- #26

Hurkyl

Staff Emeritus

Science Advisor

Gold Member

- 14,916

- 19

P.S.

Your parametric equations:

[tex]x(t)=k_1 e^t Cos[\sqrt{2}t]+k_2e^t Sin[\sqrt{2}t]-1/3[/tex]

[tex]y(t)=k_1e^t\left(-\sqrt{2}Sin(\sqrt{2}t)\right)+k_2e^t\left(\sqrt{2} Cos(\sqrt{2}t)\right)+1/3[/tex]

Can be simplified: remember your trigonometry! They're equivalent to:

[tex]x(t) = A e^t \sin(\sqrt{2} t + B) - 1/3[/tex]

[tex]y(t) = A \sqrt{2} e^t \cos(\sqrt{2} t + B) + 1/3[/tex]

where [itex]A^2 = k_1^2 + k_2^2[/itex] and [itex]\tan B = k_1 / k_2[/itex].

This gives a nice way to solve for

Your parametric equations:

[tex]x(t)=k_1 e^t Cos[\sqrt{2}t]+k_2e^t Sin[\sqrt{2}t]-1/3[/tex]

[tex]y(t)=k_1e^t\left(-\sqrt{2}Sin(\sqrt{2}t)\right)+k_2e^t\left(\sqrt{2} Cos(\sqrt{2}t)\right)+1/3[/tex]

Can be simplified: remember your trigonometry! They're equivalent to:

[tex]x(t) = A e^t \sin(\sqrt{2} t + B) - 1/3[/tex]

[tex]y(t) = A \sqrt{2} e^t \cos(\sqrt{2} t + B) + 1/3[/tex]

where [itex]A^2 = k_1^2 + k_2^2[/itex] and [itex]\tan B = k_1 / k_2[/itex].

This gives a nice way to solve for

*t*as -B + Log(f(x, y)) where*f*is a linear polynomial. You ought to be able to reduce it to a nonparametric equation, but I haven't worked out the details yet.
Last edited: