exam in a couple hours... vector field question

Oct 2009
24
0
I dont thoroughly understand this... I know I shouldve seeked more help on campus but I got wrapped up in studying for my other exams.

given
dx/dt = (-1/2 3/2 ) (x)
dy/dt (3/2 -1/2 ) (y)

first told to sketch vectors on line y = 0

-1/2x + 3/2y -1/2x
3/2 x -1/2y = 3/2x

How can I tell slope and direction of vectors from this?

Here is the link with this problem on it (example 2)

http://www.math.umass.edu/~norman/331_10/Notes/m331lastnotes.pdf
 

Jester

MHF Helper
Dec 2008
2,470
1,255
Conway AR
I dont thoroughly understand this... I know I shouldve seeked more help on campus but I got wrapped up in studying for my other exams.

given
dx/dt = (-1/2 3/2 ) (x)
dy/dt (3/2 -1/2 ) (y)

first told to sketch vectors on line y = 0

-1/2x + 3/2y -1/2x
3/2 x -1/2y = 3/2x

How can I tell slope and direction of vectors from this?

Here is the link with this problem on it (example 2)

http://www.math.umass.edu/~norman/331_10/Notes/m331lastnotes.pdf
On the line \(\displaystyle y = 0\) then

\(\displaystyle
\frac{dx}{dt} = - \frac{1}{2}x\; \text{and} \; \frac{dy}{dt} = \frac{3}{2}x
\).

Picking values of \(\displaystyle x\) say \(\displaystyle x = -2, -1, 0, 1, 2 \) will give both \(\displaystyle \frac{dx}{dt}\) and \(\displaystyle \frac{dy}{dt}\) or dividing \(\displaystyle \frac{dy}{dx}\).