# exam in a couple hours... vector field question

• May 11th 2010, 06:06 AM
plopony
exam in a couple hours... vector field question
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/33...1lastnotes.pdf
• May 11th 2010, 06:36 AM
Jester
Quote:

Originally Posted by plopony
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/33...1lastnotes.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 $\displaystyle \frac{dy}{dx}$.
• May 11th 2010, 07:11 AM
plopony
thanks for the help =)
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