Solving a Graph Problem: Velocities and Equations

  • Thread starter Thread starter Michaelbk
  • Start date Start date
  • Tags Tags
    Graph
Michaelbk
Messages
6
Reaction score
0
Hi people, i was wondering if anyone might be able to see the problem with my graph:
http://img82.imageshack.us/my.php?image=grafur3.jpg
There must be something wrong, since the velocity drops so fast, and i was thinking that they should all start at v0, but this isn't the case.
(The definitions and equation should be in the picture with the graph)
In this picture you will see how i made the equation (maybe there's something i missed):
http://img393.imageshack.us/my.php?image=eqfmyx6.png
 
Physics news on Phys.org
Michaelbk said:
Hi people, i was wondering if anyone might be able to see the problem with my graph:
http://img82.imageshack.us/my.php?image=grafur3.jpg
There must be something wrong, since the velocity drops so fast, and i was thinking that they should all start at v0, but this isn't the case.
(The definitions and equation should be in the picture with the graph)
In this picture you will see how i made the equation (maybe there's something i missed):
http://img393.imageshack.us/my.php?image=eqfmyx6.png

I don't see why you would think "they should all start at v0" since you have replaced v0 by vr, vh, and vs, three different values.
 
Thread 'Direction Fields and Isoclines'
I sketched the isoclines for $$ m=-1,0,1,2 $$. Since both $$ \frac{dy}{dx} $$ and $$ D_{y} \frac{dy}{dx} $$ are continuous on the square region R defined by $$ -4\leq x \leq 4, -4 \leq y \leq 4 $$ the existence and uniqueness theorem guarantees that if we pick a point in the interior that lies on an isocline there will be a unique differentiable function (solution) passing through that point. I understand that a solution exists but I unsure how to actually sketch it. For example, consider a...
Back
Top