Solución
T
(
t
)
=
1
4
t
2
+
2
⟨
1
,
2
t
,
1
⟩
\bold{T}(t) = \frac{1}{\sqrt{4t^2+2}}\lang 1, 2t, 1\rang
T
(
t
)
=
4
t
2
+
2
1
⟨
1
,
2
t
,
1
⟩