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)
You need to show me the model, so I can answer the question.
