d/dz=-2z?
\(div \ \vec F = <∂_x, ∂_y, ∂_z>\bullet \)
so doing the first bit
/( = -y^2 + ... + ... /)
You have the right idea, but you're taking different derivatives. For y - xy^2, the i component, you're taking the derivative with respect to x, so we treat y as a constant. So can you see how it would instead by -y^2, for the first part?
Then you take the derivative with respect to y for xy + z^2. What would that be, treating x and z as constants?
yes i see that...-y^2+2z-y^2?
Well, if we have xy + z^2 and we're treating x and z as constants when taking the derivative of the expression with respect to y, we go term by term. d/dy of xy = x, right? If we imagine x is like 6, we know d/dy of 6y = 6, so it's a similar sort of thing. Similarly, d/dy of z^2 is just 0, right?
oh oki yes i agree
Cool. So what about the last term then?
Awesome! Looks good to me. So if we just add together everything, we've got the divergence of the vector field. After that, we sub in x = y = z = 1 (which is the easy part of course).
-2y^2+2x...-2(1)^2+2(x)...is is this?
Yep, but you can plug x = 1 in too, so you should end up with 0 (unless if I've made some mistake... it's so easy to make mistakes with all this algebra!)
its zero... whats the purpose of calculating it ? what is it was 1? does that matter
It's a little bit hard for me to explain, but you can imagine it like a whole bunch of vectors either coming out of or going into a point, I guess. It has to do with sources and sinks.
would the process be different is the number wasnt zero?
Nope, same process. I'm not sure if the significance of a divergence's value would be assessed in your course, but whether or not the curl is 0 is usually something assessed (and has to do with conservative fields).
Do you think you can give the curl a go? The determinant notation might make it look daunting but it's just a cross product (which is annoying in itself, I guess).
can i do it then show you what i did
Go for it
from here how do i simplify
You have the right general gist, but where you have, for example, d/dy - (xz-zy^2), it should be multiplication instead. So, you're actually finding the derivative of xz-zy^2 with respect to y in that instance.
So, for starters, you would find d/dy of F3 and subtract d/dz of F2, and that's your i component.
The little crosses really help me, and I imagine a bigger cross for the j component (although I just did curved arrows instead so there aren't too many overlapping arrows).
does that mean that d/dx is muplicated by (xy-zy^2) also
Do you mean xz - zy^2? If yes, then yeah.
i take the derivate of dy (xz-y^2)...d/dz (xy+z^2).... 2y-2z?
for the i component
Isn't it d/dy (xz - zy^2)?
Your d/dz part is correct.
im a bit confused for the j component...i thought it was 1-0
Your i component looks right! The j component is d/dz (F1) - d/dx (F3). Does that help?
why d/dz-d/dx...isnt d/dx-d/dz?
Nope, it's the opposite way to how the i and k components are calculated; they go from left to right, in a sense, while the j goes from right to left. Was my diagram a bit too sloppy?
no now i undertsand it! thank you
Ok sweet, you're welcome!
k component looks good, but I think it's (0-z)j