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<math>-\cos(\frac{\pi}{2}+ x)=\sin x = x -\frac{1}{3!}x ^3 +\dots \approx x</math> | <math>-\cos(\frac{\pi}{2}+ x)=\sin x = x -\frac{1}{3!}x ^3 +\dots \approx x</math> | ||
| + | (when x is small) | ||
to realize that the linearized system at the critical point (pi/2,0) | to realize that the linearized system at the critical point (pi/2,0) | ||
| − | has a 1 times y<sub>1</sub> in that spot. | + | has a 1 times y<sub>1</sub> in that spot. I think it's easier to use the |
Jacobian matrix the way I demonstrated in class to see this. | Jacobian matrix the way I demonstrated in class to see this. | ||
Revision as of 09:18, 21 September 2013
Homework 5 collaboration area
Question from Ryan Russon
For problems p.159: 4,7,11 are we supposed to accompany each solution with a sketch of the what is happening at the critical points or are we fine just stating what is happening at those points based on the eigenvalues of the linearized system? My confusion is stemming from the answer in the back of the book for #11 which says, "Use -cos(+- 1/2...)," you get the picture.
Answer from Steve Bell :
Those problems don't seem to ask for a sketch, so don't bother. (If you wanted to practice for the exam, testing yourself to see if you would know how to draw a sketch if you had to would be therapeutic.)
That cryptic remark about the trig identity can be used like so
$ -\cos(\frac{\pi}{2}+ x)=\sin x = x -\frac{1}{3!}x ^3 +\dots \approx x $
(when x is small) to realize that the linearized system at the critical point (pi/2,0) has a 1 times y1 in that spot. I think it's easier to use the Jacobian matrix the way I demonstrated in class to see this.
