Tangent Line at t=0 for Parametric Curve (8sin(5t)-3, 8cos(5t)+3, 3t-4)

[alchal]

Homework Statement

If the parametric equations of a curve C are (x= 8sin(5 t) -3, y=8cos(5 t) + 3, z=3 t -4),
what are the parametric equations of the tangent line to this curve at t=0 ?

The Attempt at a Solution

x(t)=8sin(5t)-3
y(t)=8cos(5t)+3
z(t)=3t-4

x'(t)=40cos(5t)
y'(t)=-40sin(5t)
z'(t)=3

x'(0)=40
y'(0)=0
z'(0)=3

not even sure if I am doing this right...

 

[Rasalhague]

You're doing alright so far. A parametric equation for the tangent line will take the form a + hb, where a is a position vector for the point on your original curve at t = 0, h is a scalar, namely the parameter for this new equation describing the tangent line, and b is a displacement vector tangent to your original curve at t = 0, for example the derivative you found: (40,0,3).