Toronto Math Forum

MAT244-2014F => MAT244 Math--Lectures => Topic started by: Hyunmin Jung on September 15, 2014, 01:12:24 AM

Title: Graphing direction field in general
Post by: Hyunmin Jung on September 15, 2014, 01:12:24 AM
It's been a long time since I have taken past math courses so I am not really sure what's going on in some of the content.

My question is for 1.1.31

Okay, I drawn a direction field but I am not sure where I go from there.

What exactly does the solution mean?? Are the solution just slope of the y in general?

there is not a diverging point where vector changes direction (since in this question direction goes all over the place except when it is fixed sign when t is negative) so I guess it does not diverge in this question?

How can you tell where there is asymptote? Do you have set the slope to be 0 before deriving the asymptote?? If so, Why??
By factoring it does come up to be + or - root of (2t-1) But I am assuming because the sign is all negative where t < 0 so there is only one asymptote at + root of (2t-1)? Is my logic correct?

Thanks in advance
Title: Re: Graphing direction field in general
Post by: Victor Ivrii on September 15, 2014, 09:44:41 AM
It's been a long time since I have taken past math courses so I am not really sure what's going on in some of the content.

My question is for 1.1.31

Okay, I drawn a direction field but I am not sure where I go from there.

What exactly does the solution mean?? Are the solution just slope of the y in general?
Solution of the equation is the function $y(t)$ satisfying equation.
Quote
there is not a diverging point where vector changes direction (since in this question direction goes all over the place except when it is fixed sign when t is negative) so I guess it does not diverge in this question?

How can you tell where there is asymptote? Do you have set the slope to be 0 before deriving the asymptote?? If so, Why??
By factoring it does come up to be + or - root of (2t-1) But I am assuming because the sign is all negative where t < 0 so there is only one asymptote at + root of (2t-1)? Is my logic correct?

Thanks in advance
There is no asymptote at all.