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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?

Answer :

horaciocrotte

Answer:

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)

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