1.
x=cos(t)
y=sin(t)
z=t^2+1
(1 0 1)
x'(t) = -sint
y'(t) = cost
z'(t) = 2t
r'(t) = [-sint cost 2t]
r(t) = [cos(0) sin(0) (1)]
= [1 0 1]
r'(t) = [-sin(0) cos(0) 0]
= [0 1 0]
--> [1 0 1]+t[0 1 0]
x(t) = 1
y(t) = 1t
z(t) = 1


2. 
r'(t) = ti+2j+(t+1)k
r(0)=i+2j+3k
integrate r'(t)
= [t^2/2 2t t^2/2+t + C]
r(0) becomes [0 0 0]+c, which 
makes C = [1 2 3]
insert back into original integrated
equation:
[t^2/2+1 2t+2 t^2/2+t+3]