Question

What is the tangent and normal unit vector of (3t,(4-t),5t)

Answer 1

unit Tangent is < velocity > / magnitude of velocity

Answer 2

can you find the velocity vector?

Answer 3

< 3, 1, 5> then over sqrt(35) for unit tangent but lost on normal

Answer 4

<3, -1, 5> i guess actually

Answer 5

normal is zero here... velocity is in a straight line, so no curvature means no centripetal acceleration...

Answer 6

remember the normal vector point along the radius of curvature... if velocity is always 'straight' then there's no curvature (no centripetal direction)

Answer 7

ok, how would I find it if it did have one say <3cos(t), 4sin(t), 5t>

Answer 8

whoops.

Answer 9

if r(t) = <3cos(t), 4sin(t), 5t> ?
then find T as usual < V >/|V|
then differentiate < T > and divide by |dT/dt|

Answer 10

then differentiate < T > and divide by |dT/dt|
so < dT/dt > /| dT/dt|

sorry, I put "magnitude of T" in the denominator initially; it should be magnitude of dT/dt

Answer 11

ok? questions?

Answer 12

ok thank you

Answer 13

sure:)