Zela101
  • Zela101
Find curl F
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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Zela101
  • Zela101
Question: \(\large F(x, y, z) = x^2i − 2j + yzk\) \(\large div~F=\frac{\partial}{\partial x}(x^2)+\frac{\partial}{\partial y}(-2)+\frac{\partial}{\partial z}(yz)\) \(\large div~F=2x+0+y\) \(\large div~F=2x+y\)
Zela101
  • Zela101
|dw:1449641454777:dw|
Zela101
  • Zela101
\(curl~ F=4y~i+(yx^2-2xyz)~j-4x~k=<4y,yx^2-2xyz,-4x>\)

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More answers

Zela101
  • Zela101
The book has a different answer for curl F
jim_thompson5910
  • jim_thompson5910
what answer does the book have?
jim_thompson5910
  • jim_thompson5910
maybe they're equivalent
anonymous
  • anonymous
Try doing it like this: \[\huge \begin{vmatrix} i & j & k\\ \partial x & \partial y & \partial z\\ F_x & F_y & F_z \end{vmatrix}\] Fx, Fy, and Fz are the components of the vector
anonymous
  • anonymous
Not sure if you'll get the same or different answer
Zela101
  • Zela101
The answer of the book is curl F = zi.
Zela101
  • Zela101
I have another question. I have two vectors given, F and G. The question asks me to find ∇•(F x G). Which vector should i use the gradient for?
anonymous
  • anonymous
I believe you'd have to take the cross product of F and G which will yield another vector. And then you dot that with the del operator.
jim_thompson5910
  • jim_thompson5910
@Zela101 can you post a screenshot of the full problem? The first problem you posted
Zela101
  • Zela101
Ok
jim_thompson5910
  • jim_thompson5910
I agree, you find vector h = f X g and then find the gradient of h
anonymous
  • anonymous
Or you could recognize that it's a triple product
Zela101
  • Zela101
anonymous
  • anonymous
Not sure which one would be easier or quicker. Probably the triple product but it might turn out ugly. Should get the same thing though
Zela101
  • Zela101
So H=FxG Then ∇• H ?
anonymous
  • anonymous
Yep! And here is the det for the triple product:\[\huge \begin{vmatrix} \partial x & \partial y & \partial z\\ F_x & F_y & F_z\\ G_x & G_y & G_z \end{vmatrix}\]
jim_thompson5910
  • jim_thompson5910
I'm not sure why your book is saying curl(F) = zi. That's very strange. Maybe there's a typo or mixup
Zela101
  • Zela101
Alright, thanks @jim_thompson5910 :) Thanks @cshrix ! Alright, i'll go over the nabla and work on it..
anonymous
  • anonymous
Sounds good X)
Zela101
  • Zela101
That's the answer to this question. I got the Div F correct but the curl F didn't match..
anonymous
  • anonymous
Did you try using the determinant I set up for you (my first post)?
Zela101
  • Zela101
Yes, that's what i did.
anonymous
  • anonymous
hmm, give me a minute
jim_thompson5910
  • jim_thompson5910
@Zela101 I see what went wrong with the curl
anonymous
  • anonymous
I got so the book answer is correct
anonymous
  • anonymous
I can type it out for you if you'd like
Zela101
  • Zela101
\(\large F(x, y, z) = x^2i − 2j + yzk\) \(\large F(x, y, z) = Mi − Nj + Pk\) \(\Large \begin{vmatrix} i & j & k\\ \frac{\partial}{\partial x}& \frac{\partial}{\partial y} & \frac{\partial}{\partial z}\\ M & N & P \end{vmatrix}\)
Zela101
  • Zela101
Type it out please.
anonymous
  • anonymous
Sure :) Give me a minute to get it typed up
Zela101
  • Zela101
I'm probably setting up the cross product incorrectly.
anonymous
  • anonymous
\[\large =<\partial y[yz]-\partial z[-2],-\partial x[yz]-\partial z[x^2],\partial x[-2]-\partial y[x^2]>\]
anonymous
  • anonymous
When you evaluate the partial derivatives of the j and k components, you'll see that they're all 0. Taking the partial with respect to x of (yz) is 0! The partial WRT z of (x^2) is 0 and so on and so forth.
anonymous
  • anonymous
Except for the i component, where the partial WRT y of (yz) is just z! (treating z as a constant)
anonymous
  • anonymous
WRT = with respect to
Zela101
  • Zela101
Oh, so we don't take into account the div F when setting up the cross product?
anonymous
  • anonymous
No we have to set up the determinant that I set up in my first post X)
Zela101
  • Zela101
I see :O
anonymous
  • anonymous
This one: \[\huge \begin{vmatrix} i & j & k\\ \partial x & \partial y & \partial z\\ F_x & F_y & F_z \end{vmatrix}\]
anonymous
  • anonymous
X)
Zela101
  • Zela101
Isn't the symbol \(F_x\) is a different way of writing partials though?
anonymous
  • anonymous
Also I made a sign error here: \[\large =<\partial y[yz]-\partial z[-2],-\partial x[yz] \color{red}{+}\partial z[x^2],\partial x[-2]-\partial y[x^2]>\]But it doesn't really matter anyway
jim_thompson5910
  • jim_thompson5910
\[\huge \begin{array}{|ccc|cc} i & j & k & i & j\\ \partial x & \partial y & \partial z & \partial x & \partial y\\ P & Q & R & P & Q\end{array} \] Notice we have this diagonal (in red) \[\huge \begin{array}{|ccc|cc} \color{red}{i} & j & k & i & j\\ \partial x & \color{red}{\partial y} & \partial z & \partial x & \partial y\\ P & Q & \color{red}{R} & P & Q\end{array} \] that means we're going to multiply i, \(\Large \partial y\) and R. But what's really going on is we're applying the partial derivative with respect to y on the function R(x,y,z) = yz this happens with all of the diagonals (more or less) @CShrix is basically saying the same thing
anonymous
  • anonymous
Yeah but sometimes it's also the notation for the x component of a vector.. bit confusing
jim_thompson5910
  • jim_thompson5910
http://www.math.harvard.edu/archive/21a_spring_09/PDF/13-05-curl-and-divergence.pdf
Zela101
  • Zela101
The book has the same thing, i read it wrong xD I was thinking i will replace the partials with what i got for div F
Zela101
  • Zela101
Thanks, you guys are helpful. Much love <33
anonymous
  • anonymous
Ahh, nope! The del operators are part of the cross product X)
anonymous
  • anonymous
You are very welcome :)
Zela101
  • Zela101
Take care. Thanks again!
anonymous
  • anonymous
Yes! and you're welcome! X) You can either take the cross product first and then the dot product, or you can use the triple product determinant that i also set up for you.
Zela101
  • Zela101
\(∇ · (F × G) = ∇ · (−(z + 4y^2)i + (4xy + 2xz)j + (2xy − x)k) = 4x\)

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