The position vector for a particle moving on a helix is

c (t) = (cos (t), sin (t), t^2).

(a) Find the speed of the particle at time t0 = 4pi.

(b) Find a parametrization for the tangent line to c (t) at t0 = 4pi.

(c) Where will this line intersect the xy plane?

a) 25.15

b)

x = 1

y = t

z = (4pi) ^2 + t * (8pi) = 4pi (4pi + 2t)

c) (x, y) = (1, - 2pi)

Step-by-step explanation:

a)

First lets calculate the velocity, that is, the derivative of c (t) with respect to t:

v (t) = (-sin (t), cos (t), 2t)

The velocity at t0=4pi is:

v (4pi) = (0, 1, 8pi)

And the speed will be:

s (4pi) = √ (0^2+1^2 + (8pi) ^2) = 25.15

b)

The tangent line to c (t) at t0 = 4pi has the parametric form:

(x, y, z) = c (4pi) + t*v (4pi)

Since

c (4pi) = (1, 0, (4pi) ^2)

The tangent curve has the following components:

x = 1

y = t

z = (4pi) ^2 + t * (8pi) = 4pi (4pi + 2t)

c)

The intersection with the xy plane will occurr when z = 0

This happens at:

t1 = - 2pi

Therefore, the intersection will occur at:

(x, y) = (1, - 2pi)