Question

Medal*** Please help
Find the divergence and the curl of the vectorfield
F(x,y,z)= (y-xy^2)i+(xy+z^2)j+(xz-zy^2)k at the point (x,y,z)= (1,1,1)

Answer 1

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.

Answer 2

please help me out with the formulas im very new to this topic

Answer 3

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.

Answer 4

derivative of d/dx= 1-2y? d/dy=2z? d/dz=-2z?

Answer 5

\(\vec F(x,y,z)= <y-xy^2, \ xy+z^2, \ xz-zy^2>\)
\(div \ \vec F = <∂_x, ∂_y, ∂_z>\bullet <y-xy^2, \ xy+z^2, \ xz-zy^2>\)

so doing the first bit

/( = -y^2 + ... + ... /)

Answer 6

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?

Answer 7

Then you take the derivative with respect to y for xy + z^2. What would that be, treating x and z as constants?

Answer 8

yes i see that...-y^2+2z-y^2?

Answer 9

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?

Answer 10

oh oki yes i agree

Answer 11

Cool. So what about the last term then?

Answer 12

x-y^2

Answer 13

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).

Answer 14

-2y^2+2x...-2(1)^2+2(x)...is is this?

Answer 15

Here 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.

Answer 16

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!)

Answer 17

its zero... whats the purpose of calculating it ? what is it was 1? does that matter

Answer 18

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.

Answer 19

would the process be different is the number wasnt zero?

Answer 20

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).

Answer 21

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).

Answer 22

can i do it then show you what i did

Answer 23

Go for it

Answer 24

from here how do i simplify

Answer 25

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.

Answer 26

Also, 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 cross-products in general, I guess).

Answer 27

So, for starters, you would find d/dy of F3 and subtract d/dz of F2, and that's your i component.

Answer 28

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).

Answer 29

does that mean that d/dx is muplicated by (xy-zy^2) also

Answer 30

Do you mean xz - zy^2? If yes, then yeah.

Answer 31

i take the derivate of dy (xz-y^2)...d/dz (xy+z^2).... 2y-2z?

Answer 32

for the i component

Answer 33

Isn't it d/dy (xz - zy^2)?

Answer 34

Your d/dz part is correct.

Answer 35

-2yz-2z (i)

Answer 36

im a bit confused for the j component...i thought it was 1-0

Answer 37

Your i component looks right!

The j component is d/dz (F1) - d/dx (F3). Does that help?

Answer 38

why d/dz-d/dx...isnt d/dx-d/dz?

Answer 39

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?

Answer 40

no now i undertsand it! thank you

Answer 41

Ok sweet, you're welcome!

Answer 42

(z-0)j
(y-(1-2xy)k

Answer 43

k component looks good, but I think it's (0-z)j