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?

Accepted Solution

Answer:a) 25.15b)    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) = 25.15b)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 = 1y = tz = (4pi)^2 + t *(8pi) = 4pi(4pi + 2t)c)The intersection with the xy plane will occurr when z = 0This happens at:   t1 = -2piTherefore, the intersection will occur at:(x,y) = (1, -2pi)