MHB Another field lines of 3D vector functions question

brunette15
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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? :)
 
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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
 
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