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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)

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  1. Tommynaut
    • one year ago
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    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
    • one year ago
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    please help me out with the formulas im very new to this topic

  3. Tommynaut
    • one year ago
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    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
    • one year ago
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    derivative of d/dx= 1-2y? d/dy=2z? d/dz=-2z?

  5. IrishBoy123
    • one year ago
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    \(\vec F(x,y,z)= <y-xy^2, \ xy+z^2, \ xz-zy^2>\) \(div \ \vec F = <∂_x, ∂_y, ∂_z>\bullet <y-xy^2, \ xy+z^2, \ xz-zy^2>\) so doing the first bit /( = -y^2 + ... + ... /)

  6. Tommynaut
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    yes i see that...-y^2+2z-y^2?

  9. Tommynaut
    • one year ago
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    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
    • one year ago
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    oh oki yes i agree

  11. Tommynaut
    • one year ago
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    Cool. So what about the last term then?

  12. anonymous
    • one year ago
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    x-y^2

  13. Tommynaut
    • one year ago
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    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
    • one year ago
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    -2y^2+2x...-2(1)^2+2(x)...is is this?

  15. Tommynaut
    • one year ago
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    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.

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  16. Tommynaut
    • one year ago
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    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
    • one year ago
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    its zero... whats the purpose of calculating it ? what is it was 1? does that matter

  18. Tommynaut
    • one year ago
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    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
    • one year ago
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    would the process be different is the number wasnt zero?

  20. Tommynaut
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    can i do it then show you what i did

  23. Tommynaut
    • one year ago
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    Go for it

  24. anonymous
    • one year ago
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  25. anonymous
    • one year ago
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    from here how do i simplify

  26. Tommynaut
    • one year ago
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    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
    • one year ago
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    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).

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  28. Tommynaut
    • one year ago
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    So, for starters, you would find d/dy of F3 and subtract d/dz of F2, and that's your i component.

  29. Tommynaut
    • one year ago
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    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
    • one year ago
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    does that mean that d/dx is muplicated by (xy-zy^2) also

  31. Tommynaut
    • one year ago
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    Do you mean xz - zy^2? If yes, then yeah.

  32. anonymous
    • one year ago
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    i take the derivate of dy (xz-y^2)...d/dz (xy+z^2).... 2y-2z?

  33. anonymous
    • one year ago
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    for the i component

  34. Tommynaut
    • one year ago
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    Isn't it d/dy (xz - zy^2)?

  35. Tommynaut
    • one year ago
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    Your d/dz part is correct.

  36. anonymous
    • one year ago
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    -2yz-2z (i)

  37. anonymous
    • one year ago
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    im a bit confused for the j component...i thought it was 1-0

  38. Tommynaut
    • one year ago
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    Your i component looks right! The j component is d/dz (F1) - d/dx (F3). Does that help?

  39. anonymous
    • one year ago
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    why d/dz-d/dx...isnt d/dx-d/dz?

  40. Tommynaut
    • one year ago
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    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
    • one year ago
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    no now i undertsand it! thank you

  42. Tommynaut
    • one year ago
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    Ok sweet, you're welcome!

  43. anonymous
    • one year ago
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    (z-0)j (y-(1-2xy)k

  44. Tommynaut
    • one year ago
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    k component looks good, but I think it's (0-z)j

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