Question

How do you find the parametric equations for the tangent line to the curve `x=t^4−1`, `y=t^2+1`, `z=t^3` at the point (15, 5, 8)?

Answer 1

`vec r = ((15),(5),(8)) + lambda ((32),(1),(6))`

Explanation:

for `vec r = ((t^4 - 1),(t^2 +1),(t^3)) = ((15),(5),(8))`, we can see that `t = 4`

the tangent vector (or velocity vector) is the first derivative:

`vec r'\_(t = 4) = ((4t^3),(2t),(3t^2))_(t = 4) = ((4(4^3)),(2(4)),(3(4^2))) = ((32),(1),(6))`

so for generalised line `vec r = vec r_o + lambda vec r'` we have here

`vec r = ((15),(5),(8)) + lambda ((32),(1),(6))`