Two unit tangent vectors for parametric equation

[medicalphysicsguy]

Greetings,

A curve is given parametrically. Find two unit tangent vectors to C at P.

\(\displaystyle x=e^{2t} \hspace{12} y=e^{-t} \hspace{12} z=t^2+4 \hspace{12} P(1,1,4)\)

The book does not prepare us for this one. I can take the three derivatives and get parametric equations for the tangent line fine. That would give me:

\(\displaystyle x'=2e^{2t} \hspace{12} y'=-e^{-t} \hspace{12} z'=2t \hspace{12}\)

t=0 so that would give me:

\(\displaystyle x'=2 \hspace{12} y'=-1 \hspace{12} z'=0\)

for a final parametric equation of the tangent line of:

\(\displaystyle x = 1 + 2t \hspace{12} y = 1 - t \hspace{12} z = 4\)

but I do NOT understand this "two unit tangent vectors" business.

Thanks for any help!!

-mpg

 

[pka]

A curve is given parametrically. Find two unit tangent vectors to C at P.
\(\displaystyle x=e^{2t} \hspace{12} y=e^{-t} \hspace{12} z=t^2+4 \hspace{12} P(1,1,4)\)
\(\displaystyle x'=2e^{2t} \hspace{12} y'=-e^{-t} \hspace{12} z'=2t \hspace{12}\)
t=0 so that would give me:
\(\displaystyle x'=2 \hspace{12} y'=-1 \hspace{12} z'=0\)
for a final parametric equation of the tangent line of:
\(\displaystyle x = 1 + 2t \hspace{12} y = 1 - t \hspace{12} z = 4\)
but I do NOT understand this "two unit tangent vectors" business.

The vector \(\displaystyle v=<2,-1,0>\) is a tangent vector at \(\displaystyle (1,1,4)\).

But you want two unit tangents, so \(\displaystyle \pm\dfrac{v}{\|v\|}\).