Find a unit tangent vector to r(t)=e^(2t)i+e^(-1)j+(t^2+4)k at (1,1,4).
8 answers
To get the point (1,1,4) it appears there's an error. t=0 produces (1,1/e,4) as written. Is the coefficient of j e^(-t)?
oh yes it is.
r(t)=e^(2t)i+e^(-t)j+(t^2+4)k at (1,1,4)
Cool. Now just differentiate term by term:
r'(t) = 2e^(2t)i - e^(-t)j + 2tk
r'(1) = 2e²i - 1/ej + 2k
∥r'∥ = √(4e^4 + 1 + 4)
So, the unit vector at (1,1,4) = 1/√(4e^4 + 1 + 4) 2e²i - 1/ej + 2k
r'(t) = 2e^(2t)i - e^(-t)j + 2tk
r'(1) = 2e²i - 1/ej + 2k
∥r'∥ = √(4e^4 + 1 + 4)
So, the unit vector at (1,1,4) = 1/√(4e^4 + 1 + 4) 2e²i - 1/ej + 2k
Rats. t=0
r'(0) = (2e²,-1,4)
you take it from there.
r'(0) = (2e²,-1,4)
you take it from there.
RATS!
r'(0) = (2,-1,0)
unit vector = 1/√5 (2,-1,0)
r'(0) = (2,-1,0)
unit vector = 1/√5 (2,-1,0)
how does 1/e equal to 1? ∥r'∥ = √(4e^4 + 1 + 4)
oh ok, now i get it.