# Toronto Math Forum

## MAT244--2020F => MAT244--Lectures & Home Assignments => Chapter 2 => Topic started by: Suheng Yao on September 15, 2020, 12:24:18 PM

Title: Lecture 0201 question
Post by: Suheng Yao 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?
Title: Re: Lecture 0201 question
Post by: Victor Ivrii 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.
Title: Re: Lecture 0201 question
Post by: Suheng Yao 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.
Title: Re: Lecture 0201 question
Post by: Victor Ivrii 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
Title: Re: Lecture 0201 question
Post by: Jianfeng Huang 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? :)
Title: Re: Lecture 0201 question
Post by: Victor Ivrii on September 17, 2020, 12:34:30 PM
We use $x$ and $y$ because we can exclude $t$ but neither $x$ nor $y$