- #1

- 99

- 0

x(t) = _________

y(t) = _________

z(t) = _________

r'(t) = <-sin(t), cos(t), 1>

r'(0) = <0,1,1>

my answer:

x = cos(-4pi/6) + 0t

y = sin(-4pi/6) +1t

z = -4pi/6 +t

the only part i got correct is the z, anyone know what im doing wrong?

- Thread starter Whatupdoc
- Start date

- #1

- 99

- 0

x(t) = _________

y(t) = _________

z(t) = _________

r'(t) = <-sin(t), cos(t), 1>

r'(0) = <0,1,1>

my answer:

x = cos(-4pi/6) + 0t

y = sin(-4pi/6) +1t

z = -4pi/6 +t

the only part i got correct is the z, anyone know what im doing wrong?

- #2

StatusX

Homework Helper

- 2,564

- 1

You want the derivative at the tangent point, not t=0.

- #3

- 99

- 0

yea your right, i dont know what i was thinking, thanks

- Replies
- 3

- Views
- 940

- Last Post

- Replies
- 6

- Views
- 2K

- Replies
- 3

- Views
- 6K

- Replies
- 3

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 13

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 8

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 4K

- Last Post

- Replies
- 6

- Views
- 8K