Questão 2 (2,0 pontos) Considere a curva
\gamma(t)=\left(3 \cos \left(\frac{t}{5}\right), 3 \operatorname{sen}\left(\frac{t}{5}\right), \frac{4 t}{5}\right)
a) Determine os vetores tangente e normal unitários T(t) e N(t).
b) Calcule a curvatura de \gamma.

Passo 1

O enunciado dá a curva paramétrica
\gamma(t)=\bigl(3\cos(\tfrac{t}{5}),\,3\sin(\tfrac{t}{5}),\,\tfrac{4t}{5}\bigr)
e pede
(a) o vetor tangente unitário T(t) e o vetor normal unitário N(t),
(b) a curvatura \kappa.
Para isso precisamos calcular derivadas de \gamma, normas de vetores e usar
T(t)=\frac{\gamma'(t)}{\|\gamma'(t)\|},\quad N(t)=\frac{T'(t)}{\|T'(t)\|},\quad \kappa=\frac{\|T'(t)\|}{\|\gamma'(t)\|}.

Passo 2

Calculemos \gamma'(t) componente a componente:
\gamma'(t) =\Bigl(-\tfrac{3}{5}\sin(\tfrac{t}{5}),\, \tfrac{3}{5}\cos(\tfrac{t}{5}),\, \tfrac{4}{5}\Bigr).
A norma é
\|\gamma'(t)\| =\sqrt{\Bigl(\tfrac{3}{5}\Bigr)^{2}\bigl(\sin^{2}+\cos^{2}\bigr) +\Bigl(\tfrac{4}{5}\Bigr)^{2}} =\sqrt{\tfrac{9}{25}+\tfrac{16}{25}} =\sqrt{1} =1.

Passo 3

Como \|\gamma'(t)\|=1, o vetor tangente unitário é
T(t)=\gamma'(t) =\Bigl(-\tfrac{3}{5}\sin(\tfrac{t}{5}),\, \tfrac{3}{5}\cos(\tfrac{t}{5}),\, \tfrac{4}{5}\Bigr).

Passo 4

Derivando T(t) obtemos
T'(t) =\Bigl(-\tfrac{3}{25}\cos(\tfrac{t}{5}),\, -\tfrac{3}{25}\sin(\tfrac{t}{5}),\,0\Bigr),
cuja norma é
\|T'(t)\| =\sqrt{\Bigl(\tfrac{3}{25}\Bigr)^{2}\bigl(\cos^{2}+\sin^{2}\bigr)} =\frac{3}{25}.
Como \|\gamma'(t)\|=1, a curvatura segue
\kappa=\frac{\|T'(t)\|}{\|\gamma'(t)\|} =\frac{3/25}{1} =\frac{3}{25}.

Passo 5

O vetor normal unitário é
N(t)=\frac{T'(t)}{\|T'(t)\|} =\frac{1}{3/25} \Bigl(-\tfrac{3}{25}\cos(\tfrac{t}{5}),\, -\tfrac{3}{25}\sin(\tfrac{t}{5}),\,0\Bigr) =\bigl(-\cos(\tfrac{t}{5}),\,-\sin(\tfrac{t}{5}),\,0\bigr).

Resposta

(a)
T(t)=\Bigl(-\tfrac{3}{5}\sin(\tfrac{t}{5}),\, \tfrac{3}{5}\cos(\tfrac{t}{5}),\, \tfrac{4}{5}\Bigr), \qquad N(t)=\bigl(-\cos(\tfrac{t}{5}),\,-\sin(\tfrac{t}{5}),\,0\bigr).
(b)
\kappa=\frac{3}{25}.

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