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$begingroup$

I am starting with describing the streamlines of velocity fields.

Let's work out the following example so as to see where I get stuck:

$$v(x, y, z) = x hati + y hatj - x hatk$$

Now we set up the differential equation for the field lines:

$$fracdxx = fracdyy = - fracdzx$$

From here on I do not understand why we do what we do; the solution states:

Thus, $z + x = C_1$, $y = C_2 x$. The streamlines are straight halflines emanating from the z-axis
and perpendicular to the vector $hati + hatk$

Please explain why.

vector-analysis vector-fields

share|cite|improve this question

asked Mar 13 at 18:57

JD_PMJD_PM

16111

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $dfracdxx=-dfracdzximplies dx+dz=0implies x+z=C_1$. You can get the second equation by equating $dfracdxx$ and $dfracdyy$ and integrating in a similar manner.
  $endgroup$
  – Shubham Johri
  Mar 13 at 19:02
  

  
  
  
  

add a comment | 

1

1

$begingroup$

I am starting with describing the streamlines of velocity fields.

Let's work out the following example so as to see where I get stuck:

$$v(x, y, z) = x hati + y hatj - x hatk$$

Now we set up the differential equation for the field lines:

$$fracdxx = fracdyy = - fracdzx$$

From here on I do not understand why we do what we do; the solution states:

Thus, $z + x = C_1$, $y = C_2 x$. The streamlines are straight halflines emanating from the z-axis
and perpendicular to the vector $hati + hatk$

Please explain why.

vector-analysis vector-fields

share|cite|improve this question

asked Mar 13 at 18:57

JD_PMJD_PM

16111

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $dfracdxx=-dfracdzximplies dx+dz=0implies x+z=C_1$. You can get the second equation by equating $dfracdxx$ and $dfracdyy$ and integrating in a similar manner.
  $endgroup$
  – Shubham Johri
  Mar 13 at 19:02
  

  
  
  
  

add a comment | 

$begingroup$

I am starting with describing the streamlines of velocity fields.

Let's work out the following example so as to see where I get stuck:

$$v(x, y, z) = x hati + y hatj - x hatk$$

Now we set up the differential equation for the field lines:

$$fracdxx = fracdyy = - fracdzx$$

From here on I do not understand why we do what we do; the solution states:

Thus, $z + x = C_1$, $y = C_2 x$. The streamlines are straight halflines emanating from the z-axis
and perpendicular to the vector $hati + hatk$

Please explain why.

vector-analysis vector-fields

share|cite|improve this question

asked Mar 13 at 18:57

JD_PMJD_PM

16111

$endgroup$

share|cite|improve this question

asked Mar 13 at 18:57

JD_PMJD_PM

16111

share|cite|improve this question

asked Mar 13 at 18:57

JD_PMJD_PM

16111

asked Mar 13 at 18:57

JD_PMJD_PM

16111

asked Mar 13 at 18:57

JD_PMJD_PM

16111