anonymous
  • anonymous
How do I draw a vector field by hand? F(x,y,z)=-y k
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
@phi @Zarkon @inkyvoyd
klimenkov
  • klimenkov
\(\vec{F}(x,y,z)=-y\cdot\vec{k}\)
anonymous
  • anonymous
Yes

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

klimenkov
  • klimenkov
|dw:1359238654831:dw|
anonymous
  • anonymous
I'm not sure what you did there?
abb0t
  • abb0t
I'm not sure what she did there either.
klimenkov
  • klimenkov
This field doesn't depent on \(x\) and \(z\). That means that there will be the same vectors in the planes \(y=\text{const}\). \(\vec{k}\) is a vector that has length = 1 and is collinear to \(z\)-axis. I drew 3 planes \(y=\text{const}\) for different values of this constant.
anonymous
  • anonymous
So, say (1,0,0).. it would be the zero vector, so wouldn't it just be a point and not a line of length 1?
klimenkov
  • klimenkov
A vector field is a vector function. Its argument is a vector (point) and its value is also a vector. For example if you take a point (1,0,0), thats mean that \(x=1,y=z=0.\). So, \(\vec{F}(1,0,0)=0\cdot\vec{k}=\vec{0}\). For every vector that has y-coordinate = 0, it will be zero vector. y=0 is a plane. Every vector in this plane will be zero-vector. If you take (0,1,0), then \(\vec{F}(0,1,0)=1\cdot\vec{k}=\vec{k}\). Every vector in the plane y=1 will be \(\vec{k}\). And so on.
anonymous
  • anonymous
I get it!! Thank you so much! So everything with a negative y value would be going up and everything with a positive y value would be going down?
klimenkov
  • klimenkov
Yes. Sorry, my fault. I have lost the "-" sign. I'm happy that you've got it. You are welcome.
anonymous
  • anonymous
Thank you!

Looking for something else?

Not the answer you are looking for? Search for more explanations.