MHB Another field lines of 3D vector functions question

Click For Summary
To find the field lines of the 3D vector function F(x, y, z) = yi - xj + k, the user derived the equations dx/dt = y, dy/dt = -x, and dz/dt = 1. They computed dy/dx = -x/y, leading to the equation y^2 + x^2 = c, which represents a circle. The user expressed uncertainty about relating y^2 + x^2 = c to the parameter t and sought assistance with the mathematical representation. A response suggested using the parametric form of a circle to substitute into the equations to verify the solution.
brunette15
Messages
58
Reaction score
0
I am trying to find the field lines of the 3D vector function F(x, y, z) = yi − xj +k.

I began by finding dx/dt =y, dy/dt = -x, dz/dt = 1.

From here I computed dy/dx = -x/y, and hence y^2 + x^2 = c.

For dz/dt = 1, I found that z = t + C, where C is a constant.

I am unsure where to go from here however, and how to write y^2 + x^2 = c in terms of t. I am guessing as this is a formula for a circle we use sin^2 + cos^2 = 1, however I am unsure how to show this mathematically.

Can anyone please help me complete this question? :)
 
Physics news on Phys.org
brunette15 said:
I am trying to find the field lines of the 3D vector function F(x, y, z) = yi − xj +k.

I began by finding dx/dt =y, dy/dt = -x, dz/dt = 1.

From here I computed dy/dx = -x/y, and hence y^2 + x^2 = c.

For dz/dt = 1, I found that z = t + C, where C is a constant.

I am unsure where to go from here however, and how to write y^2 + x^2 = c in terms of t. I am guessing as this is a formula for a circle we use sin^2 + cos^2 = 1, however I am unsure how to show this mathematically.

Can anyone please help me complete this question? :)

Hi brunette15! (Smile)

The parametric form of a circle is $(x(t), y(t)) = (r\cos(\omega t-\phi_0), r\sin(\omega t-\phi_0))$.
If we substitute that in $\d x t =y$, $\d y t = -x$, and $y^2 + x^2 = c$, we can find $r, \omega$ and $\phi_0$, and verify that it is indeed a solution.
 
I like Serena said:
Hi brunette15! (Smile)

The parametric form of a circle is $(x(t), y(t)) = (r\cos(\omega t-\phi_0), r\sin(\omega t-\phi_0))$.
If we substitute that in $\d x t =y$, $\d y t = -x$, and $y^2 + x^2 = c$, we can find $r, \omega$ and $\phi_0$, and verify that it is indeed a solution.

I see! Thankyou! :D
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
5K
  • · Replies 1 ·
Replies
1
Views
514
  • · Replies 3 ·
Replies
3
Views
963
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K