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anonymous
 one year ago
Medal*** Please help
Find the divergence and the curl of the vectorfield
F(x,y,z)= (yxy^2)i+(xy+z^2)j+(xzzy^2)k at the point (x,y,z)= (1,1,1)
anonymous
 one year ago
Medal*** Please help Find the divergence and the curl of the vectorfield F(x,y,z)= (yxy^2)i+(xy+z^2)j+(xzzy^2)k at the point (x,y,z)= (1,1,1)

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Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3How 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0please help me out with the formulas im very new to this topic

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Sure. 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0derivative of d/dx= 12y? d/dy=2z? d/dz=2z?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.2\(\vec F(x,y,z)= <yxy^2, \ xy+z^2, \ xzzy^2>\) \(div \ \vec F = <∂_x, ∂_y, ∂_z>\bullet <yxy^2, \ xy+z^2, \ xzzy^2>\) so doing the first bit /( = y^2 + ... + ... /)

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3You 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?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Then you take the derivative with respect to y for xy + z^2. What would that be, treating x and z as constants?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes i see that...y^2+2zy^2?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Well, 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?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Cool. So what about the last term then?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Awesome! 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).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.02y^2+2x...2(1)^2+2(x)...is is this?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Here I've got the 3 fundamental formulas from vector calculus (there are a lot more!). We use the bottom two for the question you're doing.

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Yep, 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!)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its zero... whats the purpose of calculating it ? what is it was 1? does that matter

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3It'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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would the process be different is the number wasnt zero?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Nope, 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).

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Do 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).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0can i do it then show you what i did

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0from here how do i simplify

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3You have the right general gist, but where you have, for example, d/dy  (xzzy^2), it should be multiplication instead. So, you're actually finding the derivative of xzzy^2 with respect to y in that instance.

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Also, I don't know if you'll find this useful but I made this image just now of what I do mentally when I find curls (or crossproducts in general, I guess).

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3So, for starters, you would find d/dy of F3 and subtract d/dz of F2, and that's your i component.

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3The 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).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0does that mean that d/dx is muplicated by (xyzy^2) also

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Do you mean xz  zy^2? If yes, then yeah.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i take the derivate of dy (xzy^2)...d/dz (xy+z^2).... 2y2z?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Isn't it d/dy (xz  zy^2)?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Your d/dz part is correct.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0im a bit confused for the j component...i thought it was 10

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Your i component looks right! The j component is d/dz (F1)  d/dx (F3). Does that help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0why d/dzd/dx...isnt d/dxd/dz?

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Nope, 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?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0no now i undertsand it! thank you

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3Ok sweet, you're welcome!

Tommynaut
 one year ago
Best ResponseYou've already chosen the best response.3k component looks good, but I think it's (0z)j
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