Question

HEEEEEEEEEELLLLLLLLLLPPPPPPPPPP!!!!!!!!!!!!!!!!!!! Medal will be given if fully answered:
C: r(t)=(4sin(3t))i+(2t)j+(4cos(3t))k, sketch and indicate orientation of C for 0 <= t <= 2pi

Answer 1

you can write ≤ as <=

Answer 2

notice if you ignore the j-component, you have
4 sin(3t) i + 4 cos(3t) k
that will describe a circle of radius 4 in the x-z plane
(it will make one full circle from t=0 to 2pi/3 )

but at the same time, the y component is increasing at a rate of 2t

Answer 3

Is that enough info ?

Answer 4

So 4sin(t)+4cos(t) is the same thing as 4sin(3t)+4cos(3t) as a circle? I thought its not allowed to have a 3 for t.

Answer 5

yes, it's still a circle, although the speed at the curve is traced out will be different

Answer 6

the "3" in front of the t can be thought of as the speed

Answer 7

Ok so for the whole equation is it a helix?

Answer 8

no, it's like a spring

Answer 9

yes, a helix

now you need to figure out which way it turns

Answer 10

How is it a spring? I thought it was a circle increasing by 2t?

Answer 11

Do I plug in 0 <= t <=2pi to find which way it goes?

Answer 12

it's a spring that lies parallel to the y-axis

Answer 13

What is the orientation then sorry I don't know how to use t to get the orientation.

Answer 14

orientation in 3D is a bit difficult to describe because i really depends on where your reference is. if you view the graph from the xz plane, it's traced couterclockwise

Answer 15

Oh I figured it out. Could you help me with one more problem?

Answer 16

The mathematical term is helix (a spring is a mechanical thing)