Question

r(t) =cos(pi*t)i +sin(pi*t)j +2tk
I already found the velocity of the particle when it's height is =6, which is
v(3)=(-pi)j+2k

c) When the particle has height 6, it leaves the helix and moves along the tangent line at the constant velocity found in part (b). Find a vector parametric equation for the position of the particle (in terms of the original parameter t) as it moves along this tangent line.

Answer 1

if its moving along the tangent line, what is the equation of the tangent line?

Answer 2

isnt it just the component derivatives, anchored to the point at t=3?

Answer 3

x(t) = r(3) + v(3)t

Answer 4

yeah, like that :)

Answer 5

oh and the time is 0 when the particle is at height 6

Answer 6

r(t) =cos(pi*t)i +sin(pi*t)j +2tk

r = <x,y,z>

r' = <x',y',z'> = tangent vector

L =

x = x(t) + x'(t)n
y = y(t) + y'(t)n
z = z(t) + z'(t)n

Answer 7

Yes, I found by doing eqn of tangent line: L(t) = r(3) +r'(3)t, which is what ingenuus has written b/c r'(t)=v(t), but my answer is wrong. I came up with L(t)=(-1,0,6) + (0, -pi, 2)t so I don't know why this is incorrect. I have to write it in vector form b/c the space where I enter the answer is not set up to write it in parametric (b/c it's webwork).

Answer 8

is the answer (-1,-pi(t-3),t+3) ?
maybe it's because of the sentence "in terms of the original parameter t "

Answer 9

no, it didn't work either.