Asked by mando

Find a unit tangent vector to r(t)=e^(2t)i+e^(-1)j+(t^2+4)k at (1,1,4).
Answered by Steve
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 <b>j</b> e^(-t)?
Answered by mando
oh yes it is.
Answered by mando
r(t)=e^(2t)i+e^(-t)j+(t^2+4)k at (1,1,4)
Answered by Steve
Cool. Now just differentiate term by term:

r'(t) = 2e^(2t)<b>i</b> - e^(-t)<b>j</b> + 2t<b>k</b>
r'(1) = 2e²<b>i</b> - 1/e<b>j</b> + 2<b>k</b>
∥r'∥ = √(4e^4 + 1 + 4)

So, the unit vector at (1,1,4) = 1/√(4e^4 + 1 + 4) 2e²<b>i</b> - 1/e<b>j</b> + 2<b>k</b>
Answered by Steve
Rats. t=0

r'(0) = (2e²,-1,4)

you take it from there.
Answered by Steve
RATS!
r'(0) = (2,-1,0)

unit vector = 1/√5 (2,-1,0)
Answered by mando
how does 1/e equal to 1? ∥r'∥ = √(4e^4 + 1 + 4)
Answered by mando
oh ok, now i get it.
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