Representing a curve with a vector valued function

dietcookie
Messages
15
Reaction score
0

Homework Statement


See attached image


Homework Equations





The Attempt at a Solution



Why use x=2sint??

Would it be incorrect to use the x=2cost (versus the given x=2sint)? My professor instructed us to not use the given parameters in the book and to come up with our own, and I would of used x=2cost which would of changed the resulting vector-valued function (Resulting in r = 2Cos(t)i+2Sin(t)j+4Cos2tk)

Thanks.
 

Attachments

  • se12a01061.gif
    se12a01061.gif
    13.3 KB · Views: 538
Physics news on Phys.org
Either one is fine. The difference between the two is also the same thing as changing the parameter t into pi/2-t. Do you see why?
 
Last edited:
I understand the identity sin(t)=cos[(pi/2)-t], but I really don't see why using either parameters x=2sint or x=2cost is equivalent. The book wants me to use x=2sint, wouldn't the equivalent then be actually x=2cos([(pi/2)-t]) ? Thanks
 
dietcookie said:
I understand the identity sin(t)=cos[(pi/2)-t], but I really don't see why using either parameters x=2sint or x=2cost is equivalent. The book wants me to use x=2sint, wouldn't the equivalent then be actually x=2cos([(pi/2)-t]) ? Thanks

If r(t) is a curve then r(a-t) represents the same curve. It goes through the same points, just at different values of t.
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
8K
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 16 ·
Replies
16
Views
4K