MHB What is the locus of points where $F_x = 1$ and $|F_x| = 2$?

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The discussion revolves around finding the locus of points for the vector field F given by F=0.4(y-2x)a_{x}-(200/(x^2+y^2+z^2))a_{z}. The user successfully evaluated |F| at the point P(-4,3,5) and found the unit vector direction of |F|. For the condition F_x = 1, the locus is described by the plane equation y - 2x = 2.5. In contrast, |F_x| = 2 results in two lines represented by the equations y - 2x = 5 and y - 2x = -5. The discussion clarifies the distinction between the two conditions, emphasizing the difference in the number of resulting lines.
Drain Brain
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Given the vector field $F=0.4(y-2x)a_{x}-(\frac{200}{x^2+y^2+z^2})a_{z}$ :
1. evaluate $|F|$ at $P(-4,3,5)$;
2. Find unit vector specifying the direction of $|F|$ at P.
3. Describe the locus of all points for which $ F_{x}=1; |F_{x}|=2$

I managed to solve the 1 and 2

By substituting the value of x and y to the vector field I obtain
G
$F=4.4a_{x}-4a{z}$
$|F|=5.95$

$a_{p}=\frac{F}{|F|}=0.740a_{x}-0.673a_{z}$

Can you help me with the last question.
 
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Hi Drain Brain,

What is the meaning of $a_x$ and $a_z$? I'm asking because although $F$ is supposed to be a vector field, the question you seek deals with the equation $F_x = 1$, which is not a vector equation of three dimensions.
 
Those are the unit vectors in the cartesian coordinate.
 
Ok, then when you wrote $ F_x = 1$ in part 3, did you mean $|F_x| = 1$?
 
No. It's just as it is.
 
So $ F_x $ represents the $ x $- component of $ F $, not the derivative of $ F $ with respect to $ x$?
 
Euge said:
So $ F_x $ represents the $ x $- component of $ F $, not the derivative of $ F $ with respect to $ x$?

Yes.
 
Alright. Then $F_x = 1$ is equivalent to $0.4(y - 2x) = 1$, i.e., $y - 2x = 2.5$. So the locus of points satisfying $F_x = 1$ is the plane $y - 2x = 2.5$. Can you find the locus for $|F_x| = 2$?
 
Euge said:
Alright. Then $F_x = 1$ is equivalent to $0.4(y - 2x) = 1$, i.e., $y - 2x = 2.5$. So the locus of points satisfying $F_x = 1$ is the plane $y - 2x = 2.5$. Can you find the locus for $|F_x| = 2$?

Hi euge!

I just want to know what's the difference when we find the locus of points for $F_x = 1$ and $|F_x| = 2$? The absolute value confuses me.
 
  • #10
The difference is that the latter gives two lines, whereas the former gives one line. Have you tried working them out?
 
  • #11
Euge said:
The difference is that the latter gives two lines, whereas the former gives one line. Have you tried working them out?

my answer for $|F|=2$ are

$y-2x=5$ and $y-2x=-5$ are these correct?
 
  • #12
They're correct!
 

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