 Author Topic: Lecture 0201 question  (Read 523 times)

Suheng Yao Lecture 0201 question
« on: September 15, 2020, 12:24:18 PM »
This is the last question from yesterday's lecture 0201 section. I still don't understand why there is a negative sign on the right side of the equation?

Victor Ivrii Re: Lecture 0201 question
« Reply #1 on: September 15, 2020, 01:11:04 PM »
Indeed, there should be no $-$ on the right, unless I change $b-y$ to $y-b$ (which I intended to to but doid not). I updated handout, it is just a single place as on the next frame everything is right.

Suheng Yao Re: Lecture 0201 question
« Reply #2 on: September 15, 2020, 02:23:46 PM »
Thanks, prof. Also, on the next slide, I feel confused about why does f(x) have a single minimum at x=a and have equilibrium at x=a and y=b? I really don't get the idea here.

Victor Ivrii Re: Lecture 0201 question
« Reply #3 on: September 16, 2020, 03:47:54 AM »
You can investigate $f(x)$ as in Calculus I.

Also because $x=x(t)$ and $y=y(t)$ and $x=a,y=b$ is a constant solution (equilibrium). Look at the picture on the next slide. We excluded $t$ from our analysis but it does not mean that it had gone

Jianfeng Huang

• Newbie
• • Posts: 1
• Karma: 0 Re: Lecture 0201 question
« Reply #4 on: September 16, 2020, 11:20:50 AM »
I was also wondering how to extent the concept of direction field in this case, as both x and y are functions of t. Why the direction field shows the relationship between x and y? Is it because equations do not include t explicitly, and theoretically we can have 3-dimensional direction field? Victor Ivrii Re: Lecture 0201 question
« Reply #5 on: September 17, 2020, 12:34:30 PM »
We use $x$ and $y$ because we can exclude $t$ but neither $x$ nor $y$