## anonymous one year ago 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)

1. Tommynaut

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.

2. anonymous

3. Tommynaut

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.

4. anonymous

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

5. IrishBoy123

$$\vec F(x,y,z)= <y-xy^2, \ xy+z^2, \ xz-zy^2>$$ $$div \ \vec F = <∂_x, ∂_y, ∂_z>\bullet <y-xy^2, \ xy+z^2, \ xz-zy^2>$$ so doing the first bit /( = -y^2 + ... + ... /)

6. Tommynaut

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?

7. Tommynaut

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

8. anonymous

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

9. Tommynaut

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?

10. anonymous

oh oki yes i agree

11. Tommynaut

Cool. So what about the last term then?

12. anonymous

x-y^2

13. Tommynaut

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).

14. anonymous

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

15. Tommynaut

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.

16. Tommynaut

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!)

17. anonymous

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

18. Tommynaut

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.

19. anonymous

would the process be different is the number wasnt zero?

20. Tommynaut

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).

21. Tommynaut

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).

22. anonymous

can i do it then show you what i did

23. Tommynaut

Go for it

24. anonymous

25. anonymous

from here how do i simplify

26. Tommynaut

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.

27. Tommynaut

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).

28. Tommynaut

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

29. Tommynaut

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).

30. anonymous

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

31. Tommynaut

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

32. anonymous

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

33. anonymous

for the i component

34. Tommynaut

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

35. Tommynaut

36. anonymous

-2yz-2z (i)

37. anonymous

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

38. Tommynaut

Your i component looks right! The j component is d/dz (F1) - d/dx (F3). Does that help?

39. anonymous

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

40. Tommynaut

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?

41. anonymous

no now i undertsand it! thank you

42. Tommynaut

Ok sweet, you're welcome!

43. anonymous

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

44. Tommynaut

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